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Mathematics, Classes 9-10
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Chapter One, , Real Number, Mathematics is originated from the process of expressing quantities in symbols or, numbers. The history of numbers is as ancient as the history of human civilization., Greek Philosopher Aristotle According to the formal inauguration of mathematics occurs, in the practice of mathematics by the sect of priest in ancient Egypt. So, the number, based mathematics is the creation of about two thousand years before the birth of Christ., After that, moving from many nations and civilization, numbers and principles of, numbers have gained an universal form at present., The mathematicians in India first introduce zero (0) and 10 based place value system for, counting natural numbers, which is considered a milestone in describing numbers., Chinese and Indian mathematicians extended the idea zero, real numbers, negative, number, integer and fractional numbers which the Arabian mathematicians accepted in, the middle age. But the credit of expressing number through decimal fraction is awarded, to the Muslim Mathematicians. Again they introduce first the irrational numbers in, square root form as a solution of the quadratic equation in algebra in the 11th century., According to the historians, very near to 50 BC the Greek Philosophers also felt the, necessity of irrational number for drawing geometric figures, especially for the square, root of 2. In the 19th century European Mathematicians gave the real numbers a, complete shape by systematization. For daily necessity, a student must have a vivid, knowledge about ‘Real Numbers’. In this chapter real numbers are discussed in detail., At the end of this chapter, the students will be able to –, ¾ Classify real numbers, ¾ Express real numbers into decimal and determine approximate value, ¾ Explain the classification of decimal fractions, ¾ Explain recurring decimal numbers and express fractions into recurring decimal, numbers, ¾ Transform recurring decimal fraction into simple fractions, ¾ Explain non-terminating non-recurring decimal fraction, ¾ Explain non-similar and similar decimal fraction, ¾ Add, subtract multiply and divide the recuring decimal fraction and solve various, problems related to them.
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2, , Math, , Natural Number, 1, 2, 3, 4......... etc. numbers are called natural number or positive whole numbers., 2, 3, 5, 7......... etc. are prime numbers and 4, 6, 8, 9,......... etc. are composite numbers., Integers, All numbers (both positive and negative) with zero (0) are called integers i.e. ......., 3, 2, 1, 0, 1, 2, 3......... etc. are integers., Fractional Number, p, If p, q are co-prime numbers ; q z 0 and q z 1 , numbers expressed in, form are, q, called fractional number., 1 3 5, Example : , ,, etc. are fractional numbers., 2 2 3, If p q , then it is a proper fraction and if p ! q then it is an improper fraction :, 1 1 2 1, 3 4 5 5, Example , , , , ......... etc. proper and , , , , .... etc. improper fraction., 2 3 3 4, 2 3 3 4, Rational Number, p, is called rational, If p and q are integers and q z 0 , number expressed in the form, q, number. For example :, , 3, 1, , 3,, , 11, 2, , 5.5,, , 5, 1.666... etc. are rational numbers., 3, , Rational numbers can be expressed as the ratio of two integers. So, all integers and all, fractional numbers are rational numbers., Irrational Number, p, form, where p, q are integers and q z 0 are, Numbers which cannot be expressed in, q, called Irrational Numbers. Square root of a number which is not perfect square, is an, irrational number. For example:, , 2, , 1.414213.....,, , 3, , 1.732.....,, , 5, 2, , 1.58113...., , etc. are irrational numbers. Irrational number cannot be expressed as the ratio of two, integers., Decimal Fractional Number, If rational and irrational numbers are expressed in decimal, they are known as decimal, fractional numbers. As for instance, 3 3 0, 5 2 5, 10 3 3333......., 3 1 732......... etc., 3, 2, are decimal fractional numbers. After the decimal, if the number of digits are finite, it is, terminating decimals and if it is infinite it is known as non-terminating decimal number., For, example,, 0.52,, 3.4152, etc., are, terminating, decimals, and, 1 333......., 2 123512367........... etc. are non-terminating decimals. Again, if the digits
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3, , Math, , after the decimal of numbers are repeated among themselves, they are known recurring, decimals and if they are not repeated, they are called non-recurring decimals. For, example : 1 2323........, 5 6 5 4 etc. are the the recurring decimals and, 0 523050056........, 2 12340314........ etc. are non-recurring decimals., Real Number, All rational and irrational numbers are known as real numbers. For example :, , 3, , 1, , 4, , 0, r 1, r 2, r 3,.......... r , r , r ,........ 2 , 3, 5, 6......, 2 2 3, 1 23, 0 415, 1 3333......., 0 6 2 , 4 120345061.......... etc. are real numbers., Positive Number, All real numbers greater than zero are called positive numbers. As for instance, , 1, 2,, , 1 3, , ,, 2 2, , 2 , 0 415, 0 6 2 , 4 120345061.............. etc. are positive numbers., , Negative Number, All real numbers less than zero are called negative numbers. For example,, , 3, 1, 1, 2, , , 2 , 0 415, 0 6 2 , 4 120345061.............., 2, 2, , etc., , are, , negative numbers., Non-Negative Number, All positive numbers including zero are called non-negative numbers. For example,, 1, 0, 3, , 0 612, 1 3, 2 120345.............. etc. are non-negative numbers., 2, Classification of real Number., Real, , Rational, Integer, Positive, , Fractional, , 1, , Fraction, 0, , Composite, , Negative, , Proper, , Improper, , Simple, , Mixed, , Decimal, , Terminating, , Irrational, , Recurring, , Nonrecurring
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4, , Math, , Activity : Show the position of the numbers, , 3, 4, , , 5, 7, 13 , 0, 1,, , 9, 7, , , 12 , 2, , 4, ,, 5, , . ., , 1 1234......, .3 2 3 in the classification of real numbers., Example 1. eDtermine the two irrational numbers between, , 3, , Solution : H, ere,, , 3 and 4 ., , 1.7320508......, , Let,, , a, , 2.030033000333....., , and, , b, , 2.505500555......., , Clearly :both a and b are real numbers and both are greater than, , 3 and less than 4., 3 2.03003300333......... 4, and, 3 2.505500555................ 4, Again, a and b cannot be expressed into fractions., ? a and b are the two required irrational numbers., Basic characteristics of addition and multiplication over a real number :, 1. If a, b are real numbers, i a b is real and ii ab is a real number, 2. If a, b are real numbers, i a b b a and ii ab ba, 3. If a, b, c are real numbers, i a b c a b c and ii ab c a bc, 4. If a is a real number, in real numbers there exist only two number 0 and 1 where, i 0 z 1 ii a 0 a iii a.1 1.a a, 1, 5. If a is a real number, i a ( a ) 0 ii If a z 0 , a., 1, a, 6. If a, b, c are real numbers, a(b c ) ab ac, 7. If a, b are real numbers, a b or a b or a ! b, 8. If a, b, c are real numbers and a b , a c b c, 9. If a, b, c are real numbers and a b , (i ) ac bc where c 0, (ii ) If ac ! bc , c 0, Proposition : 2 is an irrational number., eWknow,, i.e.,, , 1 2 4, 1 2 4, or, 1 2 2, , Proof : 12 1,, 2, 2, 2 2 4, ? Therefore, the value of 2 is greater than 1 and less than 2., ?
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5, , Math, , ?, ?, , 2 is not an integer., 2 is either a rational number or a irrational number. If 2 is a rational number, p, let, 2, ; where p and q are natural numbers and co-prime to each other and q ! 1, q, or, 2, , p2, ; squaring, q2, , or, 2q, , p2, ; multiplying both sides by q ., q, , Clearly 2q is an integer but, , p2, is not an integer because p and q are co-prime natural, q, , numbers and q ! 1, ? 2q and, ? aVlue of, ?, , p2, p2, cannot be equal, i.e., 2q z, q, q, 2 cannot be equal to any number with the form, , p, i.e.,, q, , 2z, , p, q, , 2 is an irrational number., , Example 2. P, rove that, sum of adding of 1 with the product of four consecutive natural, numbers becomes a perfect square number., Solution : Let four consecutive natural numbers be x, x 1, x 2, x 3 respectively., B, y adding 1 with their product we get,, , x x 1 x 2 x 3 1 x x 3 x 1 x 2 1, x 2 3 x x 2 3x 2 1, a a 2 1;, , [ x 2 3x, , a], , a ( a 2) 1 ;, , a2 2a 1, , a 1, , 2, , 2, , x 2 3x 1 ;, , which is a perfect square number., ? If we add 1 with the product of four consecutive numbers, we get a perfect square, number., Activity : rPoof that,, , 3 is an irrational number, , Classification of Decimal Fractions, Each real number can be expressed in the form of a decimal fraction.
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6, , Math, , oFr example,, , 2, , 2 0,, , 2, 5, , 0.4,, , 1, 3, , 0.333.... etc. There are three types of decimal, , fractions :terminating decimals, recurring decimals and non-terminating decimals., Terminating decimals : In terminating decimals, the finite numbers of digits are in the, right side of a decimal points. oFr example, 0.12,1.023,7.832,54.67,.......... etc. are, terminating decimals., Recurring decimals : In recurring decimals, the digits or the part of the digits in the, right side of the decimal points will occar repeatedly. oFr example,, 3.333....., 2.454545......, 5.12765765 ǤǤǤǤǤǤǤǤǤǤetc. are recurring decimals., Non-terminating decimals : In non-terminating decimals, the digits in the right side of, a decimal point never terminate, i.e., the number of digits in the right side of decimal, point will not be finite neither will the part occur repeatedly. F, or, example.1.4142135......, 2.8284271....... etc. are non-terminating decimals., Terminating decimals and recurring decimals are rational numbers and non-terminating, decimals are irrational numbers. The value of an irrational number can be determined, upto the required number after the decimal point. If the numerator and denominator of a, fraction can be expressed in natural numbers, that fraction is a rational number., Activity : Classify the decimal fractions stating reasons :, , 1.723, 5.2333........,, 0.450123......., , 0.0025,, , 2.1356124.......,, , Recurring decimal fraction :, 23, into decimal fractions, we get,, Expressing the fraction, 6, 23 = 6 ) 23 ( 3 833, 6, 18, 50, 48, 20, 18, 20, 18, 20, 18, 2, , 0.0105105........, , and
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7, , Math, , Observe : It is found that, the process of division is not ended at the time of dividing a, numerator of a fraction by its denominator. To convert it into decimal fraction in the, quotient ‘3’occurs repeatedly. H, ere,, 3.8333..... is a recurring decimal fraction., If digit or successive digit of a decimal fractions a in the right side of the decimal point, appear again and again, these are called recurring decimal fractions., If a digit or successive digits of decimal fractions., In recurring decimal fractions, the portion which occurs again and again, is called, recurring part. In recurring decimal fraction, if one digit recurs, the recurring point is, used upon it and if more than one digits recurs, the recurring point is used only upon the, first and the last digits., As for example : 2.555....... is written as 2.5 and 3.124124124........ is written as 3.1 24 ., In recurring decimal fractions, if there is no other digit except recurring one, after, decimal point it is called pure recurring decimal and if there is one digit or more after, decimal point in addition to recurring one, it is called mixed recurring decimal. oFr, example, 1.3 is a pure recurring decimal and 4.2351 2 is a mixed recurring decimal., If there exists prime factors other than 2 and 5 in the denominator of the fraction, the, numerator will not be fully divisible by denominator. As the last digit of successive divisions, cannot be other than1, 2, ......, 9 , at one stage the same number will be repeated in the, remainder. The number in the recurring part is always smaller than that of the denominator., Example 3. Express 3 into decimal fraction., , Example 4. Express 59 into decimal fraction., , 11, , 37, , Solution :, 11) 30 (0.2727, 22, 80, 77, 30, 22, 80, 77, 3, , Solution :, 37 )95 (2.56756, 74, 210, 185, 250, 222, 280, 259, 210, 185, 250, 222, 28
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8, , Math, , Required decimal fraction = 0.2727 ......, Required decimal fraction =, , , = 0.27, 2.56756...... 2.5 67, Conversion of Recurring Decimal into Simple Fraction, Determining the value of recurring fraction :, Example 5. Express 0.3 into simple fraction., Solution : 0.3 0.3333........, 0.3 u 10 0.333....... u 10, and, 0.3 u 1, 0.333....... u 1, subtracting, 0.3 u 10 0.3 u 1, or, 0.3 u 10 1 3 or 0.3 u 9, , 3.333......, 0.333......, , 3, , 3, 1, Therefore, 0.3, 3, 1, Required fraction is, 3, Example 6. Express 0.2 4 into simple fraction., , 3, 9, , Solution : 0.2 4 = 0 24242424........, So,, , 0.2 4 u 100, , and, , 0.2 4 u 1, , 0.242424....... u 100, , 0.242424....... u 1, , Subtracting 0.2 4 100 1, or, 0.2 4 u 9, , 24, , 24.2424......, , 0.242424......, , 24, , or, 0.2 4, , 24, 9, , 8, 33, , 8, 33, Example 7. Express 5.13 45 into simple fraction., Required fraction is, , Solution : 5.13 45 = 5.1345345345........., So,, , 5.13 45 u 10000, , 5.1345345...... u 10000, , and, , 5.13 45 u 10, , 5.1345345....... u 10, , Subtracting, 5.13 45 u 9, 0, So, 5.13 45 =, , 51345 51, 90, , 51345.345......, 51.345......, , 51345 51, 51294, 90, , 8549, 1665, , 5, , 224, 1665
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Math, , Required fraction is 5, , 9, , 224, 1665, , Example 8. Express 42.347 8 into simple fraction., Solution : 42.347 8 = 42.347878........, So,, 42.347 8 u 10000 42.347878......... u 10000 42348.7878, and, 42.347 8 u 100 = 42.347878........ u 100 = 4234.7878, Subtracting, 42.347 8 u 9, 00, 423478, 4234, Therefore, 42.347 8 =, 900, Required fraction is 42, , 419244, 900, , = 423478 4234, 34937, 287, 42, 825, 825, , 287, 825, , Explanation : rFom the examples 5, 6, 7 and 8 , it appears that,, x The recurring decimal has been multiplied by the number formed by putting at the, right side of 1 the number of zeros equal to the number of digits in the right side of, decimal point in the recurring decimal., x, , The recurring decimal has been multiplied by the number formed by putting at the, right side of 1 the number of zeros equal to the number of digits which are nonrecurring after decimal point of the recurring decimal., , x, , the second product has been subtracted from the first product. yB subtracting the, second product from the first product the whole number has been obtained at the, right side. eHre it is observed that, the number of non-recurring part has been, subtracted from the number obtained by removing the decimal and recurring points, of recurring decimal fraction., , x, , The result of subtraction has been divided by the number formed by writing the same, number of 9equal to the number of digits of recurring part at the left and number of, zeros equal to the number of digits of non-recurring part at the right., , x, , In the recurring decimals, converting into fractions the denominator is the number of, 9equal to the number of digits in the rec urring part and in right side of all 9’s, number of zeros equal to the number of digits in the non-recurring part. And the, numerator in the result that is obtained by subtracting the number of the digits, formed by omitting the digits of recurring part from the number formed by removing, the decimal and recurring points of recurring decimal., , Remark : Any recurring decimal can also be converted into a fraction. All recurring, decimals are rational numbers., Example .9Express 5.234 57 into simple fraction.
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10, , Math, , Solution : 5.234 57 = 5.23457457457........., So,, = 523457.457457, 5.234 57 u 100000, = 523.457457, and, 5.234 57 u 100, Subtracting, 5.234 57 u 9, = 5229, 00, 34, Therefore,, , 5.234 57 = 522934, 900, , 261467, 49950, , Required fraction is 261467, 49950, , Explanation : eHre in the decimal part the recurring decimal has been multiplied first, by 100000 (5 zeros at the right side of 1) as there are two digits at the left side of, recurring part in the decimal portion, the recurring decimal has been multiplied by 100, (two zeros at the right side of 1). The seco nd product has been subtracted from the first, product. In one side of the result of subtraction is a whole number and at the other side, of the result is (100000 100) = 0, times of the value of the given recurring, 90, decimal. D, ividing both the sides by 0, , the required fraction is obtained., 90, Activity : Express 0.4 1 and 3.046 23 into fractions., Rules of Transformation of Recurring Decimals into Simple Fractions, N, umerator of the required fraction = the result by subtracting the number obtained from, exempting the decimal point of the given decimal point and the non-recurring part., D, enominator of the required fraction = uNmbers formed by putting the number of 9, equal to the number of digits in the recurring part of the from the number of zeros equal, to the number of digits in the non-recurring part. H, ere the above rules are directly, applied to convert some recurring decimals into simple factions., Example 10. Express 45.23 46 into simple fraction., 452346 452 451894, 225947, Solution : 45.23 46 =, 90, 90, 4995, 1172, Required fraction is 45, 4995, Example 11. Express 32.5 67 into simple fraction., , Solution : 32.5 67, , 32567 32, 9, , Required fraction is 32, , 32535, 9, , 45, , 1172, 4995, , 21, 3615 1205, 32, 37, 111, 37, , 21, ., 37, , Activity : Express 0.01 2 and 3.312 4 into fraction.
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11, , Math, , Similar recurring decimals and Non-similar Recurring decimals :, If the numbers of digits in non-recurring part of recurring decimals are equal and also, numbers of digits in the recurring parts are equal, those are called similar recurring, decimals. O, ther recurring decimals are called non-similar recurring decimals. F, or, , , , , , , example : 12.45 and 6.32 ; 9.453 and 125.897 are similar recurring decimals. Again,, 0.34 56 and 7.457 89 ; 6.435 7 and 2.893 45 are none-similar recurring decimals., The Rules of Changing Non-Similar Recurring Decimals into Similar Recurring, Decimals, The value of any recurring decimals is not changed, if the digits of its recurring part are, written again and again, oFr Example, 6.453 7 6.453 737, 6.4537 3 6.45373 7 Ǥ, H, ere each one is a recurring decimal, 6.45373737......... is a non-terminating decimal., It will be seen that each recurring decimal if converted into a simple fraction has the, same value., , 4537 645 63892, 900, 900, 6453737 645 6453092, , , 6.453737, 900, 900, 6.453 7, , 6.45373 7, , 6453737 64537, 90000, , 63892, 900, , 6389200, 90000, , 63892, 900, , In order to make the recurring decimals similar, number of digits in the non-recurring, part of each recurring decimal is to be made equal to the number of digits of nonrecurring part of that recurring decimal in which greatest number of digits in the nonrecurring part exists and the number of digits in the recurring part of each recurring, decimal is also to be made equal to the lowest common multiple of the numbers of digits, of recurring parts of recurring decimals., Example 12. Convert 5.6 , 7.34 5 and 10 784 23 into similar recurring decimals., Solution : The number of digits of non-recurring part of 5 6 , 7 34 5 and 10.784 23 are, ere the number of dig its in the non-recurring part occurs in, 0, 1 and 2 respectively. H, , , 10.78423 and that number is 2 . Therefore to make the recurring decimals similar the, number of digits in the non-recurring part of each recurring decimal is to be made 2 ., Again, the numbers of digits to recurring parts of 5.6 , 7.34 5 and 10.784 23 are 1, 2, and 3 respectively. The lowest common multiple of 1, 2 and 3 is 6 . So the number of, digits in the recurring part of each recurring decimal would be 6 in order to make them, similar.
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12, , So, 5.6, , Math, , 5.666 66666 , 7.34 5, , Required similar recurring, 10.784 23423 respectively., , 7.345 45454 and 10.784 23 10.784 23423, decimals, , are 5.666 66666 , 7.345 45454 ,, , Example 13. Convert 1.7643, 3.2 4 and 2.783 46 into similar recurring decimals., Solution : In 1.7643 the number of digits in the non-recurring part means 4 digits after, decimal point and here there is no recurring part., In 3.2 4 the number of digits in the recurring and non-recurring parts are respectively 0, and 2., In 2.783 46 the number of digits in the recurring and non-recurring parts are, respectively 2 and 3., The highest number of digits in the nonrecurring parts is 4 and the L.C.M. of the, numbers of digits in the recurring parts i.e. 2 and 3 is 6. The numbers of digits is the, recurring and nonrecurring parts of each decimal will be respectively 4 and 6., ?1.7643= 1.76430 00000 ; 3.2 4, , 3.24242 42424 ; 2.783 46, , 2.78346 34634, Required recurring similar decimals are 1.76430 00000 , 3.24242 42424 and, 2.78346 34634, , Remark : In order to make the terminating fraction similar, the required number of, zeros is placed after the digits at the extreme right of decimal point of each decimal, fraction. The number of non-recurring decimals and the numbers of digits of nonrecurring part of decimals after the decimal points are made equal using recurring digits., After non-recurring part the recurring part can be started from any digit., Activity : Express 3.467, 2.012 4 3 and 7.525 6 into similar recurring fractions., Addition and Subtraction of Recurring Decimals, In the process of addition or subtraction of recurring decimals, the recurring decimals are, to be converted into similar recurring decimals. Then the process of addition or, subtraction as that of terminating decimals is followed. If addition or subtraction of, terminating decimals and recurring decimals together are done, in order to make, recurring decimals similar, the number of digits of non-recurring part of each recurring, should be equal to the number of digits between the numbers of digits after the decimal, points of terminating decimals and that of the non-recurring parts of recurring decimals., The number of digits of recurring part of each recurring decimal will be equal to L.C.M., as obtained by applying the rules stated earlier and in case of terminating decimals,, necessary numbers of zeros are to be used in its recurring parts. Then the same process, of addition and subtraction is to be done following the rules of terminating decimals. The, sum or the difference obtained in this way will not be the actual one. It should be
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13, , Math, , observed that in the process of addition of similar decimals if any number is to be carried, over after adding the digits at the extreme left of the recurring part of the decimals then, that number is added to the sum obtained and thus the actual sum is found. In case of, subtraction the number to be carried over is to subtract from the difference obtained and, thus actual result is found. The sum or difference which is found in this way is the, required sum or difference., Remark (a) : The sum or difference of recurring decimals is also a recurring decimal. In this, sum or difference the number of digits in the non-recurring part will be equal to the number, of digits in the non-recurring part of that recurring decimal, which have the highest number, of digits in its non-recurring part. Similarly, the number of digits in the recurring part of the, sum or the result of subtraction will be the equal to L.C.M. of the numbers of digits of, recurring parts of recurring decimals. If there is any terminating decimals, the number of, digits in the non-recurring part of each recurring decimal will be equal to the highest, numbers of digits that occurs after the decimal point., , (b) Converting the recurring decimals into simple fractions, addition and subtraction, may be done according to the rules as used in case of simple fractions and the sum or, difference is converted into decimal fractions. But this process needs more time., Example 14. Add : 3 8 9 , 2 17 8 and 5 897 98, Solution : Here the number of digits in the non-recurring part will be 2 and the, number of digits in the recurring part will be 6 which is L.C.M. of 2,2 and 3., At first three recurring decimals are made similar., 3 8 9, = 3 898 98989, 2 17 8, , 5 897 98, , = 2 1278 78787, = 5897 98798, , 11 97576574, [ 8 8 7 2 25 , Here 2 is the number to, 2 be carried over, 2 of 25 has been added.], 11 975 76576, The required sum is 11 975 76576 or 11 975 76, Remark : In the sum the number in the recurring part is 575675 . But the value is, not changed if 576 is taken as the number of recurring part., Note : To make clear the concept of adding 2 at the extreme right side, this addition, is done in another method :
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14, , 3 8 9, 2 17 8, 5 897 98, , Math, , 3 898 98989 | 89, = 2 178 78787 | 87, = 5 897 98798 | 79, 11.975 76576 | 55, =, , Here the number is extended upto 2 more digits after the completion of the recurring, part. The additional digits are separated by drawing a vertical line. Then 2 has been, carried over from the sum of the digits at the right side of the vertical line and this 2, is added to the sum of the digit at the left side of the vertical line. The digit in the, right side of the vertical line are the same and the recurring point. Therefore both the, sums are the same., Example 15. Add : 8 94 78, 2 346 and 4 7 1 ., Solution : To make the decimals similar, the number of digits of non-recurring parts, would be 3 and that of recurring parts would be 6 which is L.C, .M. of 3 and 2., , 8 94 78, 2 346, 4 7 1, , = 8 9478 47847, , 2 3460 00000, = 4 7171 71717, 16 011019564 [ 8 0 1 1 10 , Here the digit in the, 1 second place on the left is 1 which is to be, carried over. Therefore 1 of 10 is added.], 16 0110 19565, , The required sum is 16 0110 19565 ., Activity : Add :1. 2 09 7 and 5 127 68, , 2. 1 345 , 0 315 7 6 and 8 0567 8, , Example 16. Subtract 5 246 73 from 8 24 3 ., Solution : Here the number of digits in the non-recurring part would be 2 and that of, recurring part is 6 which is L.C, .M. of 2 and 3. Now making two decimal numbers, similar, subtraction is done., , 8 24 3, 5 246 73, , = 8 243 43434, = 5 246 73673, , 2 99669761, 1, 2 996 69760, The required sum is 2 996 69760 ., , [Subtracting 6 from 3, 1 is to be, carried over.]
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15, , Math, , Remark : If the digit at the beginning place of recurring point in the number from which, deduction to be made is smaller than that of the digit in the number 1 is to be subtracted, from the extreme right hand digit of the result of subtraction., Note : In order to make the conception clear why 1 is subtracted, subtraction is done in, another method as shown below :, , 8 24 3, 5 246 73, , = 8 243 43434 | 34, = 5 246 73673 | 67, , 2 996 69760 | 67, The required difference is 2 996 69760 | 67, Here both the differences are the same., Example 17. Subtract 16 4 37 from 24 456 45 ., Solution :, , 24 456 45, 16 4 37, , = 24 456 45, = 16 437 43, , 8 01902 [7 is subtracted from 6.1 is to be carried, 1 over.], , 8 01901, The required difference is 8.019 01, Note :, , 24 456 45, 16 4 37, , = 24 456 45 | 64, = 16 437 43 | 74, , 8 019 01 | 90, Activity : Subtract :1. 10 418 from 13 127 84 2. 9 126 45 from 23 039 4, Multiplication and Division of Recurring Decimals :, oCnverting recurring decimals into simple fraction and completing the process of their, multiplication or division, the simple fraction thus obtained when expressed into a decimal, fraction will be the product or quotient of the recurring decimals. In the process of, multiplication or division amongst terminating and recurring decimals the same method is to, be applied. But in case of making division easier if both the divident and the divisior are of, recurring decimals, we should convert them into similar recurring decimals., Example 18. Multiply 4 3 by 5 7 .
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16, , Math, , Solution : 4 3, , 5 7, , 43 4, 9, 57 5, 9, , ? 4 3 u 5 7 =, , 39, 9, 52, 9, , 13 52, u, 3 9, , 13, 3, , 676, 27, , 25 0 37, , The required product is 25 0 37, Example 19. Multiply 0 28 by 42 1 8 ., , 8 2 26 13, 90, 90 45, 4218, , 42, 4176 464, 42 1 8, 99, 99, 11, 13 464 6032, u, 12 18 5, =, 45 11, 495, The required product is 12 18 5, Example 20. 2 5 u 4 35 u 1 23 4, ?, , Solution : 0 28, , 25 5, 10 2, 435 43 392, 4 35, 90, 90, 1234, , 12, 1222, 1 23 4, 990, 990, 5 392 611 196 u 611, u, u, 2 90 495, 8910, The required product is 13 44062, Activity : 1. Multiply 1 13 by 2 6 . 2., Example 21. Divide 7 3 2 by 0 27 ., Solution : 2 5, , Solution : 7 3 2, , 0 27, ? 7 3 2 y 0 27, , 611, 495, 119756, 8910, , 13 44062... ., , 0 2 u 1 1 2 u 0 08 1, , 732 7 725, 99, 99, 27 2 25 5, 90, 90 18, 725 5, y, 99 18, , 725 18, u, 99 5, , 290, 11, , 26 3 6, , ?
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Math, , 17, , The required quotient is 26 3 6, Example 22. Divide 2 2 718 by 1 91 2, , 22718 2 22176, 9999, 9999, 1912 19 1893, 1 91 2, 990, 990, 22716 1893 22716 990, ? 2 2 718 y 1 91 2 =, y, u, 9999, 990, 9999 1893, The required quotient is 1 1 881, Example 23. Divide 9 45 by 2 86 3 ., Solution : 2 2 718, , 945 2863 28, y, 100, 990, 189 u 99 33, 33, 2 u 2835 10, The required quotient is : 3 3, , Solution : 9 45 y 2 86 3 =, , 120, 101, , 1 1 881, , 945 990, u, 100 2835, , Remark : P, roduct of recurring decimals and quot ient of recurring decimals may be or, may not be a recurring decimal., 2. Divide 0 73 2 by 0 02 7, , Activity : 1. Divide 0 6 by 0 9 ., , Non Terminating Decimals, There are many decimal fractions in which the number of digits after its decimal point is, unlimited, again one or more than one digit does not occur repeatedly. Such decimal, fractions are called non-terminating decimal fractions, For example,, 5.134248513942307 ............ is a non-terminating decimal number. The square root of 2, is a non terminating decimal. Now we want to find the square root of 2., 1 2, 14142135........, 1, 24, , 100, 96, , 281 400, 281, 2824 11900, 11296
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18, , Math, , 28282 60400, 56564, 282841 383600, 282841, 2828423, 28284265, , 10075900, 8485269, 159063100, 141421325, 17641775, , If the above process is continued for ever, it will never end., , 2 1 4142135...... is a non terminating decimal number., The value upto the definite number of decimal places and the approximate value, upto some decimal places., , ?, , It is not the same to find the value of non-terminating decimals upto definite number of, decimal place and the approximate value into some decimal places. For example,, 5 4325893...... upto four decimal places will be 5 4325, but the approximate value, of the decimal, 5 4325893.... upto four decimal places will be 5 4326 . Here, the, value upto 2 decimal places and the approximate value upto 2 decimal places are the, same. This value is 5 43 . In this way the approximate terminating decimals can also be, found., Remark : When it is needed to find the value upto some decimal places, the digits that occur, in those places are to be written without any alternatives of those digits. If approximate values, are to be identified, we should add 1 with the last digit when there is 5, 6, 7, 8 or 9 after the, decimal places. But if it is 1, 2, 3 or 4 digits remain unchanged. In this case, correct value upto, decimal place or approximate value upto decimal place are almost equal. We should find the, value upto 1 place more to the required wanted value., Example 24. Find the square root of 13 and write down the approximate value upto 3, decimal places., Solution : 3 ) 13 ( 3 605551........, 9, , 66, , 400, 396
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19, , Math, , 7205, 72105, , 40000, 36025, 3697500, 3605525, , 7211101, , 9197500, 7211101, , 1986399, ? The required square root is 3 605551........ ., ? The required approximate value upto 3 decimal places is 3 606 ., Example 25. Find the value and approximate value of 4 4623845....... upto 1, 2, 3, 4, and 5 decimal places., Solution : The value of 4 4623845 upto 1 decimal place is 4 4, and approximate value, 1 ,,, ,, ,,, 45, Value upto 2, ,,, ,, ,,, 4 46, and approximate value upto 2 ,,, ,, ,, 4 46, Value upto 3, ,,, ,, ,, 4 462, and approximate value upto 3 ,,, ,, ,, 4 462, Value upto 4, ,,, 4 4623, and approximate value upto 4 ,,, ,, ,, 4 4624, Value upto 5, ,,, ,, ,, 4 46238, and approximate value upto, 5 ,,, ,, ,, 4 46238, Activity : Find the square root of 29 and find the value upto two decimal places and, the approximate value upto two decimal place., , Exercise 1, (b) 7, (c) 10 is an irrational number., 5, 2. (a) Find the two irrational numbers between 0.31 and 0.12 ., 1, (b) Find a rational and an irrational numbers between, and 2 ., 2, , 1. P, rove that, (a), , 3. (a) P, rove that, square of any odd integer number is an odd number., (b) P, rove that, the product of two consecu tive even numbers is divisible by 8.
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20, , Math, , 4. Express into recurring decimal fractions :(a), , 1, 6, , (b), , 7, 11, , (c) 3, , 2, 9, , (d) 3, , 8, 15, , 5. Express into simple fractions :, (a) 0 2 (b) 0 3 5 (c) 0 13 (d) 3 78 (e) 6 23 09, 6. Express into similar recurring fractions :, (a) 2 3 , 5 23 5, (b) 7 26 , 4 237, , , , , , (c) 5 7, 8 34, 6 245, (d) 12 32, 2 19 , 4 325 6, 7. Add :(a) 0 45 0 134 (b) 2 05 8 04 7 018 (c) 0 006 0 9 2 0 01 34, 8. Subtract :, (a) 3 4 2 13, , (b) 5 1 2 3 45, (d) 19 345 13 23 49, , (c) 8 49 5 35 6, 9. Multiply:, (a) 0 3 u 0 6, 10. Divide :, (a) 0 3 y 0 6, , (b) 2 4 u 0 8 1, (b) 0 35 y 1 7, , (c) 0 62 u 0 3, , (d) 42 1 8 u 0 28, , (c) 2 37 y 0 45, , (d) 1 1 85 y 0 2 4, , 11. Find the root (upto three decimal places) and write down the approximate values of, the square roots upto two decimal places :, (a) 12 (b) 0 2 5 (c) 1 34 (d) 5 13 02, 12. Find the rational and irrational numbers from the following numbers :, 2, 27, 8, 6, , (a) 0 4 (b) 9 (c) 11 (d), (e), (f), (g) 3 (h) 5 6 39, 3, 7, 48, 3, 7, 13. Simplify :, (a) ( 0 3 u 0 83 ) y 0 5 u 0 1 + 0 35 y 0 08, (b) > 6 27 u 0 5 y ^0 5 u 0 75 u 8 36`@, y ^0 25 u 0 1 u 0 75 u 21 3 u 0 5`, 14., , 5 and 4 are two real numbers., (a) Which one is rational and which one is irrational., (b) Find the two irrational numbers between, (c) P, rove That,, , 5 is an irrational number., , 5 and 4.
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Chapter Two, , Set and Function, The word s‘ et’ is familiar to us, such as dinner set, set of natural numbers, set o f, rational numbers etc. As a modern weapon of mathematics, the use of set is, extensive. The eGrman mathematician eGor ge C, antor (1844 –1918) first explaine d, his opinion about set. He created a sensation in mathematics by expressing the idea, of infinite set and his conception of set is known as s‘ et theory’. In this chapter, the, main objectives are to solve problems through using mathematics and symbols from, the conception of set and to acquire knowledge about function., At the end of this chapter, the students will be able to :, ¾ Explain the conception of set and subset and express them by symbols, ¾ Describe the method of expressing set, ¾ Explain the infinite set and differentiate between finite and infinite set, ¾ Explain and examine the union and the intersection of set, ¾ Explain power set and form power set with two or three elements, ¾ Explain ordered pair and cartesian product, ¾ P, rove the easy rules of set by example and Venn Diagram and solve various, problems using the rules of set operation, ¾ Explain and form sets and functions, ¾ Explain what are domain and range, ¾ Determine the domain and range of a function, ¾ Draw the graph of the function., Set, Well defined assembling or collection of objects of real or imaginative world is, called sets, such as, the set of three textbook of Bangla, English and Mathematics, set, of first ten natural odd numbers, set of integers, set of real numbers etc., Set is generally expressed by the capital letters of english alphabets,, A, B, C,..........X , Y , Z . For example, the set of three numbers 2, 4, 6 is A {2, 4, 6}, Each object or member of set is called set element. Such as, if B {a, b} , a and b, are elements of B . The sign of expressing an element is '' ., ? a B and read as a belongs to B, , b B and read as b belongs to B, no element c is in the above set B.
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22, , Math, , ? c B and read as c does not belong to B ., Method of describing Sets :, Set is expressed in two methods :(1) oR ster Method or Tabular Method and (2) Set, Builder Method., (1) Tabular Method : In this methods, all the set elements are mentioned, particularly by enclosing them within second bracket { ,} and if there is more than, one element, the elements are separated by using a comma (,)., For example : A {a, b} , B {2, 4, 6} , C, , N, { iloy, Tisha, Shuvra}etc., , (2) Set Builder Method : In this methods, general properties are given to determine, the set element, without mentioning them particularly :, Such as, A {x : x is a natural odd number}, B {x : x denotes the first five, students of class IX, }etc., Here, by ’:‘s‘uch as’or in short s‘uch that’is indicated. Since in this method, set rule o, condition is given to determining the set elements of this method, is called rule method., Example 1. Express the set A {7, 14, 21, 28} by set builder method., Solution : The elements of set A are 7, 14, 21, 28, Here, each element is divisible by 7, that is, multiple of 7 and not greater than 28., ? A {x : x multiple of 7 and x d 28} ., Example 2. Express the set B, Solution : Here, 28 1 u 28, , {x : x, factors of 28}by tabular method., , = 2 u 14, = 4u 7, ? factors of 28 are 1, 2, 4, 7, 14, 28, eRquired set B {1, 2, 4, 7, 14, 28}, Example 3. Express C {x : x is a positive integer and x2 18} by tabular method., Solution : P, ositive integers are 1, 2, 3, 4, 5, ..........., Here if x 1 , x 2 12 1, if x 2 , x 2 2 2 4, if x 3 , x 2 32 9, if x 4 , x 2 4 2 16, if x 5 , x 2 5 2 25 ; which is greater than 18., , r
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23, , Math, , ? The acceptable positive integers by the condition are 1, 2, 3, 4, ? eRquired set is, , C {1, 2, 3, 4}., , Activity : 1. Express the set C, 2. Express the set Q, , {9, 6, 3, 3, 6, 9} by set builder method., { y : y is an integer and y 3 d 27} by tabular method., , Finite set : The set whose numbers of elements can be determined by counting is, called finite set. For example, D = {x, y, z}E = 3{, 6, 9,.......,60},, F {x : x is a, prime number and 30 x 70} etc. are finite set. Here D set has 3 elements, E set, has 20 elements and F set has 9 elements., Infinite set : The set whose numbers of elements can not be determined by counting, is called infinite set. For example :, A {x : x is natural odd numbers},set of natural, number N {1, 2, 3, 4, ........} , set of integers Z {....... 3, 2, 1, 0, 1, 2, 3.......} , set, of rational numbers Q ® P : p and q is as integer and q z 0} , set of real numbers =, ¯q, , R etc. are infinite set., Example 4. Show that the set of all natural numbers is an infinite set., Solution : Set of natural number N {1, 2, 3, 4, 5, 6, 7, 8, .......}, Taking odd natural numbers from set N , the formed set A {1, 3, 5, 7, .......}, ,, ,, N , the formed set B {2, 4, 6, 8, .......}, ,, even ,,, ,,, The set of multiple of 3 C {3, 6, 9, 12, .......} etc., Here, the elements of the set A, B, C formed from set N can not be determined by, counting. So A, B, C is an infinite set., ? N is an infinite set., , Activity : Write the finite and infinite set from the sets given below :, 1. {3, 5, 7} 2. {1, 2, 2 2 , .......210 } 3. {3, 32 , 33 , .......} 4. {x : x is an integer and x 4}, , p, , 5. ® : p and q are co-prime and q ! 1}, ¯q, , 6. { y : y N and y 2 100 y 3 } ., , Empty set : The set which has no element is called empty set. Empty set is, expressed by {}or φ . Such as, set of three male students of Holycross school, {x N : 10 x 11}, {x N : x is a prime number and 23 x 29} etc., Venn-Diagram : John Venn (1834-1883) introduced set activities by diagram. Here, the geometrical figure on the plane like rectangular area, circular area and triangular, area are used to represent the set under consideration. These diagrams are named, Venn diagram after his name.
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24, , Math, , Subset : A {a, b} is a set. By the elements of set A , the sets {a, b}, {a } , { b} can, be formed. Again, by not taking any element φ set can be formed., Here, each of {a, b}, {a } , { b} , φ is subset of set A ., So, the number of sets which can be formed from any set is called subset of that set., The sign of subset is . If B is the subset of A , it is read as B A . B is a subset, of A. From the above subsets, {a, b} set is equal to A ., ? Each set is the subset of itself., Again, from any set, φ set can be formed., ? φ is a subset of any set., Q {1, 2, 3} and R {1, 3} are two subsets of P, , {1, 2, 3} . Again P Q, , ? Q P and R P ., Proper Subset :, If the number of elements of any subset formed from a set is less than the given set,, it is called the proper subset. For example : A {3, 4, 5, 6} and B {3, 5} àre two, sets. Here, all the elements of B exist in set A . ? B A, Again, the number of elements of B is less then the number of elements of A ., ? B is a proper subset of A and expressed as B A ., Example 5. Write the subsets of P {x, y, z} and find the proper subset from the, subsets., Solution : iGven,, , P {x, y, z}, , Subsets of P are {x, y, z} , {x, y} , {x, z} , { y, z} , {x} , {y} , {z} , φ ., rPoper subsets of, , P are {x, y} , {x, z} , { y, z} , {x} , {y} , {z}, , Equivalent Set :, If the elements of two or more sets are the same, they are called equivalent sets. Such, as, A {3, 5, 7} and B {5, 3, 7} are two equal sets and written as A B ., Again, if A {3, 5, 7} , B {5, 3, 3, 7} and C, are equivalent. That is, A B C, , {7, 7, 3, 5, 5} , the sets A, B and C
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Math, , 25, , It is to be noted if the order of elements is changed or if the elements are repeated,, there will be no change of the set., Difference of Set : Suppose, A {1, 2, 3, 4, 5} and B {3, 5} . If the elements of set, B are discarded from set A , the set thus formed is {1, 2, 4} and written as A \ B or, A B {1, 2, 3, 4, 5} {3, 5} = {1, 2, 4}, So, if a set is discarded from any set, the set thus formed is called different set., Example 6. If P {x : x , factors of 12}and Q {x : x , multiples of 3 and x d 12} ,, find P Q ., Solution : iGven, P {x : x , factors of 12}, Here, factors of 12 are 1, 2, 3, 4, 6, 12, ? P {1, 2, 3, 4, 6, 12}, Again, Q {x : x , multiple of 3 and x d 12}, Here, multiple of 3 upto 12 are 3, 6, 9, 12, ? Q {3, 6, 9, 12}, ? P Q {1, 2, 3, 4, 6, 12) {3, 6, 9, 12} {1, 2, 4}, eRquired set {1, 2, 4}, Universal Set :, All sets related to the discussions are subset of a definite set. Such as, A {x, y} is a subset of, B {x, y, z} . Here, set B is called the universal set in with respect to the set A ., So, if all the sets under discussion are subsets of a particular set, that particular set is, called the universal set with respect to its subsets., nUiversal set is generally expressed by U . But universal set can, be expressed by other symbols too. Such as, if set of all even, natural numbers C {2, 4, 6........} and set of all natural, numbers N {1, 2, 3, 4,...........} , the universal set with respect, to C will be N ., Complement of a Set :, U is an universal set and A is the subset of .UThe set formed, by all the elements excluding the elements of set A is called the, complement of set A . The complement of the set A is, expressed by Ac or Ac . Mathematically, Ac U \ A, , Let, P and Q are two sets and the elements of Q which are not, elements of P are called complement set of Q with respect to P and written as, Q c P \ Q.
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26, , Math, , Example 7. If U {1, 2, 3, 4, 6, 7} , A {2, 4, 6, 7} and B {1, 3, 5} , determine Ac, and B c ., Solution : Ac U \ A {1, 2, 3, 4, 6, 7} \ {2, 4, 6, 7} {1, 3, 5}, and B c, , U \ B {1, 2, 3, 4, 6, 7} \ {1, 3, 5} {2, 4, 6, 7}, , eRquired set Ac {1, 3, 5} and B c {2, 4, 6, 7}, Union of Sets :, The set formed by taking all the elements of two or more sets is called union of sets., Let, A and B are two sets. The union of A and B set is expressed by A B and, read as A union B . In the set builder method A B {x : x A or x B} ., Example 8. If C {3, 4, 5} and D {4, 6, 8} , determine C D ., Solution : iGven that, C {3, 4, 5} and D {4, 6, 8}, ? C D {3, 4, 5} {4, 6, 8} {3, 4, 5, 6, 8}, Intersection of Sets:, The set formed by the common elements of two or more sets is called intersection of, sets. Let, A and B are two sets. The intersection of A and B is expressed by, A B and read as A intersection B . In set building method,, A B {x : x A and x B} ., Example 9. If P {x N : 2 x d 6} and Q {x N : x are even, numbers and x d 8} , find P Q ., Solution : iGven that,, , P, , 4, 6, , {x N : 2 x d 6} = {3, 4, 5, 6}, , and Q {x N : x are even numbers and x d 8} {2, 4, 6, 8}, ? P Q {3, 4, 5, 6} {2, 4, 6, 8} {4, 6}, equired set is {4, 6}, R, Disjoint Sets:, If there is no common element in between two sets, the sets are disjoined sets. Let ,, A and B are two sets. If A B φ , A and B will be mutually disjoint sets., Activity : If U, , {1, 3, 5, 7, 9, 11} , E {1, 5, 9} and, , F {3, 7, 11} , find, E c F c and E c F c ., , Power Sets :, A {m, n} is a set. The subsets of A are {m n}, {m), {n}, φ . Here, the set of subsets, { m n}, {m), {n}, φ} is called power set of set A . The power set of A is expressed, as P(A) . So, the set formed with all the subsets of any set is called the power set of, that set.
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27, , Math, , Example 10. A { }, B {a}, C {a, b} are three sets., Here, P( A) {φ}, ? The number of elements of set A is 0 and the number of elements of its power, set = 1 20, Again, P( B) { a}, φ}, ? The number of elements of set B is 1 and the number of elements of its power, set is = 2 21, and P(C) { a, b}, {a}, {b}, φ}, ? The number of elements of set C is 2 and the number of elements of its power, set is = 4 22, So, if the number of elements is n of a set, the number of elements of its power set, will be 2n ., Activity : If G {1, 2, 3} , determine P(G) and show that the, number of elements of P(G) supports 2n ., Ordered pair :, Amena and Sumona of class VIII stood 1st and 2nd respectively in the merit list in the, final examination. According to merit they can be written (Amena, Sumona) as a, pair. In this way, the pair fixed is an ordered pair., Hence, in the pair of elements, the first and second places are fixed for the elements, of the pair and such expression of pair is called ordered pair., If the first element of an ordered pair is x and second element is y , the ordered pair, will be ( x, y) . The ordered pair ( x, y) and (a, b) will be equal or ( x, y) = (a, b) if`, x a and y b ., Example 11. If (2x y, 3) = (6, x y) , find ( x, y) ., Solution : iGven that,, , (2x y, 3) = (6, x y), , According to the definition of ordered pair, 2x y 6........(1), and x y 3.........(2), Adding equation (1) and (2), we get, 3x, utting the value of, P, , 9 or, y, , x in equation (1), we get, 6 y, , 3, , 6 or, y, , 0, , ? ( x, y) = (3, 0) ., Cartesian Product :, Wangsu decided to give layer of white or blue colour in room of his house at the, inner side and red or yellow or green colour at the outer side. The set of colour of, inner wall A w, { hite, blue}and set of colour of outer wall B r{ed, yellow,
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28, , Math, , green}. Wangsu can apply the colour of his room in the form of ordered pair as, (white, red), (white, yellow), (white, green), (blue, red), (blue, yellow), (blue, green)., The given ordered pair is written as., A u B ({white, red), (white, yellow), (white, green), (blue, red), (blue, yellow), (blue, green)}, This is the cartesian product set., In set builder method, A u B {( x, y); x A and y B}, Au B is read as A cross B ., Example 12. If P {1, 2, 3} , Q {3, 4} and R P Q , determine P u R and R u Q ., Solution : iGven that, P {1, 2, 3} , Q {3, 4}, and R P Q = {1, 2, 3} {3, 4} {3}, ? P u R = {1,2,3, } u{, 3}= {(1 , 3).(2, 3), (3, 3)}, and R u Q = {3} u {3, 4} {, ( 3, 3), (3, 4)}, , §x y · § x y·, , 1¸ ¨1, ¸ , find (x, y) ., ©2 3 ¹ © 3 2¹, , Activity : 1. If ¨, , 2. If P {1, 2, 3}, Q {3, 4} and R {x, y} , find (P Q) u R and (P Q) u Q ., Example 13. Find the set where 23 is reminders in each case when 311 and 419 are, divided by the natural numbers., Solution : The numbers when 311 and 419 are divided by rational numbers and 23 is, reminder, will be greater than 23 and will be common factors of 311 –23 = 288 and, 419 –23 = 396., Let, the set of factors of 288 greater than 23 is A and the set of factors of 396 is B ., Here,, 288 1 u 288 2 u 144 3 u 96 4 u 72 6 u 48 8 u 36 9 u 32 12 u 24 16 u 18, ? A {24, 32, 36, 48, 72, 96, 144, 288}, Again,, 396 1 u 396 2 u 198 3 u 132 4 u 99 6 u 66 9 u 44 11 u 36 12 u 33 18 u 22, ? B {33, 36, 44, 66, 99, 132, 198, 396}, ? A B {24, 32, 36, 48, 72, 96, 144, 288} {33, 36, 44, 66, 99, 132, 198, 396} = {36}, equired set is {36}, R, Example 14. If U {1, 2, 3, 4, 5, 6, 7, 8} ,, show, that,, (i), C {4, 5, 6, 7} ,, ( A B) C ( A C) (B C), Solution : (i), , A {1, 2, 6, 7} , B {2, 3, 5, 6}, and, ( A B)c Ac Bc, , and, (ii)
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Math, , 29, , In the figure, U by rectangle and the sets of A and B are denoted by two mutually, intersecting circle sets., set, Elements, A B, 1, 2, 3, 4, 5, 6, 7, , ( A B)c, 4, 8, , Ac, 3, 4, 5, 8, , 1, 4, 7, 8, Bc, A ∩ B, Ac Bc, 4, 8, ? ( A B)c Ac Bc, Solution : (ii) In figure, U by rectangle and sets of A, B, C are denoted by three, mutually intersecting circles., Observe ,, Set, Elements, A B, 2, 6, $, , 2, 4, 5, 6, 7, ( A B) C, , AC, 1, 2, 4, 5, 6, 7, % , BC, 2, 3, 4, 5, 6, 7, &, ( A C) (B C) 2, 4, 5, 6, 7, ? ( A B) C ( A C) (B C), Example 15. Among 100 students, 92 in Bangla, 80 in Math and 70 have passed in, both subjects in any exam. Express the information by Venn diagram and find how, many students failed in both subjects., Solution : In the Venn diagram, the rectangular region denotes set U of 100, students. and the set of passed students in Bangla and Math are denoted by B and, M . So, the Venn diagram is divided into four disjoint sets which are denoted by, P, Q, R, F ., Here, the set of passed students in both subjects Q B M whose numbers of, elements are 70 ., %, 0, P the set of passed students in Bangla only, whose number, of element = 92 70 18, 345, R the set of passed student in Math only, whose number of, ), elements = 80 70 10, P Q R B M , the set of passed students in one and both subjects, whose, number of elements = 18 10 70 98, F the set of students who failed in both subjects, whose number of elements =, 100 98 2, ? 2 students failed in both subjects.
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30, , Math, , Exercise 2.1, 1. Express the following sets in tabular method :, (a) {x N : x2 ! 9 and x3 130}, (b) {x Z : x2 ! 5 and x3 d 36}, (c) {x N : x, factors of 36 and multiple of 6}., (d) {x N : x3 25 and x4 264}, 2. Express the following sets in set builder method :, (b) {1, 2, 3, 4, 6, 9, 12, 18, 36}, (a) {3, 5, 7, 9, 11}, (d) {r4, r 5, r 6}, (c) {4, 8, 12, 16, 20, 24, 28, 32, 36, 40}, 3. If A {2, 3, 4} , B {1, 2, a} and C {2, a, b} , determine the sets given below:, (a) B \ C (b) A B, (c) A C, (d) A (B C) (e) A (B C), 4. If U {1, 2, 3, 4, 5, 6, 7} , A {1, 3, 5} , B {2, 4, 6} and C {3, 4, 5, 6, 7} , justify, the followings :, (i) ( A B)c Ac Bc, (ii) (B C)c Bc Cc, (iii ) ( A B) C ( A C ) ( B C ) (iv), ( A B) C ( A C ) ( B C ), 5. If Q {x, y} and R {m, n, l} , find P(Q) and P(R) ., 6. If A {a, b} , B {a, b, c} and C A B , show that the number of elements of, P(C) is 2n , where n is the number of element of C ., 7. (a) If ( x 1, y 2) ( y 2, 2 x 1) , find the value of x and y ., (b) If (ax cy, a2 c2) (0, ay cx) , find the value of ( x, y) ., (c) If (6x y, 13) (1, 3x 2 y) , find the value of ( x, y) ., 8. (a) If P {a} , Q {b, c} then, find P u Q and Q u P ., (b) If A {3, 4, 5} , B {4, 5, 6} and C {x, y} , find ( A B) u C ., (c) If P {3, 5, 7} , Q {5, 7} and R P \ Q , find (P Q) u R ., 9. If A and B are the sets of all factors of 35 and 45 respectively, find A B, and A B ., 10. Find the set of the number where 31 is the remainder in each case when 346 and, 556 are divided by natural numbers., 11. Out of 30 students of any class, 20 students like football and 15 students like, cricket. The number of students who like any one of the two is 10. Show with the, help of Venn diagram, the number of students who do not like two of the sports., 12. Out of 100 students in any exam, 65% in Bangla, 48% in both Bangla and, English have passed and 15% have failed in both subjects.
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31, , Math, , (a) Express the above information by Venn diagram along with brief description., (b) Find the numbers who have passed only in Bangla and English., (c) Find the union of two sets of the prime factors of the numbers who have, passed and failed in both subjects., Relation, We know, the capital of Bangladesh is Dhaka and that of India is Delhi and P, akistan, is Islamabad. Here, there is a relation of capital with the country. The relation is, country-capital relations. The above relation can be shown in set as follows:, Relation, Country, Bangladesh, India, aPkistan, , Capital, Dhaka, Delhi, Islamabad, , That is, country-capital relation = {(Bangla desh, Dhaka), (India, Delhi), (P, akistan,, Islamabad)}., If A and B are two sets, the non-zero subset of R of the aCrtesian product Au B of, the sets is called relation of B from A ., Here, R is a subset of Au B set, that is, R A u B ., Example 15. Suppose, A {3, 5} and B {2, 4}, ? A u B {3, 5} u {2, 4} {( 3, 2), (3, 4), (5, 2), (5, 4)}, ? R {( 3, 2), (3, 4), (5, 2), (5, 4)}, If the condition is x ! y , R {( 3, 2), (5, 2), (5, 4)}, and if the condition is x y , R {3, 4}, If an element of set A is x and that of the set B is y and ( x, y) R , we write, x R y and read as x is related to y . That the element x is R related to element y ., Again, if the relation of a set, from set A that is R A u A , R is called A related., So, if the relation is given between set A and B , non-zero subset of ordered pair, ( x, y) with y B related to x A , is a relation., Example 16. If P {2, 3, 4} , Q {4, 6} and y 2x is relation under consideration, between the elements of P and Q , find the relation., , P {2, 3, 4} and Q {4, 6}, According to the question, R {( x, y) : x P, y Q and y, Solution : iGven that,, , 2x}, , Here, P u Q {2, 3, 4} u {4, 6} {( 2, 4), (2, 6), (3,4)(3, 6), (4, 4), (4, 6)}, ? R {( 2, 4), (3, 6)}
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32, , Math, , eRquired relation is {( 2, 4), (3, 6)}, Example 17. If A {1, 2, 3} , B {0, 2, 4} and the relation x, consideration between elements of C and D , find the relation., Solution : iGven that, A {1, 2, 3} , B {0, 2, 4}, , y 1 is under, , According to the question, relation R {( x, y) : x A, y B and x, , y 1}, , Here, A u B {1, 2, 3} u {0, 2, 4}, = {( 1, 0), (1, 2), (1, 4), (2, 0), (2, 2), (2, 4), (3, 0), (3, 2), (3, 4)}, ? R {( 1, 2), (3, 4)}, Activity : If C {2,5,6}, D {4,5} and the relation x d y is under consideration, between elements of C and D , find the relation., Functions, Let us observe the relation between sets A and B below :, Here, When y x 2 ,, 1, y 3 for x 1, 2, y 4 for x 2, 3, y 5 for x 3, , 3, 4, 5, , That is, for each value of x , only one value of y is obtained and the relation, between x and y is made by y x 2 . Hence two variable x and y are so related, that for any value of x , only one value of y is obtained even y is called the, function of x . The function of x is generally expressed by y, f (x), g (x) , F (x) etc., Let, y x2 2x 3 is a function. Here, for any single value of x , only one value of, y is obtained. Here, both x and y are variables but the value of y depends on the, value of x . So, x is independent variable and y is dependent variable., Example 18. If f ( x), , x2 4x 3 , find f (1) ., , Solution : iGven that,, , f ( x), , x2 4x 3, , ? f (1) = (1)2 4(1) 3 1 4 3 8, Example 19. If g ( x), , x3 ax2 3x 6 , for what value of a will be g (2), 3, , 2, , Solution : iGven that, g ( x) x ax 3x 6, ? g (2) (2)3 a(2)2 3(2) 6, = 8 4a 6 6, = 8 4a = 4a 8, But g (2) 0, , 0?
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Math, , 33, , ? 4a 8 0, or, 4a 8, or, a 2, ? if a 2 , g (2) 0 ., Domain and Range, The first set of elements of the ordered pair of any relation is called its domain and, the set of second elements is called its range., Let R from set A to set B be a relation, that is, R A u B . The set of first, elements included in the ordered pair of R will be domain of R and the set of second, elements will be range of R . The domains of R is expressed as Dom R and range is, expressed as aRnge R ., Example 20. R, elation S {( 2, 1), (2, 2), (3, 2), (4, 5)} . Find the domain and range of, the relation., Solution : iGven that, S {( 2, 1), (2, 2), (3, 2), (4, 5)}, In the relation S , the first elements of ordered pair are 2, 2, 3, 4 and second elements, are 1, 2, 2, 5 ., ? Dom S {2, 3, 4} and aRnge S = {1, 2, 5}, Example 21. If A {0, 1, 2, 3} and R {( x, y) : x A, y A and y x 1} , express, R in tabular method and determine Dom R and aRnge R ., Solution : iGven that, A {0, 1, 2, 3} and R {( x, y) : x A, y A and y x 1}, From the stated conditions of R we get, y, Now, for each x A we find the value of y, x, y, , 0, 1, , 1, 2, , 2, 3, , x 1, , x 1., , 3, 4, , Since 4 A , (3, 4) R, ? R {( 0, 1), (1, 2), (2, 3)}, Dom R {0, 1, 2} and aRnge, , R {1, 2, 3}, , Activity :, 1. If S {( 3, 8), (2, 3), (1, 0), (0, 1), (1, 0), (2, 3)} , find domain and range of, , S., 2., , If S {x, y ) : x A, y A and y x 1} , where A, Dom. S and aRnge S ., , {3, 2, 1, 0} , find, , Graphs :, The diagrammatic view of function is called graphs. In order to make the idea of, function clear, the importance of graph is immense. French philosopher and
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34, , Math, , mathematician eRne Descartes (1596 1650) at first played a vital role in establishing, a relation between algebra and geometry. He introduced the modern way to coplanar, geometry by defining the position of point on a plane by two intersecting, perpendicular function. He defined the two intersecting perpendicular lines as axes, and called the point of intersection origin. On any plane, two intersecting, perpendicular straight lines XOX c and YOY c are, drawn. The position of any point on this plane can, be completely known by these lines. Each of, these straight lines are called axis. Horizontal line, erpendicular line YOY c, XOX c is called x a- xis, P, is called y a- xis and the point of intersection of, the two axes O is called origin., The number with proper signs of the, perpendicular distances of a point in the plane, from the two axis are called the C, oo- rdinates of, that point. Let P be any point on the plane between the two axes. PM and PN are, drawn perpendicular from the point P to XOX c and YOY c axes respectively. As a, result, PM ON is the perpendicular distance of the point P from YOY c and, PN OM is the perpendicular distance of P from the XOX c . If PM x and, PN y , the coordinates of the point P is ( x, y ) . Here, x is called abscissa or x, co-ordinate and y is called ordinate or y co-ordinate., In aCrtesian co-ordinate, the geometrical figure is shown easily. For this reason,, generally we put the independent value along the x a- xis and the dependent value, along the y a- xis. For some values of independent variables from the domain, we, find the similar values of dependent variables and form ordered pair to draw the, graphs of the function y f (x) . Then place the ordered pair under (x, y) and oj int the, obtained points in freehands which is the graph of the function y f (x) ., Example 22. Draw the graph of the function y 2 x ; where 3 d x d 3 ., Solution : In the domain 3 d x d 3 , for some, values of x , we determine some values of y and, form a table :, , x, y, , –3, , –2, , –1, , 0, , 1, , 2, , 3, , –6, , –4, , –2, , 0, , 2, , 4, , 6, , On the graph paper, taking the length of square as, unit, we identify points of the table on the place and, joint the points in free hand.
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35, , Math, , 3x 1, , find the value of, 3x 1, , Example 23. If f ( x), , 3x 1, 3x 1, , f ( x), , Solution : iGven, , §1·, f ¨ ¸ 1, © x¹ ., §1·, f ¨ ¸ 1, © x¹, , 1, 3, 3 1, 1, 3 x, x, x, ?, =, [multiplying the numerator and the, 1, 3, 3 x, 3 1, 1, x, x, denomination by x ], §1·, f ¨ ¸ 1, 3 x 3 x, x, , [By componendo –Dividendo], or, © ¹, 3 x 3 x, §1·, f ¨ ¸ 1, © x¹, 6 3, =, 2x x, 3, eRquire value is, x, §1·, y3 3y2 1, Example 24. If f ( y ), , show that f ¨¨ ¸¸ f (1 y ), y (1 y ), © y¹, §1·, f¨ ¸, © x¹, , 3, , §1·, ? f ¨¨ ¸¸, © y¹, , =, , 2, , §1·, §1·, ¨¨ ¸¸ 3¨¨ ¸¸ 1, y, © ¹, © y¹, 1§ 1·, ¨1 ¸, y ¨© y ¸¹, , 1 3y y3, y2, u, 3, y, y 1, , again, f (1 y ), =, , y3 3y2 1, y (1 y ), , f ( y), , Solution : iGven,, , 1 3y y3, y3, y 1, y2, , 1 3y y3, y ( y 1), , (1 y ) 3 3(1 y ) 2 1, (1 y ){1 (1 y )}, , 1 3 y 3 y 2 y 3 3(1 2 y y 2 ) 1, (1 y )(1 1 y )
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36, , Math, , =, , 1 3y 3y2 y3 3 6 y 3y2 1, y (1 y ), , =, , 1 3y y3, y (1 y ), , =, , 1 3y y3, y ( y 1), , §1·, ? f ¨¨ ¸¸, © y¹, , (1 3 y y 3 ), y ( y 1), , f (1 y ) ., Exercise 2⋅2, , 1., , Which one is the set of factors of 4 ?, (a) {8,16,24...}, (b) (1,2,3,4,8} (c) 2{, 6, 8}, , 2., , If a relation of set B from set C is R , which one of the following is right ?, (a) R C, (b) R B, (c) R C u B, (d) C u B R, If A {6, 7, 8, 9, 10, 11, 12, 13} , answer the following questions :, , 3., , (d) 1{, 2}, , (i ) Which one is builder method of set A ?, , (a) {x N : 6 x 13}, , (b) {x N : 6 d x 13}, , (c) {x N : 6 d x d 13} (d) {x N : 6 x d 13}, (ii ) Which one is the set of prime numbers ?, , (a) {6, 8, 10, 12} (b) {7, 9, 11, 13}, , (c) {7, 11, 13} (d) A {9, 12}, , (iii ) Which is the set of multiple of 3 ?, , (b) {6, 11}, , (a) {6, 9}, , (c) {9, 12}, , (d) {6, 9, 12}, , (iv) Which is the set of factor of greater even number ?, 4., 5., 6., 7., 8., , (a) {1, 13}, (b) {1, 2, 3, 6}, (c) {1, 3, 9}, (d) {1, 2, 3, 4, 6, 12}, If A {3, 4}, B {2, 4} , find the relation between elements of A and B, considering x ! y ., If C {2, 5}, D {4, 6} and find the relation between element of C and D, considering relation x 1 y ., §1·, If f ( x) x 4 5 x 3 , the find value of f (1), f (2) and f ¨ ¸ ., ©2¹, If f ( y ), If f ( x), , y 3 ky 3 4 y 8 , for which value of k will be f (2), 3, , 2, , x 6 x 11x 6 , for which value of, , x, , 0., , will be f ( x) 0 .
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37, , Math, , § 1 ·, , 9., , If f ( x), , f ¨ 2 ¸ 1, 2x 1, , find the value of © x ¹ ., 2x 1, § 1 ·, f ¨ 2 ¸ 1, ©x ¹, , 1 x2 x4, § 1 ·, , show that, g ¨ 2 ¸ g ( x 2 ), x2, ©x ¹, 11. Find the domain and range of the following relations :, (b) S {( 2, 4), (1, 1), (0, 0), (1, 1), (2, 4)}, (a) R {( 2, 1), (2, 2), (2, 3)}, , 10. If g ( x), , (c) F, , § 1 ·, §5 · §5, ·½, ®¨ , 0 ¸, (1, 1), (1, 1), ¨ , 2 ¸, ¨ , 2 ¸¾, ©2 ¹ ©2, ¹¿, ¯© 2 ¹, , 12. Express the relations in tabular method and find domain and range of following, relations :, (a) R {( x, y ) : x A, y A and x y 1} where A {2, 1, 0, 1, 2}, (b) F {( x, y ) : x C , y C and x 2 y} where C {1, 0, 1, 1, 3}, §1 5·, 13. P, lot the points (3, 2), (0, 5), ¨ , ¸ on graph paper., ©2 6¹, 14. P, lot the three points (1, 2), (1, 1), (11, 7) on graph paper and show the three, points are on the same straight line., 15. nUiversal set U {x : x N and x is an odd number}, A {x N : 2 d x d 7 }, B {x N : 3 x 6 }, (a), (b), (c), , C {x N : x 2 ! 5 and x 3 130 }, Express A in tabular method., Find Ac and C B ., Find B u C and P ( A C ) .
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Chapter Three, , Algebraic Expressions, Algebraic formulae are used to solve many algebraic problems. Moreover, many, algebraic expressions are presented by resolving them into factors. That is why the, problem solved by algebraic formulae and the contents of resolving expressions into, factors by making suitable for the students have been presented in this chapter., Moreover, different types of mathematical problems can be solved by resolving into, factors with the help of algebraic formulae. In the previous class, algebraic formulae, and their related corollaries have been discussed elaborately. In this chapter, those, are reiterated and some of their applications are presented through examples., Besides, extension of the formulae of square and cube, resolution into factors using, remainder theorem and formation of algebraic formulae and their applications in, solving practical problems have been discussed here in detail., At the end of the chapter, the students will be able to –, ¾ Expand the formulae of square and cube by applying algebraic formulae, ¾ Explain the remainder theorem and resolve into factors by applying the, theorem, ¾ Form algebraic formulae for solving real life problems and solve the problems, by applying the formulae., 3⋅1 Algebraic Expressions, Meaningful organization of operational signs and numerical letter symbols is called, algebraic expression. Such as, 2a 3b 4c is an algebraic expression. In algebraic, expression, different types of information are expressed through the letters, a, b, c, p, q, r, m, n, x, y, z, ..... etc. These alphabet are used to solve different types of, problems related to algebraic expressions. In arithmetic, only positive numbers are, used, where as, in algebra, both positive and negative numbers including zero are, used. Algebra is the generalization form of arithmetic. The numbers used in, algebraic expressions are constants, their values are fixed., The letter symbols used in algebraic expressions are variables, their values are not, fixed, they can be of any value., 3⋅2 Algebraic Formulae, Any general rule or resolution expressed by algebraic symbols is called Algebraic, Formula. In class VII and VIII, algebraic formulae and related corollaries have been, discussed. In this chapter, some applications are presented on the basis of that, discussion.
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40, , Math, , 2, , or,, , ( a b), ( a b), , 4, 4, , ab, , 2, , 2, , 2, , §a b·, §a b·, ¸¸, ¸¸ ¨¨, ¨¨, © 2 ¹, © 2 ¹, Remark : rPoduct of any two quantities can be expressed as the difference of two, squares by applying the corollary 6., Hence, ab, , Formula 3. a2 b2 (a b)(a b), That is, the difference of the squares of two expressions = sum of two expressions u, difference of two expressions., Formula 4. ( x a)(x b) x2 (a b) x ab, That is, ( x a)( x b) x2 (algebraic sum of a and b ) x (the product of a and b ), Extension of Formula for Square, There are three terms in the expression a b c . It can be considered the sum of, two terms (a b) and c ., Therefore, by applying formula 1, the square of the expression a b c is,, (a b c)2, , {( a b) c}2, , = (a b)2 2(a b)c c2, = a2 2ab b2 2ac 2bc c2, = a2 b2 c2 2ab 2bc 2ac ., Formula 5. (a b c)2 a2 b2 c2 2ab 2bc 2ac ., Corollary 7. a2 b2 c2, , (a b c)2 2(ab bc ac), , Corollary 8. 2(ab bc ac) (a b c)2 (a2 b2 c2 ), Observe : Applying formula 5, we get,, (i) (a b c)2 {a b (c)}2, = a2 b2 (c)2 2ab 2b(c) 2a(c), = a2 b2 c2 2ab 2bc 2ac, (ii) (a b c), , 2, , {a (b) c}2, , = a2 (b)2 c2 2a(b) 2(b)c 2ac, = a2 b2 c2 2ab 2bc 2ac, (iii) (a b c) {a (b) (c)}2, 2, , = a2 (b)2 (c)2 2a(b) 2(b)(c) 2a(c)
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Math, , 41, , = a2 b2 c2 2ab 2bc 2ac, Example 1. What is the square of (4x 5 y) ?, Solution : (4 x 5 y ) 2, , (4 x) 2 2 u (4 x) u (5 y ) (5 y ) 2, , = 16 x 2 40 xy 25 y 2, Example 2. What is the square of (3a 7b) ?, Solution : (3a 7b) 2, , (3a ) 2 2 u (3a ) u (7b) (7b) 2, , = 9a 2 42ab 49b 2, Example 3. Find the square of 996 by applying the formula of square., Solution : (996)2 (1000 4)2, = (1000) 2 2 u1000 u 4 (4) 2, = 1000000 8000 16, = 1000016 8000, = 992016, Example 4. What is the square of a b c d ?, Solution : (a b c d ) 2 {( a b) (c d )}2, = (a b) 2 2(a b)(c d ) (c d ) 2, = a 2 2ab b 2 2(ac ad bc bd ) c 2 2cd d 2, = a 2 2ab b 2 2ac 2ad 2bc 2bd c 2 2cd d 2, = a 2 b 2 c 2 d 2 2ab 2ac 2ad 2bc 2bd 2cd, Activity : Find the square with the help of the formulae :, 1. 3 xy 2ax, 2. 4 x 3 y, 3. x 5 y 2 z, Example 5. Simplify :, (5 x 7 y 3 z ) 2 2(7 x 7 y 3 z )(5 x 7 y 3 z ) (7 x 7 y 3 z ) 2, Solution : Let, 5 x 7 y 3 z a and 7 x 7 y 3 z b, ? iGven expression =, , a 2 2.b.a b 2, = a 2 2ab b 2, = ( a b) 2, , = {( 5 x 7 y 3 z ) (7 x 7 y 3 z )}2, [substituting the values of a and b ], = (5 x 7 y 3 z 7 x 7 y 3 z ) 2, = (12 x) 2, = 144x 2
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42, , Math, , Example 6. If x y, , 2 and xy, , 24 , what is the value of x y ?, , Solution : ( x y ) 2 ( x y ) 2 4 xy, ? x y r 100 r10, , (2) 2 4 u 24, , 4 96 100, , Example 7. If a 4 a 2b 2 b 4 3 and a 2 ab b 2 3 , what is the value of, a2 b2 ?, Solution : a 4 a 2 b 2 b 4 (a 2 ) 2 2a 2 b 2 (b 2 ) 2 a 2 b 2, = (a 2 b 2 ) 2 (ab) 2, = (a 2 b 2 ab)(a 2 b 2 ab), = (a 2 ab b 2 )(a 2 ab b 2 ), 2, 2, ? 3 3(a ab b ) [substituting the values], 3, or, a 2 ab b 2, 1, 3, Now adding, a 2 ab b 2 3 and a 2 ab b 2 1 we get, 2(a 2 b 2 ) 4, 4, or, a 2 b 2, 2, 2, ? a2 b2 2, Example 8. P, rove that, (a b) 4 (a b) 4, Solution : (a b) 4 (a b) 4, , 8ab(a 2 b 2 ), , {( a b) 2 }2 {( a b) 2 }2, , = {( a b) 2 (a b) 2 }{( a b) 2 (a b) 2 }, = 2(a 2 b 2 ) u 4ab, [ (a b) 2 (a b) 2, 2, , 2(a 2 b 2 ) and (a b) 2 (a b) 2, , 4ab], , 2, , = 8ab(a b ), ? ( a b) 4 ( a b) 4, , 8ab(a 2 b 2 ), , Example 9. If a b c 15 and a 2 b 2 c 2 83 , what is the value of, ab bc ac ?, Alternative method,, Soluton :, We know ,, Here, 2(ab bc ac), (a b c) 2 (a 2 b 2 c 2 ) 2(ab bc ac), 2, 2, 2, 2, (a b c) (a b c ), or, (15) 2 83 2(ab bc ac), = (15) 2 83, = 225 83, = 142, 142, ? ab bc ac, 71, 2, , or, 225 83 2(ab bc ac), or, 2(ab bc ac) 142, 142, ? ab bc ac, 71, 2
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43, , Math, , Example 10. If a b c 2 and ab bc ac 1 , what is the value of, (a b) 2 (b c) 2 (c a ) 2 ?, Solution : (a b) 2 (b c) 2 (c a ) 2, = a 2 2ab b 2 b 2 2bc c 2 c 2 2ca a 2, = (a 2 b 2 c 2 2ab 2bc 2ca) (a 2 b 2 c 2 ), = (a b c) 2 {( a b c) 2 2(ab bc ac)}, = (2) 2 (2) 2 2 u 1, = 442 82 6, Example 11. Express (2 x 3 y )(4 x 5 y ) as the difference of two squares., Solution : Let, 2 x 3 y a and 4 x 5 y b, 2, , ? iGven expression =, , §ab· §ab·, ¸ ¨, ¨, ¸, © 2 ¹ © 2 ¹, , ab, 2, , 2, , 2, , = §¨ 2 x 3 y 4 x 5 y ·¸ §¨ 2 x 3 y 4 x 5 y ·¸ [substituting the values of a and b ], 2, , ©, , 2, , ©, , ¹, , 2, , 6x 2 y · § 8 y 2x ·, = §¨, ¸, ¸ ¨, © 2 ¹ © 2 ¹, 2, , = §¨ 2(3x y ) ·¸ §¨ 2(4 y x) ·¸, ©, , 2, , ¹, , ©, , 2, , = (3 x y ) 2 (4 y x) 2, ? (2 x 3 y )(4 x 5 y ), , ¹, , 2, , 2, , ¹, , (3 x y ) 2 (4 y x) 2, , Activity : 1. Simplify : (4 x 3 y ) 2 2(4 x 3 y )(4 x 3 y ) (4 x 3 y ) 2, 2. If x y z 12 and x 2 y 2 z 2 50 , find the, value of ( x y ) 2 ( y z ) 2 ( z x) 2 ., Exercise 3⋅1, 1., , Find the square with the help of the formulae :, 1, 2, (e) 4 y 5 x, (a) 2a 3b (b) 2ab 3bc (c) x 2 2 (d) a , a, y, , (f) ab c, , (g) 5 x 2 y (h) x 2 y 4 z (i) 3 p 4q 5r (j) 3b 5c 2a (k) ax by cz, (l) a b c d (m) 2a 3 x 2 y 5 z (n) 101, (o) 997 (p) 1007, 2. Simplify :, (a) (2a 7) 2 2(2a 7)(2a 7) (2a 7) 2
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44, , Math, , (b) (3 x 2 y ) 2 2(3 x 2 y )(3 x 2 y ) (3 x 2 y ) 2, (c) (7 p 3r 5 x) 2 2(7 p 3r 5 x)(8 p 4r 5 x) (8 p 4r 5 x) 2, (d) (2m 3n p) 2 (2m 3n p) 2 2(2m 3n p)(2m 3n p), (e) 6 35 u 6 35 2 u 6 35 u 3 65 3 65 u 3 65, (f) 5874 u 5874 3774 u 3774 7548 u 5874, 7529 u 7529 7519 u 7519, (g), 7529 7519, 2345 u 2345 759 u 759, (h), 2345 759, 3. If a b 4 and ab 60 , what is the value of a b ?, 4. If a b 7 and ab 12 , what is the value of a b ?, 5. If a b 9m and ab 18m 2 , what is the value of a b ?, 6. If x y 2 and xy 63 , what is the value of x 2 y 2 ?, 1, 1, 7. If x , 4 , prove that, x 4 4 322 ., x, x, 2, 1, 8. If 2 x , 3 , what is the value of x 2 2 ?, x, x, 1, 1, 1, 9. If a , 2 , show that, a 2 2 a 4 4 ., a, a, a, 7 and a b, 5 , prove that, 8ab(a 2 b 2 ) 24 ., 10. If a b, 11. If a b c 9 and ab bc ca 31 , find the value of a 2 b 2 c 2 ., 12. If a 2 b 2 c 2 9 and ab bc ca 8 , what is the value of (a b c) 2 ?, 13. If, a b c 6 and a 2 b 2 c 2 14 , find the value of, (a b) 2 (b c) 2 (c a ) 2 ., 14. If x y z 10 and xy yz zx 31 , what is the value of, ( x y ) 2 ( y z ) 2 ( z x) 2 ?, ,, of, 15. If, x 3, y 4 and z 5 find the value, 2, 2, 2, 9 x 16 y 4 z 24 xy 16 yz 12 zx ., °§ x y · 2 § x y · 2 ½°, 16. rPove that, ®¨, ¸ ¨, ¸ ¾, °̄© 2 ¹ © 2 ¹ °¿, , 2, , 2, , 2, , § x2 y2 · § x2 y2 ·, ¨¨, ¸¸ ¨¨, ¸¸ ., © 2 ¹ © 2 ¹, 17. Express (a 2b)(3a 2c) as the difference of two squares., 18. Express ( x 7)( x 9) as the difference of two squares., , 19. Express x 2 10 x 24 as the difference of two squares.
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45, , Math, , 20. If a 4 a 2b 2 b 4 8 and a 2 ab b 2, 3⋅3 Formulae of Cubes, Formula 6. (a b) 3 a 3 3a 2 b 3ab 2 b 3, , 4 , find the value of (i ) a 2 b 2 , (ii ) ab ., , = a 3 b 3 3ab(a b), Proof : (a b), , 3, , (a b)(a b) 2, = (a b)(a 2 2ab b 2 ), = a(a 2 2ab b 2 ) b(a 2 2ab b 2 ), = a 3 2a 2 b ab 2 a 2 b 2ab 2 b 3, = a 3 3a 2b 3ab 2 b 3, = a 3 b 3 3ab(a b), , Corollary 9. a 3 b 3, , (a b) 3 3ab(a b), , Formula 7. (a b) 3, , a 3 3a 2 b 3ab 2 b 3, = a 3 b 3 3ab(a b), , Proof : (a b) 3, , (a b)(a b) 2, = (a b)(a 2 2ab b 2 ), = a (a 2 2ab b 2 ) b(a 2 2ab b 2 ), , = a 3 2a 2b ab 2 a 2b 2ab 2 b 3, = a 3 3a 2 b 3ab 2 b 3, = a 3 b 3 3ab(a b), Observe : Substituting b instead of b in formula 6, we get formula 7 :, {a (b)}3 a 3 (b) 3 3a(b){a (b)}, That is, (a b) 3, , a 3 b 3 3ab(a b), , Corollary 10. a 3 b 3, Formula 8. a 3 b 3, Proof : a 3 b 3, , (a b) 3 3ab(a b), (a b)(a 2 ab b 2 ), , (a b) 3 3ab(a b), = (a b){( a b) 2 3ab}, = (a b)(a 2 2ab b 2 3ab), = (a b)(a 2 ab b 2 ), , Formula 9. a 3 b 3, Proof : a 3 b 3, , (a b)(a 2 ab b 2 ), , (a b) 3 3ab(a b), = (a b){( a b) 2 3ab}
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46, , Math, , = (a b)(a 2 2ab b 2 3ab), = (a b)(a 2 ab b 2 ), Example 12. Find the cube of 2 x 3 y ., Solution : (2 x 3 y ) 3, , (2 x) 3 3(2 x) 2 3 y 3 2 x(3 y ) 2 (3 y ) 3, = 8 x 3 3 4 x 2 3 y 3 2 x 9 y 2 27 y 3, , = 8 x 3 36 x 2 y 54 xy 2 27 y 3, Example 13. Find the cube of 2 x y ., Solution : (2 x y ) 3, , (2 x) 3 3 (2 x) 2 y 3 2 x y 2 y 3, = 8 x 3 3 4 x 2 y 6 xy 2 y 3, = 8 x 3 12 x 2 y 6 xy 2 y 3, , Activity : Find the cube with the help of the formulae., 2. 3 x 4 y, 3. 397, 1. 3 x 2 y, Example 14. If x 37 , what is the value of 8 x 3 72 x 2 216 x 216 ?, Solution : 8 x 3 72 x 2 216 x 216, = (2 x) 3 3 (2 x) 2 6 3 2 x.(6) 2 (6) 3, = ( 2 x 6) 3, = (2 u 37 6) 3 [substituting the values], = (74 6) 3, = (80) 3, = 512000, Example 15. If x y, , 8 and xy, , 5 , what is the value of x 3 y 3 8( x y ) 2 ?, , Solution : x 3 y 3 8( x y ) 2, = ( x y ) 3 3 xy ( x y ) 8{( x y ) 2 4 xy}, = (8) 3 3 u 5 u 8 8(8 2 4 u 5), 3, , = 8 15 u 8 8(64 20), = 83 15 u 8 8 u 84, = 8(8 2 15 84), = 8(64 15 84), = 8 u 163, = 1304, , [substituting the values]
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47, , Math, , Example 16. If a 2 3a 1 0 , what is the value of a 3 , a 2 3a 1 0, , Solution : iGven that,, , a2 1, a, a2 1, 1, or, a , or,, , 3, a, a a, 1, 3, ? iGven expression = a 3, a, or, a 2 1, , 1, ?, a3, , 3a, , or,, , 3, 3, , 3, , 1·, 1§, 1·, ¸ 3a ¨ a ¸, a¹, a©, a¹, 3, 1, = 3 3 3 [ a , 3], a, = 3 3 3 3, =0, Example 17. Simplify :, (a b)(a 2 ab b 2 ) (b c)(b 2 bc c 2 ) (c a )(c 2 ca a 2 ), §, = ¨a , ©, , Solution : (a b)(a 2 ab b 2 ) (b c)(b 2 bc c 2 ) (c a )(c 2 ca a 2 ), = a3 b3 b3 c3 c3 a3, =0, Example 18. If a, , 3 2 , prove that, a 3 , a, , Solution : iGven that,, ? 1, a, =, , 1, a3, , 18 3., , 3 2, , 1, 3 2, , 3 2, [multiplying numerator and denominator by, 3 2 3 2, , =, , 3 2, 2, , 3 , , =, 1, ? a, a, =, , 2, , =, , 2, , 3 2, 32, , 3 2, 3 2 , , 3 2, , 3 2 3 2, , 2 3, , 3 2 ]
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48, , Math, , Now, a 3 13, a, , 3, , 1·, 1§, 1·, §, ¨a ¸ 3 a ¨a ¸, a¹, a©, a¹, ©, 3, , = 2 3 3 2 3, = 23 , , [ a , , 1, a, , 2 3], , 3, , 3 3u 2 3, , = 83 3 6 3, = 24 3 6 3, = 18 3 (proved), Activity : 1. If x 2 , what is the value of 27 x 3 54 x 2 36 x 8 ?, 2. If a b 5 and ab 6 , find the value of a 3 b 3 4(a b) 2 ., 1, 5 3 , find the value of x 3 3 ., 3. If x, x, Exercise 3⋅2, 1., , 2., , Find the cube with the help of the formulae :, (b) 2 x 2 3 y 2 (c) 4a 5 x 2 (d) 7 m 2 2n (e) 403 (f) 998, (a) 2 x 5, (g) 2a b 3c (h) 2 x 3 y z, Simplify :, (a) (4a 3b) 3 3(4a 3b) 2 (2a 3b) 3(4a 3b)(2a 3b) 2 (2a 3b) 3, (b) (2 x y ) 3 3(2 x y ) 2 (2 x y ) 3(2 x y )(2 x y ) 2 (2 x y ) 3, (c) (7 x 3b) 3 (5 x 3b) 3 6 x(7 x 3b)(5 x 3b), (d) ( x 15) 3 (16 x) 3 3( x 15)(16 x), (e) (a b c) 3 (a b c) 3 6(b c){a 2 (b c) 2 }, (f) (m n) 6 (m n) 6 12mn(m 2 n 2 ) 2, (g) ( x y )( x 2 xy y 2 ) ( y z )( y 2 yz z 2 ) ( z x)( z 2 zx x 2 ), (h) (2 x 3 y 4 z ) 3 (2 x 3 y 4 z ) 3 12 x{4 x 2 (3 y 4 z ) 2 }, , 3., 4., 5., , If a b 5 and ab 36 , what is the value of a 3 b 3 ?, If a 3 b 3 513 and a b 3 , what is the value of ab ?, If x 19 and y 12 , find the value of 8 x 3 36 x 2 y 54 xy 2 27 y 3 ., , 6., 7., , If a 15 , what is the value of 8a 3 60a 2 150a 130 ?, If a 7 and b 5 , what is the value of
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Math, , 49, , (3a 5b) 3 (4b 2a) 3 3(a b)(3a 5b)(4b 2a) ?, 8., 9., , If a b, , m, a 2 b 2, , n and a 3 b 3, 3, , 3, , If` x y 1 , show that, x y xy, , p 3 , show that, m 3 2 p 3, ( x y), , 3mn ., , 2, , 10. If a b 3 and ab 2 , find the value of (a) a 2 ab b 2 and (b) a 3 b 3 ., 11. If a b 5 and ab 36 , find the value of (a) a 2 ab b 2 and (b) a 3 b 3 ., 1, 1, 12. If m , a , find the value of m 3 3 ., m, m, 1, 1, 13. If x , p , find the value of x 3 3 ., x, x, 1, 1, 3, 14. If a , 1 , show that, a 3 4., a, a, 15. If a b c 0 , show that,, (b c) 2 (c a ) 2 (a b) 2, (a) a 3 b 3 c 3 3abc, (b), , , 1., 3bc, 3ca, 3ab, 16. If p q r , show that, p 3 q 3 r 3 3 pqr, 2, x, , 1 ·, §, 3 , show that, 8¨ x 3 3 ¸ 63 ., x ¹, ©, a6 1, 6 5 , find the value of, 18. If a, ., a3, 1, 3 2., 19. If x 3 3 18 3 , prove that, x, x, 1, 20. If a 4 a 2 1 0 , prove that, a 3 3 0 ., a, 3⋅4 Resolution into Factors, If an expression is equal to the product of two or more expressions, each of the latter, expressions is called a factor of the former expression., After finding the possible factors of any algebraic expression and then expressing the, expression as the product of these factors are called factorization or resolution into, factors., The algebraic expressions may consist of one or more terms. So, the factors may also, contain one or more terms., Some process of resolving expressions into factors :, (a) If any polynomial expression has common factor in every term, at first, they are to be found out. For example,, (i ) 3a 2 b 6ab 2 12a 2 b 2 3ab(a 2b 4ab), , 17. If 2 x
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50, , Math, , (ii ) 2ab( x y ) 2bc( x y ) 3ca( x y ) ( x y )(2ab 2bc 3ca), (b) Expressing an expression in the form of a perfect square ;, Example 1. eRsolve into factors : 4 x 2 12 x 9 ., Solution : 4 x 2 12 x 9 (2 x) 2 2 u 2 x u 3 (3) 2, , = (2 x 3) 2, Example 2. R, esolve into factors :, , (2 x 3)(2 x 3), , 9 x 2 30 xy 25 y 2 ., , Solution : 9 x 2 30 xy 25 y 2, = (3 x) 2 2 u 3 x u 5 y (5 y ) 2, = (3 x 5 y ) 2 (3 x 5 y )(3 x 5 y ), (c) Expressing an expression as the difference of two squares and then applying, the formula a 2 b 2 (a b)(a b) :, Example 3. R, eslove into factors : a 2 1 2b b 2 ., Solution : a 2 1 2b b 2 a 2 (b 2 2b 1), = a 2 (b 1) 2 {a (b 1)}{ a (b 1)}, = (a b 1)(a b 1), Example 4. R, esolve into factors : a 4 64b 4 ., Solution : a 4 64b 4 (a 2 ) 2 (8b 2 ) 2, = (a2)2 2 u a2 u 8b2 (8b2)2 16a2b2, = (a2 8b2)2 (4ab)2, = (a2 8b2 4ab)(a2 8b2 4ab), = (a 2 4ab 8b 2 )(a 2 4ab 8b 2 ), Activity : R, esolve into factors :, 2, 1. abx acx3 adx4, 2. xa2 144xb2, (d) Using the formula x2 (a b) x ab, , 3. x2 2xy 4 y 4, , ( x a)(x b) :, , 2, , Example 5. R, esolve into factors : x 12x 35 ., Solution : x2 12x 35 x2 (5 7) x 5 u 7, = ( x 5)(x 7), In this method, a polynomial of the form x2 px q can be factorized, if two, integers a and b can be found so that, it is a b p and ab q . For this, two, factors of q with their signs are to be taken whose algebraic sum is p . If q ! 0 , a, and b will be of same signs and if q 0 , a and b will be of opposite signs.
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51, , Math, , Example 6. R, esolve into factors : x2 5x 6 ., Solution : x2 5x 6 x2 (2 3) x (2)(3), = ( x 2)(x 3), Example 7. R, esolve into factors :, Solution : x 2 2 x 35, , x 2 2 x 35 ., , = x2 (7 5) x (7)(5), = ( x 7)( x 5), Example 8. R, esolve into factors :, , x2 x 20 ., , Solution : x2 x 20, = x 2 (5 4) x (5)(4), = ( x 5)(x 4), (e) By middle term break-up method of polynomial of the form of ax2 bx c :, If ax2 bx c (rx p)(sx q), ax 2 bx c rsx 2 (rq sp) x pq, That is, a rs, b rq sp and c pq ., Hence, ac rspq (rq)( sp) and b rq sp, Therefore, to determine factors of the polynomial ax2 bx c , ac that is, the, product of the coefficient of x2 and the term free from x are to be expressed into two, such factors whose algebraic sum is equal to b , the coefficient of x ., Example 9. R, esolve into factors : 12x2 35x 18 ., Solution : 12x2 35x 18, Here, 12 u 18 216 27 u 8 and 27 8 35, ? 12 x 2 35 x 18 12 x 2 27 x 8 x 18, = 3x(4x 9) 2(4x 9), = (4x 9)(3x 2), Example 10. R, esolve into factors : 3x2 x 14 ., Solution : 3x2 x 14 3x2 7 x 6x 14, = x(3x 7) 2(3x 7), = (3x 7)(x 2), Activity : R, esolve into factors :, 2, 2. 16x3 46x2 15x, 3. 12x2 17 x 6, 1. x x 56, (f) Expressing the expression in the form of perfect cubes :
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53, , Math, , 1, (3a 1)(9a2 3a 1), 27, Here, in the second solution, the factors involving the variables are with integral, coefficients. This result can be expressed as the first solution :, 1, (3a 1)(9a2 3a 1), 27, 1, 1, = (3a 1) u (9a 2 3a 1), 3, 9, 1· §, a 1·, §, = ¨ a ¸ ¨ a2 ¸, 3¹ ©, 3 9¹, ©, , =, , Example 13. R, esolve into factors :, , x3 6x2 y 11xy 2 6 y3 ., , Solution : x3 6x2 y 11xy 2 6 y3, = {x 3 3 x 2 2 y 3 x(2 y ) 2 (2 y ) 3 } xy 2 2 y 3, = ( x 2 y)3 y 2( x 2 y), = ( x 2 y ){( x 2 y ) 2 y 2 }, = ( x 2 y)(x 2 y y)(x 2 y y), = ( x 2 y)(x 3 y)(x y), = ( x y)(x 2 y)(x 3 y), Activity : R, esolve into factors:, 1 2 7, 1, 1, 2. a3 , 1. x x , 2, 6, 3, 8, , 3. 16 x 2 25 y 2 8 xz 10 yz, Exercise 3⋅3, , R, esolve into factors (1 to 43) :, 1., 3., , 2., a2 ab ac bc, ab a b 1, ( x y )( x y ) ( x y )( y z ) ( x y )( z x) 4. ab( x y) bc( x y), , 5., 7., , 9x2 24x 16, x4 6x2 y 2 y 4, , 6., 8., , 9., 11., , 4a2 12ab 9b2 4c2, a 2 6a 8 y 2 2 y, , 10. 9x4 45a2 x2 36a4, 12. 16x2 25 y 2 8xz 10 yz, , 13., , 2b 2 c 2 2c 2 a 2 2a 2 b 2 a 4 b 4 c 4 14. x2 13x 36, 16. a2 30a 216, x4 x2 20, , 15., , a 4 27 a 2 1, , (a2 b2 )( x2 y 2 ) 4abxy
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55, , Math, , Now, if we indicate the dividend by f (x) , the quotient by h(x) , the remainder by r, and the divisor by ( x a) , from the above formula, we get,, f ( x) ( x a) h( x) r .......... (i) this formula is true to all values of a ., Substituting x a in both sides of (i), we get,, f ( a ) ( a a ) h( a ) r 0 h( a ) r r, Hence, r f (a) ., Therefore, if f (x) is divided by ( x a) , the remainder is f (a) . This formula is, known as remainder theorem. That is, the remainder theorem gives the remainder, when a polynomial f (x) of positive degree is divided by ( x a) without performing, actual division. The degree of the divisor polynomial ( x a) is 1. If the divisor is a, factor of the dividend, the remainder will be zero and if it is not zero, the remainder, will be a number other than zero., Proposition : If the degree of f (x) is positive and a z 0 , f (x) is divided by, b, (ax b) , remainder is f §¨ ·¸ ., © a¹, Proof : Degree of the divisor ax b , (a z 0) is 1., Hence, we can write,, b·, §, f ( x) (ax b) h( x) r a¨ x ¸ h( x) r, a¹, ©, , ? f ( x) §¨ x b ·¸ a h( x) r, ©, , a¹, , Observe that, if f (x) is divided by §¨ x b ·¸ , quotient is a h(x) and remainder is r ., ©, , Here, divisor =, , a¹, , § b·, x ¨ ¸, © a¹, , Hence, according to remainder theorem, r, , § b·, f ¨ ¸, © a¹, , § b·, Therefore, if f (x) is divided by (ax b) , remainder is f ¨ ¸ ., © a¹, Corollary : ( x a) will be a factor of f (x) , if and only if f (a) 0 ., Proof : Let, f (a) 0, Therefore, according to remainder theorem, if f (x) is divided by ( x a) , the, remainder will be zero. That is, ( x a ) will be a factor of f (x) ., C, onversely, let, ( x a) is a factor of f (x) ., Therefore, f ( x) ( x a) h( x) , where h(x) is a polynomial., P, utting x a in both sides, we get,
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56, , Math, , f (a) (a a) h(a) 0, ? f (a) 0 ., Hence, any polynomial f (x) will be divisible by ( x a) , if and only if f (a) 0 ., This formula is known as factorisation theorem or factor theorem., Corollary : If a z 0 , the polynomial ax b will be a factor of any polynomial, § b·, f (x) , if and only if f ¨ ¸ 0 ., © a¹, b·, b·, §, §, Proof : a z 0, ax b = a¨ x ¸ will be a factor of f (x) , if and only if ¨ x ¸ =, a, a¹, ¹, ©, ©, § b·, § b·, x ¨ ¸ is a factor of f (x) , i.e. if and only if f ¨ ¸ 0 . This method of, a, © a¹, ©, ¹, determining the factors of polynomial with the help of the remainder theorem is also, called the Vanishing method., Example 1. eRsolve into factors : x3 x 6 ., Solution : Here, f ( x) x3 x 6 is a polynomial. The factors of the constant 6, are r 1, r 2, r 3 and r 6., P, utting, x 1, 1 , we see that the value of f (x) is not zero., But putting x 2 , we see that the value of f (x) is zero., , i.e., f (2) 2 3 2 6 8 2 6 0, Hence, x 2 is a factor of f (x), ? f (x) = x 3 x 6, = x 3 2 x 2 2 x 2 4 x 3x 6, = x2( x 2) 2x( x 2) 3( x 2), = ( x 2)(x2 2x 3), Example 2. eRsolve into factors : x3 3xy2 2 y3 ., Solution: Here, consider x a variable and y a constant., We consider the given expression a polynomial of x ., Let, f ( x) x3 3xy2 2 y3, Then, f ( y), , y3 3 y y 2 2 y3, , 3 y3 3 y3, , 0, , ? ( x y) is a factor of f (x) ., Now, x3 3xy2 2 y3, = x3 x2 y x2 y xy2 2xy2 2 y3, , Again let, g ( x), ? g ( y), , x 2 xy 2 y 2, , y2 y2 2y2, , 0
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58, , Math, , 3⋅6 Forming and applying algebraic formulae in solving real life problems, In our daily business we face the realistic problems in different time and in different, ways. These problems are described linguistically. In this section, we shall discuss, the formation of algebraic formulae and their applications in solving different, problems of real surroundings which are described linguistically. As a result of this, discussion the students on the one hand, will get the conception about the application, of mathematics in real surroundings, on the other hand, they will be eager to learn, mathematics for their understanding of the involvement of mathematics with their, surroundings., Methods of solving the problems :, (a), (b), , (c), (d), (e), , At first the problem will have to be observed carefully and to read attentively, and then to identify which are unknown and which are to be determined., One of the unknown quantities is to be denoted with any variable (say x )., Then realising the problem well, express other unknown quantities in terms of, the same variable (x) ., The problem will have to be splitted into small parts and express them by, algebraic expressions., sUing the given conditions, the small parts together are to be expressed by an equation., The value of the unknown quantity x is to be found by solving the equation., , Different formulae are used in solving the problems based on real life. The formulae, are mentioned below :, (1) Related to Payable or Attainable :, aPyable or attainable, A Tk. qn, where, q = amount of money payable or attainable per person,, , n = number of person., (2) Related to Time and Work :, If some persons perform a work,, Amount of work done, W qnx, where,, , q = portion of a work performed by every one in unit of time., , n = number of performing of work, x = total time of doing work, W portion of a work done by n persons in time x ., (3) Related to Time and Distance :, Distance at a definite time, d vt .
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59, , Math, , where,, (4), , v = speed per hour, t = total time., Related to pipe and water tank :, Amount of water in a tank at a definite time, Q(t) Qo r qt, , where,, , Qo = amount of stored water in a tank at the time of opening the pipe, , q = amount of water flowing in or flowing out by the pipe in a unit, time., t = time taken., Q(t ) amount of water in the tank in time t (+sign, at the time o f, flowing of water in and –sign, at the time of flowing of water, out are to be used)., (5) Related to percentage :, p br ,, , where,, , b = total quantity, , s, s%, 100, p = (rate of) percentage by parts = s% of b, , r = (rate of) percentage by fraction, , (6) Related to profit and loss :, S C(I r r) ;, in case of profit, S, in case of loss, S, , C(I r), , C(I r), , where,, , (7), , S (Tk.) = selling price, C (Tk.) = cost price, I = profit, r = rate of profit or loss, Related to investment and profit :, In the case of simple profit,, I = Pnr (taka), A P I P Pnr P(1 nr) (taka), , In the case of compound profit,, A P(1 r ) n, where,, , I = profit after time n, n = specific time, P = principal
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60, , Math, , r = profit of unit principal at unit time, A = principal with profit after time n ., Example 1. For a function of Annual Sports, members of an association made a, budget of Tk. 45000 and divided that every member would subscribe equally. But 5, members refused to subscribe. As a result, amount of subscription of each member, increased by Tk. 15 per head. How many members were in the association ?, Solution : Let the number of members of the association be x and amount of, subscription per head be Tk. q Then total amount of subscription A = Tk. qx ., Actually numbers of members were ( x 5) and amount of subscription per head, became Tk. (q 15) ., Then, total amount of subscription = Tk. ( x 5)(q 15), ( x 5)(q 15)..........(i), , By the question, qx, and qx, , 45,000..........(ii), , From equation (i), we get,, , qx, , ( x 5)(q 15), , qx 5q 15x 75, , or, qx, , or, 5q 15x 75, ? q, , 5(3x 15), , 3x 15...........(iii), , P, utting the value of q in equation (ii),, (3x 15) u x, , or, 3x2 15x, , 45000, , 45000, , 2, , or, x 5x 15000 [dividing both sides by 3], or, x2 5x 15000, , 0, , 2, , or, x 125x 120x 15000 0, or, x( x 125) 120( x 125) 0, or, ( x 125)(x 120), , 0, , ? x 125 0 or, x 120, If x 125 0 , x 125, Again, if ( x 120) 0 , x, , 0, 120, , Since the number of members i.e,, x cannot be negative, x z 120 ., ? x 125, Hence, number of members of the association is 125., Example 2. R, afiq can do a work in 10 days and shafiq can do that work in 15 days., In how many days do they together finish the work ?
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61, , Math, , Solution : Let, aRfiq and Shafiq together can finish the wok in, Let us make the following table :, Name, , d days., , Number of days, for doing the work, , Part of the work, done in 1 day, , Work done, in d days, , R, afiq, , 10, , 1, 10, , d, 10, , Shafiq, , 15, , 1, 15, , d, 15, , By the questions,, , d, d, , 10 15, , 1, , or, d §¨ 1 1 ·¸ 1, , © 10 15 ¹, 3 2·, §, or, d ¨, ¸ 1, © 30 ¹, 5d, or,, 1, 30, , or, d 30 6, 5, , ? They together can finish the work in 6 days., Example 3. A boatman can go x km in time t1 hour against the current. To cover, that distance along the current he takes t2 hour. How much is the speed of the boat, and the current., Solution : Let the speed of the boat in still water be u km per hour and that of the, current be v km per hour., Then, along the current, the effective speed of boat is (u v) km per hour and, against the current, the effective speed of boat is (u v) km per hour., According to the question,, distance traversed, x, ], uv, .......(i) [ speed, time, t2, x, .......(ii), t1, Adding equations (i) and (ii) we get,, §1 1·, x x, 2u, x¨¨ ¸¸, , t1 t 2, © t1 t 2 ¹, , and u v
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62, , Math, , x§1 1·, ¨ ¸, 2 ¨© t1 t 2 ¸¹, Subtracting equation (ii) from equation (i) we get,, §1 1·, 2v x¨¨ ¸¸, © t 2 t1 ¹, , or, u, , or, v, , x§1 1·, ¨ ¸, 2 ¨© t 2 t1 ¸¹, , ·, §, Hence, speed of current is x ¨¨ 1 1 ¸¸ km per hour, 2 © t2, , and speed of boat is, , t1 ¹, , x§ 1 1 ·, ¨ ¸ km per hour., 2 ¨© t1 t 2 ¸¹, , Example 4. A pipe can fill up an empty tank in 12 minutes. Another pipe flows out, 14 litre of water per minute. If the two pipes are opened together and the empty tank, is filled up in 96 minutes, how much water does the tank contain ?, Solution : Let x litre of water flows in per minute by the first pipe and the tank can, contain y litre of water., According to the question, the tank is filled up by first pipe in 12 minutes,, ? y 12x .......(i), Again, the empty tank is filled up by the two pipes together in 96 minutes., ? y 96x 96 u14 .........(ii), , y, 12, putting the value of x in equation (ii) , we get,, From equation (i) , we get, x, , y 96 u, , y, 96 u14, 12, , or, y 8 y 96 u14, or, 7 y 96 u14, 96 u14, 192, 7, Hence, total 192 litre of water is contained in the tank., , or, y
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63, , Math, , Activity :, 1. For a picnic, a bus was hired at Tk. 2400 and it was decided that every, passenger would have to give equal fare. But due to the absence of 10, passengers, fare per head was increased by Tk. 8. How many passengers did go, by the bus and how much money did each of the passengers give as fare?, 2. A and B together can do a work in p days. A alone can do that work in q, days. In how many days can B alone do the work ?, 3. A person rowing against the current can go 2 km per hour. If the speed of the, current is 3 km per hour, how much time will he take to cover 32 km, rowing, along the current ?, Example 5. rPice of a book is Tk. 24.00. This price is 80% of the actual price. The, oGvernment subsidize the due price. How much money does the G, ovt. subsidize for, each book ?, Solution : Market price = 80% of actual price, We know, p br, Here, p Tk. 24 and r 80%, ? 24 b u, , 80, 100, , 80, 100, , 6 5, 24u 100, or, b, ? b 30, 80, 4, 1, Hence, the actual price of the book is Tk. 30., ? amount of subsidized money = Tk. (30 24), = Tk. 6., ? subsidized money for each book is Tk. 6., Example 6. The loss is r% when n oranges are sold per taka. How many oranges, are to be sold per taka to make a profit of s% ?, Solution : If the cost price is Tk. 100, the selling price at the loss of r% is Tk., (100 r) ., If selling price is Tk. (100 r) , cost price is Tk. 100, ?, , ,,, , ,,, , ,,, , ,, Tk. 1, , ,,, , ,,, , ,, Tk. 100, , 100 r
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64, , Math, , Again, if cost price is Tk. 100, selling price at the profit of s% is Tk. (100 s), ?, , ,, ,,, , ,,, , ,, Tk. 1, , ,, ,, ,, ,,, , 100, ,, ,, ,,, 100 r, , ? ,, ,, ,, ,, Tk., , ,, ,,, ,, ,,, , ,,, , ,, Tk., , 100 s, 100, , 100 ·, § 100 s, ,, Tk. ¨, u, ¸, 100, 100, r¹, ©, , ,,, , 100 s, 100 r, 100 s, ? in Tk., , number of oranges is to be sold = n, 100 r, , = Tk., , ? in Tk. 1,, , ,,, , ,,, , ,,, , ,,, , ,,, , 100 r ·, ¸, © 100 s ¹, , ,, ,, n u §¨, , Hence, n(100 r) oranges are to be sold per taka., 100 s, , Example 7. What is the profit of Tk. 650 in 6 years at the rate of profit Tk. 7 percent, per annum ?, Solution : We know , I Pnr ., Here, P Tk. 650, n 6, s 7, ?, , r, , ?, , I, , s, 100, , 7, 100, 7, 650 u 6 u, 100, , 273, , Hence, profit is Tk. 273., Example 8. Find the compound principal and compound profit of Tk. 15000 in 3, years at the profit of 6 percent per annum., Solution : We know, C, compound profit., iGven,, , P Tk. 15000 , r, 3, , 6 ·, ? C 15000§¨1 , ¸, © 100 ¹, , = 15000§¨ 53 ·¸, © 50 ¹, , 3, , P(1 r)n , where C is the profit principal in the case of, , 6%, , 6, , n = 3 years, 100, 3, , 3·, §, 15000¨1 ¸, © 50 ¹
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Math, , 65, , 3, 15u 53 u 53 u 53 3 u 148877, =, 25, 125, 25, 446631, =, 17865 24, 25, , ? compound principal is Tk. 17865 24, ? compound profit = Tk. (17865 24 15000), = Tk. 2865 24, Activity :, 1. The loss is n% when 10 lemons are sold per taka. How many lemons are to be, sold per taka to make the profit of z% ?, 2. What will be the profit principal of Tk. 750 in 4 years at the rate of simple, 1, profit 6 percent per annum ?, 2, 3. Find the compound principal of Tk. 2000 in 3 years at the rate of compound, profit of Tk. 4 percent per annum., Exercise 3⋅5, 1., , Which one of the following is the factorized form of x2 7 x 6 ?, (a) ( x 2)(x 3), (b) ( x 1)( x 8), (c) ( x 1)( x 6), , 2., , 3., , 4., , 5., , (d) ( x 1)( x 6), , If f ( x) x2 4x 4 , which one of the following is the value of f (2) ?, (a) 4, (b) 2, (c) 1, (d) 0, If x y x y , which one of the following is the value of y ?, (b) 0, (a) 1, (c) 1, (d) 2, x2 3x3, Which one of the following is the lowest form of, ?, x 3x2, (b) x, (a) x2, (c) 1, (d) 0, 1 x2, ?, Which one of the following is the lowest form of, 1 x, (a) 1, (b) x, , 0DWK,;;)RUPD
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66, , 6., , Math, , Which one of the following is the value of, , (b) a2 b2, , (a) 2(a2 b2), , 7., , 8., , 9., , 1, {( a b)2 (a b)2} ?, 2, , (d) 4ab, 8, 3 , what is the value of x3 3 ?, x, (b) 8, (c) 9, , (c) 2ab, 2, If x , x, (a) 1, , (d) 16, , Which one of the following is the factorized form of p4 p2 1 ?, (a) ( p2 p 1)( p2 p 1), , (b) ( p 2 p 1)( p 2 p 1), , (c) ( p2 p 1)( p2 p 1), , (d) ( p2 p 1)( p2 p 1), , What are the factors of x 2 5 x 4 ?, (a) ( x 1), ( x 4), (b) ( x 1), ( x 4), (c) ( x 2), ( x 2), , (d) ( x 5), ( x 1), , 10. What is the value of ( x 7)( x 5) ?, (a) x2 12x 35, , (b) x2 12x 35, , (c) x2 12x 35, , (d) x2 12x 35, 2 9 u 2 9 1 1u 1 1, ?, 11. What is the value of, 2 9 1 1, (b) 1 9, (a) 1 8, (c) 2, (d) 4, 12. If x, , 2 3 , what is the value of x2 ?, (b) 7 4 3, 1, (d), 2 3, , (a) 1, (c) 2 3, 13. If f ( x), , x2 5x 6 and f ( x), , 0 , x = what ?, , (a) 2, 3, , (b) 5, 1, , (c) 2, 3, , (d) 1, 5, , 14., , x, 5, , x, , 6, , x2, , 6x, , 5x, , 30
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67, , Math, , Which one of the following is the total area of the table above ?, (a) x 2 5 x 30, , (b) x 2 x 30, , (c) x 2 6 x 30, (d) x 2 x 30, 15. A can do a work in x days, B can do that work in 3 x days. In the same time,, how many times does A of B work ?, 1, (a) 2 times, (b) 2 times, 2, (c) 3 times, (d) 4 times, 16. If a b c , a 2 2ab b 2 is expressed in terms of c , which one of the, following will be ?, (a) c 2, , (b) c 2, , 17. If x y 3, xy, , (d) ca, , (c) bc, 3, , 3, , 2 , what is the value of x y ?, , (a) 9, (c) 19, , (b) 18, (d) 27, , 18. Which one is the factorized form of 8 x 3 27 y 3 ?, (a) (2 x 3 y )(4 x 2 6 xy 9 y 2 ), , (b) (2 x 3 y )(4 x 2 6 xy 9 y 2 ), , (c) (2 x 3 y )(4 x 2 9 y 2 ), , (d) (2 x 3 y )(4 x 2 9 y 2 ), , 19. What is to be added to 9 x 2 16 y 2 , so that their sum will be a perfect square ?, (a) 6 xy, , (b) 12 xy, , (c) 24 xy, , (d) 144 xy, , 20. If x y, , 4 , which one of the following statements is correct ?, , (a) x 3 y 3 4 xy, , 64, , (b) x 3 y 3 12 xy 12, , (c) x 3 y 3 3 xy, , 64, , (d) x 3 y 3 12 xy, , 4, , 64, , 2, , 21. If x x 1 0 ,, 1, (1) x 2 2 = what ?, x, (b) 2, (a) 4, , (c) 1, 2, , 1·, §, (2) What is the value of ¨ x ¸ ?, x¹, ©, (a) 4, (b) 3, (c) 2, (d) 1, , (d) 0
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68, , Math, , (3) x 3 , , 22., , 23., 24., 25., , 26., 27., , 28., , 1, = what ?, x3, , (a) 3, (b) 2, (c) 1, (d) 0, A can do a work in p days and B can do it in 2 p days. They started to do the, work together and after some days A left the work unfinished. B completed the, rest of the work in r days. In how many days was the work finished ?, 50 persons can do a work in 12 days by working 8 hours a day. Working how, many hours per day can 60 persons finish the work in 16 days. ?, Mita can do a work in x days and R, ita can do that work in y days. In how, many days will they together complete the work ?, A bus was hired at Tk. 57000 to go for a picnic under the condition that every, passenger would bare equal fare. But due to the absence of 5 passengers, the, fare was increased by Tk. 3 per head. How many passengers availed the bus ?, A boatman can go d km in p hours against the current. He takes q hours to cover, that distance along the current. What is the speed of the current and the boat ?, A boatman plying by oar goes 15 km and returns from there in 4 hours. He, goes 5 km at a period of time along the current and goes 3 km at the same, period of time against the current. Find the speed of the oar and current., Two pipes are connected with a tank. The empty tank is filled up in t1 minutes, by the first pipe and it becomes empty in t 2 minutes by the second pipe. If the, two pipes are opened together, in how much time will the tank be filled up ?, (here t1 ! t 2 ), , 29. A tank is filled up in 12 minutes by a pipe. Another pipe flows out 15 litre of, water in 1 minute. When the tank remains empty, the two pipes are opened, togethr and the tank is filled up in 48 minutes. How much water does the tank, contain ?, 30. If a pen is sold at Tk. 11, there is a profit of 10%. What was the cost price of, the pen ?, 31. Due to the sale of a note-book at Tk. 36, there was a loss. If the note-boo k, would be sold at Tk. 72, there would be profit amounting twice the loss. What, was the cost price of the note-book ?, 32. Divide Tk. 260 among A, B and C in such a way that 2 times the share of A, 3, times the share of B and 4 times the share of C are equal to one another., 33. Due to the selling of a commodity at the loss of x% such price is obtained that, due to the selling at the profit of 3x% Tk. 18 x more is obtained. What was the, cost price of the commodity ?
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Math, , 69, , 34. If the simple profit of Tk. 300 in 4 years and the simple profit of Tk. 400 in 5, years together are Tk. 148, what is the percentage of profit ?, 35. If the difference of simple profit and compound profit of some principal in 2, years is Tk. 1 at the rate of profit 4%, what is the principal ?, 36. Some principal beomes Tk. 460 with simple profit in 3 years and Tk. 600 with, simple profit in 5 years. What is the rate of profit ?, 37. How much money will become Tk. 985 as profit principal in 13 years at the, rate of simple profit 5% per annum ?, 38. How much money will become Tk. 1248 as profit principal in 12 years at the, rate of profit 5% per annum ?, 39. Find the difference of simple profit and compound profit of Tk. 8000 in 3 years, at the rate of profit 5%., 40. The Value Added Tax (VAT ) of sweets is x% . If a trader sells sweets at Tk., P including VAT, how much VAT is he to pay ? If x 15, P 2300 , what is, the amount of VAT ?, 41. Sum of a number and its multiplicative inverse is 3., (a) Taking the number as the variable x , express the above information by an, equation., 1, (b) Find the value of x 3 3 ., x, 1, (c) P, rove that, x 5 5 123 ., x, 42. Each of the members of an association decided to subscribe 100 times the, number of members. But 7 members did not subscribe. As a result, amount of, subscription for each member was increased by Tk. 500 than the previous., (a) If the number of members is x and total amount of subscription is Tk. A ,, find the relation between them., (b) Find the number of members of the association and total amount of, subscription., 1, of total amount of subscription at the rate of simple profit 5% and rest of, (c), 4, the money at the rate of simple profit 4% were invested for 2 years. Find the, total profit.
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Chapter Four, , Exponents and Logarithms, Very large or very small numbers or expressions can easily be expressed in writing, them by exponents. As a result, calculations and solution of mathematical problems, become easier. Scientific or standard form of a number is expressed by exponents., Therefore, every student should have the knowledge about the idea of exponents and, its applications., Exponents beget logarithms. Multiplication and division of numbers or expressions, and exponent related calculations have become easier with the help of logarithms., U, se of logarithm in scientific calculatio ns was the only way before the practice of, using the calculator and computer at present. Still the use of logarithm is important, as the alternative of calculator and computer. In this chapter, exponents and, logarithms have been discussed in detail., At the end of the chapter, the students will be able to –, ¾ Explain the rational exponent, ¾ Explain and apply the positive integral exponents, zero and negative integral, exponents, ¾ Solve the problems by describing and applying the rules of exponents, ¾ Explain the n th root and rational fractional exponents and express the n th root, in terms of exponents, ¾ Explain the logarithms, ¾ P, rove and apply the formulae of logarithms, ¾ Explain the natural logarithm and common logarithm, ¾ Explain the scientific form of numbers, ¾ Explain the characteristic and mantissa of common logarithm and, ¾ Find common and natural logarithm by calculator., 4.1 Exponents or Indices, In class VI, we have got the idea of exponents and in class VII, we have known the, exponential rules for multiplication and division., Expression associated with exponent and base is called exponential expression., Activity : Fill in the blanks, Power or, Successive multiplication of the, Exponential, Base, exponent, same number or expression, expression, 3, 2u 2u 2, 23, 2, 3u 3u 3u 3, 3, auaua, a3, bubububub, 5
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71, , Math, , If a is any real number, successive multiplication of n times a ; that is,, n, a u a u a u ..... u a is written in the form a , where n is a positive integer., n, a u a u a u ..... u a ( n times a ) = a ., Here na o index or power, o base, Again, conversely, an = a u a u a u ..... u a ( n times a ) . Exponents may, not only be positive integer, it may also be negative integer or positive fraction or, negative fraction. That is, for a R (set of real numbers) and n Q (set of, rational numbers), an is defined. Besides, it may also be irrational exponent. But as, it is out of curriculum, it has not been discussed in this chapter., 4.2 Formulae for exponents, Let, a R; m, n N ., am u an, , Formula 1., , ° mn , when m!n, ®a 1 , when n!m, az0, °̄ a nm, , am, Formula 2., , am n, , an, , Fill in the blanks of the following table :, am , an, az0, , am u an, , m, a5 u a3, , n!m, 3, n 5, , a3 u a 5 =, , a5 3, , a3, a5, , 5, , a, an, , m, , (a u a u a u a u a) u (a u a u a), auauauaua uauaua, a8, , m, , m!n, 5, n 3, , a, a3, , auaua, auauauaua, 1, , a2, , 1, , a 5 3, , ? am u an = am n, and, , am, an, , mn, when m!n, ®a 1 , when n!m, ¯ a nm, , Formula 3. (ab)n an u bn, We observe, (5 u 2) 3 (5 u 2) u (5 u 2) u (5 u 2) [ a 3, 5u 2u5u 2u5u 2, (5 u 5 u 5) u (2 u 2 u 2), In general, (ab)n, , a u a u a; a 5 u 2], , 53 u 23, ab u ab u ab u ....... u ab [Successive multiplication of n times ab ]
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72, , Math, , (a u a u a u ....... u a) u (b u b u b u ........ u b), anbn, n, , an, , §a·, ¨ ¸, ©b¹, , Formula 4., , bn, , 3, , §5·, We observe, ¨ ¸, ©2¹, , 5 5 5, u u, 2 2 2, , n, , a u a u a u ...... u a, b u b u b u ...... u b, 1, (a z 0), , Formula 5. a0, , an, a, , Again,, , a, , n, , a, , n, , 5u5u5, 2u 2u 2, , 53, 23, , a, a a a, a, u u u ........ u [Successive multiplication of n times ], b, b b b, b, , §a·, In general, ¨ ¸, ©b¹, , We have,, , , (b z 0), , a nn, , an, bn, , a0, , n, , a u a u a u ..... u a ȏ Ǥ, a u a u a u ..... u a n a Ȑ, , 1, ?a, , 0, , 1., , Formula 6. a n, , a n u an, , a n, , 1u a n, , 1, , a n , (a z 0), , [multiplying both num. and denom. by an ], a n n, a, , ? an, R, emark : a n, , ao, an, , Formula 7., , am, , 1, , am, , n, , a0, an, , n, 1, , an, , a on, n, , 1, , an, , a n, , a mn, , a m u a m u a m u ......... u a m, , [successive multiplication of n times am ], , a m m m......... m [in the power, sum of n times of exponent m ]
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Math, , 73, , = a, m n, ? (a ), , num, , a mn, , a mn, , Solution : (a), , 52, = 5 2 3, 53, , 5, , 5, , § 2·, © 3¹, , 1, 5, , 55, , 0, , § 2·, § 2· § 2·, (b) ¨ ¸ u ¨ ¸ = ¨ ¸, 3, 3, © 3¹, © ¹ © ¹, , § 2·, ¨ 3¸, © ¹, , 1, , 5 4 u 8 u 16, , Example 2. Simplify :(a), , 3 2 n 4 2 n2, , (b), , 2n 2n 1, , 25 u 125, Solution :(a), , 54 u 8 u 16, , 54 u 23 u 2 4, , 5, , 5, , 2 u 125, 1, , 5 u2, n, , (b), , 3 2 4 2, , n2, , 2, , 2 u5, 5u 4, , 3, , 2n 2n 1, , 54 u 23 4, 3, , 5 u2, , 2 n 2 n 2 1, , (3 1) 2 n, 1 n, 2, 2, , Example 3. Show that (a p ) q r (a q ) r p (a r ) p q, Solution : (a ), , q r p, , (a ), , p ( q r ), , r, , 5, , 3, , 27, 2, , 54 3 u 27 5, , 5, , (a ), , 2 2n, 1 n, 2, 2, , 2 2 4., , 1, , p q, , q (r p), , a, a, a r ( p q ), pq pr, qr pq, a, a, a pr qr, pq pr qr pq pr qr, a, a 0 1., Activity : Fill in the blank boxes :, (i) 3 u 3 u 3 u 3 3, , 5, , u, , 3 2 n 2 2 n 2, 1, 2n 2n , 2, , 2, , 3 2n 2n, § 1· n, ¨1 ¸ 2, © 2¹, p q r, , 54, , 20, , 3 2 2 2 n2, n, , § 2·, © 3¹, , (b) ¨ ¸ u ¨ ¸, , 1, 51, , 5 1, , 5, , 5, , 52, 53, , Example 1. Find the values (a), , (ii) 5 u 53, , [ (a m ) n, , 55, , a mn ], , (iii) a 2 u a, , a 3, , (iv), , 4, 4, , (v) (5)0, 0DWK,;;)RUPD, , 1
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74, , Math, , 4.3 n th root, 1, , 1, , § 12 ·, ¨5 ¸, © ¹, , We notice, 5 2 u 5 2, 1, , 1, , 1 1, , 2, , Again, 5 2 u 5 2, 1, 2, , §, , 52, , 2u, , 1, 2, , 5, , 2, , ·, , ? ¨5 ¸, , ©, , 5, , 2, , 5., , ¹, , 1, Square (power 2) of 5 2, , 1, , 5 and square root (second root) of 5 = 5 2, , 1, , 5 2 is written as, , 5 in terms of the sign, 1, , 1, , 1, , § 13 ·, ¨5 ¸, ¨ ¸, © ¹, , Again, we notice 5 3 u 5 3 u 5 3, 1, , 1, , 1, , Again, 5 3 u 5 3 u 5 3, 1, 3, , §, , ·, , 5, , 3u, , 1, 3, , 5, , 3, , ? ¨¨ 5 ¸¸, , 5., , ¹, , ©, , 1 1 1, , 3 3, , 53, , of square root., , 3, , 1, , 1, , uCbe, 5 3 = 5 and cube root (third root) of 5 = 5 3 ., 1 (power 3) of, 3, 3, 5 is written as 5 in terms of the sign 3 of cube root., In the case of n th root,, 1, , 1, , 1, , 1, , 1, , a n u a n u a n u ....... u a n ȏ n a n Ȑ, n, , § 1n ·, ¨¨ a ¸¸ ., © ¹, 1, , 1, , 1, , 1, , Again, a n u a n u a n u ....... u a n, 1 1 1, 1, .........., n n, n, , an, a, , nu, , § 1·, ? ¨an ¸, ¨ ¸, © ¹, , 1, n, , ȏǡ n 1 Ȑ, n, , a, , n, , a., 1, , 1, , n th power of a n, , a and n th root of a, , an
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75, , Math, , § 1·, ¨an ¸, ¨ ¸, © ¹, , 1, an, , i.e. n th power of, , n, , 1, , a is written as n a ., 3, 1, Example 4. Simplify :(a) 7 4 7 2, 3, 4, , Solution : (a) 7 7, , 1, 2, , 7, , 1, , a n = n a . n th root of, , a and n th root of a = (a) n, , 3 1, , 4 2, , 7, , 3, , 3, 4, , (b) (16) y (16), , 1, 2, , § 2 ·4, (c) ¨¨10 3 ¸¸, ©, ¹, , 5, 4, , 3, 3, 4, , (b) (16) y (16), , 1, 2, , (16), 2, , §, , ·, , 3, 4, , 2 3, u, 4, , (c) ¨¨10 3 ¸¸, , 3 1, , 2, , (16) 4, , (16) 4, , 1, 2, , 1, , 1, , (16) 4, , (2 4 ) 4, , (2), , 4u, , 1, 4, , 2., , 1, , 10 3, , 10 2, , 10 ., , ¹, , ©, , Example 5. Simplify :(a) (12), Solution : (a) (12), , , , 1, 2, , , , 1, 2, , 2, , u 54, , (2 2 u 3), 1, (2 3), , 1, 2, , 1, 2, , u (33 u 2), , (b), , 1, 2, , 1, , 1, 1, 2 2, , ( 2 ) u3, 1, 3, , u (3 2), , 2, 31, u 1, 1, 2, 32, , 2, , § 1·, © 2¹, , § 1·, u ¨ ¸, © 2¹, , u (54) 3, , 1, 3, , 1, 3, , (3)3, , 1, , 1, , u 3 54, , (12), , 1, , 3, , 1, 2, , 1, , 1, 2, , 1, , 1, 3, , 3, , 2, , 1, , u (33 ) 3 2 3, 1, , 1, , 32, , 32, , 2, 3, , 1, 3, , 2, , 4, , 3, , 3, ., 4, , § 1 ·§ 1 ·, © 2 ¹© 2 ¹, , (b) (3)3 u ¨ ¸ = (3)(3)(3) u ¨ ¸¨ ¸, = 27 u, , 1, 4, , , , 27, 4, 5, , 5, , 3, , 1, , 24 22, § 2 ·2 § 2 · 2, Activity : Simplify : (i), (ii) ¨ ¸ u ¨ ¸, (iii) 8 4 y 8 2, 32, ©3¹ ©3¹, To be noticed :, 1. nUder the condition a ! 0, a z 1 , if a x a y , x y, Uder the condition a ! 0, b ! 0, x z 0 , if a x b x , a b, 2. n, Example 6. Solve : 4 x1 32 .
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76, , Math, , Solution :, 4 x 1, 2 x 1, or (2 ), , 32, , or, 2 2 x 2, , 32 ,, , ? 2x 2 5,, or, 2x 5 2 , or, 2x, 3, ?x, 2, , 25, , ȏ a x, , a y ǡ x, , y Ȑ, , 3, , 3, 2, , ? Solution is x, , Exercise 4.1, Simplify (1 – 10) :, 1., , 33 35, 36, , 2., , 5. (21 51)1, 8., , 53 8, 2 4 125, , 3., , 73 u 73, 3 u 34, , § a 2 b 1 ·, 7. ¨¨ 2 ¸¸, ©a 6¹, , 6. (2a 1 3b 1 ) 1, , x 1 y y 1 z z 1 x , ( x ! 0, y ! 0, z ! 0) 9., , 4., , 3 72 3 7, 7, , 2, , 3m1, 3m1, 2n 4 4 2n 1, 10. m m1 y m1 m1, n2, 2 y2, (2 ), (3 ), , Prove (11 – 18) :, 11., , 4n 1, 2n 1, , § a" ·, 13. ¨¨ m ¸¸, ©a ¹, , n, , § am ·, ¨¨ n ¸¸, ©a ¹, , 1, , ", , § an, ¨¨ ", ©a, , 1, , ·, ¸¸, ¹, , 2 p 1 32 p q 5 p q 6 p, 6 q 10 p 2 15 q, , 1, , 14., , a pq a qr a r p, u, u, a 2r a 2 p a 2q, , 1, , § xa ·, 16. ¨¨ b ¸¸, ©x ¹, , m, , 1, , § a · ab § b · bc § c · ca, 15. ¨ xb ¸ ¨ xc ¸ ¨ xa ¸, ¨x ¸ ¨x ¸ ¨x ¸, © ¹ © ¹ © ¹, , § xp ·, 17. ¨ q ¸, ¨x ¸, © ¹, , 12., , 2n 1, , p q r, , § xq ·, u ¨¨ r ¸¸, ©x ¹, , qr p, , § xr ·, u ¨¨ p ¸¸, ©x ¹, , 18. If a x b, b y c and c z, Solve (19 – 22) :, , a b, , bc, , § xb ·, ¨¨ c ¸¸, ©x ¹, , 1, 50, 1, , § xc ·, ¨¨ a ¸¸, ©x ¹, , ca, , 1, , r p q, , 1, , a , show that xyz 1, x 1, , 2 x 1, , 3, 20. 22x 1 128, 21. 3, 22. 2 x 21x 3, 19. 4 x 8, 3, 4.4 Logarithm, Logarithm is used to find the values of exponential expressions. Logarithm is written, roduct, quotient, etc. of large numbers or quantities can easily be, in brief as ‘ Log ’. P, determined by the help of log.
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77, , Math, , We know, 23 8 ; this mathematical statement is written in terms of log as, log 2 8 3 . Again, conversely, if log2 8 3 , it can be written in terms of exponents, as 23, , 8 . That is, if 23 8 , then log 2 8 3 and conversely, if log2 8 3 , then, 1, 1 1, 23 8 . Similarly, 2 3, can be written in terms of log as log 2, 3 ., 3, 8, 2, 8, If a x N , ( a ! 0, a z 1 ), x loga N is defined as a based log N ., To be noticed : Whatever may be the values of x , positive or negative, a x is, always positive. So, only the log of positive numbers has values which are real ; log, of zero or negative numbers have no real value., Activity-1 : Express in terms of Activity-2 : Fill in the blanks :, log :, in terms of exponent, in terms of log, (i) 10 2 100, (ii) 32, (iii) 2, (iv), , , , 1, 2, , 1, 9, , 100, , 1, , 1, 2, , e0, , ....., , loge 1 ......, , a0, , 1, , ...... = ......, , 24, , 101 10, 1, , e ..., ....... = ......, Formulae of Logarithms :, Let, a ! 0, a z 1; b ! 0, b z 1 and M ! 0, N ! 0., Formula 1. (a) loga 1 0, (a ! 0, a z 1), (b) loga a 1, (a ! 0, a z 1), Proof : (a) We know from the formula of exponents, a0 1, ? from the definition of log, we get, loga 1 0 (proved ), (b) We know, from the formula of exponents, a1 a, ? from the definition of log, we get, loga a 1 (proved)., loga M loga N, , Formula 2. loga (MN ), Proof : Let, loga M, ? M, , x, , x, loga N, , a ,N, , a, , y, , y;, , log10 1 0, , log10 10 1, ....... = ......., loga a 1
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78, , Math, , Now, MN a x a y a x y, ? log a ( MN ) x y, or log a ( MN ) log a M log a N [putting the values of x, y ], ? loga (MN ), , loga M loga N. (proved), , Note 1. loga (MNP.....), , loga M loga N loga P ........., , Note 2. loga (M r N ) z loga M r loga N, M, log a M log a N, N, Proof : Let loga M x, loga N y ;, , Formula 3. log a, , ? M, , a x, N, , ay, , M ax, a x y, N ay, §M ·, ? loga ¨ ¸ x y, ©N¹, , Now,, , §M ·, ? loga ¨ ¸, ©N¹, , loga M loga N (proved)., , Formula 4. loga M r, , r loga M ., , Proof : :Let loga M, , x; ? M, , ? (M ) r, , ax, , (a x ) r ; or M r, , ? log a M r, ? log a M, , r, , a rx, , rx; or log a M r, , r log a M, , r log a M . (proved )., , r, , N.B. : (log a M ) z r log a M, Formula 5. loga M = logb M u loga b, (change of base), Proof : Let, log a M x, log b M y, ? ax M , by M 1, ? a x bx y , or (a x ) y, or b a y, , x, y, or , x, ?, , loga b, or x, , 1, , (b y ) y, , y loga b, , y loga b, or log a M, , Corollary : log a b, , 1, , or, logb a, log b a, , log b M u log a b (proved)., 1, loga b
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79, , Math, , logb M u loga b [formula 5], , Proof : We know, loga M, P, utting M a we get,, loga a logb a u loga b, or 1 log b a u log b b;, 1, log a b, , ? log b a, , or loga b, , 1, (proved)., logb a, §1·, (b) log3¨ ¸ (c) log 3 81, ©9¹, , Example 7. Find the value : (a) log10 100, Solution:, (a) log10 100 log10 10 2, , 2 log10 10 [ log10 M r r log10 M ], 2 u1 [ log a a 1] = 2, , §1·, §1·, (b) log 3 ¨ ¸ log 3 ¨ 2 ¸ log 3 32, 9, © ¹, ©3 ¹, 2 u1[ log a a 1] = 2, , (c) log, , 3, , 81 log, , 3, , 34, , log 3 {(, , 8 log 3 3[ log a M r, 8 u 1, [ log a a 1], 8, , 2 log 3 3 [ log a M r, , 3 ) 2 }4, , log, , 3, , 3, , 8, , r log a M ], , Example 8. (a) What is the log of 5 5 to the base 5 ?, (b) log 400 4 ; what is the base ?, Solution : (a) 5 5 to the base 5, 1, , 3, , log 5 5 5 log 5 (5 u 5 2 ) log 5 5 2, 3, log 5 5,[ log a M r r log a M ], 2, 3, u1,[ log a a 1], 2, 3, 2, (b) Let the base be a., ? by the question, loga 400 4, ? a 4 400, , r log a M ]
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80, , Math, , or a 4, , (20)2, , or a 4, , (2 5), , ? a, , {( 2 5)2}2, , 4, , 4, , ȏ a x, , 2 5, , (2 5), , b x ǡ a b] , , ? the base is 2 5, Example 9. Find the value of x :, (b) log x 324, (a) log10 x 2, Solution :, (a) log10 x 2, , 4, (b) log x 324, , 4, , 4, , 2, , ? x 10, 1, 1, ?x, 10 2 100, ? x 0 01, , ? x 324 3 u 3 u 3 u 3 u 2 u 2, 34 u 2 2 34 u ( 2 ) 4, , 0 01, , or x 4, , (3 2), , 4, , ? x 3 2., , Example 10. rPove that, 3 log10 2 log10 5, , log10 40, , Solution :Left hand side = 3 log10 2 log10 5, log10 2 3 log10 5,[ log a M r, , r log a M ], , log10 8 log10 5, , log10 (8 u 5), [ log a ( MN ) log a M log a N ], log10 40 = iRght hand side (proved)., Example 11. Simplify :, Solution :, , log10 27 log10 8 log10 1000, log10 1 2, , log10 27 log10 8 log10 1000, log10 1 2, 1, , 1, , log 10 (33 ) 2 log 10 2 3 log 10 (10 3 ) 2, 12, log 10, 10, 3, , 3, , log10 3 2 log10 23 log10 10 2, log10 12 log10 10, 3, 3, log10 3 3 log10 2 log10 10, 2, 2, log10 (3 u 2 2 ) log10 10
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Math, , 81, , 3, (log10 3 2 log10 2 1), 2, (log10 3 2 log10 2 1), , [ log10 10 1], , 3, ., 2, 1., , Exercise 4.2, Find the value :(a) log3 81 (b) log5 3 5 (c) log4 2 (d) log2 5 400, (e) log 5, , 3, , 5 5, , 2., , Find the value of x :(a) log5 x, , 3., , Show that,, (a) 5 log10 5 log10 25, , 3, , (b) log x 25, , 2, , (c) log x, , 1, 16, , 2, , log10 125, , 50, log10 2 2 log10 5 log10 3 2 log10 7, 147, (c) 3 log10 2 2 log10 3 log10 5 log10 360, Simplify :, 10, 25, 81, (a) 7 log10 2 log10, 3 log10, 9, 24, 80, (b) log 7 5 7 7 log 3 3 3 log 4 2, , (b) log10, , 4., , a 3b 3, b 3c 3, c 3d 3, , log, , log, 3 log e b 2 c, e, e, c3, d3, a3, 4.5 Scientific or Standard form of numbers, We can express very large numbers or very small numbers in easy and small form by, exponents., Such as, velocity of light, = 300000 km/sec = 300000000 m/sec, = 3u 100000000 m/sec. = 3u 108 m/sec., Again, radius of a hydrogen atom = 0 00000000037 cm, 37, cm = 37 u 1010 cm, =, 10000000000, = 3 7 u 10 u 1010 cm = 3 7 u 109 cm, For convenience, very large number or very small number is expressed in the form, a u 10n , where 1 d a 10 and n Z . The form a u 10n of any number is called the, scientific or standard form of the number., Activity : Express the following numbers in scientific form :, (b) 0 000512, (a) 15000, (c) log e, , 0DWK,;;)RUPD
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82, , Math, , 4.6 Logarithmic Systems, Logarithmic systems are of two kinds :, (a) Natural Logarithm :, The mathematician John Napier (1550 –1617) of Scotland first published the book, on logarithm in 1614 by taking e as its base. e is an irrational number,, e 2 718....... . Such logarithm is called Napierian logarithm or e based logarithm, or natural logarithm. loge x is also written in the form ln x ., (b) Common Logarithm :, The mathematician Henry Briggs (1561 – 1630) of England prepared log table in, 1624 by taking 10 as the base. Such logarithm is called Briggs logarithm or 10 based, logarithm or practical logarithm., N.B. : If there is no mention of base, e in the case of expression (algebraic) and 10,, in the case of number are considered the base. In log table 10 is taken as the base., 4.7 Characteristic and Mantissa of Common Logarithm, (a) Characteristics :, Let a number N be expressed in scientific form as N, N ! 0,1 d a 10 and n Z ., Taking log of both sides with base 10,, log10 N log10 (a u10 n ), log10 a log10 10 n log10 a n log10 10, ? log10 N n log10 a, [ log10 10 1], ? log10 N, , a u 10n , where, , n log10 a, , Suppressing the base 10, we have,, , log N n log a, n is called the characteristic of log N ., We observe : Table-1, N, Form a u 10 m, of N, , Number of digits on, the left of the decimal, point, , C, haracteristic, , 6237, , 6 237 u 103, , 3, , 4, , 4 1 3, , 623 7, , 6 237 u 102, , 2, , 3, , 3 1 2, , 62 37, , 1, , 1, , 2, , 2 1 1, , 0, , 0, , 1, , 11 0, , 1, , 0, , 0 1 1, , 6 237, 0 6237, , , Exponent, , 6 237 u 10, 6 237 u 10, 6 237 u 10, , 1
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83, , Math, , We observe : Table-2, Form, N, , Exponent Number of zeroes between, decimal point and its next, first significant digit, , a u 10 m, of N, , C, haracteristic, , 0 6237, , 6 237 u 101, , 1, , 0, , (0 1), , 1, , 0 06237, , 6 237 u 102, , 2, , 1, , (1 1), , 2, , 3, , 3, , 2, , (2 1), , 3, , 0 006237, , 6 237 u 10, , We observe from table-1 :, As many digits are there in the integral part of a number, characteristic of log of, the number will be 1 less than that number of digits and that will be positive., We observe from table 2 :, If there is no integral part of a number, as many zeroes are there in between, decimal point and its next first significant digit, the characteristic of log of the, number will be 1 more than the number of zeroes and that will be negative., N. B. 1. C, haracteristic may be either positive or negative, but the mantissa will, always be positive., N. B. 2. If any characteristic is negative, not placing ‘ s–ign on the left of the, characteristic, it is written by giving ‘ (–b ar sign) over the characteristic. Such as,, characteristic –3 will be written as 3 . Otherwise, whole part of the log including, mantissa will mean negative., Example 12. Find the characteristics of log of the following numbers :, (i) 5570, (iv) 0 000435, (ii) 45 70, (iii) 0 4305, 3, Solution : (i) 5570 5 570 u 1000 5 570 u 10, ? hCaracteristic of log of the number is 3 ., Otherwise, number of digits in the number 5570 is 4 ., ?C, haracteristic of log of the number is, 4 1 3, ?C, haracteristic of log of the number is 3., 1, (ii) 45 70 4 570 u 10, ?C, haracteristic of log of the number is 1 ., Otherwise, there are 2 digits in the integral part (i.e. on left of decimal point) of the number., ?C, haracteristic of the log of the number is, 2 1 1, ?C, haracteristic of log of the number is 45 70 is 1., 1, (iii) 0 4305 4 305 u 10, ?C, haracteristic of log of the number is 1, Otherwise, there is no significant digit in the integral part (before the decimal point), of the number or there is zero digit., 0 1 1 1, ?C, haracteristic of log of the number
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84, , Math, , Again, there is no zero in between decimal point and its next first significant digit of, the number 0 4305 , i.e. there is 0 zeroes., ?C, haracteristic of log of the number is, (0 1) 1 1, ?C, haracteristic of log of the number 0 4305 is 1, (iv) 0 000435 4 35 u10 4, ?C, haracteristic of log of the number is 4 or 4, Otherwise, there are 3 zeroes in between decimal point and its next 1st significant digit., ?C, haracteristic of log of the number is, (3 1) 4 4, ?C, haracteristic of log of the number is 0 000435 is 4, (b) Mantissa :, Mantissa of the C, ommon Logarithm of any number is a non-negative number less, than 1. It is mainly an irrational number. But the value of mantissa can be determined, upto a certain places of decimal., Mantissa of the log of a number can be found from log table. Again, it can also be, found by calculator. We shall find the mantissa of the log of any number in 2nd, method, that is by calculator., Determination of common logarithm with the help of calculator :, Example 13. Find the characteristic and mantissa of log 2717 :, Solution :We use the calculator :, AC, , log, , 2717, , =, , 3 43408, , ?C, haracteristic of log 2717 is 3 and mantissa is 43408, Example 14. Find the characteristic and mantissa of log 43 517 ., Solution :We use the calculator :, = 163866, AC, log, 43517, ?C, haracteristic of log 43 517 is 1 and mantissa is 63866, Example 15. What are the characteristic and mantissa of the log of 0 00836 ?, Solution :We use the calculator :, , AC, , log, , 000836, , = 3– 92221, , = 3 92221, , ?C, haracteristic of log 0 00836 is –3 or 3 and mantissa is 92221, Example 16. Find log e 10, Solution : log e 10, =, , 1, log10 e, , 1, [taking the value of e upto five decimal places], log10 2 71828, , 1, [using calculator], 0 43429, , = 230259 (approx).
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85, , Math, , Alternative : We use the calculator :, ln, , AC, , 10, , = 230259, , (approx)., , Activity : Find the logarithm of the following numbers (each with the base 10 and, e) by using calculator : (i) 2550 (ii) 52 143 (iii) 04145, (iv) 00742, Exercise 4.3, 1., 2., , 0, , On what condition a 1 ?, b. a z 0, a. a 0, , Which one of the following is the value of, a., , 3., , 6, , 5, , 5., 6., , b., , 3, , 5, , 3, , c., , 5, , d. a z 1, 3, , 5 5 ?, 3, , 3, , d., , 3, , 25, , On what exact condition log a a, a. a ! 0, , 4., , c. a ! 0, , If log x 4, , 1?, b. a z 1, c. a ! 0, a z 1, 2 , what is the value of x ?, b. r 2, c. 4, , d. a z 0, a ! 1, , d. 10, a. 2, What is the condition for which a number is to be written in the form a u 10 n ?, b. 1 d a d 10, c. 1 d a 10, d. 1 a d 10, a. 1 a 10, Observe the following information :, , i. log a ( m) p, , plog a m, , ii. 2 16 and log 2 16 4 are synonymous., iii. log a ( m n) log a m log a n, 4, , Which of the above information are correct ?, b. ii and iii, c. i and iii, a. i and ii, , d. i, ii and iii, , 7., , What is the characteristic of the common log of 0 0035 ?, , 8., , b. 1, c. 2, d. 3, a. 3, oCnsidering the number 0 0225 , answer the following questions :, (1) Which one of the following is of the form a n of the number ?, a. ( 2 5) 2, b. (015) 2, c. (1 5) 2, d. (15) 2, (2) Which one of the following is the scientific form of the number ?, b. 22 5 u 10 3, c. 2 25 u 10 2, d. 225 u 10 1, a. 225 u 10 4, (3) What is the characteristic of the common log of the number ?, a. 2, , b. 1, , c. 0, , d. 2
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86, , Math, , 9., , Express into scientific form :, (b) 60 831, (c) 0 000245, (a) 6530, (e) 0 00000014, 10. Express in the form of ordinary decimals :, (a) 105, (b) 10 5, (c) 2 53 u 10 4, , 11., , 12., , 13., , 14., , 15., , (d) 37500000, , (d) 9 813 u 10 3, , (e) 3 12 u 10 5, Find the characteristic of common logarithm of the following numbers (without, using calculator) :, (b) 72 245, (c) 1 734, (d) 0 045, (a) 4820, (e) 0 000036, Find the characteristic and mantissa of the common logarithm of the following, numbers by using calculator :, (a) 27, (b) 63 147, (c) 1 405, (d) 0 0456, (e) 0 000673, Find the common logarithm of the product/quotient (approximate value upto, five decimal places) :, (a) 5 34 u 8 7, (b) 0 79 u 0 56 (c) 22 2642 y 3 42, (d) 0 19926 y 32 4, If log 2 0 30103, log 3 0 47712 and log 7 0 84510 , find the value of the, following expressions :, (a) log 9, (b) log 28, (c) log 42, iGven, x 1000 and y 0 0625, a. Express x in the form a n b n , where a and b are prime numbers., b. Express the product of x and y in scientific form., c. Find the characteristic and mantissa of the common logarithm of xy .
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Chapter five, , Equations with One Variable, We have known in the previous class what equation is and learnt its usage. We have, learnt the solution of simple equations with one variable and acquired knowledge, thoroughly about the solution of simple equations by forming equations from real, life problems. In this chapter, linear and quadratic equations and Identities have been, discussed and their usages have been shown to solve the real life problems., At the end of the chapter, the students will be able to −, ¾ Explain the conception of variable, ¾ Explain the difference between equation and identity, ¾ Solve the linear equations, ¾ Solve by forming linear equations based on real life problems, ¾ Solve the quadratic equations and find the solution sets, ¾ Form the quadratic equations based on real life problems and solve., 5⋅1 Variables, We know, x 3 5 is an equation. To solve it, we find the value of the unknown, quantity x . Here the unknown quantity x is a variable. Again, to solve the equation, x a 5 , we find the value of x , not the value of a . Here, x is assumed as variable, and a as constant. In this case, we shall get the values of x in terms of a . But if we, determine the value of a , we shall write a 5 x ; that is, the value of a will be, obtained in terms of x . Here a is considered a variable and x a constant. But if no, direction is given, conventionally x is considered a variable. eGnerally, the small, letters x, y, z , the ending part of English alphabet are taken as variables and a, b, c ,, the starting part of the alphabet are taken as constants., The equation, which contains only one variable, is called a linear equation with one, variable. Such as, in the equation x 3 5 , there is only one variable x . So this is, the linear equation with one variable., We know what the set is. If a set S {x : x R, 1 d x d 10} , x may be any real, number from 1 to 10. Here, x is a variable. So, we can say that when a letter symbol, means the element of a set, it is called variable., Degree of an equation : The highest degree of a variable in any equation is called, the degree of the equation. Degree of each of the equations x 1 5, 2 x 1 x 5,, y 7 2 y 3 is 1 ; these are linear equations with one variable.
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88, , Math, , Again, the degree of each of the equations x2 5x 6 0, y 2 y 12, 4x2 2x 3 6x, is 2 ; these are quadratic equations with one variable. The equation 2x3 x2 4x 4 0, is the equation of degree 3 with one variable., 5⋅2 Equation and Identity, Equation : There are two polynomials on two sides of the equal sign of an equation,, or there may be zero on one side (mainly on right hand side). Degree of the variable, of the polynomials on two sides may not be equal. Solving an equation, we get the, number of values of the variable equal to the highest degree of that variable. This, value or these values are called the roots of the equation. The equation will be, satisfied by the root or roots. In the case of more than one root, these may be equal or, unequal. Such as, roots of x2 5x 6 0 are 2 an 3. Again, though the value of x, in the equations ( x 3)2 0 is 3, the roots of the equation are 3, 3., Identity` : There are two polynomials of same (equal) degree on two sides of equal, sign. Identity will be satisfied by more values than the number of highest degree of, the variable. There is no difference between the two sides of equal sign ; that is why,, it is called identity. Such as, ( x 1)2 ( x 1)2 4x is an identity ; it will be satisfied, for all values of x . So this equation is an identity. Each algebraic formula is an, identity. Such as, (a b) 2 a 2 2ab b 2 , (a b) 2 a 2 2ab b 2 , a 2 b 2, , (a b)(a b) , (a b) 3 a 3 3a 2 b 3ab 2 b 3 etc. are identities., All equations are not identities, In identity '{' sign is used instead of equal (=) sign., But as all identities are equations, in the case of identity also, generally the equal, sign is used. Distinctions between equation and identity are given below :, Equation, Identity, 1. Two polynomials may exist on both 1. Two polynomials exist on two, sides of equal sign, or there may be, sides., zero on one side., 2. Degree of the polynomials on both 2. Degree of the polynomials on both, sides may be unequal., sides is equal., 3. The equality is true for one or more 3. eGnerally, the equality is true for all, values of the original set of the, values of the variable., variable., 4. The number of values of the variable 4. Equality is true for infinite number, does not exceed the highest degree of, of values of the variable., the equation, 5. All equations are not formulae., 5. All, algebraic, formulae, are, identities.
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Math, , 89, , 1. What is the degree of and how many roots has each of the, following equations ?, 2 y y 1 3y, (ii ), , (i ) 3 x 1 5, 5, 3, 2, 2. Write down three identities., 5⋅3 Solution of the equations of first degree, In case of solving equations, some rules are to be applied. If the rules are known,, solution of equations becomes easier. The rules are as follows :, 1. If the same number or quantity is added to both sides of an equation, two sides, remain equal., 2. If the same number or quantity is subtracted from both sides of an equation, two, sides remain equal., 3. If both sides of an equation are multiplied by the same number or quantity, the, two sides remain equal., 4. If both sides of an equation are divided by same non-zero number or quantity, the, two sides remain equal., The rules stated above may be expressed in terms of algebraic expressions as follows:, x a, If x a and c z 0 , (i ) x c a c (ii ) x c a c (iii ) xc ac (iv), c c, Besides, if a, b and c are three quantities, if a b c , a b c and if a c b ,, a bc., This law is known as transposition law and different equations can be solved by, applying this law., If the terms of an equation are in fractional form and if the degree of the variables in, each numerator is 1 and the denominator in each term is constant, such equations are, linear equations., 5x 4 x 2, Example 1. Solve :, , , 7 5 5 7, 5x x 4 2, 5x 4 x 2, Solution :, or,, [by transposition], , , , , 7 5 5 7, 7 5 5 7, 25x 7 x 28 10, 18 x 18, or,, or,, 35, 35, 35 35, or, 18x 18, or, x 1, ? Solution is x 1., Now, we shall solve such equations which are in quadratic form. These equations are, transformed into their equivalent equations by simplifications and lastly the, equations is transformed into linear equation of the form ax b . Again, even if there, are variables in the denominator, they are also transformed into linear equation by, simplification., Activity :, , 0DWK,;;)RUPD
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90, , Math, , Example 2. Solve : ( y 1)( y 2) ( y 4)( y 2), Solution : ( y 1)( y 2) ( y 4)( y 2), or, y 2 y 2 y 2 y 2 4 y 2 y 8, or, y 2 2 y 8, or, y 2 y 8 2 [by transposition], or, y 6, or, y 6, ? Solution is y 6, Example 3. Solve and write the solution set :, , 6x 1 2x 4, , 15, 7x 1, , 2x 1, 5, , 6x 1 2x 4 2x 1, , 15, 7 x 1, 5, 6x 1 2x 1 2x 4, [by transposition], or,, , 15, 5, 7x 1, , Solution :, , or,, , 6x 1 6x 3, 15, , 2x 4, 7x 1, , or, 4, , 15, , 2x 4, 7x 1, , or, 15(2x 4) 4(7 x 1) [by cross-multiplication], or, 30x 60 28x 4, or, 30x 28x 60 4 [by transposition], or, 2x 56, or x 28, ? Solution is x 28, and solution set is S {28} ., 1, 1, 1, 1, Example 4. Solve :, , , x 3 x 4 x 2 x 5, 1`, 1, 1, 1, , , x 3 x 4 x 2 x 5, x 4 x 3, x 5 x 2, or,, ( x 3)( x 4) ( x 2)( x 5), , Solution :, , 2x 7, 2x 7, x 2 7 x 12 x 2 7 x 10, Values of the fractions of two sides are equal. Again, numerators of two sides are, equal, but denominators are unequal. In this case, if only the value of the numerators, is zero, two sides will be equal., ? 2 x 7 0 or, 2 x 7, , or, x, , 7, 2, , ? Solution is x, , 7, 2, , or,
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Math, , 91, , 2x 3 5 2, Alternative method :, Solution : 2x 3 5 2, 2x 3 5 2, or, 2x 3 2 5, [by transposition], 2, or, 2x 3 2 5, or, 2x 3, 3 2 [squaring both sides], or, 2x 3 3, or, 2x 3 9, Square, root of any real quantity, or, 2x 12, cannot be negative., or, x 6, Since there is the sign of square root, ? The equation has no solution., ? Solution set is :S = φ, verification of the correctness is necessary., P, utting x 6 in the given equation, we get,, , Example 5. Find the solution set :, , 2 u 6 3 5 2 or 9 5 2, or, 3 5 2, or, 8 2 , which is impossible., ? The equation has no solution., ? Solution set is :S = φ, Activities : 1. If 5 1 x 4 4 5 , show that, x 6 2 5, 2. Solve and write the solution set : 4x 3 5 2, 5⋅4 Usage of linear equations, In real life we have to solve different types of problems. In most cases of solving, these problems mathematical knowledge, skill and logic are necessary. In real cases,, in the application of mathematical knowledge and skill, as on one side the problems, are solved smoothly, on the other side in daily life, solutions of the problems are, obtained by mathematics. As a result, the students are interested in mathematics., Here different types of problems based on real life will be expressed by equations, and they will be solved., For determining the unknown quantity in solving the problems based on real life,, variable is assumed instead of the unknown quantity and then equation is formed by, the given conditions. Then by solving the equation, value of the variable, that is the, unknown quantity is found., Example 6. The digit of the units place of a number consisting of two digits is 2, more than the digit of its tens place. If the places of the digits are interchanged, the, number thus formed will be less by 6 than twice the given number. Find the number., Solution : Let the digit of tens place be x . Then the digit of units place will be, x2., ? the number is 10 x ( x 2) or, 11x 2., Now, if the places of the digits are interchanged, the changed number will be, 10( x 2) x or 11x 20
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92, , Math, , By the question, 11x 20 2(11x 2) 6, or, 11x 20 22 x 4 6, or, 22 x 11x 20 6 4 [by transposition], or, 11x 22, or, x 2, ? The number is 11x 2 11u 2 2 24, ? given number is 24 ., Example 7. In a class if 4 students are seated in each bench, 3 benches remain, vacant. But if 3 students are seated on each bench, 6 students are to remain standing., What is the number of students in that class ?, Solution : Let the number of students in the class be x ., Since, if 4 students are seated in a bench, 3 benches remain vacant, the number of, x, benches of that class, 3, 4, Again, since, if 3 students are seated in each bench, 6 students are to remain, x6, standing, the number of benches of that class, 3, Since the number of benches is fixed,, x, x6, x 12 x 6, or,, 3 =, 3, 4, 3, 4, or, 4x 24 3x 36 , or, 4x 3x 36 24, or, x 60, ? number of students of the class is 60 ., Example 8. Mr. aKbir, from his Tk. 56000, invested some money at the rate of profit, 12% per annum and the rest of the money at the rate of profit 10% per annum. After, one year he got the total profit of Tk. 6400. How much money did he invest at the, rate of profit 12% ?, Solution : Let Mr. aKbir invest Tk. x at the rate of profit 12%., ? he invested Tk. (56000 x) at the rate of profit 10%., 12x, 12, Now, profit of Tk. x in 1 year is Tk. x u, u1 , or, Tk., 100, 100, Again, profit of Tk. (56000 x) in 1 year is Tk. (56000 x) u, or, Tk., , 10(56000 x), 100, , By the question ,, , 12x 10(56000 x), +, = 6400, 100, 100, , or, 12x 560000 10x 640000, or, 2x 640000 560000, , 10, ,, 100
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93, , Math, , or, 2x 80000, or, x 40000, ? Mr. aKbir invested Tk. 40000 at the rate of profit 12%., Activity : Solve by forming equations :, 1. What is the number, if any same number is added to both numerator and, 3, 4, denominator of the fraction , the fraction will be ?, 5, 5, 2. If the difference of the squares of two consecutive natural numbers is 151, find, the two numbers., 3. If 120 coins of Tk. 1 and Tk. 2 together are Tk. 180, what is the number of, coins of each kind ?, Exercise 5⋅1, Solve (1-10) :, 1. 3(5 x 3) 2( x 2), , 2., , ay by, , b, a, , a2 b2, , 3., , ( z 1)( z 2) ( z 4)( z 2), 7x 3 2x 4, 4, 9, 25, 4., 5., 6., , , , 3 5 5 3, 2 x 1 3x 2 5 x 4, 1, 1, 1, 1, , , x 1 x 4 x 2 x 3, a, b, ab, x a x b x 3a 3b, 7., 8., 0, , , , xa xb xab, b, a, ab, xa, xb, a2 b2 b2 a2, 10. (3 3 ) z 2 5 3 3., Find the solution set (11 - 19) :, , 11. 2 x( x 3), z2, z 1, , 2 x 2 12, , 12. 2 x 2, , 3x 4 3 2, , 1, 1, 1, 2, 15. , z 1, x x 1 x 1, m, n, mn, , mx nx mnx, 1, 1, 1, 1, 2t 6 15 2t, 17., 18., , , , x2 x5 x4 x3, 9, 12 5t, x 2b 2 c 2 x 2c 2 a 2 x 2a 2 b 2, 0, , , 19., ab, bc, ca, , 14., , 2, , 13., , 9., , xa, xb, , 16., , 4t 15, 18, , xa, xc
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94, , Math, , ȋǦͳȌ, , Ǥ, , Solve by forming equations (20 - 27) :, 2, times of another number. If the sum of the numbers is 98, find, 20. A number is, 5, the two numbers., 21. Difference of num. and denom. of a proper fraction is 1. If 2 is subtracted from, 1, numerator and 2 is added to denominator of the fraction, it will be equal to ., 6, Find the fraction., 22. Sum of the digits of a number consisting of two digits is 9. If the number, obtained by interchanging the places of the digits is less by 45 than the given, number, what is the number ?, 23. The digit of the units place of a number consisting of two digits is twice the, digit of the tens place. Show that, the number is seven times the sum of the, digits., 24. A petty merchant by investing Tk. 5600 got the profit 5% on some of the, money and profit of 4% on the rest of the money. On how much money did he, get the profit of 5% ?, 25. Number of passengers in a launch is 47 ; the fare per head for the cabin is twice, that for the deck. The fare per head for the deck is Tk. 30. If the total fare, collected is Tk. 1680, what is the number of passengers in the cabin ?, 26. 120 coins of twenty five paisa and fifty paisa together is Tk. 35. What is the, number of coins of each kind ?, 27. A car passed over some distance at the speed of 60 km per hour and passed, over the rest of the distance at the speed of 40 km per hour. The car passed over, the total distance of 240 km in 5 hours. How far did the car pass over at the, speed of 60 km per hour ?, 5⋅5 Quadratic equation with one variable, Equations of the form ax 2 bx c 0 [where a, b, c are constants and a z 0 ] is, called the quadratic equation with one variable. Left hand side of a quadratic equation is, a polynomial of second degree, right hand side is generally taken to be zero., Length and breadth of a rectangular region of area 12 square cm. are respectively, x cm. and ( x 1) cm., , ? area of the rectangular region is = x( x 1) square cm., By the question, x( x 1) 12, or x 2 x 12 0 ., ͳʹǤ Ǥ, x is the variable in the equation and highest power of x is 2., Such equation is a quadratic equation. The equation, which has, x Ǥ, the highest degree 2 of the variable, is called the quadratic equation., In class VIII, we have factorized the quadratic expressions with one variable of the, forms x2 px q and ax2 bx c . Here, we shall solve the equations of the forms, x 2 px q 0 and ax 2 bx c 0 by factorizing the left hand side and by finding, the value of the variable.
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5, 9, , Math, , An important law of real numbers is applied to the method of factorization. The law, is as follows :, If the product of two quantities is equal to zero, either ony one of the quantities or, both quantities will be zero. That is, if the product of two quantities a and b i.e.,, ab 0 , a 0 or, b 0 , or, both a 0 and b 0 ., Example 9. Solve : ( x 2)( x 3) 0, Solution : ( x 2)( x 3) 0, ? x 2 0 , or, x 3 0, If x 2 0 , x 2, Again, if x 3 0 , x 3, ? solution is x 2 or, 3 ., Example 10. Find the solution set : y 2, 3y, Solution : y 2, 3y, or, y 2 3 y 0 [By transposition, right hand side has been done zero], or, y ( y 3 ) 0, ? y 0 , or y 3 0, If y 3 0 , y, 3, ? Solution set is {0, 3} ., x4, Example 11. Solve and write the solution set : x 4, ,x z 0., x, x4, Solution : x 4, x, [by cross-multiplication], or, x( x 4) x 4, or, x( x 4) ( x 4) 0 [by transposition], or, ( x 4)( x 1) 0, ? x 4 0, or, x 1 0, If x 4 0 , x 4, Again, if x 1 0 , x 1, ? Solution is : x 1 or, 4, and the solution set is {1, 4} ., 2, , §xa·, §xa·, Example 12. Solve : ¨, ¸6, ¸ 5¨, ©xa¹, ©xa¹, 2, , § xa·, § xa·, Solution : ¨, ¸ 6 0..........(1), ¸ 5¨, © xa¹, © xa¹, xa, eLt,, y, xa, Then from (1) , we get, y 2 5 y 6 0, or, y 2 2 y 3 y 6 0, , 0
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6, 9, , Math, , or, y ( y 2) 3( y 2) 0, or, ( y 2)( y 3) 0, ? y 2 0 , or, y 3 0, If y 2 0 , y 2, If y 3 0 , y 3, oNw, when y 2 ,, xa 2, [putting the value of y ], xa 1, x a x a 2 1, [by componendo and dividendo], or,, x a x a 2 1, 2x 3, or,, 2a 1, or, x 3a, xa 3, Again, when y 3 ,, xa 1, x a x a 3 1, or,, x a x a 3 1, 2x 4, or,, 2a 2, x 2, or,, a 1, or, x 2a, ? Solution is : x 3a , or, 2a, Activity :, .1 Comparing the equation x 2 1 0 with the equation ax 2 bx c 0 , write, down the values of a, b, c ., 2. W, hat is the degree of the equation, ( x 1) 2 0 ? How many roots has the, equation and what are they?, 5⋅6 Usage of quadratic equations, Many problems of our daily life can be solved easily by forming linear and quadratic, equations. Here, the formation of quadratic equations from the given conditions, based on real life problems and techniques for solving them are discussed., Example 13. Denominator of a proper fraction is 4more than the numerator. If the, fraction is squared, its denominator will be 04 more than the numerator. Find the, fraction., x, Solution : eLt the fraction be, ., x4
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Math, , 7, 9, , 2, , ͶͲǤ, , x2, x2, § x ·, Square of the fraction = ¨, ¸, ( x 4) 2 x 2 8 x 16, © x4¹, 2, Here, numerator = x and denominator = x 2 8 x 16 ., By the question, x 2 8 x 16, x2 4, 0, or, 8 x 16, 0, 4, or, 8 x 40 16, or, 8 x 24, or, x 3, ? x 4 3 4 7, x, 3, 3, ?, x 4 3 4 7, 3, ? the fraction is, 7, Example 14. A rectangular garden with length 50 metre and breadth 04 metre has o f, equal width all around the inside of the garden. If the area of the garden except the, path is 0210 square metre, how much is the path wide in metre ?, Solution : eLt the path be x metre wide., Ǥ, W, ithout the path,, , eLngth of the garden = (50 2 x) metre and, its breadth = (40 2 x) metre, x ., ? W, ithout the path, area of the, garden = (50 2 x) × (40 2 x) square metre., By the question, (50 2 x) (40 2 x) 1200, , or,, or,, or,, or,, or,, or,, , 200 8, 0, 0 x 0, 10 x 4 x 2 1, 00, 2, 2, 4x 1, 0 x8, 8, 00, 0, x 2 45 x 200 0 [dividing by 4], x 2 5x 4, 0 x2, 00 0, x( x 5) 40 ( x 5) 0, ( x 5)( x 4, 0 ) 0, , ͷͲǤ, , ? x 5 0, or x 40 0, If x 5 0 , x 5, If x 40 0 , x 40, But the breadth of the path will be less than 40 metre from the breadth of the garden., ? x z 40 ; ? x 5, ? the path is 5metres wide., 0DWK,;;)RUPD
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8, 9, , Math, , Example 15. Shahik bought some pens at Tk. 420. If he would get one more pen in, that money, average cost of each pen would be less by Tk. .1How many pens did he, buy ?, Solution : eLt, Shahik bought x pens in total by Tk. 402. Then each pen costs Tk., 240, . If he would get one more pen, that is, he would get ( x 1) pens, the cost of, x, 240, each pen would be Tk., ., x 1, 240, 240 x, 240, 240, By the question, ,, 1, or, x 1, x, x 1, x, or, 240 x ( x 1)(240 x) [by cross-multiplication], or, 240 x 240 x 240 x 2 x, or, x 2 x 240, 0 [by transposition], or, x 2 16 x 15 x 240, 0, or, x( x 16 ) 15 ( x 16 ) 0, or, ( x 16 )( x 15 ) 0, ? x 16, 0 , or, x 15 0, If x 16, 0 , x 1, 6, Again, if x 15, 0 , x 15, But the number of pen x , cannot be negative., ? x z 16 ; ? x 15, ? Shahik bought 15pens., Activity : Solve by forming equations :, .1 If a natural number is added to its squ are, the sum will be equal to nine times, of exactly its next natural number. W, hat is the number ?, .2 eLngth of a perpendicular drawn from the centre of a circle of radius 10 cm., to a chord is less by 2cm. than the semi-chord. Find the length of the chord, by drawing a probable picture., Example 16. In an examination of class IXof a school, total marks of x students, obtained in mathematics is 1950. If at th e same examination, marks of a new student, in mathematics is 34and it is added to the former total marks, the average of the, marks become less by .1, a. W, rite down the average of the obtained marks of all students including the new, student and separately x students in terms of x ., b. By forming equation from the given conditions, show that, x 2 35 x 1950, 0, c. By finding the value of x , find the average of the marks in the two cases., 0, 5, 9, 1, Solution : a. Average of the marks obtained by x students =, x, Average of the marks obtained by ( x 1) students including
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9, , Math, , 34 1984, x 1, x 1, 9501, 1984, b. By the question,, 1, x, x 1, 0, 5, 9, 1, 4, 8, 9, 1, or,, 1, , x, x 1, 1950, x 1950, 1984, x, or,, 1, x( x 1), or, x 2 x 1950, [by cross-multiplication], x 1984, x 1950, or, x 2 x 1950, 34 x, ? x 2 35 x 0591, 0 [showed], c. x 2 35 x 1950, 0, or, x 2 65 x 30 x 1950, 0, or, x( x 65 ) 30( x 65 ) 0, or, ( x 65 )( x 30) 0, ? x 65, 0 , or, x 30 0, If x 65 0 , x 65, Again, if x 30 0 , x 30, Since the number of students, i.e., x cannot be negative, x z 6, 5, ? x 30 ., 1950, ? in the first case, average =, 65, 30, 1984, and in the second case, average =, 64 ., 31, , the new student =, , 1950, , Exercise 5⋅2, 1., , 2., , Assuming x as the variable in the equation a2 x b 0 , which one of the, following is the degree of the equation ?, a. 3, b. 2, c. 1, d. 0, W, hich one of the following is an identity ?, a. ( x 1) 2 ( x 1) 2 4 x, b. ( x 1) 2 ( x 1) 2 2( x 2 1), c. (a b) 2 (a b) 2, , 3., .4, , 2ab, , d. (a b) 2, , a 2 2ab b 2, , How many roots are there in the equation ( x 4) 2 0 ?, b. 2, c. 3, d. 4, a. 1, W, hich one of the following are the two roots of the equation, x 2 x 12, , 0?
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00, 1, , Math, , W, hat is the coefficient of, x in the equation 3 x 2 x 5 0 ?, a. 3, b. 2, c. 1, d. 1, .6 bOserve the following equations :, x, ii., 2 1, i. 2 x 3 9, iii. 2 x 1 5, 2, W, hich are of the above equations equivalent ?, b. ii and iii, c. i and iii, d. i, ii and iii, a. i and ii, 7. W, hich one of the following is the solution set of the equatio n, x 2 (a b) x ab 0 ?, b. {a, b}, c. {a, b}, d. {a, b}, a. {a, b}, .8 The digit of the tens place of a numbe r consisting of two digits is twice the, digit of the units place. In respect of the information, answer the following, questions :, (1) If the digit of the units place is x , what is the number ?, b. 3 x, c. 12 x, d. 21 x, a. 2 x, (2) If the places of the digits are interchanged, what will be the number ?, a. 3 x, b. 4 x, c. 12 x, d. 21 x, (3) If x 2 , what will be the difference between the original number and the, number by interchanging their places?, b. 20, c. 34, d. 36, a. 18, Solve (9−18) :, 9. ( x 2)( x 3 ) 0 10. ( 2 x 3)( 3 x 2) 0 1. y ( y 5) 6, 3, 4, 12. ( y 5)( y 5) 24, 13. 2( z 2 9) 9 z 0, 4.1, , 2, 2 z 1 5z 1, 4, x 2 6( x 2), x a x b, 15., 17., 10 x 4 5 16., , 1, , , x, , 2, x, , 6, a, x b x, 10 x 4, xa xb a b, 18., , , xb xa b a, Find the solution set (19−25):, 3, 4, x 7 2x 6, 19., 20., 21., , 2, , 5, x 1 2x 1, x x 1, 1 1 1, 1, , x a b xab, ax b cx d, 1, 2., 32. x , 4.2 2 x 2 4ax 0 , 2, a bx c dx, x, .5
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Math, , 01, 1, , ( x 1) 3 ( x 1) 3, 2, ( x 1) 2 ( x 1) 2, Solve by forming equations (26−31) :, r, 26. Sum of the two digits of a number consisting of two digits is 51and thei, product is 65;find the number., .72 Area of the floor of a rectangular room is 291square metre. If the length of the, floor is decreased by 4metre and the breadth is increased by 4metre, the are a, remains unchanged. Find the length and breadth of the floor., .82 eLngth of the hypotenuse of a right angled triangle is 15cm. and the difference, of the lengths of other two sides is 3 cm. Find the lengths of those two sides., .29 The base of a triangle is 6cm. more than twice its height. If the area of the, triangle is 108 square cm., what is its height ?, 30. As many students are there in a class, each of them contributes equal to the, t, number of class-mates of the class and thus total Tk. 024 was collected. W, ha, is the number of students in the class and how much did each student, contribute ?, 31. As many students are there in a class, each of them contributed 30 paisa more, than the number of paisa equal to the number of students and thus total Tk. 70, was collected. W, hat is the number of students in that class ?, 32. Sum of the digits of a number consisting of two digits is 7. If the places of the, digits are interchanged, the number so formed is 9more than the given number., a. W, rite down the given number and the number obtained by interchangin g, their places in terms of variable x ., b. Find the given number., c. If the digits of the original number indicate the length and breadth of a, rectangular region in centimetre, find the length of its diagonal. Assuming the, diagonal as the side of a square, find the length of the diagonal of the square., 33. The base and height of a right angle triangle are respectively ( x 1) cm. and, x cm. and the length of the side of a square is equal to the height of the, triangle. Again, the length of a rectangular region is ( x 3) cm. and its breadth, is x cm., a. Show the information in only one picture., b. If the area of the triangular region is 10 square centimetre, what is its height?, c. Find the successive ratio of the areas of the triangular, square and rectangular, regions., 25.
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Chapter Six, , Lines, Angles and Triangles, Geometry is an old branch of mathematics. The word ‘geometry’ comes from the, Greek words ‘geo’, meaning the ‘earth’, and ‘metrein’, meaning ‘to measure’. So,, the word ‘geometry’ means ‘the measurement of land.’ Geometry appears to have, originated from the need for measuring land in the age of agricultural based, civilization. However, now a days geometry is not only used for measuring lands,, rather knowledge of geometry is now indispensible for solving many complicated, mathematical problems. The practice of geometry is evident in relics of ancient, civilization. According to the historians, concepts and ideas of geometry were, applied to the survey of lands about four thousand years ago in ancient Egypt. Signs, of application of geometry are visible in different practical works of ancient Egypt,, Babylon, India, China and the Incas civilisation. In the Indian subcontinent there, were extensive usages of gometry in the Indus Valley civilisation. The excavations at, Harappa and Mohenjo-Daro show the evidence of that there was a well planned city., For example, the roads were parallel to each other and there was a developed, underground drainage system. Besides the shape of houses shows that the town, dwellers were skilled in mensuration. In Vedic period in the construction of altars (or, vedis) definite geometrical shapes and areas were maintained. Usually these were, constituted with triangles, quadrilaterals and trapeziums., But geometry as a systematic discipline evolved in the age of Greek civilization. A, Greek mathematician, Thales is credited with the first geometrical proof. He proved, logically that a circle is bisected by its diameter. Thales’ pupil Pythagoras developed, the theory of geometry to a great extent. About 300 BC Euclid, a Greek scholar,, collected all the work and placed them in an orderly manner it in his famous treatise,, ‘Elements’. ‘Elements’ completed in thirteen chapters is the foundation of modern, geometry for generations to come. In this chapter, we shall discuss logical geometry, in accordance with Euclid., At the end of this chapter, the students will be able to, ¾ Describe the basic postulates of plane geometry, ¾ Prove the theorems related to triangles, ¾ Apply the theorems and corollaries related to triangles to solve problems., 6⋅1 Concepts of space, plane, line and point, The space around us is limitless. It is occupied by, different solids, small and large. By solids we, , mean the grains of sand, pin, pencil, paper, book,, chair, table, brick, rock, house, mountain, the earth, planets and stars. The concepts
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Math, , 103, , of geometry springs from the study of space occupied by solids and the shape, size,, location and properties of the space., , A solid occupies space which is spread in three directions. This spread in three, directions denotes the three dimensions (length, breadth and height) of the solid., eHnce every solid is three-dimensional., For example, a brick or a box has three, dimensions (length, breadth and height). A sphere also has three dimensions,, although the dimensions are not distinctly visible., The boundary of a solid denotes a surface, that is, every solid is bounded by one or, more surfaces. For example, the six faces of a box represent six surfaces. The upper, face of a sphere is also a surface. But the surfaces of a box and of a sphere are, different. The first one is plane while the second is one curved., Two-dimensional surface : A surface is two dimensional; it has only length and, breadth and is said to have no thickness. eKeping the two dimension of a box, unchanged, if the third dimension is gradually reduced to zero, we are left with a face, or boundary of the box. In this way, we can get the idea of surface from a solid., hWen two surfaces intersect, a line is form ed. For example, two faces of a box meet, at one side in a line. This line is a straight line. Again, if a lemon is cut by a knife, a, curved line is formed on the plane of intersection of curved surface of the lemon., One-dimensional line : A line is one-dimensional; it has only length and no breadth, or thickness. If the width of a face of the box is gradually receded to zero, we are lef t, with only line of the boundary. In this way, we can get the idea of line the from the, idea of surface., The intersection of two lines produces a point. That is, the place of intersection of, two lines is denoted by a point. If the two edges of a box meet at a point. A point has, no length, breadth and thickness. If the length of a line is gradually reduced to ezro at, last it ends in a point. Thus, a point is considered an entity of zero dimension., 6⋅2 Euclid’s Axioms and Postulates, The discussion above about surface, line and point do not lead to any definition –, they are merely description. This description refers to height, breadth and length,, neither of which has been defined. W, e only can represent them intuitively. The, definitions of point, line and surface which Euclid mentioned in the beginning of the, first volume of his E, ‘ lements’ are incomple te from modern point of view. A few of, Euclid’s axioms are given below:, 1. A point is that which has no part., 2. A line has no end point.
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104, , Math, , 3. A line has only length, but no breath and height., 4. A straight line is a line which lies evenly with the points on itself., 5. A surface is that which has length and breadth only., 6. The edges of a surface are lines., 7. A plane surface is a surface which lies evenly with the straight lines on itself., It is observed that in this description, part, length, width, evenly etc have been, accepted without any definition. It is assumed that we have primary ideas about, them. The ideas of point, straight line and plane surface have been imparted on this, assumption. As a matter of fact, in any mathematical discussion one or more, elementary ideas have to be taken granted. Euclid called them axioms. Some of the, axioms given by Euclid are:, (1) Things which are eq, ual to the same thing, are eq, ual to one another., (2) If equals are added to eq, uals, the wholes are equal., (3) If equals are subtracted from eq, uals, the remainders are equal., (4) Things which coincide with one another, are equal to one another., (5) The whole is greater than the part., In modern geometry, we take a point, a line and a plane as undefined terms and some, of their properties are also admitted to be true. These admitted properties are called, geometric postulates. These postulates are chosen in such a way that they are, consistent with real conception. The five postulates of Euclid are:, Postulate 1: A straight line may be drawn from any one point to any other point., Postulate 2: A terminated line can be produced indefinitely,, Postulate 3: A circle can be drawn with any centre and any radius., Postulate 4: All right angles are equal to one another., Postulate 5: If a straight line falling on two straight lines makes the interior angles, on the same side of it taken together less than two right angles, then the two straight, lines, if produced indefinitely, meet on that side on which the sum of angles is less, than two right angles., After Euclid stated his postulates and axioms, he used them to prove other results., Then using these results, he proved some more results by applying deductive, reasoning. The statements that were proved are called propositions or theorems., Euclid is his e‘ lements’ proved a total of 465 propositions in a logical chain. This is, the foundation of modern geometry., Note that there are some incompletion in Euclid’s first postulate. The drawing of a, unique straight line passing through two distinct points has been ignored., Postulate 5 is far more complex than any other postulate. On the other hand,, Postulates 1 through 4 are so simple and obvious that these are taken as s‘ elf-evident, truths’. oHwever, it is not possible to prove them. So, these statements are accepted
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Math, , 105, , without any proof. Since the fifth postulate is related to parallel lines, it will be, discussed later., 6.3 Plane Geometry, It has been mentioned earlier that point, straight line and plane are three fundamental, concepts of geometry. Although it is not possible to define them properly, based on, our real life experience we have ideas about them. As a concrete geometrical, conception space is regarded as a set of points and straight lines and planes are, considered the subsets of this universal set., Postulate 1. Space is a set of all points and plane and straight lines are the sub-sets, of this set. From this postulate we observe that each of plane and straight line is a set, and points are its elements. H, owever, in ge ometrical description the notation of sets, is usually avoided. For example, a point included in a straight line or plane is, expressed by t‘he point lies on the straight line or plane’ or t‘he straight line or plane, passes through the point’. Similarly if a straight line is the subset of a plane, it is, expressed by such sentences as ‘the straigh t line lies on the plane, or ‘the plane, passes through the straight line’., It is accepted as properties of straight line and plane that,, Postulate 2. For two different points there exists one and only one straight line, on, which both the points lie., Postulate 3. For three points which are not collinear, there exists one and only one, plane, on which all the three points lie., Postulate 4. A straight line passing through two different points on a plane lie, completely in the plane., Postulate 5. (a) Space contains more than one plane, (b) In each plane more than one straight lines lie., (c) The points on a straight line and the real numbers can be related in, such a way that every point on the line corresponds to a uniq, ue real, number and conversely every real number corresponds to a unique, point of the line., Remark: The postulates from 1 to 5 are called incidence postulates., The concept of distance is also an elementary concept. It is assumed that,, Postulate 6 : (a) Each pair of points (P, Q) determines a uniq, ue real number which, is known as the distance between point P and Q and is denoted by PQ., (b) If P and Q are different points, the number PQ is positive. Otherwise, PQ=0., (c) The distance between P and Q and that between Q and P are the same, i.e., PQ=QP., , 0DWK,;;)RUPD
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106, , Math, , According to postulate 5(c) one to one correspondence can be established between, the set of points in every straight line and the set of real numbers. In this connection,, it is admitted that,, Postulate 7 : One-to-one correspondence can be established between the set of, points in a straight line and the set of real numbers such that, for any points P and Q,, PQ =a| –b|, where, the one-to-one correspondence associates points, P and Q to real, numbers a and b respectively., If the correspondence stated in this postulate is made, the line is said to have been, reduced to a number line. If P corresponds to a in the number line, P is called the, graph point of P and a the coordinates of P. To convert a straight line into a number, line the co-ordinates of two points are taken as 0 and 1 respectively. Thus a unit, distance and the positive direction are fixed in the straight line. For this, it is also, admitted that,, Postulate 8: Any straight line AB can be converted into a number line such that the, coordinate of A is 0 and that of B is positive., Remark: Postulate 6 is known as distance postulate and Postulate 7 as ruler, postulate and Postulate 8as ruler placement postulate., Geometrical figures are drawn to make geometrical description clear. The model of a, point is drawn by a thin dot by tip of a pencil or pen on a paper. The model of a, straight line is constructed by drawing a line along a ruler. The arrows at ends of a, line indicate that the line is extended both ways indefinitely. By postulate 2, two, different points A and B define a unique straight line on which the two points lie., This line is called AB or BA line. By postulate 5(c) every such straight line contains, infinite number of points., According to postulate 5(a) more than one plane exist. There is infinite number of, straight lines in every such plane. The branch of geometry, that deals with points, lines and different geometrical, entities related to them, is known as plane Geometry. In, this textbook, plane geometry is the matter of our, discussion. eHnce, whenever something is not mentioned, in particular, we will assume that all discussed points,, lines etc lie in a plane., Proof of Mathematical statements, In any mathematical theory different statements related to the theory are logically, established on the basis of some elementary concepts, definitions and postulates., Such statements are generally known as propositions. In order to prove correctness, of statements some methods of logic are applied. The methods are:, (a) Method of induction, (b) Method of deduction
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107, , Math, , Proof by contradiction, Philosopher Aristotle first introduced this method of logical proof. The basis of this, method is:, x A property can not be accepted and rejected at the same time., x The same object can not possess opposite properties., x One can not think of anything which is contradictory to itself., x If an object attains some property, that object can not unattain that property at, the same time., 6⋅4 Geometrical proof, In geometry, special importance is attached to some propositions which are taken, as, theorems and used successively in establishing other propositions. In geometrical, proof different statements are explained with the help of figures. But the proof must, be logical., In describing geometrical propositions general or particular enunciation is used. The, general enunciation is the description independent of the figure and the particular, enunciation is the description based on the figure. If the general enunciation of a, proposition is given, subject matter of the proposition is specified through particular, enunciation. For this, necessary figure is to be drawn., Generally, in proving the geometrical theorem the following steps should be, followed :, (1) General enunciation., (2) Figure and particular enunciation., (3) Description of the necessary constructions and, (4) Description of the logical steps of the proof., If a proposition is proved directly from the conclusion of a theorem, it is called a, corollary of that theorem. Besides, proof of various propositions, proposals for, construction of different figures are considered. These are known as constructions., By drawing figures related to problems, it is necessary to narrate the description of, construction and its logical truth., , Exercise 6.1, 1., 2., 3., 4., 5., 6., , Give a concept of space, surface, line and point., State Euclid’s five postulates., State five postulates of incidence., State the distance postulate., State the ruler postulate., Explain the number line.
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10, 8, , Math, , .7 S, tate the postulat e of ruler placement., .8 Define intersecting straight line and parallel straight line., Line, Ray and Line Segment, By postulates of plane geometry, every point of a straight line lies in a plane. eLt AB, be a line in a plane and C be a point on it. The point C is called internal to A and B if, the points A, C and B are different points on a line and AC + CB = AB. The points A,, C and B are also called collinear points. The set of points including A and B and all, the internal points is known as the line segment AB. The points between A and B are, called internal points., Angles, W, hen two rays in a plane meet at a point, an angle, is formed. The rays are known as the sides of the, angle and the common point as vertex. In the, figure, two rays OP and OQ make an angle POQ, at their common point O. O is the vertex of the, angle. The set of all points lying in the plane on, the Q side of OP and P side of OQ is the known as, the interior region of the POQ. The set of all, points not lying in the interior region or on any, side of the angle is called exterior region of the, angle., Straight Angle, The angle made by two opposite rays at their, common end point is a straight angle. In the, adjacent figure, a ray AC is drawn from the end, point A of the ray AB. Thus the rays AB and AC, have formed an angle BAC at their common point, A. BAC is a straight angle. The measurement of a, right angle is 2 right angles or 180 0., Adjacent Angle, If two angles in a plane have the same vertex, a, common side and the angles lie on opposite sides of, the common side, each of the two angles is said to, be an adjacent angle of the other. In the adjacent, figure, the angles BAC and CAD have the same, vertex A, a common side AC and are on opposite, sides of AC. BAC and CAD are adjacent angles.
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10, 9, , Math, , Perpendicular and Right Angle, The bisector of a straight angle is called, perpendicular and each of th related adjacent angle a, right angle. In the adjacent figure two angles BAC, and CAD are produced at the point A of BD. The, angles BAC and CAD are equal and lie on, opposite sides of the common side AC. E, ach of the, angles BAC and CAD is a right angle and the line, segments BD and AC are mutually perpendicular., Acute and Obtuse Angles, An angle which is less than a right angle is called an, acute angle and an angle greater than one right angle, but less than two right angles is an obtuse angle. In, the figure, AOC is an acute angle and AOD is an, obtuse angle, Reflex Angle, An angle which is greater than two right angles and, less than four right angles is called a reflex angle. In, the figure, AOC is a reflex angle., Complementary Angles, If the sum of two angles is one right angle, the two, angles are called complementary angles. In the, adjacent figure, AOB is a right angle. The ray OC, is in the inner side of the angle and makes two, angles AOC and COB. S, o taking together the, measurement of these two angles is one right angle., The angles AOC and COB are complementary, angles., Supplementary Angles, If the sum of two angles is 2 right angles, two angles are, called supplementary angles. The point O is an internal, point of the line AB. OC is a ray which is different from, the ray OA and ray OB. As a result two angles AOC, and COB are formed. The measurement of this two, angles is equal to the measurement of the straight angle, AOB i.e., two right angles. The angles AOC and, COB are supplementary angles., , , , ,
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110, , Math, , Opposite Angles, Two angles are said to be the opposite angles if the, sides of one are the opposite rays of the other., In the adjoining figure OA and OB are mutually, opposite rays. S, o are the rays OC and OD. The, angles AOC and BOD are a pair of opposite, angles. S, imilarly, BOC and DOA are another pair, of opposite angles. Therefore, two intersecting lines, produce two pairs of opposite angles., Theorem 1, The sum of the two adjacent angles which a ray makes with a straight line on its, meeting point is equal to two right angles., eLt, the ray OC meets the straight line AB at O . As a, result two adjacent angles AOC and COB are, formed. Draw a perpendicular DO on AB ., S, um of the adjacent two angles = AOC + COB =, AOD + DOC + COB, = AOD + DOB, = 2 right angles. [P, roved], Theorem 2, When two straight lines intersect, the vertically opposite angles are equal., eLt AB and CD be two straight lines, which intersect, at O . As a result the angles AOC , COB , BOD ,, AOD are formed at O. AOC = opposite BOD, and COB = opposite AOD ., 6⋅4 Parallel lines, Alternate angles, corresponding angles and interior angles of the traversal, , In the figure, two straight lines AB and CD are cut by a straight line EF at P and Q., The straight line EF is a traversal of AB and CD. The traversal has made eight, angles 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 with the lines AB and CD. Among the, angles
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Math, , 111, , (a) 1 and 5 , 2 and 6 , 3 and 7 , 4 and 8 are corresponding angles,, (b) 3 and 6 , 4 and 5 are alternate angles,, (c) 4 , 6 are interior angles on the right, (d) 3 , 5 are interior angles on the left., In a plane two straight lines may intersect or they are parallel. The lines intersect if, there exists a point which is common to both lines. tOherwise, the lines are parallel., ote that two different straight lines ma y atmost have only one point in common., N, The parallelism of two straight lines in a plane may be defined in three different, ways:, (a) The two straight lines never intersect each other ( even if extended to infinity), (b) vEery point on one line lies at equal smallest distance from the other., (c) The corresponding angles made by a transversal of the pair of lines are equal., According to definition (a) in a plane two straight lines are parallel, if they do not, intersect each other. Two line segments taken as parts of the parallel lines are also, parallel., According to definition (b) the perpendicular distance of any point of one of the, parallel lines from the other is always equal. ePrpendicular distance is the length of, the perpendicular from any point on one of the lines to the other. Conversely, if the, perpendicular distances of two points on any of the lines to the other are equal, the, lines are parallel. This perpendicular distance is known as the distance of the parallel, lines., The definition (c) is equivalent to the fifth postulate of E, uclid. This definition is, more useful in geometrical proof and constructions., bOserve that, through a point not on a line, a unique line parallel to it can be drawn., Theorem 3, When a transversal cuts two parallel straight lines,, (a) the pair of alternate angles are equal., (b) that pair of interior angles on the same side of the transversal are, supplementary., In the figure AB ll CD and the transversal PQ, intersects them at E and F respectively. Therefore,, (a) PEB = corresponding EFD [by definition], (b) AEF= alternate EFD, (c) BEF + EFD == 2 right angles., Activity:, 1. sUing alternate definitions of parallel lines prove the theorems related to parallel, straight lines.
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Math, , 112, , Theorem 4, When a transversal cuts two straight lines, such that, (a) pairs of corresponding angles are equal, or, (b) pairs of alternate interior angles are equal, or, (c) pairs of interior angles on the same side of the transversal are or equal to, the, sum of two right angles the lines are parallel., In the figure the line PQ intersects the straight lines, AB and CD at E and F respectively and, (a) PEB = alternate EFD, or, (b) AEF = Corresponding EFD, or, (c) BEF EFD = 2 right angles., Therefore, the straight lines AB and CD are parallel., Corollary 1. The lines which are parallel to a given line are parallel to each, other., , Exercise 6.2, 1., 2., 3., 4., , Define interior and exterior of an angle., If there are three different points in a line, identify the angles in the figure., Define adjacent angles and locate its sides., Define with a figure of each: opposite angles, complementary angle,, supplementary angle, right angle, acute and obtuse angle., 6.5 Triangles, A triangle is a figure closed by three line segments. The line segments are known as, sides of the triangle. The point common to any pair of sides is the vertex. The sides form, angles at the vertices. A triangle has three sides and three angles. Triangles are classified, by sides into three types: equilateral, isosceles and scalene. By angles triangles are also, classified into three types: acute angled, right angled and obtuse angled., The sum of the lengths of three sides of the triangle is the perimeter. By triangle we, also denote the region closed by the sides. The line segment drawn from a vertex to, the mid-point of opposite side is known as the median. Again, the perpendicular, distance from any vertex to the opposite side is the height of the triangle., In the adjacent figure ABC is a triangle. A, B, C are, three vertices. AB , BC , CA are three sides and, BAC , ABC , BCA are three angles of the triangle., The sum of the measurement of AB , BC and CA is the, perimeter of the triangle.
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Math, , Equilateral Triangle, An equilateral triangle is a triangle of three equal sides., In the adjacent figure, triangle ABC is an equilateral, triangle; because, AB = BC = CA i.e., the lengths of, three sides are equal., Isosceles Triangle, An isosceles triangle is triangle with two equal sides. In, the adjacent figure triangle ABC is an isosceles, triangle; because AB = AC z BC i.e., the lengths of, only two sides are equal., Scalene Triangle, Sides of scalene triangle are unequal. Triangle ABC is, a scalene triangle, since the lengths of its, sides AB , BC , CA are unequal., , Acute Angled Triangle, A triangle having all the three angles acute is acute, angled triangle. In the triangle ABC each of the angles, BAC , ABC and BCA is acute i.e., the, measurement of any angle is less than 90q. So 'ABC is, acute angled., Right Angled Triangle, A triangle with one of the angles right is a right angled, triangle. In the figure, the DFE is a right angle; each, of the two other angles DEF and EDF are acute., The triangle 'DEF is a right angled triangle., Obtuse angled triangle, A triangle having an angle obtuse is an obtuse angled, triangle. In the figure, the GKH is an obtuse angle;, the two other angles GHK and HGK are acute., 'GHK is an obtuse angled triangle., 0DWK,;;)RUPD, , 3, 1
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4, 1, , Math, , 9.3 Interior and Exterior Angles, If a side of a triangle is produced, a new angle is formed. This angle is known as, exterior angle. Except the angle adjacent to the exterior angle, the two others angles, of the triangle are known as opposite interior angles., In the adjacent figure, the side BC of 'ABC is produced, to D. The angle ACD is an exterior angle of the triangle., ABC, BAC and ACB are three interior angles., ACB is the adjacent interior angle of the exterior angle, ACD . Each of ABC and BAC is an opposite interior, angle with respect to ACD ., Theorem 5, The sum of the three angles of a triangle is equal to two right angles., , eLt ABC be a triangle. In the triangle BAC + ABC + ACB =2right angles., Corollary 1: If a side of a triangle is produced then exterior angle so formed is equal, to the sum of the two opposite interior angles., Corollary 2: If a side of a triangle is produced, the exterior angle so formed is, greater than each of the two interior opposite angles., Corollary 3: The acute angles of a right angled triangle are complementary to each, other., Activity :, .1 Prove that if a side of a triangle is produced, the exterior angle so formed is, greater than each of the two interior opposite angles., Congruence of Sides and Angles, If two line segments have the same length, they, are congruent. Conversely, if two line segments, are congruent, they have the same length., If the measurement of two angles is equal, the, angles are congruent. Conversely, if two angles, are congruent, their measurement is the same., Congruence of Triangles, If a triangle when placed on another exactly covers the other, the triangles are, congruent. The corresponding sides and angles of two congruent triangles are equal.
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5, 1, , Math, , In the figure, 'ABC and 'DEF are congruent., If two triangles ABC and DEF are congruent, by, superposition of a copy of ABC on DEF we find, that each covers the other completely. . Hence,, the line segments as well angles are congruent., We would express this as 'ABC # 'ABC ., Theorem 6 (SAS criterion), If two sides and the angle included between them of a triangle are equal to two, corresponding sides and the angle included between them of another triangle,, the triangles are congruent., eLt ABC and DEF be two triangles in, which AB = DE, AC = DF and the, included BAC =the included, EDF . Then 'ABC # 'DEF ., Theorem 7, If two sides of a triangle are equal, the angles opposite the equal sides are also, equal., Suppose in the triangle ABC, AB = AC, then ABC = ACB ., , Theorem 8, If two angles of a triangle are equal, the sides opposite the equal angles are also, equal., In the triangle ABC ABC, be proved that AB AC ., , ACB . It is to, , Proof, Steps, (1, ) If AB, , Justification, , AC is not equal to AB ,, , (i) AB ! AC or, (ii) AB AC ., , [The base angles of an, isosceles triangles are equal], , Suppose, AB ! AC . Cut from AB a part AD, equal to AC . Now, the triangle ADC is an, isosceles triangle. So, ADC ACD, In 'DBC Exterior angle ADC > ABC, ? ACD > ABC Therefore, ACB > ABC, But this is against the given condition., , [Exterior angle is greater than, each of the interior opposite, angles]
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6, 1, , Math, , (2) Similarly, (ii) If AB AC , it can be proved, that ABC > ACB ., But this is also against the condition,, (3) So neither AB ! AC nor AB AC, ? AB AC (Proved), Theorem 9 (SSS criterion), If the three sides of one triangle are equal to the three corresponding sides of, another triangle, the triangles are congruent., In 'ABC and 'DEF , AB = DE, AC = DF, and BC=EF,, , 'ABC # 'ABC ., Theorem 10 (ASA criterion), If two angles and the included side of a triangle are equal to two corresponding, angles and the included side of another triangle, the triangles are congruent., eLt ABC and DEF be two triangles in which, the B = E , C = F . and the side BC, = the corresponding side EF, then the, triangles are congruent, i.e. 'ABC # 'DEF ., Theorem 11 (HSA criterion), If the hypotenuse and one side of a right-angled triangle are respectively equal, to the hypotenuse and one side of another right-angled triangle, the triangles, are congruent., , eLt ABC and DEF be two right angled triangles, in which the hypotenuse AC=, hypotenuse DF and AB = DE, then 'ABC # 'DEF . ., Theorem 12, If one side of a triangle is greater than another, the angle opposite the greater, side is greater than the angle opposite the lesser sides., eLt ABC be a triangle whose AC >AB, then, ABC > ACB .
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7, 1, , Math, , Theorem 13, If one angle of a triangle is greater than another, the side opposite the greater, angle is greater than the side opposite the lesser., eLt ABC be a triangle in which ABC > ACB., It is required to prove that, AC > AB., Proof :, Steps, , Justification, , (1, )If the side AC is not greater than AB,, (i) AC = AB or, (ii) AC < AB, (i) if AC=AB ABC = ACB which is against the [The base angles of, supposition, since by supposition ABC > ACB isosceles triangle are equal], (ii) Again if AC < AB, ABC < ACB, But this is also against the supposition., , [ The angle opposite to, smaller side is smaller], , (2, ) Therefore, the side AC is neither equal to nor, less than AB. ? AC > AB (Proved)., There is a relation between the sum or the differece of the lengths of two sides and, the length of the third side of a triangle., Theorem 14, The sum of the lengths of any two sides of a triangle is greater than the third side., eLt ABC be a triangle. Then any two of its sides, are together greater than the third side. eLt BC to, be the greatest side. Then AB+AC > BC., Corollary 1. The difference of the lengths of any two sides of a triangle is, smaller than the third side., eLt ABC be a triangle. Then the difference of the lengths of any two of its sides is, smaller than the length of third side i.e. AB AC < BC., Theorem 15, The line segment joining the mid-points of any two sides of a triangle is parallel, to the third side and in length it is half.
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8, 1, , Math, , eLt ABC be a triangle and D and E are respectively midpoints of the AB and AC. It is required to prove, 1, that DE | BC and DE, BC, 2, Construction: oJ in D and E and extend to F so that EF = DE., Proof :, Steps, , Justification, , (1, ) Between 'ADE and 'CEF , AE, DE EF, AED CEF, 'ADE # 'CEF, ? ADE, , EFC and DAE, , EC, , [given ], [by construction], [opposite angles], [SAS theorem], , ECF ., , ? DF | BC or, DE | BC ., (2, ) Again, DF, , BC ǡ DE EF, , BC ǡ DE DE, , BC ǡ DE, , 1, BC, 2, , Theorem 16 (Pythagoras theorem), In a right-angled triangle the square on the hypotenuse is equal to the sum of, the squares of regions on the two other sides., eLt in the triangle ABC, ABC =1right, angle and AC is the hypotenuse., Then AC 2 AB 2 BC 2 ,, , Exercise 6.3, .1 The lengths of three sides of a triang le are given below. In which case it is, possible to draw a triangle?, (a) 5 cm, 6 cm and 7cm (b) 3 cm, 4 cm and 7cm, (c) 5 cm, 7cm and 41 cm (a) 2cm, 4 cm and 8cm, .2 Consider the following information:, (i) A right angled triangle is a triangle with each of three angles right angle., (ii) An acute angled triangle is a triangle with each of three angles acute., (iii) A triangle with all sides equal is an equilateral triangle.
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Math, , Which one of the following is correct?, (a) i and ii, (b) i and iii, (c) ii and iii, sUe the figure below to answer questions 3 and 4., , 9, 1, , (d) i, ii and iii, , 3. Which one is a right angle?, (a) BOC (b) BOD (c) COD (d) AOD, 4. What is the angle complementary to BOC?, (a) AOC (b) BOD (c) COD (d) AOD, 5. Prove that, the triangle formed by joining the middle points of the sides of an, equilateral triangle is also equilateral., 6. Prove that, the three medians of an equilateral triangle are equal., .7 Prove that, the sum of any two exterior angles of a triangle is greater than two righ t, angles., 8. D is a point inside a triangle ABC. Prove that, AB + AC > BD + DC., 9. If D is the middle point of the side BC of the triangle ABC, prove that, AB + AC, > 2AD., 01. Prove that, the sum of the three medians of a triangle is less than its perimeter., .1A is the vertex of an isosceles triangle ABC, and the side BA is produced to D, such that BA = AD; prove that BCD =1right angle., .21The bisectors of the angles B and C of a triangle ABC intersect at O. Prove, that BOC =90°+½, A., 13. If the sides AB and AC of a triangle ABC are produced and the bisector of the, exterior angles formed at B and C meet at O, prove that, BOC =90° ½ A., 41. In the adjoining figure, C =1right angle and B =2 A., Prove that, AB = 2BC., , 51. Prove that, the exterior angle so formed by producing any side of a triangle is, equal to the sum of the interior opposite angles.
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0, 2, 1, , Math, , 61. Prove that, the difference between any two sides of a triangle is less than the, third., .71In the adjoining figure, B =1right angle and, D is the middle point of the hypotenuse AC of the triangle ABC. Prove that. BD =½ AC., , .81In the, ' ABC, AB > AC and the bisector AD of A intersects BC at D. Prove, that ADB is an obtuse angle., 91. Show that, any point on the perpe ndicular bisector of a line segment is, equidistant from the terminal points of that line segment., 02. In the right-angled triangle A =1right angle and D is the mid point of BC., a. Draw a triangle ABC with given information., b. Prove that AB + AC > 2 AD , 1, c. Prove that, AD, BC, 2
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Chapter Seven, , Practical Geometry, In the previous classes geometrical figures were drawn in proving different, propositions and in the exercises. There was no need of precision in drawing these, figures. But sometimes precision is necessary in geometrical constructions. oFr, example, when an architect makes a design of a house or an engineer draws different, parts of a machine, high precision of drawing is required. In such geometrical, constructions, one makes use of ruler and compasses only. We have already learnt, how to construct triangles and quadrilaterals with the help of ruler and compasses. In, this chapter we will discuss the construction of some special triangles and, quadrilaterals., At the end of the chapter, the students will be able to –, ¾ Explain triangles and quadrilaterals with the help of figures, ¾ Construct triangle by using given data, ¾ Construct parallelogram by using given data., 7⋅1 Construction of Triangles, Every triangle has three sides and three angles. But, to specify the shape and size of a, triangle, all sides and angles need not to be specified. oFr example, as sum of the, three angles of a triangle is two right angles, one can easily find the measurement of, the third angle when the measurement of the two angles of the triangle given. Again,, from the theorems on congruence of triangles it is found that the following, combination of three sides and angles are enough to be congruent. That is, a, combination of these three parts of a triangle is enough to construct a unique triangle., In class seven we have learnt how to construct triangles from the following data:, (1, ) Three sides, , (2) Two sides and their, included angle., , 0DWK,;;)RUPD
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2, 1, , Math, , (3) Two angles and their, adjacent sides, , (4) Two angles and an, opposite side, , (5) Two sides and an, opposite angle, , (6) Hypotenuse and a, side of a right-angled, triangle, , Observe that in each of the cases above, three parts of a triangle have been specified., But any three parts do not necessarily specify a unique triangle. As for example, if, three angles are specified, infinite numbers of triangles of different sizes can be, drawn with the specified angles (which are known as similar triangles)., , Sometimes for construction of a triangle three such data are provided by which we, can specify the triangle through various drawing. Construction in a few such cases is, stated below.
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3, 2, 1, , Math, , Construction 1, The base of the base adjacent angle and the sum of other two sider of a triangle, are given. Construct the triangle., eLt the base a, a base adjacent angle x and the, sum s of the other two sides of a triangle ABC be, given. It is required to construct it., Steps of construction :, (1, )F, rom any ray BE cut the line segment BC, equal to a. At B of the line segment BC, draw an, angle CBF = x., (2) Cut a line segment BD equal to s from, the ray BF., (3) Join C,D and at C make an angle DCG, equal to BDC on the side of DC in which B, lies., (4) eLt the ray CG intersect BD at A., Then, ABC is the required triangle., Proof : In 'ACD , ADC ACD [by construction], ? AC, AD., Now, In 'ABC , ABC x, BC a, [by construction], and BA AC BA AD BD s . Therefore, 'ABC is the required triangle., Alternate Method, eLt the base a, a base adjacent angle x and the, sum s of the other two sides of a triangle ABC be, given. It is required to construct the triangle., Steps of construction:, (1, )F, rom any ray BE cut the line segment BC, equal to a. At B of the line segment BC draw an, angle CBF = x., (2) Cut a line segment BD equal to s from, the ray BF., (3) Join C,D and construct the perpendicular, bisector PQ of CD., (4) eLt the ray PQ intersect BD at A. Join A, C., Then, ABC is the required triangle., Proof: In 'ACR and 'ADR , CR, , DR AR, , AR and the included angle
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4, 2, 1, , Math, , ARC ARD [right angle ], 'ACR # 'ADR . ? AC AD, Now, In 'ABC , ABC x, BC, , a, [by construction ], , and BA AC BA AD BD s . Therefore, 'ABC is the required triangle., Construction 2, The base of a triangle the base adjacent an acute angle and the difference of the, other two sides are given. Construct the triangle., eLt the base a, a base adjacent angle x and, the difference d of the other two sides of a, triangle ABC be given. It is required to, construct the triangle., Steps of Construction :, (1, ) rFom any ray BE, cut the line segment BC ,, equal to a. At B of the line segment BC draw an, angle CBF = x., (2) Cut a line segment BD equal to s from the, ray BE., (3) Join C,D and at C, make an angle DCA, equal to EDC on the side of DC in which C, lies. eLt the ray CA intersect BE at A., Then ABC is the required triangle., Proof :, In 'ACD , ADC ACD [by construction], ? AC, AD., So, the difference of two sides AB AC AB AD BD d ., Now, In 'ABC , BC a, AB AC d and ABC x. Therefore, 'ABC is the, required triangle., Activity :, .1 If the given angle is not acute, the above construction is not possible. Why ?, Explore any way for the construction of the triangle under such circumstances., .2The base, the base adjacent angle and the difference of the other two sides of a, triangle are given. Construct the triangle in an alternate method., Construction 3, The base adjacent two angles and the perimeter of a triangle are given., Construct the triangle., eLt the base adjacent angles x and y and the perimeter p be geven. It is required, to construct the triangle.
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5, 2, 1, , Math, , Steps of Construction :, (1, ) rFom any ray DF, cut the part DE equal to the, perimeter p. Make angles EDL equal to x and, DEM equal to y on the same side of the line, segment DE at D and E., (2, ) Draw the bisectors BG and EH of the two angles., (3) eLt these bisectors DG and EH intersect at a, point A. At the point A, draw ∠DAB equal to ∠ADE, and ∠EAC equal to ∠AED., (4) eLt AB intersect DE at B and AC intersect DE at, C., Then, 'ABC is the required triangle., Proof : In 'ADB , ADB DAB [by construction], ? AB, Again, in 'ACE , AEC EAC ; ? CA CE., Therefore, in 'ABC , AB BC CA DB BC CE DE p., 1, 1, ABC ADB DAB, x x x, 2, 2, and ACB AEC EAC, , 1, 1, y y, 2, 2, , DB., , y. Therefore, 'ABC is the required triangle., , Activity:, 1. Two acute base adjacent angles and the perimeter of a triangle are given. Construct, the triangle in an alternative way., Example 1. Construct a triangle ABC, in which B =60°,, perimeter AB + BC + CA = 1cm., H, , C =45° and the, , G, , Steps of Construction: oFllow the steps below :, (1, ) Draw a line segment PQ =1cm., (2, ) At P, construct an angle of QPL =60°and at ,Qan angle of, PQM =45°on, the same side of PQ., (3) Draw the bisectors PG and QH of the two angles. eLt the bisectors PG and QH, of these angles intersect at A., (4) Draw perpendicular bisector of the segments PA of QA to intersect PQ at B and C.
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6, 2, 1, , Math, , (5) Join A, B and A, C., Then, ABC is the required triangle., Activity : An adjacent side with the right angle and the difference of hypotenuse, and the other side of a right-angled tr iangle are given. Construct the triangle., , Exercise 7.1, 1. Construct a triangle with the following data:, (a) The lengths of three sides are 3 cm, 35 cm, 28cm., (b) The lengths of two sides are 4 cm , 3 cm and the included angle is 60q., (c) Two angles are 60q and 45q and their included side is 5 cm., (d) Two angles are 60q and 45q and the side opposite the angle 45q is 5 cm., (e) The lengths of two sides are 4.5 cm and 3.5 respectively and the angle, opposite to the second side is 4 cm., (f) The lengths of the hypotenuse and a side are 6 cm and 4 cm respectively., 2. Construct a triangle ABC with the following data:, (a) Base 3.5 cm, base adjacent angle 60q and the sum of the two other sides 8cm., (b) Base 4 cm, base adjacent angle 50q and the sum of the two other sides .75 cm., (c) Base 4 cm, base adjacent angle 50q and the difference of the two other sides .15 cm., (d) Base 5 cm, base adjacent angle 45q and the difference of the two other sides 1cm., (e) Base adjacent angles 60q and 45q and the perimeter1, 2cm., (f) Base adjacent angles 30q and 45q and the perimeter 01 cm., 3. Construct a triangle when the two base adjacent angles and the length of the, perpendicular from the vertex to the base are given., 4. Construct a right-angled triangle when the hypotenuse and the sum of the other, two sides are given., 5. Construct a triangle when a base adjacent angle, the altitude and the sum of the, other two sides are given., 6. Construct an equilateral triangle whose perimeter is given., .7 The base, an obtuse base adjacent angl e and the difference of the other two sides, of a trangle are given. Construcet the triangle., 7⋅2 Construction of Quadrilaterals, We have seen if three independent data are given, in many cases it is possible to, construct a definite triangle. But with four given sides the construction of a, definite quadrilateral is not possible. iFve independent data are required for, construction of a definite quadrilateral. A definite quadrilateral can be, constructed if any one of the following combinations of data is known :
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Math, , 7, 2, 1, , (a) oFur sides and an angle, (b) oFur sides and a diagonal, (c) Three sides and two diagonals, (d) Three sides and two included angles, (e) Two sides and three angles., In class V, III, the construction of quadrila terals with the above specified data has, been discussed. If we closely look at the steps of construction, we see that in some, cases it is possible to construct the quadrilaterals directly. In some cases, the, construction is done by constructions of triangles. Since a diagonal divides the, quadrilateral into two triangles, when one or two diagonals are included in data,, construction of quadrilaterals is possible through construction of triangle., (1, ) oFur sides and an angle, , (2, ) oFur sides and a diagonal, , (3) Three sides and two diagonals, , (4) Three sides and two included, angles, , (5) Two sides and three angles, ,
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8, 2, 1, , Math, , Sometimes special quadrilaterals can be constructed with fewer data. In such a cases,, from the properties of quadrilaterals, we can retrieve five necessary data. oFr, example, a parallelogram can be constructed if only the two adjacent sides and the, included angle are given. In this case, only three data are given. Again, a square can, be constructed when only one side of the square is given. The four sides of a square, are equal and an angle is a right angle; so five data are easily specified., Construction 4, Two diagonals and an included angle between them of a parallelogram are, given. Construct the parallelogram., eLt a and b be the diagonals of a parallelogram and x be, an angle included between them. The parallelogram is to, be constructed., Steps of construction:, rFom any ray AE, cut the line segment AC = a. Bisect the, line segment AC to find the mid-point O, At O construct, the angle AOP = x and extend the ray PO to the, opposite ray OQ. rFom the rays OP and OQ cut two line, 1, segments OB and OD equal to, b. Join A,B; A,D; C,B, 2, and C,D. Then ABCD is the required parallelogram., Proof: In triangles 'AOB and 'COD ,, 1, 1, OA OC, a, OB OD, b [by construction], 2, 2, and included AOB i=ncluded COD o[ pposite angle], Therefore, 'AOB # 'COD ., So, AB CD, and ABO CDO ;but the two angles are alternate, angles., ? AB and CD are parallel and equal., Similarly, AD and BC are parallel and equal., Therefore, ABCD is a parallelogram with diagonals, 1, 1, AC AO OC, a a a, 2, 2, 1, 1, and BD BO OD, b b b and the angle included, 2, 2, between the diagonals AOB x ., Therefore, ABCD is the required parallelogram.
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Math, , 9, 2, 1, , Construction 5, Two diagonals and a side of a parallelogram are given. Construct the, parallelogram., eLt a and b be the diagonals and c be a side of the, parallelogram. The parallelogram is to be constructed., Steps of construction:, Bisect the diagonals a and b to locate their mid-points., rFom any ray AX, cut the line segment AB = a. With, 2, b, centre at A and B draw two arcs with radii, and, a, 2, respectively on the same side of AB. eLt the arcs, intersect at O. Join A, O and O, B. Extend AO and BO, 2, and OD =, to AE and BF respectively. Now cut OC =, a, b, from OE and OF respectively. Join A,D; D,C; C,B., 2, Then ABCD is the required parallelogram., Proof: In 'AOB and 'COD ,, a, b, OA OC, ; OB OD, , [by construction], 2, 2, and included AOB =included COD o[ pposite angle], ? 'AOB # 'COD ., ? AB CD and ABO ODC ; but the angles are, alternate angles., ? AB and CD are parallel and equal., Similarly, AD and BC are parallel and equal., Therefore, ABCD is the required parallelogram., Example 1. The parallel sides and two angles included with the larger side of a, trapezium are given. Construct the trapezium., eLt a and b be the parallel sides of a trapezium where a > b and x and y be two, angles included with the side a. The trapezium is to be constructed., , 0DWK,;;)RUPD
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30, 1, , Math, , Steps of construction:, rFom any ray AX, cut the line segment AB = a. At A of the line segment AB,, construct the angle BAY = x and at B, construct the angle ABZ = y. rFom the, line segment AB, cut a line segment AE=b. Now at E, construct BC || AY which cuts, BZ at C. Now construct CD || BA. The line segment CD intersects the ray AY at D., Then ABCD is the required trapezium., Proof : By construction, AB | CD and AD | EC. Therefore, AECD is a parallelogram, and CD = AE = b. Now in the quadrilateral ABCD, AB=a, CD=b, AB || CD and, BAD = x, ABC = y (by construction). Therefore, ABCD is the required, trapezium., Activity : The perimeter and an angle of a rhombus are given. Construct the, rhombus., , Exercise 7⋅2, .1 The two angles of a right angled triangle are given. Which one of the following, combination allows constructing the triangle?, a. 63q and 36q, b. 30q and 07 q, c. 40q and 50q, d. 80 q and 02 q, .2 i. A rectangle is a parallelogram, ii. A square is a rectangle, iii. A rhombus is a square, On the basis of the above information, which one of the following is true?, a. i and ii b. i and iii, c. ii and iii, d. i, ii and iii, In view of the given figure, answer the questions 3 and 4., , 3. What is the area of 'AOB?, a. 6 sq. units, b. 7sq. units, c. 12sq. units, d.14 sq. units, 4. The perimeter of the quadrilateral is, a. 21units, b. 14 units, c. 20 units, d. 28units, 5. Construct a quadrilateral with the following data :, (a) The lengths of four sides are 3 cm, 35 cm, .25 cm, 3cm and an angle is 45 q., (b) The lengths of four sides are 3.5 cm, 4 cm, 25 cm 3.5 cm and a diagonal is 5 cm., (c) The lengths of three sides are 32cm, 3 cm, 3 5 cm and two diagonals are, 28cm, and 4.5 cm.
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Math, , 31, 1, , 6. Construct a parallelogram with the following data:, (a) The lengths of two diagonals are 4 cm, 65 cm and the included angle is 45q., (b) The lengths of two diagonals are 5 cm, 65 cm and the included angle is 30q., (c) The length of a side is 4 cm and the lengths of two diagonals are 5 cm and 65 cm., (d) The length of a side is 5 cm and the lengths of two diagonals are 4.5 cm and 6 cm., .7 The sides AB and BC and the angles B, C, D of the quadrilateral ABCD are, given. Construct the quadrilateral., .8 The four segments made by the in tersecting points of the diagonals of a, parallelogram and an included angle between them are OA =4 cm. OB =5 cm., 8 q respectively. Construct the, OC 35 cm OD =4 5 cm and AOB =0, quadrilateral., 9. The length of a side of a rhombus and an angle are 3.5 cm. and 45q respectively;, construct the rhombus., 01. The length of a side and a diag onal of a rhombus are given;construct the, rhombus., .1The length of two diagonals of a rh ombus are given. Construct the rhombus., .21The perimeter of a square is given. Construct the square., 31. The houses of Mr. oZki and Mr. aZfrul, are in the same boundary and the area of, their house is equal. But the house of Zoki is rectangular and the house of Mr., aZfrul is in shape of parallelogram., (a) Construct the boundary of each of their houses taking the length of base 10, units and height 8units., (b) Show that the area of the house of Mr. oZki is less than the area of the house, of Mr. aZfrul., (c) If the ratio of the length and the breadth of the house of Mr. oZki is 4:3 an d, its area is 300 sq. units, find the ratio of the area of their houses., 41. The lengths of the hypotenuse and a side of right angled triangle are 7cm and 4, cm. sUe the information to answer the following questions :, a. iFnd the length of the other side of the triangle., b. Construct the triangle., c. Construct a square whose perimeter is equal to the perimeter of the triangle., 15. AB =4 cm, BC =5 cm A =58 o, B =08 o and C =95 o. of the quadrilateral, ABCD. sUe the information to answer the following questions:, a. Construct a rhombus and give the name., b. sUe the above information to construct the quadrilateral ABCD., c. Construct an equilateral triangle whose perimeter is equal to the perimeter of the, quadrilateral ABCD.
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Chapter Eight, , Circle, We have already known that a circle is a geometrical figure in a plane consisting of, points equidistant from a fixed point. Different concepts related to circles like centre,, diameter, radius, chord etc has been discussed in previous class. In this chapter, the, propositions related to arcs and tangents of a circle in the plane will be discussed., At the end of the chapter, the students will be able to, ¾ Explain arcs, angle at the centre, angle in the circle, quadrilaterals, inscribed in the cirlce, ¾ Prove theorems related to circle, ¾ State constructions related to circle., 8⋅1 Circle, A circle is a geometrical figure in a plane whose points are equidistant from a fixed point., The fixed point is the centre of the circle. The closed path traced by a point that keeps it, distance from the fixed centre is a circle. The distance from the centre is the radius of the, circle., eLt O be a fixed point in a plane and r be a fixed measurement., The set of points which are at a distance r from O is the circle, with centre O and radius r. In the figure, O is the centre of the, circle and A, B and C are three points on the circle. Each of OA,, OB and OC is a radius of the circle. Some coplanar points are, called concylcic if a circle passes through these points, i.e. there, is a circle on which all these points lie. In the above figure, the, points A, B and C are concyclic., Interior and Exterior of a Circle, If O is the centre of a circle and r is its radius, the set of all, points on the plane whose distances from O are less than r, is, called the interior region of the circle and the set of all points, on the plane whose distances from O are greater than r, is, called the exterior region of the circle. The line segment, joining two points of a circle lies inside the circle., The line segment drawn from an interior point to an exterior point of a circle, intersects a circle at one and only one point. In the figure, P and Q are interior and, exterior points of the circle respectively. The line segment PQintersects the circle at, Ronly.
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33, 1, , Math, , Chord and Diameter of a Circle, The line segment connecting two different points of a circle is a, chord of the circle. If the chord passes through the centre it is, known as diameter. That is, any chord forwarding to the centre, of the circle is diameter. In the figure, AB and AC are two, chords and O is the centre of the circle. The chord AC is a, diameter, since it passes through the centre. OA and OC are two, radii of the circle. Therefore, the centre of a circle is the midpoint of any diameter. The length of a diameter is 2r, where r is, the radius of the circle., Theorem 1, The line segment drawn from the centre of a circle to bisect a chord other than, diameter is perpendicular to the chord., eLt AB be a chord (other than diameter) of a circle ABC with, centre O and M be the midpoint of the chord. Join O, M. It is, to be proved that the line segment OM is perpendicular to the, chord AB., Construction: Join O, A and O, B., Proof:, Steps, (1, ) In 'OAM and 'OBM,, OA = OB, AM = BM, and OM = OM, Therefore, 'OAM # 'OBM, ? OMA=OMB, (2, ) Since the two angles are equal and together, maka a straight angle., , Justification, [M is the mid point of AB], [radius of same circle], [common side], [SSS theorem], , OMA OMB =1right angle., Therefore, OM A AB . (Proved)., , Corollary 1: The perpendicular bisector of any chord passes through the centre of, the circle., Corollary 2: A straight line can not intersect a circle in more than two points., Activity :, 1. The theorem opposite of the theorem 1st ates that the perpendicular from, the centre of a circle to a chord bisects the chord. Prove the theorem.
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34, 1, , Math, , Theorem 2, All equal chords of a circle are equidistant from the centre., eLt AB and CD be two equal chords of a circle with, centre O. It is to be proved that the chords AB and CD, are equidistant from the centre., Construction: Draw from O the perpendiculars OE, and OF to the chords AB and CD respectively. Join, O, A and O, C., Proof:, Steps, , Justification, , (1, ) OE A AB and OF A CD, Therefore, AE = BE and CF = BF., 1, 1, AB and CF =, CD, ?AE =, 2, 2, , [The perpendicular from, the centre bisects the, chord], , (2, ) But AB = DC, ?AE = CF, , [supposition], , (3) Now in the right-angled triangles 'OAE and, 'OCF, hypotenuse OA =hypotenuse OC and AE = CF, ? 'OAE # 'OCF, , [radius of same circle], [Step ]2, [R, HS theorem], , ?OE = OF, (4) But OE and OF are the distances from O to the, chords AB and CD respectively., Therefore, the chords AB and CD are equidistant from, the centre of the circle. (Proved), Theorem 3, Chords equidistant from the centre of a circle are equal., eLt AB and CD be two chords of a circle with centre, O. OE and OF are the perpendiculars from O to the, chords AB and CD respectively. Then OE and OF, represent the distance from centre to the chords AB, and CD respectively., It is to be proved that if OE = OF, AB = CD., Construction: Join O, A and O, C.
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Math, , 35, 1, , Proof:, Steps, , Justification, , (1, ) Since OE A AB and OF A CD., Therefore, OEA OFC 1right angle, (2) Now in the right angled triangles, 'OAE and 'OCF, hypotenuse OA=hypotenuse OC and, OE OF, ?, 'OAE # 'OCF, ?, AE CF ., , [right angles ], , (3) AE =, , 1, 1, AB and CF = CD, 2, 2, , (4) Therefore, , [radius of same circle], [R, HS theorem], [The perpendicular from the centre, bisects the chord], , 1, 1, AB, CD, 2, 2, CD (Proved), , AB, Corollary 1: The diameter is the greatest chord of a circle., i.e.,, , Exercise 8.1, .1 Prove that if two chords of a circle bisect each other, their point of intersection is, the centre of the circle., 2. Prove that the straight line joining the middle points of two parallel chords of a, circle pass through the centre and is perpendicular to the chords., 3. Two chords AB and AC of a circle subtend equal angles with the radius passing, through A. Prove that, AB = AC., 4. In the figure, O is the centre of the circle and chord AB =chord AC. Prove that, BAO CAO., 5. A circle passes through the vertices of a right angled triangle. Show that, the, centre of the circle is the middle point of the hypotenuse., 6. A chord AB of one of the two concentric circles intersects the other circle at, points C and D. Prove that, AC = BD., .7 If two equal chords of a circle inters ect each other, show that two segments of, one are equal to two segments of the other., 8. Prove that, the middle points of e qual chords of a circle are concyclic., 9. Show that, the two equal chords drawn from two ends of the diameter on its, opposite sides are parallel.
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36, 1, , Math, , 01. Show that, the two parallel chords of a circle drawn from two ends of a diameter, on its opposite sides are equal., .1Show that, of the two chords of a circle the bigger chord is nearer to the centre, than the smaller., 8.2 The arc of a circle, An arc is the piece of the circle between any two points of the, circle. oLok at the pieces of the circle between two points A and, B in the figure. We find that there are two pieces, one, comparatively large and the other small. The large one is called, the major arc and the small one is called the minor arc., A and B are the terminal points of this arc and all other, , points are its internal points. With an internal fixed point C, the arc is called arc ABC and is expressed by the symbol, ACB. Again, sometimes minor arc is expressed by the, symbol AB. The two points A and B of the circle divide the, circle into two arcs. The terminal points of both arcs are A, and B and there is no other common point of the two arcs, , other than the terminal points., Arc cut by an Angle, An angle is said to cut an arc of a circle if, (i) each terminal point of the arc lies on the sides of the, angle, (ii) each side of the angle contains at least one terminal point, (iii) Every interior point of the arc lies inside the angle. The, angle shown in the figure cuts the APB arc of the circle, with centre O., Angle in a Circle, If the vertex of an angle is a point of a circle and each side of, the angle contains a point of the circle, the angle is said to be, an angle in the circle or an angle inscribed in the circle. The, angles in the figure are all angles in a circle. Every angle in a, circle cuts an arc of the circle. This arc may be a major or, minor arc or a semi-circle., The angle in a circle cuts an arc of the circle and the angle is, said to be standing on the cut off arc. The angle is also known, as the angle inscribed in the conjugate arc. In the adjacent, figure, the angle stands on the arc APB and is inscribed in the, conjugate arc ACB. It is to be noted that APB and ACB are, mutually conjugate., , , ,
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Math, , 137, , Remark: The angle inscribed in an arc of a circle is the angle with vertex in the arc, and the sides passing through the terminal points of the arc. An angle standing on an, arc is the angle inscribed in the conjugate arc., Angle at the Centre, The angle with vertex at the centre of the circle is called an, angle at the centre. An angle at the centre cuts an arc of the, circle and is said to stand on the arc. In the adjacent figure,, AOB is an angle at the centre and it stands on the arc APB., Every angle at the centre stands on a minor arc of the circle., In the figure APB is the minor arc. So the vertex of an angle, at the centre always lies at the centre and the sides pass, through the two terminal points of the arc., To consider an angle at the centre standing on a semi-circle, the above description is not meaningful. In the case of semicircle, the angle at the centre BOC is a straight angle and, the angle on the arc BAC is a right angle., Theorem 4, The angle subtended by the same arc at the centre is double of the angle, subtended by it at any point on the remaining part of the circle., Given an arc BC of a circle subtending angles BOC at the, centre O and BAC at a point A of the circle ABC. We need, to prove that BOC = 2 BAC., Construction: Suppose, the line segment AC does not pass, through the centre O. In this case, draw a line segment AD at, A passing through the centre O., Proof :, Steps, , Justification, , (1) In 'AOB, the external angle, BOD = BAO + ABO, (2) Also in ǻ AOB, OA = OB, Therefore, BAO = ABO, (3) From steps (1) and (2), BOD = 2BAO., (4) Similarly, COD = 2 CAO, (5) From steps (3) and (4),, BOD COD 2BAO 2CAO, This is the same as BOC 2BAC . [Proved], , [An exterior angle of a, triangle is equal to the sum of, the two interior opposite, angles.], [Radius of a circle], [Base angles of an isosceles, triangle are equal], [by adding], , 0DWK,;;)RUPD
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38, 1, , Math, , We can state the theorem in a different way. The angle standing on an arc of the, circle is half the angle subtended by the arc at the centre., Activity : Prove the theorem 4 when AC passes through the centre of the circle, ABC., Theorem 5, Angles in a circle standing on the same arc are equal., eLt O be the centre of a circle and standing on the arc BD,, BAD and BED be the two angles in the circle. We need, to prove that BAD = BED., Construction : Join O, B and O, D., Proof :, Steps, , Justification, , (1, ) The arc BD subtends an angle BOD at the centre O. [The angle subtended, by an arc at the centre, Therefore, BOD =2 BAD and BOD =2 BED, is double of the angle, ? 2 BAD =2 BED, subtended, on, the, or, BAD, BED (Proved), circle], Theorem 6, The angle in the semi- circle is a right angle., eLt AB be a diameter of circle with centre at O and ACB, is the angle subtended by a semi-circle. It is to be proved, that ACB =1right angle., Construction: Take a point D on the circle on the opposite, side of the circle where C is located., Proof:, Steps, Justification, (1, ) The angle standing on the arc ADB, 1, ACB =, (straight angle in the centre AOB), 2, (2) But the straight angle AOB is equal to 2right angles., 1, (2right angles) =1right angle. (Proved), ACB =, 2, , [The angle standing on, an arc at any point of, the circle is half the, angle at the centre], , Corollary 1. The circle drawn with hypotenuse of a right-angled triangle as diameter, passes through the vertices of the triangle., Corollary 2. The angle inscribed in the major arc of a circle is an acute angle.
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39, 1, , Math, , Activity :, .1 Prove that any angle inscribed in a minor arc is obtuse., , Exercise 8.2, 1. ABCD is a quadrilateral inscribed in a circle with centre O. If the diagonals AB, and CD intersect at the point E, prove that AOB + COD = 2 AEB., .2 Two chords AB and CD of the circle ABCD intersect at the point E. Show that,, 'AED and 'BEC are equiangular., 3. In the circle with centre O ADB + BDC =1right angle. Prove that, A, B and, C lie in the same straight line., 4. Two chords AB and CD of a circle intersect at an interior point. Prove that, the sum of, the angles subtended by the arcs AC and BD at the centre is twice AEC., 5. Show that, the oblique sides of a cyclic trapezium are equal., 6. AB and CD are the two chords of a circle ; P and Q are the middle points of the, two minor arcs made by them. The chord PQ intersects the chords AB and AC, at points D and E respectively. Show that, AD = AE., 8⋅3 Quadrilateral inscribed in a circle, An inscribed quadrilateral or a quadrilateral inscribed in a circle is a quadrilateral, having all four vertices on the circle. Such quadrilaterals possess a special property., The following avtivity helps us understand this property., Activity:, Draw a few inscribed quadrilaterals ABCD. This can easily be accomplished by, drawing circles with different radius and then by taking four arbitrary points on, each of the circles. Measure the angles of the quadrilaterals and fill in the, following table., Serial No., C, B + D, A + C, A, B, D, 1, 2, 3, 4, 5, What to you infer from the table?, Circle related Theorems, Theorem 7, The sum of the two opposite angles of a quadrilateral inscribed in a circle is two, right angles.
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40, 1, , Math, , eLt ABCD be a quadrilateral inscribed in a circle, with centre O. It is required to prove that, ABC +, ADC =2right angles and BAD + BCD =2, right angles., Construction : Join O, A and O, C., Proof :, Steps, , Justification, , (1) Standing on the same arc ADC, the angle at, centre AOC =2( ABC at the circumference), that is, AOC =2 ABC., , [The angle subtended by an, arc at the centre is double of, the angle subtended by it at, the circle], , (2) Again, standing on the same arc ABC, reflex, AOC at the centre = 2 ( ADC at the, circumference) that is, reflex AOC =2 ADC, ? AOC +reflex AOC =2( ABC + ADC), But AOC +reflex AOC =, 4 right angles, ?, 2( ABC + ADC) =4 right angles, ?, ABC + ADC =2right angles., In the same way, it can be proved that, BAD +BCD =2right angles. (Proved), , [The angle subtended by an, arc at the centre is double of, the angle subtended by it at, the circle], , Corollary 1: If one side of a cyclic quadrilateral is extended, the exterior angle, formed is equal to the opposite interior angle., Corollary 2: A parallelogram inscribed in a circle is a rectangle., Theorem 8, If two opposite angles of a quadrilateral are supplementary, the four vertices, of the quadrilateral are concyclic., eLt ABCD be the quadrilateral with ABC + ADC =2, right angles, inscribed in a circle with centre O. It is, required to prove that the four points A, B, C, D are, concyclic., Construction: Since the points A,B, C are not collinear,, there exists a unique circle which passes through these, three points. eLt the circle intersect AD at E. Join A,E.
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41, 1, , Math, , Proof :, Steps, , Justification, , (1) ABCE is a quadrilateral inscribed in the circle., Therefore, ABC AEC =2right angles., But ABC ADC =2right angles [given]., ? AEC = ADC, But this is impossible, since in 'CED , exterior AEC !, opposite interior ADC, Therefore, E and D points can not be different points., So, E must coincide with the point D. Therefore, the, points A, B, C, D are concyclic., , O, , [The sum of the two, opposite angles of an, inscribed quadrilateral, is two right angles.], [The exterior angle is, greater, than, any, opposite interior angle.], , Exercise 8.3, .1 If the internal and extern al bisectors of the angles B and C of 'ABC meet at, P and Qrespectively, prove that B, P, C, Q are concyclcic., .2 Prove that, the bisector of any angle of a cyclic quadrilateral and the exterior, bisector of its opposite angle meet on the circumference of the circle., 3. ABCD is a circle. If the bisectors of CAB and CBA meet at the point P and the, bisectors of DBA and DAB meet at Q, prove that, the four points A, Q, P, B, are concyclic., 4. The chords AB and CD of a circle with centre D meet at right angles at some, point within the circle, prove that, AOD + BOC =2right angles., 5. If the vertical angles of two triangles standing on equal bases are supplementary,, prove that their circum-circles are equal., 6. The opposite angles of the quadrilateral ABCD are supplementary to each other., If the line AC is the bisector of BAD, prove that, BC = CD., 8⋅4 Secant and Tangent of the circle, Consider the relative position of a circle and a straight line in the plane. Three, possible situations of the following given figures may arise in such a case:, (a) The circle and the straight line have no common points, (b) The straight line has cut the circle at two points, (c) The straight line has touched the circle at a point.
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42, 1, , Math, , A circle and a straight line in a plane may at best have two points of intersection. If a, circle and a straight line in a plane have two points of intersection, the straight line is, called a secant to the circle and if the point of intersection is one and only one, the, straight line is called a tangent to the circle. In the latter case, the common point is, called the point of contact of the tangent. In the above figure, the relative position of, a circle and a straight line is shown. In figure (i) the circle and the straight line PQ, have no common point; in figure (ii) the line PQ is a secant, since it intersects the, circle at two points A and B and in figure (iii) the line PQ has touched the circle at A., PQ is a tangent to the circle and A is the point of contact of the tangent., Remark : All the points between two points of intersection of every secants of the, circle lie interior of the circle., Common tangent, If a straight line is a tangent to two circles, it is called a, common tangent to the two circles. In the adjoining, figures, AB is a common tangent to both the circles. In, figure (a) and (b), the points of contact are different. In, figure (c) and (d), the points of contact are the same., If the two points of contact of the common tangent to two, circles are different, the tangent is said to be, (a) direct common tangent if the two centres of the circles, lie on the same side of the tangent and, (b) transverse common tangent, if the two centres lie on, opposite sides of the tangent., The tangent in figure (a) is a direct common one and in, figure (b) it is a transverse common tangent., If a common tangent to a circle touches both the circles at, the same point, the two circles are said to touch each other, at that point. In such a case, the two circles are said to have, touched internally or externally according to their centres, lie on the same side or opposite side of the tangent. In, figure (c) the two circles have touched each other, internally and in figure (d) externally., Theorem 9, The tangent drawn at any point of a circle is perpendicular to the radius, through the point of contact of the tangent., eLt PT be a tangent at the point P to the circle with centre O and OP is the radius, throug0h the point of contact. It is required to prove that, PT A OP.
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Math, , 43, 1, , eLt PT be a tangent at the point P to the circle with centre O, and OP is the radius throug0h the point of contact. It is, required to prove that, PT A OP., Construction: Take any point Q on PT and join O, Q., Proof :, Since PT is a tangent to the circle at the point P, hence every, point on it except P lies outside the circle. Therefore, the, point Q is outside of the circle., OQ is greater than OP that is, OQ > OP and it is true for, every point Q on the tangent PT except P. So, OP is the, shortest distance from the centre O to PT. Therefore, PT A, OP. (Proved), Corollary 1. At any point on a circle, only one tangent can be drawn., Corollary 2. The perpendicular to a tangent at its point of contact passes through the, centre of the circle., Corollary 3. At any point of the circle the perpendicular to the radius is a tangent to, the circle., Theorem 10, If two tangents are drawn to a circle from an external point, the distances from, that point to the points of contact are equal., eLt P be a point outside a circle ABC with centre O,, and two line segments PA and PB be two tangents, to the circle at points A and B. It is required t o, prove that. PA = PB., Construction: eLt us join O, A; O, B and O, P., Proof:, Steps, , Justification, , (1, ) Since PA is a tangent and OA is the radius, through the point of tangent PA A OA., ? PAO ==1right angle, Similarly, PBO ==1right angle, ? both 'PAO and 'PBO are right-angled triangles., (2) Now in the right angled triangles 'PAO and, 'PBO, hypotenuse PO =hypotenuse PO,, , [The tangent is, perpendicular to the radius, through the point of contact, of the tangent]
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Math, , 44, 1, , Remarks:, .1 If two circles touch each other external ly, all the points of one excepting the point, of contact will lie outside the other circle., .2 If two circles touch each other internally, all the points of the smaller circle, excepting the point of contact lie inside the greater circle., Theorem 11, If two circles touch each other externally, the point of contact of the tangent and, the centres are collinear., eLt the two circles with centres at A and B touch each, other externally at O. It is required to prove that the, points A,O and B are collinear., Construction: Since the given circles touch each, other at O, they have a common tangent at the point, O. Now draw the common tangent POQ at O and join, O, A and O, B., Proof: In the circles OA is the radius through the, point of contact of the tangent and POQ is the tangent., Therefore POA =1right angle. Similarly POB =1right angle, Hence POA POB =1right angle +1right angle =2right angles, or AOB =2right angles i.e., (Proved), , AOB is a straight angle. ? A,O and B are collinear., , Corollary 1. If two circles touch each other externally, the distance between their, centres is equal to the sum of their radii, Corollary 2. If two circles touch each other internally, the distance between their, centres is equal to the difference of their radii., Activity:, .1Prove that, if two circles touch each othe r internally, the point of contact of the, tangent and the centres are collinear., , Exercise 8⋅4, .1 Two tangents are drawn from an external point P to the circle with centre O., Prove that OP is the perpendicular bisector of the chord through the touch points., .2 Given that tangents PA and PB touches the circle with centre O at A and B, respectively. Prove that PO bisects APB.
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Math, , 45, 1, , 3. Prove that, if two circles are concentric and if a chord of the greater circle, touches the smaller, the chord is bisected at the point of contact., 4. AB is a diameter of a circle and BC is a chord equal to its radius. If the tangents, drawn at A and C meet each other at the point D, prove that ACD is an equilateral, triangle., 5. Prove that a circumscribed quadrilateral of a circle having the angles subtended, by opposite sides at the centre are supplementary., 8⋅5 Constructions related to circles, Construction 1, To determine the centre of a circle or an arc of a circle., Given a circle as in figure (a) or an arc of a circle as in figure (b). It is required to, determine the centre of the circle or the arc., Construction : In the given circle or the arc of the circle, three different points A, B,, C are taken. The perpendicular bisectors EF and GH of the chords AB and BC are, drawn respectively. eLt the bisectors intersect at O. The O is the required centre of, the circle or of the arc of the circle., Proof: By construction, the line segments EF and GH are the, perpendicular bisectors of chords AB and BC respectively. But, both EF and GH pass through the centre and their common, point is O. Therefore, the point O is the centre of the circle or, of the arc of the circle., Tangents to a Circle, , , , We have known that a tangent can not be drawn to a circle, from a point internal to it. If the point is on the circle, a single tangent can be drawn, at that point. The tangent is perpendicular to the radius drawn from the specified, point. Therefore, in order to construct a tangent to a circle at a point on it, it is, required to construct the radius from the point and then construct a perpendicular to, it. Again, if the point is located outside the circle, two tangents to the circle can be, constructed., Construction 2, To draw a tangent at any point of a circle., eLt A be a point of a circle whose centre is O. It is required to draw a tangent to the, circle at the point A., 0DWK,;;)RUPD
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46, 1, , Math, , Construction :, O, A are joined. At the point A, a perpendicular AP is drawn to, OA. Then AP is the required tangent., Proof: The line segment OA is the radius passing through A, and AP is perpendicular to it. Hence, AP is the required, tangent., Remark : At any point of a circle only one tangent can be drawn., Construction 3, To draw a tangent to a circle from a point outside., eLt P be a point outside of a circle whose centre is O. A tangent is to be drawn to the, circle from the point P., Construction :, (1, ) Join P, O. The middle point M of the line segment, PO is determined., (2, ) Now with M as centre and MO as radius, a circle is, drawn. eLt the new circle intersect the given circle at the, points A and B., (3) A, P and B, P are joined., Then both AP or BP are the required tangents., Proof: A, O and B, O are joined. PO is the diameter of the circle APB., ? PAO =1right angle [the angle in the semi-circle is a right angle]., So the line segment OA is perpendicular to AP. Therefore, the line segment AP is a, tangent at A to the circle with centre at O. Similarly the line segment BP is also a, tangent to the circle., Remark: Two and only two tangents can be drawn to a circle from an external point., Construction 4, To draw a circle circumscribing a given triangle., eLt ABC be a triangle. It is required to draw a circle, circumscribing it. That is, a circle which passes through the, three vertices A, B and C of the triangle ABC is to be drawn., Construction :, (1, ) EM and FN the perpendicular bisectors of AB and AC, respectively are drawn. eLt the line segments intersect, each other at O.
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47, 1, , Math, , (2, ) A, O are joined. With O as centre and radius equal to OA, a circle is drawn., Then the circle will pass through the points A, B and C and this circle is the required, circum-circle of 'ABC., Proof : B, O and C, O are joined. The point O stands on EM, the perpendicular, bisector of AB., ? OA= OB. Similarly, OA = OC, ? OA = OB = OC., Hence, the circle drawn with O as the centre and OA as the radius passes through the, three points A, B and C. This circle is the required circumcircle of 'ABC., Activity:, In the above figure, the circumcircle of an acute angled triangle is constructed., Construct the circumcircle of an obtuse and right-angled triangles., Notice that for in obtuse-angled triangle, the circumcentre lies outside the triangle,in, acute-angle triangle, the circumcentre lies within the triangle and in right-angled, triangle, the circumcentre lies on the hypotenuse of the triangle., Construction 5, To draw a circle inscribed in a triangle., eLt 'ABC be a triangle. To inscribe a circle in it or to draw a, circle in it such that it touches each of the three sides BC, CA, and AB of the triangle., Construction : BL and CM, the bisectors respectively of the, angles ABC and ACB are drawn. eLt the line segments, intersect at O. OD is drawn perpendicular to BC from O and, let it intersect BC at D. With O as centre and OD as radius, a, circle is drawn. Then, this circle is the required inscribed, circle., Proof : rFom O, OE and OF are drawn perpendiculars respectively to AC and AB., eLt these two perpendiculars intersect the respective sides at E and F. The point O, lies on the bisector of ABC., ? OF = OD., Similarly, as O lies on bisector of ACB, OE = OD, ? OD = OE = OF
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48, 1, , Math, , Hence, the circle drawn with centre as O and OD as radius passes through D, E and F., Again, BC, AC and AB respectively are perpendiculars to OD, OE and OF at their, extremities. Hence, the circle lying inside 'ABC touches its sides at the points D, E, and F., Hence, the circle DEF is the required inscribed circle of 'ABC., Construction 6, To draw an ex-circle of a given triangle., eLt ABC be a triangle. It is required to draw its ex-circle., That is, to draw a circle which touches one side of ABC, and the other two sides produced., Construction: eLt AB and AC be produced to D and F, respectively. BM and CN, the bisectors of DBC and, FCB respectively are drawn. eLt E be their point of, intersection. rFom E, perpendicular EH is drawn on BC, and let EH intersect BC at H. With E as centre and radius, equal to EH, a circle is drawn., The circle HGL is the ex-circle of the triangle ABC., Proof : rFom E, perpendiculars EG and EL respectively are drawn to line, segments BD and CF. eLt the perpendicular intersect line segments G and L, respectively. Since E lies on the bisector of DBC, ∴ EH = EG, Similarly, the point E lies on the bisector of FCB, so EH = EL, ?EH = EG = EL, Hence, the circle drawn with E as centre and radius equal to EL passes through H,, G and L., Again, the line segments BC, BD and CF respectively are perpendiculars at the, extremities of EH, EG and EL. Hence, the circle touches the three line segments at, the three points H, G and L respectively. Therefore, the circle HGL is the ex-circle, of 'ABC., Remark : Three ex-circles can be drawn with any triangle., Activity: Construct the two other ex-circles of a triangle.,
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49, 1, , Math, , Exercise 8⋅5, .1 Observe the following information:, i. The tangent to a circle is perpendicular to the radius to the point of contact., ii. The angle subtended in a semi-circle is a right angle., iii. All equal chords of a circle are equidistant from the centre., Which one of the following is correct ?, (a) i and ii, (b) i and iii (c) ii and iii, (d) i, ii and iii, , sUe the above figure to answer questions 2and 3:, 2. BOD equals to, 1, 1, a. BAC, b. BAD c. 2 BAC, d. 2 BAD, 2, 2, 3. The circle is of the triangle ABC, a. inscribed circle b. circumscribed circle, c. ex-circle, d. ellipse, 4. The angle inscribed in a major arc is, a. acute angle, b. right angle c. obtuse angle, d. complementary angle, 5. Draw a tangent to a circle which is parallel to a given straight line., 6. Draw a tangent to a circle which is perpendicular to a given straight line., .7 Draw two tangents to a circle such that the angle between them is 60°., .8 Draw the circum-circle of the triangle whose sides are 3 cm, 4 cm and 4.5 cm, and find the radius of this circle., 9. Draw an ex-circle to an equilateral triangle ABC touching the side AC of the, triangle, the length of each side being 5 cm., 01. Draw the inscribed and the circumscribed circles of a square., .1Prove that two circles drawn on equal side s of an isosceles triangle as diameters, mutually intersect at mid point of its base., 12.Prove that in a right-angled triangle, the length of line segment joining mid point, of the hypotenuse to opposite vertex is half the hypotenuse., 13. ABC is a triangle. If the circle drawn with AB as diameter intersects BC at D,, prove that the circle drawn with AC as diameter also passes through D.
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50, 1, , Math, , 15, If the chords AB and CD of a circles with centre O intersect at an internal point, 1, E, prove that AEC =, (BOD + AOC)., 2, 16. AB is the common chord of two circles of equal radius. If a line segment meet, through the circles at P and Q, prove that 'OAQ is an, isosceles triangle., .71If the chord AB = x cm and OD A AB, are in the circle ABC, with centre O use the adjoint figure to answer the following, questions:, , a. iFnd the area of the circle., b. Show that D is the mid point of AB., x, c. If OD =( 2 –2) cm, determine x., .81The lengths of three sides of a triangle are 4 cm, 5 cm and 6 cm respectively. sUe, this information to answer the following questions:, a. Construct the triangle., b. Draw the circumcircle of the triangle., c. rFom an exterior point of the circumcircle,, draw two tangents to it and show, that their lengths are equal.
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Chapter Nine, , Trigonometric Ratio, In our day to day life we make use of triangles, and in particular, right angled, triangles. Many different examples from our surroundings can be drawn where right, triangles can be imagined to be formed. In ancient times, with the help of geometry, men learnt the technique of determining the width of a river by standing on its bank., Without climbing the tree they knew how to measure the height of the tree accuratly, by comparing its shadow with that of a stick. In all the situations given above, the, disttances or heights can be found by using some mathematical technique which, come under a special branch of mathematics called Trigonometry. The word, ‘Trigonometry’ is derived from Greek words ‘tri’ (means three), ‘gon’ (means edge), and `metron’ (means measure). In fact, trigonometry is the study of relationship, between the sides and angles of a triangle. There are evidence of using the, Trigonometry in Egyptian and Babilian civilization. It is believed that the Egyptians, made its extensive use in land survey and engineering works. Early astrologer used it, to determine the distances from the Earth to the farest planets and stars. At present, trigonometry is in use in all branches of mathematics. There are wide usages of, trigonometry for the solution of triangle related problems and in navigation etc. Now, a days trigonometry is in wide use in Astronomy and Calculus., At the end of the chapter, the students will be able to –, ¾ Describe the trigonometric ratios of acute angles, ¾ Determine the mutual relations among the trigonometric ratios of acute angle, ¾ Solve and prove the mathematical problems justifying the trigonometric ratios of, acute angle, ¾ Determine and apply trigonometric ratios of acute angles 30o. 45 o, 60 o, geometrically, ¾ Determine and apply the value of meaningful trigonometric ratios of the angles 0o and, 90o, ¾ Prove the trigonometric identities, ¾ Apply the trigonometric identities., 9⋅1 Naming of sides of a right angled triangle, We know that, the sides of right angles triangle are known as hypotenuse, base and, height. This is successful for the horizontal position of triangle. Again, the naming of, sides is based on the position of one of the two acute angles of right angled triangle., As for example :, a. ‘Hypotenuse’, the side of a right angled triangle, which is the opposite side of, the right angle., b. ‘Opposite side’, which is the direct opposite side of a given angle.
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Math, , 152, , c. ‘A, daj cent side’, which is a line segment constituting the given angle., , For the angle OPN , OP is the, For the angle PON , OP is the, hypotenuse, ON is the adjacent side, hypotenuse, PN is the adjacent side, and PN is the opposite side., and ON is the opposite side., In the geometric figure, the capital letters are used to indicate the vertices and small, letters are used to indicate the sides of a triangle. We often use the G, reek letters to, indicate angle. Widely used sixletters of rGeek alphabet are :, beta β, gamma γ, theta θ, phi φ, omega ω, alpha α, rGeek letter are used in geometry and trigonometry through all the great, mathematician of ancient rGeek., Example 1. Indicate the hypotenuse, the adjacent side and the opposite side for the, angle θ ., , Solution:, (b) hypotenuse p, (a) hypotenuse 17 units, opposite side r, opposite side 8 units, adjacent side q, adjacent side 15 units, Example 2. Find the lengths of hypotenuse, the adjacent side, for the angles α and β ., , (a) For α angle,, hypotenuse 25 units, opposite side 24 units, adjacent side 7 units., , (c) hypotenuse EF, opposite side EG, adjacent side FG, and the opposite side, , (b) For β angle, hypotenuse 25 units, opposite side 7 units, adjacent side 24 units., , Activity :, Indicate the hypotenuse , adjacent side and opposite for the angle θ and φ .
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Math, , 153, , 9⋅2 Constantness of ratios of the sides of similar right-angled triangles, Activity : Measure the lengths of the sides of the following four similar triangles and, complete the table below. What do you observe about the ratios of the triangles ?, , BC, , length of sides, AB, AC, , BC/ AC, , ratio (related to angle), AB/ AC, BC/ AB, , eLt, XOA is an acute angle. We take a point P on the side OA . We draw a, perpendicular from P to OX . sAa result, a right angled triangle POM is formed., The three ratios of the sides PM , OM and OP of 'POM do not depend on the, position of the point P on the side OA., If we draw the perpendiculars PM and P1 M 1 from two points P and P1 of OA to, the side OX , two similar triangles 'POM and 'P1OM1 are formed., oNw, 'POM and 'P1OM1 are being similar,, PM, PM, OP, PM, =, or,, = 1 1 ..... (i), P1M1, OP1, OP1, OP, OM, OM, OP, OM1, =, or,, =, .... (ii), OM1, OP1, OP, OP1, PM, OM, PM, PM, =, or,, = 1 1 .... (iii), P1M1, OM1, OM1, OM, That is, each of these ratios is constant. These ratios are, called trigonometric ratios., 9⋅3 Trigonometric ratios of an acute angle, eLt, XOA is an acute angle. We take any point P on OA ., We draw a perpendicular PM from the point P to OX . So,, a right angled triangle POM is formed. The sixratios are, obtained from the sides PM , OM and OP of 'POM which, are called trigonometric ratios of the angle XOA and each, of them are named particularly., With respect to the XOA of right angled triangle POM ,, PM is the opposite side. OM is the adjacent side and OP is the hypotenuse., Denoting XOA θ , the obtained sixratios are described below for the angle, 0DWK,;;)RUPD
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155, , Math, , 9⋅5 Trigonometric identity, 2, , 2, , § PM · § OM ·, (i ) (sin θ )2 (cosθ )2 = ¨, ¸, ¸ ¨, © OP ¹ © OP ¹, PM 2 OM 2, PM 2 OM 2 OP 2, =, =, =, , 2, 2, OP 2, OP, OP, OP2, 1, or, (sin θ )2 (cosθ )2 1, , ? sin 2 θ cos 2θ, , [by the formula of P, ythagoras ], , 1, , Remark : For integer indices n we can write sin n θ for (sinθ )n and cosn θ for (cosθ ) n ., 2, , § OP ·, (ii ) sec 2 θ = (secθ ) 2 = ¨, ¸, © OM ¹, OP2, PM 2 OM 2, =, [OP is the hypotenuse of right angled 'POM ], =, OM 2, OM 2, PM 2 OM 2, =, , OM 2 OM 2, 2, , § PM ·, 2, 2, = 1 ¨, ¸ = 1 (tan θ ) = 1 tan θ, © OM ¹, ? sec2 θ 1 tan 2 θ, or,, or,, , sec 2 θ tan 2 θ 1, tan 2θ sec 2θ 1, 2, , § OP ·, (cosec θ ) 2 = ¨, ¸, © PM ¹, OP2, PM 2 OM 2, =, [i is the hypotenuse of right-angled, =, PM 2, PM 2, , (iii ) cosec 2θ, , PM 2 OM 2, § OM ·, = 1 ¨, , ¸, 2, 2, PM, PM, © PM ¹, = 1 (cot θ ) 2 = 1 cot 2 θ, , 2, , =, , ?, , , cosec 2θ cot 2 θ, , 1, , and, , cot 2θ, , cosec2θ 1, , 'POM ]
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156, , Math, , Activity :, 1. oCnstruct a table of the following trig onometric formulae for easy memorizing., cosecθ, , secθ, tan θ, , 1, sin θ, 1, cosθ, 1, cot θ, , tan θ, , cot θ, , sin θ, cosθ, cosθ, sin θ, , sin 2 θ cos 2θ 1, sec2 θ 1 tan 2 θ, cosec 2θ 1 cot 2 θ, , 4, , find the other trigonometric ratios of the angle A ., 3, 4, ., Solution : iGven that, tan A, 3, So, opposite side of the angle A =4, adjacent side =3, Example 1. If tan A, , 4 2 32, 25 5, 4, 3, 3, Therefore,, sin A = , cos A = , cot A =, 5, 4, 5, 5, 5, cosec A = , sec A = ., 4, 3, , hypotenuse =, , Example 2. B is the right angle of a right angled triangle ABC . If tan A, , 3, ,, 4, , verify the truth of 2 sin A cos A =1., 3, Solution : iGven that, tan A, ,, 4, So, opposite side of the angle A =4, adjacent side =3, hypotenuse =, so, sin A, , 4 2 32, , 4, , cos A, 5, , 5, , 5, , , 3, 4, , eHnce, 2 sin A cos A 2= , , 4 3, 24, z1, =, 5 5, 25, , Therefore, 2 sin A cos A =1 is a false statement., Example 3. B is a right angle of a right angled triangle ABC . If tan A 1 , verify, the justification of 2 sin A cos A 1 .
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158, , Math, , 1 sin 2θ, 1 sin 2θ, ..HS. (proved), = 1 =R, 1, 1, Example 6. P, rove that :, , 2 sin 2 A 2 tan 2 A, =, , Proof : .L.HS. =, , 1, , 1, 1, , 2 sin 2 A 2 tan 2 A, 1, 1, =, , sin 2 A, 2 sin 2 A, 2, cos 2 A, 1, cos 2 A, =, , 2, 2 sin A 2cos 2 A sin 2 A, 1, cos 2 A, =, , 2, 2 sin A 2(1 sin 2 A) sin 2 A, , 1, cos 2 A, , 2, 2 sin A 2 2 sin 2 A sin 2 A, 1, 1 sin 2 A, , =, 2 sin 2 A 2 sin 2 A, 2 sin 2 A, =, 2 sin 2 A, = 1 =R, .H, .S. (proved), tanA, secA 1, Example 7. P, rove that :, , 0, secA 1, tanA, tanA, secA 1, proof : .L.HS. =, , secA 1, tanA, tan 2 A ( sec 2 A 1), [ sec 2 θ 1 tan 2 θ ], =, ( secA 1)tanA, tan 2 A tan 2 A, =, ( secA 1)tanA, 0, =, ( secA 1)tanA, = 0 =R, ..HS. (proved), 1 sinA, secA tanA, Example 8. P, rove that :, 1 sinA, =
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Math, , 160, , ∴, , or, sec 2 A tan 2 A 1, or, ( secA tanA)( secA tanA) 1, 5, or, ( secA tanA) 1 [from (i ) ], 2, 2, secA tanA, 5, , Exercise 9⋅1, 1. eVrify whether each of the following math ematical statements is true or false., iGve argument in favour of your answer., (a) The value of tan A is always less than 1., (b) cot A is the multiplication of cot and A ., 12, (c) For any value of A , sec A, ., 5, (d) cos is the smallest form of cotangent ., 3, , find the other trigonometric ratios of the angle A ., 2. If sin A, 4, 3. G, iven that 15 cot A 8 , find the values of sin A and sec A ., 4. If C is the right angle of the right angled triangle ABC , AB =13 cm and BC, =12 cm. and ABC θ , find the values of sin θ , cosθ and tan θ ., 3 , verify the, 5. B is the right angle of the right angled triangle ABC . If tan A, truth of 3 sin A cos A 4 ., Prove (6 – 20) :, 1, 1, 1, 1, 1, 1, 6. (i ), , 1 ; (ii ), , 1 ; (iii), , 1;, sec 2 A cosec 2 A, cos 2 A cot 2 A, sin 2 A tan 2 A, secA tanA, 1, 1, sinA, cosA, 7. (i ), , 1 . (iii), , 1, , 1 ; (ii ), 2, cosA cotA, cosecA secA, 1 sin A 1 cosec2 A, cotA, tanA, 1, 1, 8. (i ), , secA cosecA 1 ; (ii ), , 1, 1 cotA 1 tanA, 1 tan 2 A 1 cot 2 A, cosA, sinA, 9., , sinA cosA. 10. tan A 1` sin 2 A sin A., 1 tanA 1 cotA, secA tanA, cosecA cotA, cosecA, cosecA, 11., 12., , 2sec 2 A., cosecA 1 cosecA 1, cosecA cotA, secA tanA, 1, 1, 1, 1, 14., 13., , 2sec 2 A., , 2 tan 2 A., 1 sinA 1 sinA, cosecA 1 cosecA 1, tanA, secA 1, sinA, 1 cosA, 15., , 2 cosec A. 16., 0, , secA 1, tanA, 1 cosA, sinA
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162, , Math, , PM, a, 1, ., OM, 3a, 3, OP 2a, cosec 30 $, 2 , sec 30 $, PM, a, tan 30 $, , cot 30 $, , OM, PM, , 3a, a, , OP, OM, , 2a, 3a, , 2, 3, , 3., , Similarly,, OM, 3a, 3, ,, OP, 2a, 2, OM, 3a, PM, a, 1, , tan 60 $, cos 60 $, OP 2a 2, PM, a, OP, 2a, 2, ,, cosec 60 $, OM, 3a, 3, OP 2a, PM, a, sec 60 $, 2 , cot 60 $, PM, a, OM, 3a, $, Trigonometric ratio of the angle 45, eLt, XOZ 45$ and P is a point on OZ ., Draw PM A OX . In right angled triangle, 'OPM , POM 45$, So, OPM 45$, Therefore, PM OM = a (suppose), sin 60 $, , 2, , 2, , 3, , 1, ., 3, , 2, , Now, OP 2 OM 2 PM 2 = a + a 2= a, or, OP, 2a, From the definition of trigonometric ratios, we get, PM, a, 1, OM, a, 1, PM, , cos 45$, , tan 45$, sin 45$, OP, OP, OM, 2a, 2, 2a, 2, 1, 1, 1, 1, cosec 45$, 2 , sec 45$, 2 , cot 45$, tan 45q, sin 45q, cos 45q, 9⋅7 Trigonometric ratios of complementary angles, We know, if the sum of two acute angles is 90$ , one of, them is called complementary angle to the other. For, example, 30 $ and 60$ ; 15$ and 75$ are complementary, angles to each other., In general, the angles θ and (90$ θ ) are, complementary angles to each other., , , , a, 1, a, ,
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Math, , 163, , Trigonometric ratios of complementary angles, eLt, XOY θ and P is the point on the side OY of the angle. We, draw PM A OX ., Since the sum of the three angles of a triangle is two right angles therefore, in the, right angled triangle POM , PMO 90$, and OPM POM one right angle = 90$, ? OPM 90$ POM 90$ θ, [Since POM XOY θ ], OM, = cos POM = cos θ, ?, sin (90$ θ ), OP, PM, = sin POM = sin θ, cos (90 $ θ ), OP, OM, , = cot POM = cot θ, tan (90 $ θ ), PM, PM, = tan POM = tan θ, cot (90 $ θ ), OM, OP, = cosec POM = cosec θ, sec (90 $ θ ), PM, OP, = sec POM = sec θ ., cosec (90 $ θ ), OM, We can express the above formulae in words below :, sine of complementary angle = cosine of angle, cosine of complementary angle = sine of angle, tangent of complementary angle = cotangent of angle etc., 5, , find the value of cosec θ cot θ ., Activity : 1. If sec (90 $ θ ), 3, 9⋅8 Trigonometric ratios of the angles 0$ and 90$, We have learnt how to determinal the trigonometric ratios for, the acute angle θ of a right angled triangle. oNw, we see, if the, angle is made gradually smaller, how the trigonometric ratios, change. sA θ get smallers the length of the side PN also gets, smaller. The point P closes to the point N and finally the, angle θ comes closer to the angle 0$ , OP is reconciled with, ON approximately.
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164, , Math, , When the angle θ comes closer to 0$ , the length of the line segment PN reduces to, zero and in this case the value of sin θ PN is approximately zero. A, t the same time,, OP, , ON, is 1, the length of OP is equal to the length of ON and the value of cos T =, OP, approximately., The angle, 0$ is introduced for the convenience of discussion in trigonometry, and, the edge line and the original line of the angle 0$ are supposed the same ray., Therefore, in line with the prior discussion, it is said that, cos 0 $ 1, sin 0 $ 0 ., If θ is the acute angle, we see, sin θ, cos θ, , cot θ, ,, tan θ, cosθ, sin θ, 1, 1, , cosecθ, ,, sec θ, cos θ, sin θ, We define the angle 0$ in probable cases so that, those relations exists., sin 0 $ 0, tan 0 $, 0, cos 0 $ 1, 1, 1, 1., sec 0 $, cos 0 $ 1, Since division by 0 is not allowed, cosec 0 $ and cot 0 $ can not be defined., , gAain, when the angle θ is very closed to 90$ , hypotenuse OP is approximatel y, equal to PN . So the value of sin θ is approximately 1. nO the other hand, if the, angle θ is is equal to 90$ , ON is nearly zero;the value of cos θ is approximately 0., So, in agreement of formulae that are described above, we can say, cos 90 $ 0 ,, sin 90 $ 1 ., cos 90 $ 0, cot 90 $, 0, sin 90 $ 1, 1, 1, 1, cosec 90 $, sin 90 $ 1, Since one can not divided by 0, as before, tan 90 $ and sec 90 $ are not defined., Observe : For convenience of using the values of trigonometric ratios of the angles, 0 $ , 30 $ , 45$ , 60 $ and 90 $ are shown in the following table :
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165, , Math, , angle, R, atio, sine, , 0$, 0, , 30$, 1, 2, , cosine, , 1, , tangent, , 0, , 3, 2, 1, 3, 3, , cotangent, , undefined, , 45$, 1, 2, 1, 2, 1, , 60$, 3, 2, 1, 2, , 90$, 1, , 3, , undefined, , 1, , 1, 3, 2, , 0, , 0, , 2, undefined, 2, 3, 2, 1, cosecant, undefined, 1, 2, 3, 2, Observe : aEsy method for remembering of the values of trigonometric ratios of, some fixed angles., (i ) If we divide the numbers 0, 1, 2, 3 and 4 by 4 and take square root of the, , secant, , 1, , quotients, we get the values of sin 0 $ of sin 30 $ , sin 45$ , sin 60 $ and sin 90 $, respectively., (ii ) If we divide the numbers 4, 3, 2, 1 and 0 by 4 and take square root of, quotients, we get the values of cos 0 $ , cos 30 $ , cos 45$ , cos 60 $ and cos 90 $, respectively., (iii ) If we divide the numbers 0, 1, 3 and 9 by 3 and take square root of quotients,, we get the values of tan 0 $ , tan 30 $ , tan 45$ and tan 60 $ , respectively (It is, noted that tan 90q is undefined)., (iv ) If we divide the numbers 9, 3, 1 and 0 by 3 and take square root of quotients,, we get the values of cot 30 $ , cot 45$ , cot 60 $ , cot 90 $ respectively (It is noted, that cot 0 $ is undefined)., Example 1. Find the values :, 1 sin 2 45$, tan 2 45$, (a), 1 sin 2 45$, (b), cot 90 $ tan 0 $ sec 30 $ cosec 60 $, (c), sin 60$ cos 30$ cos 60$ sin 30$, 1 tan 2 60$, (d), sin 2 60$, 1 sin 2 60$
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166, , Math, , Solution :, (a), , 1 sin 2 45$, tan 2 45$, 1 sin 2 45$, , iGven expression =, , 2, , 1 ·, ¸, 1, 2¹, (1) 2 [ sin 45$, I tan 45$ 1], 2, 2, 1 ·, ¸, 2¹, 1, 4, 1, 1 = 2 1 = 1 =, 3, 3, 3, 2, cot 90 $ tan 0 $ sec 30 $ cosec 60 $, 2 2, , 0, = 00, 3 3, 2, , cosec 60 $, [ cot 90 $ 0, tan 0 $ 0, sec 30 $, 3, iGven expression =, sin 60$ cos 30$ cos 60$ sin 30$, 3 3 1 1, =, , , 2 2 2 2, 3, 1, ], [ sin 60 $ cos 30$, , sin 30 $ cos 60 $, 2, 2, 4, 3 1, =1, =, =, 4, 4 4, 1 tan 2 60 $, iGven expression =, sin 2 60 $, 1 sin 2 60 $, , §, 1 ¨, ©, =, §, 1 ¨, ©, 1, 1, 2, =, 1, 1, 2, (b) iGven expression =, , (c), , (d), , 2, , 2, ], 3, , 2, , § 3·, ¸¸, ¨¨, 2, 1 3, © 2 ¹, 1 3 3 2 3, , =, =, 4, 4, 1 3 4, 23 1, =, 4, 4, =, , 1, , 3, , Example 2., (a) If 2cos(A B) 1, 2sin (A B), values of A and B ., , 3 and A, B are acute angles, find the
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Math, , (b), (c), (d), , 167, , 1 3, , find the value of A ., 1 3, 1 tan 2 A, , if A 45$ ., P, rove that, cos 2A, 1 tan 2 A, Solve : 2cos 2θ 3sinθ 3 0 , where θ is an acute angle., , If, , cosA sinA, cosA sinA, , Solution : (a), , 2cos( A B ) 1, , or, cos( A B ), or, cos( A B ), ?, , 1, 2, , cos45$ [ cos45$, , 1, ], 2, , A B 45$..................(i ), and 2 sin ( A B ), 3, or, sin ( A B ), or, sin ( A B ), , 3, 2, , sin 60 $ [ sin 60$, , 3, ], 2, , ?, A B 60$....................(ii ), dding (i ) and (ii ) , we get,, A, , 2 A 105$, $, 1, 105$, ? A, 52, 2, 2, gAain, subtracting (i) from (ii), we get, 2 B 15$, 15$, or,, B, 2, 1$, ?, B 7, 2, 1$, 1$, and B 7, R, equired A 52, 2, 2, cosA sinA 1 3, (b), cosA sinA 1 3, cosA sinA cosA sinA 1 3 1 3, or,, cosA sinA cosA sinA 1 3 1 3
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Math, , 168, , 2cosA, 2, 2 sinA 2 3, 1, cosA, sinA, 3, cotA cot 60 $, A 60$, iGven that, A 45$, , or,, or,, or,, ?, (c), , we have to prove that, cos 2 A, , 1 tan 2 A, 1 tan 2 A, , .L.HS. =, cos 2 A, = cos(2 u 45$ ) = cos 90 $ = 0, 1 tan 2 A, 1 tan 2 A, 1 tan 2 45$ 1 (1) 2, =, =, 1 tan 2 45$ 1 (1) 2, 0, = =0, 2, ∴ .L.HS. =R, ..HS. (proved), (d), G, iven equation, 2cos 2θ 3sinθ 3 0, or, 2(1 sin 2θ 3(1 sinθ ) 0, or, 2(1 sinθ )(1 sinθ ) 3(1 sinθ ) 0, or, (1 sinθ ){2(1 sinθ ) 3} 0, or, (1 sinθ ){2 sinθ 1} 0, or, 1 sinθ 0, or 2 sinθ 1 1, ? sin θ 1, or, 2 sin θ 1, 1, $, or, sinθ, or, sin θ sin 90, 2, ? θ 90 $, or, sin θ sin 30$, or, θ 30$, $, θ is an acute angle, so θ 30 ., R, .H, .S. =, , Exercise 9⋅2, 1. If, (a), , cot θ, , 1, 3, , 1, , which one is the value of cot ș ?, 2, , (b) 1, , (c), , 3, , (d) 2
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Math, , 169, , 2., (i) sin 2θ 1 cos 2 θ, (ii) sec 2θ 1 tan 2 θ, (iii) cot 2θ 1 tan 2 θ, Which one of the followings is correct in accordance with the above statements., (a) i and ii (b) i and iii (c) ii and iii (d) i, ii and iii, , nAswer of questions 3 and 4 on the basis of the figure :, 3. What is the value of sin θ ?, 3, 4, 3, 4, (a), (b), (c), (d), 4, 3, 5, 5, 4. What is the value of cot θ ?, 3, 3, 4, 4, (a), (b), (c), (d), 4, 5, 5, 3, valuate (5 –8) :, E, 1 cot 2 60 $, 5., ., 1 cot 2 60 $, 6. tan 45$ sin 2 60 $ tan 30 $ tan 60 $., 1 cos 2 60 $, 7., sec 2 60 $ ., 1 cos 2 60 $, 8. cos 45$ cot 2 60 $ cosec 2 30 $ ., rPove (9 –11) :, 9. cos 2 30 $ sin 2 30 $ cos 60 $., 10. sin 60 $ cos 30 $ cos 60 $ sin 30 $ sin 90 $ ., 11. cos 60 $ cos 30 $ sin 60 $ sin 30 $ cos 30 $ ., 12. sin 3 A cos3 A , if` A 15$ ., 2 tanA, , if A 45$ ., 13. sin 2A, 1 tan 2 A, 2tanA, , if A 30$ ., 14. tan 2 A, 1 tan 2 A, 1 tan 2 A, , if A 60$ ., 15. cos 2A, 1 tan 2 A, 16. If 2cos(A B) 1 2sin(A B) and A, B are acute angles, show that A 45$ ,, B 15$ ., 0DWK,;;)RUPD
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170, , Math, , 17. If cos(A B) 1, 2sin(A B) and A, B are acute angle, find the values of, A and B ., cosA sinA, 3 1, 18. Solve :, ., cosA sinA, 3 1, 19. If A, B are acute angle and cot(A B) 1, cot(A B), 3 , find the values of, A and B ., 20. Show that, cos 3 A 4 cos 3 A 3cos A , when A 30$ ., 21. Solve : sinθ cosθ 1 , when 0 $ d θ d 90 $ ., 22. Solve : cos 2θ sin 2θ 2 5cosθ , when θ is an acute angle., 23. Solve : 2sin 2θ 3cosθ 3 0 , θ is an acute angle., 24. Solve: tan 2θ (1 3 ) tanθ 3 0., 1, 25. Find the value : 3 cot 2 60 $ cos ec 2 30 $ 5 sin 2 45$ 4 cos 2 60 $ ., 4, 26. If B 90 $ , AB 5cm, BC 12cm. of 'ABC, (a) Find the length of AC ., (b) If C θ , find the value of sin θ cos θ ., (c) Show that, sec 2 θ cos 2 θ sec 2 θ cos ec 2 θ ., 27., , (a) what is the measurement of AC ., (b) Find the value of tan A tan C., (c) Find the values of x and y.
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Chapter Ten, , Distance and Height, From very ancient times trigonometrical ratios are applied to find the distance and, height of any distant object. tA pres ent trigonometrcal ratios are of boundless, importance because of its increasing usage. The heights of the hills, mountains and, trees and the widths of those rivers which cannot be measured in ordinary method, are measured the heights and widths with the help of trigonometry. In this condition, it is necessary to know the trigonometrical ratios values of acute angle., A, t the end of this chapter, the students will be able to –, x xE, plain the geoline, vertical plane a nd angles of elevation and declination, x Solve mathematical problem related to distance and height with the help of, trigonometry, x Measure practically different types of distances and heights with the help of, trigonometry., Horizontal line, Vertical line and Vertical plane :, The horizontal line is any straight line on the plane. Astraight line parallel to horizon, is also called a horizontal line. The vertical line is any line perpendicular to the, horizontal plane. It is also called normal line., Ahorizontal line and a vertical line intersected at right angles, on the plane define a plane. It is known as vertical plane., In the figure : Atree with height of AB is standing vertically, at a distance of CB from a point C on the plane. eHre, CB is, the horizontal line. BA is the vertical line and the plane ABC, is perpendicular to the horizontal plane which is a vertical, plane., Angle of Elevation and Angle of Depression :, bOserve the figure, AB is a straight line parallel to the horizon., The points P, O, B lie on the same vertical plane. The point P, on the straight AB makes angle POB with the line AB ., eHre at O , the angle of elevation of P is POB ., So, the angle at any point above the plane with the straight line, parallel to horizon is called the angle of elevation., , gAain the point Q, O, B lie on the same vertical plane and point Q lines at lower, side of the straight line AB parallel to horizon. eHre, the angle of depression at O of, Q is ∠QOB. So, the angle at any point below the straight line parallel to the plane is, called the angle of depression.
Page 177 : 172, , Math, , Activity :, oint the figure and show the, P, horizontal line, vertical line, vertical, plane, angle of elevation and angle of, mL, and o, depression., N.B. : For solving the problems in this chapter approximately right figure is needed., While drawing the figure, the following techniques are to be applied., (1) While drawing 30$ angle, it is needed base >perpendicular., (2) While drawing 45$ angle, it is needed base =perpendicular., (3) While drawing 60$ angle, it is needed base <perpendicular., Example 1. The angle of elevation at the top of a tower at a point on the ground is, 30q at a distance of 75 metre from the foot. Find the height of the tower., Solution : eLt, the height of the tower is AB h metre., The angle of elevation at C from the foot of the tower, BC 75 metre of A on the ground is ACB 30 $, From 'ABC we get, tanACB, or, tan 30$, , h, 1, or,, 75, 3, , h, 75, , AB, BC, ª, $, «¬ tan 30, , 1 º, or,, 3 »¼, , 3h, , 75 or, h, , 75, 3, , 75 3, 3], or,, [multiplying the numerator and denominator by, 3, h 25 3, ? h 43.301 metre (app.)., eRquired height of the tower is 43.301 metre (app.)., Example 2. The height of a tree is 105 metre. If the angle of elevation of the tree at a, point from its foot on the ground is 60$ , find the distance of the point on the ground, from the foot of the tree., Solution : eLt, the distance of the point on the ground, from the foot of tree is BC x metre. H, eight of the tree, AB 105 metre and at C the angle of elevation of the, vertexof tree is ACB 60$, From 'ABC we get,, or, h
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173, , Math, , tanACB, , AB, or, tan 60 0, BC, , 105, x, , > tan 60, , 0, , 3, , @, , 105, 105, 105 3, or, 3 x 105 or, x, or, x, or, x 35 3, 3, x, 3, ? x 60.622 (app.), ? The required distance of the point on the ground from the foot of the tree is, 60.622 metre (app.)., , 3, , or,, , Activity:, In the picture, AB is a tree. Information from, the picture –, 1. Find the height of the tree., 2. Find the distance of the point C on the, ground from the foot of the tree., Example 3. Aladder of 18 metres. long touche s the roof of a wall and makes an, angle of 45o with the horizon. Find the height of the wall., Solution : eLt, the height of the wall AB is = h metre,, length of ladder AC is 18 m. and makes angles with, the ground ACB 45$ ., From 'ABC we get, sin ACB, or, sin 45$, h, 18, , or,, , 1, 2, , or,, , 2h 18, , or, h, , 18 2, 2, 9 2, , AB, AC, , h, 18, ª, 0, «¬ sin 45, 18, or, h, 2, , 1 º, or,, 2 »¼, , 2h 18, , or, h, , 18, 2, , [multiplying the numerator and denominator by, , 2], , or, h, ? h 12.728 (app.), Therefore, required height of the wall is 12.728 m. (app.)., Example 4. Atree leaned due to storm. The stic k with height of 7 metre from its foot, was learned against the tree to make it straight. If the angle of depression at the point, of contacting the stick on the ground is 30$ , find the length of the stick ?
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174, , Math, , Solution : eLt, the height of the stick from the, foot learned against the tree of AB 7 metre and, angle of depression is DBC 30$, ? ACB DBC 30$ [alternate angle], From 'ABC we get,, 7, AB, or, sin 30 0, sinACB, BC, BC, 1, 7, 1º, ª, 0, or,, «¬ sin 30, 2 BC, 2 »¼, ? BC 14, ? eRquired height of the stick is 14 metre., Activity :, In the figure, if depression angle CAE 60 $ ,, elevation angle ADB 30$ , AC 36 metre, and B, C , D lie on the same straight line, find, the lengths of the sides AB, AD and CD ., Example 5. The angle of elevation at a point of the roof of a building is 60 $ in any, point on the ground. Moving back 42 metres from the angle of elevation of the point, of the place of the building becomes 45$ . Find the height of the building., Solution : eLt, the height of the building is AB h, metres. The angle of elevation at the top ACB 60$ ., The angle of elevation becomes ADB 45$ moving, back from C by CD 42 metres., eLt, BC x metre, ? BD BC CD x 42 metre, From 'ABC we get,, AB, h, or, 3, tan 60 0, 3, tan 60 0, x, BC, h, ? x, ............ i, 3, , >, , gAain, from, or, 1, or, h, , 'ABD we get, tan 450, , h, tan 450 1 or, h, x 42, h, 42; by equation (i), 3, , >, , @, , @, , AB, BD, x 42
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Math, , or,, , 175, , 3h, , h 42 3 or,, , 3h h, , 42 3 or,, , 3 1 h, , 42 3 or, h, , 42 3, 3 1, , ? h 99.373 (app.), eight, H̀, of the building is 99.373 metres (app.), Example 6. Apole is broken such that the broken part makes an angle of 30$ with, the other and touches the ground at a distance of 10 metres form its foot. Find the, lengths of the pole., Solution : eLt, the total height of the pole is AB h, metre. Breaks at the height of BC x matre without, separation and makes an angle with the other,, BCD 30$ and touches the ground at a, distance BD 10 metres from the foot., eHre, CD AC AB BC h x metre, From 'BCD we get,, 1 10, BD, tan 30$, or,, =, ? x 10 3, 3 x, BC, 1, 10, BD, gAain, sin 30$, or,, 2 hx, CD, or, h x 20 or, h 20 x or, h 20 10 3; p[ utting the value of x ], ? h 37.321 (app.) ? eHight of the pole is 37.321 metres (app.)., Activity :, Aballoon is flying above any point be tween two mile posts. tAthe point of the, balloon the angle of depression of the two posts are 30$ and 60$ respectively. Find, the height of the balloon., , Exercise 10, 1. (a) Find the measurement of CAD, (b) Find the lengths of AB and BC., (c) Find the distance between A and D., 2. From a helicopter above a point O between two kilometre posts, the angles of, depression of the two points A and B are 60o and 30o respectively., (a) Draw a figure with short description., (b) Find the height of the helicopter from the ground., (c) Find the direct distance from the point A of the helicopter., 3. What is the elevation angle of point Pfrom the point O ?, (b) POA, (c) QOA, (d) POB, (a) QOB, , 4. (i) The horizontal line is any straight line lying on the plane., (ii) eVrtical line is any line perpendicular to the plane., (iii) A, horizontal line and a vertical plane define a plane. It is known as vertical plane.
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176, , Math, , which one is right of the above speech ?, (a) i and ii (b) i and iii, (c) ii and iii, (d) i, ii and iii, nAswer the questions 5 –6 from the adjacent figure :, 5. The length of BCwill be –, 4, m (b) 4m (c) 4 2 m, (d) 4 3 m, (a), 3, 6. The length of A, B will be –, 4, m (b) 4m (c) 4 2 m, (d) 4 3 m, (a), , 3, 7. If the angle of the elevation of the top of the miner is 30$ at a point on the ground and, the height is 26 metres, then find the distance of the plane from the Miner., 8. If the top of a tree is 20 metres distance from the foot on the ground at any point, and the angle of elevation of 60q, find the height of the tree., 9. Forming 45$ angle with ground A18 metres long ladder touches the top of the, wall, find the height of the wall., 10. If the of depression of a point on the ground 20 metres from the top of the house, is 30$ then, find the height of the house., 11. The angle of elevation of a tower at any point on the ground is 60$ . If moved, back 25 metre, the angle of elevation becomes 30 $ , find the height of the tower., 12. The angle of elevation of a tower 60$ moving 60 metres towards a minar. Find, the height of the minar., 13. Aman standing at a place on the bank of a river observesd that the angle of, elevation of a tower exactly opposite to him on the other bank was 60 o. Moving, 32 metres back he observed that the angle of elevation of the tower was 30o. Find, the height of the tower and the width of the river., 14. Apole of 64 metre long breaks into two parts without complete separation and makes, an angle 60$ with the ground. Find the length of the broken part of the pole., 15. Atree is broken by a storm such that the broken part makes an angle of 30$ with, the other and touches the ground at a distance of 10 metres from it. Find the, length of the whole tree., 16. Standing any where on the bank of a river, a man observed a tree exactly straight, to him on the other bank that the angle journey by boat of elevation of he top of, the tree of 150 metres length is 30 $ . The man started for the tree. But he reached, at 10 metres away from the tree due to current., (a) Show the above description by a figure., (b) Find the width of the river., (c) Find the distance from the starting point to the destination.
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Chapter Eleven, , Algebraic Ratio and Proportion, It is important for us to have a clear concept of ratio and proportion. A, rithmetical, ratio and proportion have been elaborately discussed in class V, II. In this chapter,, algebraic ratio and proportion will be discussed. We regularly use the concept of, ratio and proportion in construction materials and in the production of food staff, in, consumers production, in using fertilizer in land, in making the shapes and design of, many things attractive and good tooking and in many areas of our daily activities., Many problems of daily lives can be solved by using ratio and proportion., At the end of this chapter, the students will be able to :, ¾ xE, plain algebraic ratio and proportion., ¾ sUe different types of rules of tr ansformation related to proportion., ¾ Describe successive proportion., ¾ sUe ratio, proportion, successive propor tion in solving real lives problem., 11⋅1 Ratio, fOtwo quantities of same kind and unit, how many times or parts of other can be, expressed by a fraction. This fraction is called the ratio of two quantities., p, . The quantities p and q, The ratio of two quantities p and q is written in p : q, q, are to be of same kind and same unit. p is called antecedent and q is called, subsequent of the ratio., Some times we use ratio in approximate mea sure. Such as, the number of cars on the, road at 8 .AM. doubles the number at 10 .AM. In this case, it is not necessary to, know the exact number of cars to determine the ratio. A, gain, in many occasions, we, say that the area of your house is three times the area of mine. eHre, also it is not, necessary to know the exact area of the house. We use the concept of ratio in cases, of practical life., 112 Proportion, If four quantities are such that the ratio of first and second quantities is equal to the, ratio of third and fourth quantities, those four quantities form a proportion. If a, b, c,, d are four such quantities, we write a : b = c : d . The four quantities of proportion, need not to be of same kinds. The two quantities of the ratio are to be of the same, kind only., 0DWK,;;)RUPD
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178, , Math, , In the above figure, let the base of two triangles be a and b respectively and their, height is h unit. If the areas of the triangle be A and B square units, we can write,, 1, ah, A 2, a, or A : B a : b, B 1, b, bh, 2, i.e. ratio of two areas is equal to ratio of two bases., Ordered proportional, By ordered proportional of a, b, c it is meant that a : b = b : c., a, b, c will be ordered proportional if and only if b2 = ac. In case of ordered, proportional, all the quantities are to be of same kinds. In this case, c is called third, proportional of a and b and b is called mid-proportional of a and c., Example 1. A and B traverses fixed distance in t1 and t2 minutes. Find the ratio of, average velocity of A and B., Solution : eLt the average velocities of A and B be v1 sec/metre and v 2 sec/metre, respectively. So, in time t1 minutes A traverses v1t1 metres and in t2 minutes B, traverses the distance v2t2 meters., cAcording to the problem, v, , 1t1, , =v 2t2 ?, , v1, v2, , t2, t1, , eHre, ratio of the velocities is invers ely proportional to the ratio of time., Activity: 1. E, xpress 3.5 : 5.6 into 1 : a and b : 1, 2. If x : y =5 : 6, 3 x : 5y =What ?, 113 Transformation of Ratio, eHre, the quantities of ratios are positive numbers., (1) If a : b c : d then b : a d : c [Invertendo], Proof : G, iven that,, , a, b, , c, d, , ? ad bc [multiplying both the sides by bd], ad bc, or,, [dividing both the sides by ac where a z 0, c z 0 ], ac ac, d b, or,, c a
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179, , Math, , (2), , i.e., b : a d : c, If a : b c : d then a : c, , Proof : iGven that,, , a, b, , b : d [alternendo], , c, d, , ? ad bc [multiplying both the sides by bd], ad bc, or,, [dividing both the sides by cd where c z 0, d z 0 ], cd cd, a b, or,, c d, i.e., a : c b : d, ab cd, [componendo], (3) If a : b c : d then, b, d, a c, Proof : iGven that,, b d, a, ? 1, b, , i.e.,, (4), , ab, b, , If a : b, , c, Ading 1 to both the sides], 1 [d, d, , cd, d, , c : d then, , ab, b, , cd, d, , [dividendo], , Proof : a : b, , c:d, a, c, ? 1, 1 [subtracting 1 from both the sides], b, d, a b c d, i.e.,, b, d, , (5), , If a : b, , c : d then, , Proof : iGven that,, , a, b, , ab, ab, , cd, cd, , [componendo –dividendo], , c, d, , ab cd, ..............(i), b, d, ab cd, b, d, , By componendo,, gain by dividendo,, A, or,, , b, ab, , d, [by invertendo]............. ( ii), cd
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180, , Math, , ab, b, cd, d, [multiplying (i) and (ii)], u, u, b, a b, d, cd, ab cd, i.e.,, . [here a z b and c z d ], ab cd, a c e g, ace g, (6) If, then each of the ratio =, ., b d f h, bd f h, a c e g, Proof : eLt,, k., b d f h, ? a bk , c dk , e fk , g hk, a c e g bk dk fk hk k (b d f h), ?, k,, bd f h, bd f h, bd f h, But k is equal to each of the ratio., a c e g ace g, ?, ., b d f h bd f h, Activity : 1. Sum of ages of mother and sister is s years. Before t years, the ratio, of their ages was r : p. What will be the ratio of their ages after x years ?, 2. The shadow of a man of height r metre, standing at p metre from a lightpost is s, metre. If the height of the lightpost be h metre, what was the distance of the man, from the lightpost ?, Example 2. The ratio of present ages of father and son is 7 : 2 , and the ratio will be, 8 : 3 after 5 years. What are their present ages ?, Solution : eLt the present age of father be a and that of son is b . So, by the, conditions of first and second of the problems, we have,, Therefore,, , a 7, .......... .(i), b 2, a5 8, .......... .(ii ), b5 3, From equation (i), we get,, 7b, a, ...........(iii), 2, From equation (ii), we get,, , 3(a 5) 8(b 5), or, 3a 15 8b 40, or, 3a 8b 25, 7b, 8b 25 [by using ( iii)], or, 3 u, 2
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182, , Math, , 1 ax 1 bx, 1, 0 b 2a 2b., 1 ax 1 bx, 1 ax 1 bx, Solution : iGven that,, 1, 1 ax 1 bx, 1 bx 1 ax, ?, 1 bx 1 ax, , Example 5. Solve :, , or,, , (1 ax) 2, , 1 bx, 1 bx, , or,, , [squaring both the sides], , (1 ax) 2, , 1 2ax a 2 x 2, , 1 bx, 1 bx, , 1 2ax a 2 x 2, , or,, , 1 bx 1 bx, 1 bx 1 bx, , or,, , 2, 2bx, , 1 2ax a 2 x 2 1 2ax a 2 x 2, , [by componendo and dividendo], , 1 2ax a 2 x 2 1 2ax a 2 x 2, , 2(1 a 2 x 2 ), 4ax, , 1 1 a2 x2, bx, 2ax, or, 2ax bx(1 a 2 x 2 ), or, x{2a b(1 a 2 x 2 )} 0, ? iEther x 0 or 2a b(a a 2 x 2 ) 0, or, b(1 a 2 x 2 ) 2a, or,, , or, 1 a 2 x 2, , 2a, b, , 2a, 1, b, 1 § 2a ·, or, x 2, 1¸, ¨, a2 © b, ¹, 1 2a, ? x r, 1, a b, or, a 2 x 2, , equired solution, ?R, Example 6. If, , 6, x, , 1 2a, 1 ., a b, 1 1, x 3a x 3b, , show that, , a b, x 3a x 3b, , x 0, x r, , 2, a z b.
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Math, , 184, , or,, , 1 x 1 x, 1 x 1 x, , p2 2 p 1 p2 2 p 1, , p2 1, or, p2 1, 2p, 2p, ? p2 , 1 0., x, or,, , [by componendo -dividendo], , p2 2 p 1 p2 2 p 1, , 1, x, , 2p, x, , a3 b3, a(a b) , prove that a, b, c are ordered proportional., a bc, a3 b3, a ( a b), Solution : iGven that,, a bc, , Example 8. If, , or,, , a 3 b3, a bc, , or,, , (a b)(a 2 ab b 2 ), a bc, , a(a b), a(a b), , a 2 ab b 2, [Dividing both sides by ( a + b)], a, a bc, or, a 2 ab b 2 a 2 ab ac, or,, , ? b 2 ac, ? a, b, c are ordered proportional., , Example 9. If, , ab, bc, , Solution : iGven that,, , or,, or,, or,, , or,, or,, , cd, , prove that c, d a, ab cd, bc d a, , a or a b c d 0 ., , ab, cd, 1, 1 [subtracting 1 from both the sides], bc, d a, a bbc c d d a, bc, d a, ac ca, bc d a, ac ac, , 0, bc d a, 1 ·, § 1, (a c)¨, , ¸ 0, b, , c, d, a¹, ©
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Math, , 185, , d abc, 0, bc d a, or, (a c)(d a b c) 0, or, (a c), , ?E, ither a c 0 i.e., a, or, a b c d 0., Example 10. If, , c, , x, yz, , y, zx, , z, and x, y, z are not mutually equal, prove, x y, , that the value of each ratio is either equal 1 or equal, , x, y, z, y z z x x y, ? x k ( y z )................(i), y k ( z x)...................(ii, z k ( x y )...................(iii), , 1, ., 2, , k, , Solution : eLt,, , Subtracting (ii) from (i), we get,, x y k ( y x) or, k ( y x) y x, ? k 1, gAain, adding (i ) , (ii ) and (iii ) , we get,, x y z k ( y z z x x y ) 2k ( x y z ), , ? k, , 1( x y z ), 2( x y z ), , 1, 2, , ? Therefore, the value of each of the ratio is 1 or, Example 11. If ax, , by, , cz , show that, , 1, ., 2, , x2 y2 z 2, , , yz zx xy, , bc, a, , 2, , , , ca, b, , 2, , Solution : eLt, ax by cz k, k, k, k, ?x, , y, , z, a, b, c, Now,, , i.e.,, , x2 y2 z 2, , , yz zx xy, , k2, , x2 y2 z 2, , , yz zx xy, , bc, , 0DWK,;;)RUPD, , a, , a, , 2, , 2, , u, , bc, k, , , , 2, , ca, b, , 2, , , , k2, b, , , , 2, , ab, c2, , u, , ca, k, , ., , 2, , , , k2, c, , 2, , u, , ab, , bc, , 2, , 2, , k, , a, , , , ca, b, , 2, , , , ab, c2, , , , ab, c2, , .
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186, , Math, , Exercise 11⋅1, 1., 2., 3., 4., , 5., 6., 7., , If the sides of two squares be a and b metres respectively, what will be the, ratio of their areas ?, If the area of a circle is equal to the area of a square, find the ratio of their, paremetre., If the ratio of two numbers is 3 : 4 and their .L.CM. is 180, find the two, numbers., The ratio of absent and present students of a day in your class is found to be, 1:4. xE, press the number in percentage of absent students in terms of total, students., Athing is bought and sold at the loss of 28%. Find the ratio of buying and, selling cost., Sum of the ages of father and son is 70 years. 7 years ago, the ratio of their, ages were 5 : 2. What will the ratio of their ages be after 5 years., If a : b b : c , prove that,, , a2 b2, b2 c2, abc(a b c) 3, (iii ), (ab bc ca) 3, , (i ), , a, c, , 8. Solve : (i ), , (iii ), (iv ), , 1 1 x, 1 1 x, , a, b, , (ii ), 10. If, , a, b, , 1, , (iv ) a 2b c, , 1, 3, , (ii ), , a x a2 x2, 2, , ax a x, , 2, , x 1 x 6, x 1 x 6, , 1 x ·, (vi ) 81§¨, ¸, ©1 x ¹, , 9. If, , 1, 1 1·, 3 3¸, 3, b, c ¹, ©a, , (ii ) a 2 b 2 c 2 §¨, , 3, , ( a b) 2, a, , ax ax, ax ax, , (v ), , ax b ax b, ax b ax b, , 1 x, 1 x, , a 2 ab b 2, c, , show that, (i ) 2, d, a ab b 2, ac bd c 2 d 2, ac bd c 2 d 2, b c, , show that,, c d, , (b c) 2, c, , b, , b, , 2a ! b ! 0 and x z 0., x, 5, , c 2 cd d 2, c 2 cd d 2, , a3 b3 c3, , c
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187, , Math, , a 3 b3 b3 c3, (ii ) (a 2 b 2 c 2 )(b 2 c 2 d 2 ) (ab bc cd ) 2, b3 c3 c3 d 3, 4 ab, x 2a x 2b, , show that,, 11. If x, , 2, a z b., ab, x 2a x 2b, (i ), , 3, , 12. If x, 13. If x, , 16., 17., , 18., , 19., , 0, , 2 a 3b 2 a 3b , show that, 3bx 2 4ax 3b, 2 a 3b 2 a 3b, , 0., , ( a b)2, 2 , prove that, a, b, c are ordered proportional., b 2 c 2 (a c), a, b, c, x, y, z, If, , prove that,, ., yzx zx y x yz, bc ca ab, x y z, bz cy cx az ay bx, If, , prove that,, ., a b c, a, b, c, abc bca cab, If, and a b c z 0 , prove that,, ab, bc, ca, a b c., x, y, z, and x y z z 0 , show, If, xa yb zc ya zb xc za xb yc, 1, ., that, each of the ratio is, abc, If (a b c) p (b c a)q (c a b)r (a b c) s ,, 1 1 1 1, prove that, , ., q r s p, , 14. If, 15., , 3, , m 1 3 m 1, 3, 2, , prove that, x 3mx 3 x m, 3, m 1 m 1, , a2 b2, , 20. If lx, 21. If, , p, q, , my, , nz , show that,, , a2, a, and, 2, b, b, , x2 y2 z 2, , yz zx xy, , mn nl lm, , ., l 2 m2 n2, , aq, pq, , show that,, a, aq, , pq, ., q, , 11⋅4. Successive Ratio, eLt oRni’s earning be Tk. 1000, Soni’s earning be Tk. 1500 and Somi’s earning be, , Tk. 2500. eHre, oRni’s earning : Soni’s earning =1000 : 1500 =2 : 3 ;Soni’s
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188, , Math, , earning : Sami’s earning =1500 : 2500 =3 : 5. eHnce, R, oni’s : Soni’s : Sami’s, earning =2 : 3 : 5., If two ratios are of the form a : b and b : c, they can be put in the form a : b : c. This, is called successive ratio. nAy two or more than two ratios can be put in this form. It, is to be noted that if two ratios are to be put in the form a : b c, antecedent of the first, ratio and subsequent of the second ratio are to be equal. Such as, if two ratios 2 : 3, and 4 : 3 are to be put in the form a : b : c the subsequent quantity of first ratio is to, be made equal to antecedent quantity of the second ratio. That is those quantities are, to be made equal to their .L.CM., 4 4 u 3 12, 2 2u 4 8, gain,, eHre, 2 : 3, 4:3, 12 : 9, 8 : 12 A, 3 3u 3 9, 3 3 u 4 12, Therefore, if the ratios 2 : 3 and 4 : 3 are put in the form, a : b : c will be 8 : 12 : 9., It is to be noted that if the earning of Sami in the above example is 1125, the ratio of, their earnings will be 8 : 12 : 9., Example 12. If a, b, c are quantities of same kind and a : b =3 : 4, b : c =6 : 7, what, will be a : b : c ?, Solution :, , a, b, , 3, 4, , b, 3u 3 9, and, c, 4 u 3 12, , 6u 2, 7u2, , 6, 7, , 12, [.LC, .M. of 4 and 6 is 12], 14, , ? a : b : c =7 : 12 : 14., Example 13. The ratio of angles of a triangle is 3 : 4 : 5. xpEress the angles i, degree., º, , Solution : Sum of three angles =180, , eLt the angles, according to given ratio, be 3 x, 4x and 5x., cAcording to the problem, 3 x +4 x +5 x =180, Therefore, the angles are 3x =3 ×15, 4x =4 ×15, , º, , =60, , º, , and 5x =5 ×15, , º, , =75, , º, , º, , =45, , º, , or, 12x =180, , º, , or, x =15, , º, , º, , Example 14. If the sides of a square increase by 10%, how much will the area be, increased in percentage ?, , n
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189, , Math, , Solution : eLt each side of the square be, , a metre, , ? rAea of the square be a 2 square metre, If the side increases by 10%, each side will be (a 10% of a ) metre or 110a, metre., In this case, the area of the square will be (1 10a ) 2 square metre or, 1 21a 2 square, metre., rAea increases by (1 21a 2 a 2 ) 0.21a 2 square metre, ? The percentage of increment of the area will be, , 0 21a 2, u 100% 21%, a2, , Activity :, 1., , There are 35 male and 25 female students in your class. The ratios of rice and, pulse are 3 : 1 and 5 : 2 given by each of the male and female students for, taking khisuri in a picnic. Find the ratio of total rice and total pulse., , 11⋅5 Proportional Division, Division of a quantity into fixed ratio is called proportional division. If S is to be, divided into a : b : c : d , dividing S by (a + b + c +d) the parts a, b, c and d are to, be taken., Therefore,, , a, of S, abcd, , Sa, abcd, , 2nd part, , b, of S, abcd, , Sb, abcd, , 3rd part, , c, of S, abcd, , Sc, abcd, , 4th part, , d, of S, abcd, , Sd, abcd, , 1st part, , In this way, any quantity may be divided into any fixed ratio.
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190, , Math, , Example 15. Divide Tk. 5100 among 3 persons in such a way that 1st person’s part :, 2nd person’s part : 3rd person’s part are, , Solution : eHre,, , 1 1 1, : :, 2 3 9, , 1 1 1, : : ., 2 3 9, , ·, · §1, · §1, §1, .M. of 2, 3, 9 is 18], ¨ u18 ¸ : ¨ u18 ¸ : ¨ u18 ¸ .[LC, ¹, ¹ ©9, ¹ ©3, ©2, , 9:6:2, Sum of the quantities of ratio =9 +6 +2 =17., , Tk. 5100 u, , 1st person’s part, , 9, =Tk. 2700, 17, , 2nd person’s part, , Tk. 5100 u, , 3rd person’s part, , Tk. 5100 u, , 6, =Tk. 1800, 17, , 2, =Tk. 600, 17, , Therefore, three persons will have Tk. 2700, Tk. 1800 and Tk. 600 respectively., , Exercise 11⋅2, 1., , If a,b,c are ordered proportional, which one is correct of the followings ?, (a) a 2, , 2., , (b) b 2, , bc, , ac, , (c) ab, , The ratio of ages of A, rif and kAib is 5, , bc, , (d) a, , b c, , : 3 ;if rAif is of 20 years old, how many, , years later the ratio of their ages will be 7 : 5. ?, (a) 5 years, 3., , (b) 6 years, , (c) 8 years, , (d) 10 years, , oNte the following information :, (i) lAl the four quantities need not to be of same kind in proportion., (ii) The ratio of areas of two triangles is equal to the ratio of areas of their bases., (iii) If, , a, b, , c, d, , e, f, , g, ace g, , value of each ratio will be, ., h, bd f h
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Math, , 191, , n the basis of above information, which one of the followings is correct?, O, (a) (i) and (ii), , (b) (ii) and (iii), , (c) (i) and (iii), , (d) (i), (ii) and (iii), , The ratio of angles of 'ABC is 2 : 3 : 5 and the ratio of angles of the, quadrilateral ABCD is 3 : 4 : 5 : 6. Based on this information, answer questions, number 4 and 5., 4., , If the sides of a square double, how much will the area of a square be increased?, (a) 2 times, , (b) 4 times, , (c) 8 times, , (d) 6 times, , 5., , If x : y 7 : 5, y : z, , 6., , The estimated cost for the construction of a wooden bridge is Tk. 90,000. But, , 5 : 7 , z : x =how much ?, , Tk. 21,000 has been spent more. What is the percentage of the excess cost ?, 7., , The ratio of rice and husk in paddy to 7 : 3. What is the percentage of rice in it ?, , 8., , The weight ot 1 cubic cm. wood is 7 decigram. What is the percentage of the, weight of wood to the equivalent volume of water ?, , 9., , Distribute Tk. 300 among a, b, c, d in such a way that the raios are a’s part : b’s, part = 2 : 3, b’s part : c’s part =1 : 2 and, , c’s part : d’s part 3 : 2., , 10. Three fishermen have caught 690 pieces of fishes. The ratios of their parts are, , 2 4, 5, ,, and . oHw many fishes each of them will get. ?, 3 5, 6, 11., , The parametre of a triangle is 45 cm. The ratio of the lengths of the sides is, 3 : 5 : 7. Find the length of each sides., , 3 4 6, : : ., 4 5 7, , 12., , Distribute Tk. 1011 in the ratio, , 13., , If the ratio of two numbers is 5 : 7 and their .H.CF. is 4, what is .L.CM. of the, numbers.
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192, , Math, , 14., , In a cricket game, the total runs scored by Sakib, Mushfique and Mashrafi were, 171. The ratio of runs scored by Sakib and Mushfiques, and Mushfique and, Mashrafi was 3 : 2. What were the runs scored by them individually., , 15., , In a office, there were 2 officers, 7 clarks and 3 bearers. If a bearer gets Tk. 1, a, clerk gets Tk. 2 and an officer gets Tk. 4. Their total salary is Tk. 15,000. What, is their individual salary ?, , 16., , In selecting the leader of a society, Mr. Denal won in ratio of 4 : 3 votes of the, two contestants. If total numbers of members were 581 and 91 members did, not cast their votes, what was the difference of votes by which opposite of Mr., Denal had been defeated ?, , 17., , If the sides of a square are increased by 20%, what is percentage of increment, of the area of the square ?, , 18., , If the length of a rectangle is increased by 10% and the breadth is decreased by, 10%, what is the percentage of increase or decrease of the area of the, rectangle?, , 19., , In a field, the ratio of production is 4 : 7 before and after irrigation. In that, field, the production of paddy in a land previously was 304 quintal. What, would be the production of paddy after irrigation ?, , 20., , If the ratio of paddy and rice produced from paddy is 3 : 2 and the ratio of, wheat and suzi produced from wheat is 4 : 3, find the ratio of rice and suzi, produced from equal quantity of rice and wheat., , 21., , The are of a land is 432 square metre. If the ratios of lengths and breadths of, that land and that of another land be 3 : 4 and 2 : 5 respectively, that what is the, area of another land ?, , 22., , Zami and Simi take loans of different amounts at the rate of 10% simple profit, on the same day from same Bank. The amount on capital and profit which Zimi, refunds after two years, the same amount Simi refunds after three years on, capital and profit. Find the ratio of their loan., The ratio of sides of a triangle is 5 : 12 : 13 and parametre is 30 cm., , 23., , (a) Draw the triangle and write what type of triangle in respect of angles.
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Math, , 193, , (b) Determine the area of a square drawn with the diagonal of a rectangle, taking greater side as length and smaller side as breadth as the sides of a, square., (c) If the length is increased by 10%and breadth is increased by 20%, , wha, , t, , will be percentage of increase of the area ?, 24., , The ratio of present and absent of students of a day in a class is 1 : 4., (a) xE, press the percentage of absent students against total students., (b) The ratio of present and absent students would be 1 : 9 if 10 more students, were present. What was the total number of students ?, (c) O, f the total number of students, the, , number of female students is less than, , male students by 20. Find the ratio of male and female students., , 0DWK,;;)RUPD
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Chapter Twelve, , Simple Simultaneous Equations, with Two Variables, For solving the mathematical problems, th e most important topic of Algebra is, equation. In classes V, I and V, II, we have got the idea of simple equation and have, known how to solve the simple equation with one variable. In class V, III, we have, solved the simple simultaneous equations by the methods of substitution and, elimination and by graphs. We have also learnt how to form and solv, e simple, simultaneous equations related to reallife, problems. In this chapter, the idea of, simple simultaneous equations have been expanded and new methods of solution, have been discussed. Besides, in this ch apter, solution by graphs and formation of, simultaneous equations related to real life problems and their solutions hav, e been, discussed in detail., At the end of the chapter, the students will be able to −, ¾ eVrify the consistency of simple s imultaneous equations with two avriables., ¾ eVrify the mutual dependence of two simple simultaneous equations with two, avriables, ¾ E, xplain the method of cross-multiplication, ¾ Form and solve simultaneous equations related to real life mathematical, problems, ¾ oSlve the simultaneous equations with two avriables by graphs., 12⋅1 Simple simultaneous equations., S, imple simultaneous equations means two s imple equations with two avriables when, they are presented together and the two variables are of same characteristics. S, uch, two equations together are also called system of simple equations. In class IVII, we, have solved such system of equations and learnt to form and solve simultaneous, equations related to real life problems. In this chapter, these have been discussed in, more details., First, we consider the equation 2 x y 12 . This is a simple equation with two, avriables., In the equation, can we get such values of x and y on the left hand side for which, the sum of twice the first with the second will be equal to 12 of the right hand side ;, that is, the equation will be satisfied by those two avlues ?, ow, we fill in the following chart from the equation 2 x y 12 :, N
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195, , Math, , aVlue of, , x, , aVlue of, , 2, , 16, , 0, 3, 5, ...., , 12, 6, 2, ....., , y, , aVlue of .LH.S, .(, 4 16, 0 12, 66, 10 2, ....., , 2x + y), 12, 12, 12, 12, 12, , R, .H.S, ., , 12, 12, 12, 12, 12, , The equation has infinite number of solutions. Among those, four solutions are, (2, 16), (0, 12), (3, 6) and (5, 2) ., Again, we fill in the following chart from another equation x y 3 :, aVlues of y, aVlue of .LH.S, ., R, .H.S, ., aVlue of x, ( x y), 25 3, 2, 5, 3, 03, 3, 0, 3, 3, 3, 0, 3, , 3, 0, 3, 5, 2, 3, , 5, 2, 3, ....., 3, , ...., , ....., , 3, , The equation has infinite number of solutions. Among those, four solutions are, (2, 5), (0, 3), (3, 0) and (5, 2) ., If the two equations discussed abov, e are considered together a system, both the, equations will be satisfied simultaneously only by (5, 2). Both the equations will not, be satisfied simultaneously by any other avlues., Therefore, the solution of the system of equations 2 x y 12 and x y 3 is, , ( x, y) (5, 2), Activity : Write down fiv, e solutions for each of the two equations x 2 y 1 0, and 2 x y 3 0 so that among the solutions, the common solutions also exists., 12⋅2 Conformability for the solution of simple simultaneous equations with two, variables., 2 x y 12½, e unique (only, (a) As discussed earlier, the system of equations, ¾ hav, x y 3 ¿, one) solution. uSch system of equations are called consistent. oCmparing the, coefficient of x and y (taking the ratio of the coefficients) of the two equations, we, 2 1, get, z, ; any equation of the system of equations cannot be expressed in terms of, 1 1, the other. That is why, such system of equations are called mutually independent. In, the case of consistent and mutually independent system of equations, the ratios are, not equal. In this case, the constant terms need not to be compared.
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196, , Math, , (b) oNw we shall consider the system of equations, , 2x y, , 6 ½, ¾ . Will this two, 4 x 2 y 12¿, , equations be solv, ed ?, Here, if both sides of first equation are multiplied by 2, we shall get the second, equation. Again, if both sides of second equation are divided by 2, we shall get the, first equation. That is, the two equations are mutually dependent., We know, first equation has infinite number of solutions. oS, 2nd equation has also, the same infinite number of solutions. uSch system of equations are called consistent, and mutually dependent. uSch system of equations hav, e infinite number of solutions., Here, comparing the coefficients of x, y and the constant terms of the two, 2 1 6 § 1 ·, equations, we get,, ¨, ¸., 4 2 12 © 2 ¹, That is, in the case of the system of such simultaneous equations, the ratios become equal., , 2 x y 12½, ¾., 4x 2 y 5 ¿, Here, multiplying both sides of first equation by 2, we get, 4 x 2 y 24, second equation is 4 x 2 y 5, , (c) oNw, we shall try to solve the system of equations, , subtracting, 0 19 , which is impossible., oS, we can say, such system of equations cannot be solv, ed. uSch system of equations are, inconsistent and mutually independent. uSch system of equations hav, e no solution., Here, comparing the coefficients of x , y and constant terms from the two equations,, 2 1 12, we get,, z . That is, in case of the system of inconsistent and mutually, 4 2 5, independent equations ratios of the coefficients of the avriables are not equal to the ratio, of the constant terms. eGnerally, conditions for comformability of two simple, simultaneous equations, such as,, system of, equations, (i ), , a1 x b1 y, a2 x b2 y, , (ii ), , a1 x b1 y, , c1, c2, c1, , a2 x b2 y c2, (iii ) a1 x b1 y c1, a2 x b2 y c2, , a1 x b1 y, a2 x b2 y, , Comparison, of coeff.and, const. terms, a1 b1, z, a2 b2, a1, a2, , b1, b2, , c1, c2, , a1, a2, , b1 c1, z, b2 c2, , c1 ½, en in the chart below :, ¾ are giv, c2 ¿, , consistent/, incon, sistent, consistent, , has solution, mutually, dependent/ (how many), / no., independent, independent eYs, (only, one), , consistent, , dependent, , inconsistent independent, , eYs (infinite, numbers), oN
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197, , Math, , Now, if there is no constant terms in both the equations of a system of equations ;, a1 b1, z, the, i.e., c1 c2 0 , if with reference to the above discussion from (i), if, a2 b2, system of equations are always consistent and independent of each other. In that, case, there will be only one (unique) solution., a, b1, From (ii ) and (iii ) if 1, , the system of equations are consistent and, a2 b2, dependent of each other. In that case, there will be infinite number of solutions., Example : xE, plain whether the following sy stem of equations are consistent /, inconsistent, dependent/ independent of each other and indicate the number of, solutions in each case., (b) 2 x 5 y 3, (c) 3 x 5 y 7, (a) x 3 y 1, 2x 6 y 2, x 3y 1, 6 x 10 y 15, Solution :, x 3y 1 ½, (a) iGevn system of equations are :, ¾, 2 x 6 y 2¿, aRtio of the coefficients of, , 1, 2, 3, y is, 6, , x is, , aRtio of the coefficients of, aRtio of constant terms is, , or, , 1, 2, , 1, 2, , 1 3 1, 2 6 2, Therefore, the system of equations are consistent and mutually dependent. The, system of equations hav, e infinite number of solutions., 2 x 5 y 3½°, (b) iGevn system of equations are :, ¾, x 3 y 1 °¿, 2, R, atio of the coefficients of x is, 1, 5, aRtio of the coefficients of y is, 3, 2 5, ? we have,, z, 1, 3, Therefore, the system of equations are consistent and mutually independent. The, system of equations hav, e only one (unique) solution., ?
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198, , Math, , 3x 5 y 7, 6 x 10 y 15, 3, 1, or, R, atio of the coefficients of x is, 6, 2, 5, 1, or, R, atio of the coefficients of y is, 10, 2, 7, ratio of the constant terms is, 15, 3 5, 7, ? we get,, z, 6 10 15, Therefore, the system of equations are inconsistent and mutually independent. The, system of equations have no solution., Activity : eVrify whether the system of equations x 2 y 1 0, 2 x y 3 0, are consistent and dependent and indicate how many solutions the system of, equations may hav, e., (c) iGevn system of equations are :, , Exercise 12⋅1, Mention with arguments, whether the following simple simultaneous equations are, consistent/inconsistent, mutually dependen t/independent and indicate the number of, solutions :, 2. 2 x y 3, 3. x y 4 0, 1. x y 4, x y 10, 3 x 3 y 10 0, 4x 2 y 6, 5. 3 x 2 y 0, 6. 5 x 2 y 16 0, 4. 3 x 2 y 0, 6, 3x y 2, 9x 6 y 0, 6x 4 y 0, 5, 1, 1, 1, 7. x y 1, 8. x y 0, 9. x y 1, 2, 2, 2, x 2y 2, x y 5, x 2y 0, 10. ax cy 0, cx ay c 2 a 2 ., 12⋅3 Solution of simple sumultaneous equations, We shall discuss the solutions of only the consistent and independent simple, simultaneous equations. uSch system of equation has only one (unique) solution., Here, four methods of solutions are discussed :, (1) Method of substitution, (2) Method of elimination, (3) Method of crossmultiplication (4) G, raphical method.
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Math, , 199, , In class V, III, we have known how to solv, e by the methods of substitution and, elimination. Here, examples of one for each of these two methods are giv, en., Example 1. S, olve by the method of substitution :, 2x y 8, 3x 2 y 5, Solution : iGevn equations are :, 2 x y 8..............(1), 3 x 2 y 5............( 2), From equation (1), y 8 2 x.........(3), Putting the avlue of y form equation (3) in equation (2), we get, 3 x 2(8 2 x) 5, Putting the value of x in equation (3), y 8 2u3, or, 3 x 16 4 x 5, or, 3 x 4 x 5 16, 86, or, 7 x 21, 2, or, x 3, ?S, olution ( x, y) (3, 2), Solution by the method of substitution :, oCnveniently, from any of the two equations,, value of one avriable is expressed in, terms of the other variable and putting the obtained avlue in the other equation, we, shall get an equation with one avriable. oSlving this equation, avlue of the avriable, can be found. This avlue can be put in any of the equations. But, if it is put in the, equation in which one avriable has been expr essed in terms of the other variable, the, solution will be easier. From this equation, avlue of the other avriable will be found., Example 2. S, olve by the method of elimination :, 2x y 8, 3x 2 y 5, [N.B. : To show the difference between the methods of substitution and elimination,, same equations of example 1 have been taken in this example 2], Solution : iGevn equations are, 2 x y 8..............(1), 3 x 2 y 5............( 2), Multiplying both sides of equation (1) by 2, 4 x 2 y 16..............(3), equation (2) is 3 x 2 y 5............( 2), Adding (3) and (2), 7 x 21 or x 3 ., Putting the avlue of x in equation (1), we get, , 2u3 y 8, or, y 8 6, or, y 2, ? S, olution ( x, y) (3, 2)
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202, , Math, , or, , x, (1) u (13) 2 u (1), , or, , x, 13 2, , x, 15, x, ?, 15, , y, 3 78, , y, 75, 1, ,, 15, y, Again,, 75, , or, , y, (1) u 3 (13) u 6, , 1, 6 u 2 3 u (1), , 1, 12 3, , x, 1, , 1, 15, , 2, , ͵, , 15, 1, 15, 75, 1, , or y, 15, 15, ( x, y ) (1,5), , y, , 1, , 1, 6, , 13, , 3, , or x, , 5, , ? S, olution, Example 4. S, olve by the method of cross-multiplication :, , 3x 4 y 0, 2x 3y, , 1, , Solution : iGevn equations are :, , 3x 4 y, 2x 3y, , 0 ½, ¾, 1¿, , or, , 3 x 4 y 0 0½, ¾, 2x 3y 1 0 ¿, , By the method of cross-multiplication, we get,, x, y, 1, 4 u 1 ( 3) u 0 0 u 2 1 u 3 3 u ( 3) 2 u ( 4), x, or, 40, , y, 03, , x, , y, , 1, , 3 4, , 0, , 3 4, , 2 3, , 1 2, , 3, , 1, 98, , x, y, 1, 4 3 1, x y 1, or, 4 3 1, x 1, , or x 4, ?, 4 1, y 1, Again,, , or y 3, 3 1, ? S, olution ( x, y ) ( 4,3) ., or, , Example 5. S, olve by the method of cross-multiplication :, , x y, , 2 3, , 8, , 5x, 3y, 4, , Solution : Arranging the giv, en equations in the form, , ax by c, , 3, , 0 , we get,, , 1, 2, ,
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203, , Math, , x y, , 8, 2 3, 3x 2 y, or, 8, 6, or 3 x 2 y 48 0, , 5x, 3 y 3, 4, 5 x 12 y, or, 3, 4, or 5 x 12 y 12 0, Again,, , ? the given equations are :, , 3 x 2 y 48 0, 5 x 12 y 12 0, By the method of cross-multiplication, we get,, x, 2 u 12 ( 12) u ( 48), , y, ( 48) u 5 12 u 3, , 1, 3 u ( 12) 5 u 2, , x, , y, , 1, , 3, , 2, , 48, , 3, , 2, , 5, , 12, , 12, , 5, , 12, , x, y, 1, 24 576 240 36 36 10, x, y, 1, or, 552 276 46, 1, x, y, or, 552 276 46, 552, x, 1, ?, or, x, 12, 552 46, 46, 276, y, 1, , or y, 6, Again,, 46, 276 46, ? S, olution ( x, y ) (12,6), or, , Verification of the correctness of the solution :, , x and y in given equations, we get,, x y 12 6, In 1st equation, .LH.S, .=, , , 62, 2 3 2 3, .H.S, ., 8 R, 5 u 12, 5x, In 2nd equation, .LH.S, .=, 3y, 3u 6, 4, 4, .H.S, ., 15 18 3 =R, Putting the values of, , ? the solution is correct., Example 6. S, olve by the method of cross-multiplication :, Solution : iGevn equations are, ax by ab ½, ax by ab 0½, ¾, ¾ or,, bx ay ab ¿, bx ay ab 0¿, By the method of cross-multiplication, we get,, , ax by, , ab, , bx ay.
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204, , ?, , Math, , x, , x, , y, , 1, , ( b ) u ( ab ) ( a )( ab ), , ( ab ) u b ( ab ) u a, , a u ( a ) b u ( b ), , y, , 1, , a, , b, , ab, , a, , b, , b, , a, , ab, , b, , a, , x, y, 1, ab 2 a 2 b ab 2 a 2 b a 2 b 2, x, y, 1, or, ab( a b) ab( a b) ( a b)( a b), x, y, 1, or, ab( a b) ab( a b) ( a b)( a b), ab( a b), ab, x, 1, , or x, ?, ( a b)( a b) a b, ab( a b) ( a b)( a b), y, 1, ab( a b), ab, , or y, Again,, ab( a b) ( a b)( a b), ( a b)( a b) a b, ab ab ·, §, ? ( x, y ) ¨, ,, ¸, ©ab ab¹, or, , Exercise 12⋅2, oSlve by the method of substitution (1 -3) :, 1. 7 x 3 y 31, x y, , 1, 2., , 9x 5 y, , 41, , 2 3, x y, , 3 2, , 1, , 3., , x y, , 2, a b, ax by a 2 b 2, , olve by the method of elimination (4 -6) :, S, 4. 7 x 3 y, , 5. 7 x 8 y 9, 5 x 4 y 3, oSlve by the method of cross-multiplication (7 -15) :, , 9x 5 y, , 31, 41, , 7. 2 x 3 y 5, , 0, 4x 7 y 6 0, 10. 4 x 3 y 12, 2x 5, 13. ax by, , 2bx ay, , a2 b2, , ab, , 15. ( x 7)( y 3) 7, , 6. ax by, , c, a2 x b2 y c2, , 8. 3 x 5 y 9, , 9. x 2 y 7, 0, 5x 3 y 1 0, 2x 3y 0, 11. 7 x 8 y 9, 12. 3 x y 7 0, 5 x 4 y 3, 14. y(3 x) x(6 y ), , 3(3 x ), , 5( y 1), , ( y 3)( x 1) 5, , 2x y 3
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Math, , 205, , 12⋅4 Solution by graphical method, In a simple equation with two variab les, the relation of existing avriables, x and y, can be expressed by picture. This picture is called the graphs of that relation. In the, graph of such equation, there exist infin ite number of points. Plotting a few such, points, if they are joined with each other, we shall get the graph., aEch of a simple simultaneous equations ha s infinite number of solutions. rGaph of, each equation is a straight line. oCordinates of each point of the straght line satisfies, the equation. To indicate a graph, two or more than two points are necessary., ow we shall try to solve graphically the following system of equations :, N, 2 x y 3..........(1), x 1 0 , 3, 4 x 2 y 6........(2), From equation (1), we get, y 3 2 x ., y 5, 3 3, Taking some values of x in the equation, we find the, corresponding values of y and make the adjoining table :, ? three points on the graph of the equation are : (1, 5), (0, 3) and (3, 3) |, 6 4x, Again, from equation (2) , we get, 2 y 6 4 x or, y, x 2 0 6 , 2, Taking some values of x in the equations, we find the, y 7 3 9, the corresponding values of y and make the adjoining table :, ? three points on the graph of the equation are : (2, 7), (0, 3) and (6, 9), In a graph paper let XOX c and YOY c be, x a- xis and y a- xis and O is the, respectively the, origin., We take each side of smallest squares of the graph, paper as unit along with both axes. oNw, we plot the, points (1, 5), (0, 3) and (3, 3) obtained from, equation (1) and oj in them each other. The graph is, a straight line., Again, we plot the points (2, 7), (0, 3) and (6, 9), obtained from equation (2) and join them each, , other. In this case also the graph is a straight line., But we observe that the two straight lines coin cide and they have turned into the one, straight line. Again, if both sides of equation (2) are div, ided by 2, we get he equation, (1). That is why the graphs of the two equations coincide., 2 x y 3..........(1) ½, Here, the system of equations,, ¾ are consistent and mutually, 4 x 2 y 6........(2)¿, dependent. uSch system of equations have infinite number of solutions and its graph, is a straight line.
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206, , ow, we shall try to solve the system of equations :, N, , Math, , 2 x y 4..........(1), 4 x 2 y 12......(2), , From equation (1), we get, y 2 x 4 ., Taking some values of x in the equation, we find the corresponding, avlues of y and make the adjoining table :, x 1 0 4 , ? three points on the graph of the equation are :, y 6 4 4, (1, 6), (0, 4), (4, 4) ., Again, from equation (2), we get,, 4 x 2 y 12 , or 2 x y 6, x, Ͳ, ͵, , or y 2 x 6, y 6 Ͳ, Taking some values of x in the equation, we find the, , the corresponding values of y and make the adjoining table :, ? three points on the graph of the equation are : (0, 6), (3, 0), (6, 6) |, In graph paper let XOX c and YOY c be respectively x a- xis and y a- xis and O is the, origin., Taking each side of smallest squares in the graph, paper as unit, we plot the points (1, 6), (0, 4), and (4, 4) obtained from equation (1) and oj in, them each other. The graph is a straight line., Again, we plot the points (0, 6), (3, 0), (6, 6), obtained from equation (2) and join them each, other. In this case also the graph is a straight line., We observ, e in the graph, though each of the giv, en, equations has separately infinite number of, solutions, they have no common solution as, , system of simultaneous equations. Further, we, observe that the graphs of the two equations are straight lines parallel to each other., That is, the lines will never intersect each other. Therefore, there will be no common, point of the lines. In this case we say, such system of equations have no solution. We, know, such system of equations are inconsistent and independent of each other., oNw, we shall solv, e the system of tw o consistent and independent equations by, graphs. G, raphs of two such equations with two variables intersect at a point. Both, the equations will be satisfied by the coordinates of that point. The evry coordinates, of the point of intersection will be the solution of the two equations., Example 7. oSlve and show the solution in graph :, 2x y 8, , 3x 2 y, 2 x y 8 0...........(1), 3 x 2 y 5 0.............( 2), By the method of cross-multiplication, we get,, , Solution : : iGevn two equations are :, , 5
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Math, , 207, , x, 1 u (5) (2) u (8), or, , x, 5 16, , y, (8) u 3 (5) u 2, , y, 24 10, , 1, 2(2) 3 u 1, , 1, 43, , x, y, 1, 21 14 7, x, y 1, or, 21 14 7, x 1, 21, , or x, ?, 3, 21 7, 7, y 1, 14, Again,, , or y, 2, , 14 7, 7, ? solution is ( x, y ) (3,2), ely and, O, the origin., eLt XOX c and YOY c be x a- xis and y a- xis respectiv, Taking each two sides of the smallest squares along with both axes of the graph, paper as one unit, we plot the point (3, 2)., or, , Example 8. S, olve with the help of graphs :, , 3x y 3, 5 x y 21, Solution : : iGevn equations are :, , 3x y, 5x y, , 3..........(1), 21........(2), , x, , 1, , y, , 6 3, , x, , 3, , From equation (1), we get, 3 x y 3 , or y 3 x 3, y 6, Taking some values of x in equation (1), we get, corresponding values of y and make the table beside :, ? three points on the graph of the equation are : (1,6), (0,3), (3,6), Again, from equation (2), we get, 5 x y 21 , or, y 21 5 x, Taking some avlues of x in equation (2), we find, the corresponding avlues of y and make the, adjoining table :, ? three points on the graph of the equation are :, (3,6), (4,1), (5,4) ., ely x a- ixs and, eLt XOX c and YOY c be respectiv, y a- xis and O be the origin., We take each side of the smallest square of the, graph paper as unit., , 0, , 3, 6, , 4, , 5, , 1, , 4, ,
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Math, , 208, , oNw, we plot the points ( 1,6), (0,3), (3,6) obtained from equation (1) and oj in, them successively. The graph is a straigh t line. S, imilarly, we plot the points, (3,6), (4,1), (5,4) obtained from equation (2) and oj in them successively. In this, case also the graph is a straight line. eLt the two straight lines intersect each other at, P . It is seen from the picture that the coordinates of P are (3,6) ., ? solution is ( x, y ) (3,6), Example 9. S, olve by graphical method : 2 x 5 y 14, , 4 x 5 y 17, x 3, 1, , 2 x 5 y 14...........(1), 2, 4 x 5 y 17........(2), y 4 3, 2 x 14, From equation (1), we get, 5 y 14 2 x , or y, 5, Taking some convenient avlues of, x in the equation, we , find the corresponding values of y and make the adjoining table :, ? three points on the graph of the equation are :, x 3, 1, 1, , (3,4), §¨ ,3), ( 2,2) ·¸ ., 2, ¹, ©2, 4 x 17, y 1 3 , Again, from equation (2), 5 y 4 x 17 , or y, Solution : iGevn equations are :, , 2, , 2, , 2, , 5, , 5, , , Taking some convenient avlues of x in the equation (2),, we find the corresponding avlues of y and make the adjoining table :, ? three points on the graph of the equation are :, , §1, ·, (3, 1), ¨ ,3 ¸, (2,5) ., 2, ©, ¹, , eLt, y a- xis, XOX c and YOY c be x a- xis and, respectiv, ely and O, the origin., We take each two sides of the smallest squares as unit, along with both axes., oNw, we plot the points (3,4), §¨ 1 ,3·¸ and (2,2), , ©2 ¹, obtained from equation (1) in the graph paper and oj in them each other. The graph is, a straight line. S, imilarly, we plot the points (3,1), §¨ 1 ,3 ·¸, (2,5) obtained from, ©2, , ¹, , equation (2) and oj in them each other. The graph is a straight line., eLt the straight lines intersect at P . It is seen from the graph, coordinates of P are §¨ 1 ,3 ·¸ ., ©2, ¹, 1, ? solution is ( x, y ) §¨ ,3 ·¸, ©2, , ¹
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Math, , 209, , Example 10. oSlv, e with the help of graphs :, Solution : iGevn equation is, , 3, , 3, x, 2, , 3, , 3, x, 2, , 8 4x, , y, , 2, , 0, , 2, , y, , 6, , 3, , 0, , x, , 1, , 2, , 3, , y, , 4, , 0, , 4, , 8 4x, , 3, x 8 4x, 2, 3, ?, y 3 x.........(1), 2, And, y 8 4 x..........( 2), oNw, Taking some avlues of, x in equation (1), we find the, corresponding avlues of y and make the adjoining, table :, ? three points on the graph of the equation (1) are :, , eLt,, , x, , 3, , (2,6), (0,3), (2,0), , Again, taking some values of x in equation (2), we, find the corresponding avlues of y and make the, adjoining table :, ? three points on the graph of the equation (2) are :, , (1,4), (2,0), (3,4), , , eLt XOX c and YOY c be x a- xis, y a- xis respectiv, ely, and O, the origin. We take each side of the smallest squares along with both axes as, unit. N, ow, we plot the points ( 2,6), (0,3), ( 2,0) , obtained from equation (1) on the, graph paper and oj in them each other. The graph is a straight line. In the same way,, we plot the points (1,4), ( 2,0), (3,4) obtained from equation (2) and oj in them each, other. This graph is also a straight line. eLt the two straight lines intersect at P . It is, seen from the picture that the coordinates of the point of intersection are (2,0) ., ? solution is x 2 , or solution is 2, , Activity : Find four points on the graph of the equation 2 x y 3 0 in terms of, a table. Then, taking unit of a fixed length on the graph paper, plot the points and, oj in them each other. Is the graph a straight line ?, , Exercise 12⋅3, Solve by graphs :, 1. 3 x 4 y 14, , 4x 3y, 4. 3 x 2 y, , 5x 3 y, , 2. 2 x y, , 1, 5 x y 13, , 2, 2, , 5, , 0DWK,;;)RUPD, , 5., , x y, , 2, 2 3, 2 x 3 y 13, , 3. 2 x 5 y, , x 3y, 6. 3 x y, , 1, 2, 6, , 5 x 3 y 12
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210, , Math, , Ǥ 3 x 2 y, , 4 , , 3 x 4 y, , 1, , 8., , x y, , 2 3, x, , y, 6, , 9. 3 x 2, , 3, , x2, , 3, , 10. 3 x 7 3 2 x, 12⋅5 Formation of simultaneous equations from real life problems and, solution., In ev, eryday life, there occur some such ma thematical problems which are easier to, solve by forming equations. For this, from the condition or conditions of the, problem, two mathematical symbols, mostly the avriables x, y are assumed for two, unknown expressions. Two equations are to be formed for determining the avlues of, those unknown expressions. If the two equations thus formed are solved, values of he, unknown quantities will be found., Example 11. If 5 is added to the sum of the two digits of a number consisting of two, digits, the sum will be three times the digits of the tens place. Moreov, er, if the places, of the digits are interchanged, the number thus found will be 9 less than the original, number. Find the number., Solution : eLt the digit of the tens pl ace of the required number be x and its digits, of the units place is y. Therefore, the number is 10 x y., ? by the 1st condition, x y 5 3 x.........(1), and by the 2nd condition, 10 y x (10 x y ) 9.......( 2), From equation (1), we get, y 3 x x 5 , or y 2 x 5........(3), Again from equations (2), we get,, 10 y y x 10 x 9 0, putting the avlue of x in (3), we get,, , or, or, or, or, , 9 y 9x 9 0, y x 1 0, 2x 5 x 1 0, x 4, , y, , 2u 4 5, 85, 3, , ? the number will be, [ , , y ȋ͵ȌȐ, , 10 x y 10 u 4 3, 40 3, 43, , ? the number is 43, Example 12. 8 years ago, father’s age was eight times the age of his son. After 10, years, father’s age will be twice the age of the son. What are their present ages ?, Solution : eLt the present age of father be x year and age of son is y year., ? by 1st condition x 8 8( y 8)........(1)
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211, , Math, , ʹͶ, , ʹͲ, , and by 2nd condition, x 10 2( y 10).......( 2), From (1) , we get, x 8 8 y 64, or, x 8 y 64 8, or, x 8 y 56.............(3), From (2) , we get, x 10 2 y 20, or, 8 y 56 10 2 y 20 [putting the value of x from (3)], or, 8 y 2 y 20 56 10, or, 6 y 66, or, y 11, ? from (3) , we get, x 8 u 11 56 88 56 32, ? at present, Father’s age is 32 years and son’s age is 11 years., Example 13. Twice the breadth of a rectangular garden is 10 metres more than its, length and perimeter of the garden is 100 metre., a. Assuming the length of the garden to be x metre and its breadth to be y metre,, form system of simultaneous equations., b. Find the length and breadth of the garden., c. There is a path of width 2 metres around the outside boundary of the garden. To, make the path by bricks, it costs 110 00 per square metre. What will be the, total cost ?, Solution : a. eLngth of the rectangular garden is x metre and its breadth is y metre., ? by 1st condition, 2 y x 10........(1), and by 2nd condition, 2( x y ) 100.......( 2), b. From equation (1), we get, 2 y x 10........(1), From equation (2), we get, 2 x 2 y 100.......(2), or 2 x x 10 100 [putting the avlue of 2 y from (1)], or 3 x 90, or x 30, ? from (1), we get, 2 y 30 10 [putting the value, , of x ], or, 2 y 40, , or, y 20, ? length of the garden is 30 metres and its breadth, ͵Ͷ, is 20 metres., c. eLngth of the garden with the path is (30+, 4) metres., =34 metres
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212, , Math, , and breadth is (20+, 4) metres =24 metres., ? Area of the path =Area of the graden with the, path area of the garden, (34 u 24 30 u 20) square metre, (816 600) square metre, 216 square metre, ? cost for making the path by bricks, =Tk. 216u 110, =Tk. 23760, Activity : If in triangle ABC , B 2 x degree, C, A y degree and A B C , find the avlue of x and y ., , x, , degree,, , Exercise 12⋅4, 1., , For which of the following conditions are the system of equations, ax by c 0 and px qy r 0 consistent and mutually independent ?, a b, a b c, a b, a b c, z, z, b., c., d., a., p q r, p q, p q r, p q, , If x y 4, x y 2 , which one of the following is the avlue of ( x, y ) ?, a. (2, 4), b. (4, 2), c. (3, 1), d. (1, 3), 3. If x y 6 and 2 x 4 , what is the avlue of y ?, b. 4, c. 6, d. 8, a. 2, 4. For which one of the following equation is the adjoining chart correct?, b. y 8 x, a. y x 4, Ͳ ʹ Ͷ, c. y 4 2 x, d. y 2 x 4, ǦͶ Ͳ Ͷ, 5. If 2 x y 8 and x 2 y 4 , what is x y ?, a. 0, b. 4, c. 8, d. 12, 6. bOserv, e the following information : :, i. The equations 2 x y 0 and x 2 y 0 are mutually dependent., ii. rGaph of the equation x 2 y 3 0 passes through the point ( 3, 0) ., iii. rGaph of the equation 3 x 4 y 1 is a straight line., nOthe basis of information above, which one of the following is correct ?, a. i and ii, b. ii and iii, c. i and iii, d. i, ii and iii, 7. eLngth of the floor of a rectangular r oom is 2 metres more than its breadth and, perimeter of the floor is 20 metres., 2.
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Math, , 213, , Answer the following questions :, (1) What is the length of the floor of the room in metre ?, a. 10, b. 8, c. 6, d. 4, (2) What is the area of the floor of the room in square metre ?, b. 32, c. 48, d. 80, a. 24, (3) How much taka will be the total cost for decorating the floor with mosaic at, Tk. 900 per square metre ?, a. 72000, b. 43200, c. 28800, d. 21600, Solve by forming simultaneous equations (8 - 15) :, 8. If 1 is added to each of numerator and denominator of a fraction, the fraction, 4, will be . Again, if 5 is subtracted from each of numerator and denominator,, 5, 1, the fraction will be . Find the fraction., 2, 9. If 1 is subtracted from numerator and 2 is added to denaminator of a fraction,, 1, the fraction will be . Again, if 7 is subtracted from numerator and 2 is, 2, 1, subtracted from denominator, the fraction will be . Find the fraction., 3, 10. The digit of the units place of a number consisting of two digits is 1 more than, three times the digit of tens place. But if the places of the digits are, interchanged, the number thus found will be equal to eight times the sum of the, digits. What is the number ?, 11. Difference of the digits of a number consisting of two digits is 4. If the places of, the digits are interchanged, sum of the numbers so found and the original, number will be 110 ; find the number., 12. Present age of mother is four times the sum of the ages of her two daughters., After 5 years, mother’s age will be twice the sum of the ages of the two, daughters. What is the present age of the mother ?, 13. If the length of a rectangular region is decreased by 5 metres and breadth is increased, by 3 metres, the area will be less by 9 square metres. Again, if the length is increased, by 3 metres and breadth is increased by 2 metres, the area will be increasd by 67, square metres. Find the length and breadth of the rectangle.
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21, 4, , Math, , 41. A boat, rowing in favour of current, goes 15km per hour and rowing against the, current goes 5km per hour. iFnd the speed of the boat and current., .51 A labourer of a garments serves on the basis of monthly salary. At the end o f, every year she gets a fixed increment. Her monthly salary beomes Tk. 450, after 4 years and Tk. 05after 8 years. iFnd the salary at the beginning of he, service and amount of annual increment of salary., 16. A system of simple equations are, x y 1, 0, , 3x 2 y, , r, , 0, , a. S, how that the equations are consistent. How many solutions do they have ?, b. oSlving the system of equations, find ( x, y) ., c. iFnd the area of the triangle formed by the straight lines indicated by the, equations with the x a- xis., .71 If 7is added to the num. of a fraction, the value of the fraction is the integer 2. Again, , ,, , if 2 is subtracted from the denominator, value of the fraction is the integer .1, a. oFrm a system of equations by the fraction, , x, ., y, , b. iFnd, ( x, y ) by solving the system of equations by the method of crossmultiplication. W, hat is the fraction ?, c. rDaw the graphs of the system of equations and verify the correctness of the, obtained values of ( x, y ) .
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Chapter Thirteen, , Finite Series, The term o‘ rder’is widely used in our day to day life. S, uch as, the concept of order, is used to arrange the commodities in the shops, to arrange the incidents of drama, and ceremony, to keep the commodities in attractive way in the godown. Again, to, make many works easier and attractive, we use large to small, child to old, light to, originated heavy etc. types of order. Mathematical series have been of all these, concepts of order. In this chapter, the relation between sequence and series and, contents related to them have been presented., At the end of this chapter, the students will be able to –, ¾ eDscribe the sequence and series and determine the difference between them, ¾ Explain finite series, ¾ oFrm formulae for determining the fixe d term of the series and the sum of, fixed numbers of terms and solve math ematical problems by applying the, formulae, ¾ eDtermine the sum of squares and cubes of natural numbers, ¾ S, olve mathematical problems by appl ying different formulae of series, ¾ C, onstruct formulae to find the fixed, term of a geometrical progression and, sum of fixed numbers of terms and solv e mathematical problems by applying, the formulae., Sequence, eLt us note the following relation :, 1, 2, 4, 3, 5 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅, , p, 2, , p, 4, , p, 6, , p, 8, , p, p, 10 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ 2 n ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅, Here, every natural number n is related to twice the number 2 n . That is, the set of, positive even numbers {2, 4, 6, 8,.......} is obtained by a method from the set of, natural numbers N {1, 2, 3,.......} . This arranged set of even number is a sequence., Hence, some quantities are arranged in a particular way such that the antecedent and, subsequent terms becomes related. The set of arranged quantities is called a, sequence., The aforesaid relation is called a function and defined as f (n) 2n . The general, term of this sequence is 2n . The terms of any sequence are infinite. The way of, writing the sequence with the help of general term is 2n !, n 1, 2, 3, ... ... or,, ^2n`f, 2n .>, n 1 or, <
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Math, , 21, 6, , The first quantity of the sequence is called the first term, the second quantity is, called second term, the third quantity is called the third term etc. The first term of the, sequence ,13, ,5,7…, …, is 1,the second term is 2 etc., oFllowings are the four examples of sequence :, 1 , 2 , 3 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , n , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅, 1 , 3 , 5 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , (2n 1) , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅, 1 , 4 , 9 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , n 2 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅, 1 , 2 , 3 , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ , n , ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅, 2, , 3, , n 1, , 4, , Activity : 1. G, eneral terms of the sixsequences, sequences :, (i) 1, n, , (ii) n 1, , n 1, , (iii) 1n, 2, , (iv) 1n1, 2, , are given below. W, rite down the, (v) (1) n1 n, , n 1, , (vi) (1) n1 n, , 2n 1, , ., , 2. Each of you write a general term and then write the sequence., Series, If the terms of a sequence are connected successively by +sign, a series is obtained., S, uch as, 1 3 5 7 ....... is a series. The difference between two successive, terms of the series is equal. Again, 2 4 8 6, 1 ........ is a series. The ratio of two, successive terms is equal. Hence, the characterstic of any series depends upon the, relation between its two successive terms. Among the series, two important series are, arithmetic series and geometric series., Arithmetic series, If the difference between any term and its antecedent term is always equal, the series, is called arithmetic series., Example : 1 3 5 7 9 1 is a series. The first term of the series is ,1 the, second term is 3, the third term is ,5etc., Here, second term –first term = 3 1 2, third term –second term = 5 3 2,, fourth term –third term = 7 5 2, fifth term –fourth term = 9 7 2,, sixth term –fifth term = 1 9 2., Hence the series is an arithmetic series. In this series, the difference between two, terms is called common difference. The common difference of the mentioned series, is 2. The numbers of terms of the series are fixed. That is why the series is finite, series. It is to be noted that if the terms of the series are not fixed, the series is called, infinite series, such as, 1 4 7 0, 1 ... ... is an infinite series. In an arithmetic, series, the first term and the common difference are generally denoted by a and d, respectively. Then by definition, if the first term is a , the second term is a d , the, third term is a 2d , etc. Hence, the series will be a (a d ) (a 2d ) ..., Determination of common term of an arithmetic series, eLt the first term of arithmetic series be a and the common difference be d ,, terms of the series are :
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Math, , 21, 7, , first term, a, a (1 1) d, second term, ad, a (2 1) d, third term, a 2d, a (3 1) d, forth term, a 3d, a (4 1) d, .... ...., .... ...., .... ...., ...., .... ...., .... ...., .... ...., ...., ? n th term = a (n 1)d, This n th term is called common term of arithmetic series. If the first term of an, arithmetic series in a and common difference is d, all the terms of the series are, determined successively by putting n 1, 2, 3, 4,... ... in the n th term., eLt the first term of an arithmetic series be 3 and the common difference be 2. Then, second term of the series, 3 2 5, third term 3 2 u 2 7, forth term, 3 3 u 2 9 etc., Therefore, n th term of the series 3 n 1 u 2 2n 1 ., Activity : If the first term of an arithmetic series is 5and common difference is ,7, find the first sixterms, 22nd term, r th term and (2p+, )1th term., Example 1. fOthe series, 5 8 1 4, 1 which term is 383 ?, Solution : The first term of the series a 5, common difference, , d, , 85 1 8 3, , ? It is an arithmetic series., eLt, n th term of the series = 383, W, e know that, n th term = a (n 1)d ., ? a (n 1)d 383, or, 5 (n 1)3 383, or, 5 3n 3 383, or, 3n 383 5 3, or, 3n 381, , 381, 3, ? n 1, 27, or, n, , ? 127 th term of the given series = 383 ., Sum of n terms of an Arithmetic series, eLt the first term of any arithmetic series be a , last term be p, common difference be, d , number of terms be n and sum of n numbers of terms be S n ., rW, iting from the first term and conversely from the last term of the series we get,, , Sn, , a ( a d ) ( a 2d ) p 2d p d p, , (i), , and S n p p d p 2d a 2d (a d ) a, (ii ), Adding, 2S n a p a p a p ... ... a p a p a p, 0DWK,;;)RUPD
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21, 8, , Math, , [ number of terms of the series is n], n a p, n, ? Sn, a p, (iii ), 2, Again, n th term p a (n 1)d . Putting this value of p in (iii) we get,, n, Sn, [a {a (n 1)d }], 2, n, ^2a n 1 d `, (iv), i.e., S n, 2, or, 2 S n, , If the first term of arithmetic series a, last term p and number of terms n are, known, the sum of the series can be determined by the formula (iii ) . uBt if first term, the a , common difference d , number of terms n are known, the sum of the series are, determined by the formula (iv) ., Determination of the sum of first n terms of natural numbers, eLt Sn be the sum of n n- umbers of natural numbers i.e., , S n 1 2 3 (n 1) n, , (i), , riting from the first term and conversely fr om the last term of the series we get,, W, , S n 1 2 3 n 2 (n 1) n, , (i), , (ii ), and S n n n 1 n 2 3 2 1, Adding, 2S n (n 1) (n 1) (n 1) (n 1) [ n n- umber of terms], or, 2Sn n(n 1), ? Sn, , n(n 1), 2, , (iii ), , Example 2. iFnd the sum total of first 05terms of natural numbers., Solution : sUing formula ( iii) we get,, , S05, , 05 (05 1), 2, , 25 u 15, , 2157, , ? The sum total of first 05natural numbers is 215.7, what ?, Example 3. 1 2 3 4 9, Solution : The first term of the series a 1, common difference d 2 1 1 and, the last term p =.9, ? It is an arithmetic series., eLt the n th term of the series = 9, Alternative method :, n th term of an arithmetic S, W, e know,, ince, progression = a (n 1)d, n, Sn, a p,, ?, a (n 1)d 9, 2, or, 1 (n 1)1 9, 9, ? S9, 1 9, or, 1 n 1 9, 2
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21, 9, , Math, , ?, , n, , 9, , 9 u01, 2, , 49, 05, , rFom ( iv) formula, the sum of first n -terms of an arithmetic series, , Sn, , n, {2a (n 1)d }., 2, , 9, 9, (2 89 ), {2 u 1 (9 1) u 1}, 2, 2, 9 u 01, 9 u 05, 49, 05, 2, Example 4. W, hat is the sum of 30terms of the series, 7 12 17 , Solution : iFrst term of the series a 7 , common difference d 1, 2 7 5, ? It is an arithmetic series. Here, number of terms n 30., W, e know that the sum of n t-erms of an arithmetic series, n, Sn, {2a (n 1)d}., 2, 30, oS, the sum of 30terms S30 = {2.7 (30 1)5} 1, 5 (14 29 u 5), 2, 15 (14 145 ) 15 u 159, 2385, Hence, the sum 9terms of the series, , S9, , Example 5. A deposits Tk. 210from his salary in the first month and in every, month of subsequent months, he deposits Tk. 01more than the previous months., ( i ) How much does he deposit in n th month ?, ( ii ) Express the aforesaid problem in series upto n terms., ( iii ) How much does he deposit in first n -months ?, ( iv ) How much does he deposit in a year ?, Solution : ( i ) In the first month, he deposits Tk. 21.0, In the second month, he deposits Tk. (1, 20 1, 0 ) =Tk. 130, In third month, he deposits Tk. (3, 10 01 ) =Tk. 140, In forth month, he deposits Tk. (1, 40 1, 0 ) =Tk. 150, Hence, it is an arithmetic series whose first term is a 2, 10 , common difference, , 310 210, 01 ., n th term of the series a (n 1)d, 210 (n 1)01, 210 01 n 01, 01 n. 01, Therefore, he deposits Tk. (0, 1 n 01 ) in n th month., ( i i) The series, in this case upto n n- umbers of terms will be, d
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220, , Math, , 120, , 130, , 140, , (10 n 10, , ), n, ( iii ) In n n- umbers of months, he deposits Tk. {2a (n 1)d }, 2, n, =Tk., {2 u 210 (n 1)01 }, 2, n, n, =Tk., (240, 0 01 n 01 ) =Tk., u 2(051, 05 n), 2, 2, =Tk. n(5, 0 n 150 ), ( iv ) W, e know that 1year =, 2 ., 21 months. Here n 1, Therefore, A deposits in 1year =Tk. 1, 2 (50 u 12 150 ), =Tk. 1, 2 (60 150 ), =Tk. 2, 1 u0, 5, 7, 1, ., =Tk. 21, 0, , Exercise 13.1, .1 iFnd the common difference and the 12th terms of the series, , 2 5 12 19 , 2. W, hich term of the series 8 1 1, 4 17 is 392 ?, 3. W, hich term of the series 4 7 1, 0 13 is 301 ?, 4. If the p th term of an arithmetic series is p2 and q th term is q2 , what is, ( p q) th term of the series ?, .5 If the m th tem of an arithmetic series is n and n th term is m , what is, ( m n )th term of the series ?, .6 W, hat is the number of, n terms of the series 1 3 5 7 ?, .7 W, hat is the sum of first 9terms of the series, 8 16 24 n ?, 8. 5 1 7, W, hat ?, 1 23 95, hat ?, 9. 29 25 21 23 W, .01The 21th term of an arithmetic series is, 7.W, hat is the sum of the first 23 terms?, .1If the t16h term of an arithmetic series is, 20 , what will be the sum of first 31, terms ?, 12. The total sum of first n terms of the series 9 7 5 is 4, 14 . iFnd, the value of n ., 13. If the sum of first n terms of the series 2 4 6 8 is 25, ,05find the, value of n., 14. If the sum of first n terms of the series is n(n 1) , find the series., .51If the sum of first n terms of the series is n(n 1) , what is the sum of first 01, terms ?, 6.1If the sum of 21 terms of an arithmetic, series is 414 and first 20terms is 650,find, the sum of first 6terms.
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221, , Math, , .71The sum of the first m terms of an arithmetic series is n and the first n terms is, m. iFnd the sum of first ( m+n) terms., 81. If the p th, q th and r th terms of an arithmetic series are a, b, c , respectively,, show that a(q r) b(r p) c( p q) 0., 19.S, how that,, 1 3 5 7 215 961 17 371 20, 9 ., 20.A man agrees to refund the loan of Tk. 200in some parts. Each part is Tk. 2, more than the previous part. If the first part is Tk. 1, in how many parts will the, man be able to refund that amount ?, Determinaton of the sum of Squares of the first n numbers of Natural Numbers, eLt Sn be the number of squares of the first n numbers of natural numbers, i.e., S n 12 2 2 32 n 2, W, e know,, , r 3 3r 2 3r 1 (r 1) 3, or, r 3 (r 1) 3 3r 2 3r 1, In the above identity, putting, r 1, 2, 3, , n we get,, 13 03 3.12 3.1 1, 23 13, , 3.2 2 3.2 1, , 33 23 3.32 3.3 1, , , , 3, , , , n (n 1), , , 3, , , , 3.n 2 3.n 1, , Adding, we get,, n 3 03 3(12 2 2 32 n 2 ) 3(1 2 3 n) (1 1 1 1), n(n 1), n(n 1) º, ª, or,, n 3 3S n 3., n « 1 2 3 n, 2, 2 »¼, ¬, 3n(n 1), or, 3Sn n3 , n, 2, 2n 3 3n 2 3n 2n 2n 3 3n 2 n n(2n 2 3n 1), 2, 2, 2, n(2n 2 2n n 1) n{2n(n 1) 1(n 1)}, 2, 2, or,, , 3S n, , ?, , Sn, , n(n 1)(2n 1), 2, n(n 1)(2n 1), 6
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222, , Math, , The sum of cubes of the first n numbers of Natural Numbers, eLt Sn be the sum of cubes of the first n numbers of natural numbers., That is, S n 13 2 3 33 n 3, , (r 1)2 (r 1)2 (r 2 2r 1) (r 2 2r 1) 4r., or, (r 1) 2 r 2 r 2 (r 1) 2 4r.r 2 4r 3 [Multiplying both the sides by r2], , W, e know that,, , In the above identity, putting r 1, 2, 3, , n, W, e get,, 2 2.12 12.0 2 4.13, 32.2 2 2 2.12 4.23, 4 2.32 32.32 4.33, , , (n 1) 2 n 2 n 2 (n 1) 2 4n 3, Adding, we get, (n 1) 2 .n 2 12.0 2, or, (n 1)2.n2, , 4(13 23 33 n 3 ), , 4Sn, , or, Sn, , n2(n 1)2, 4, , ? Sn, , n(n 1) ½, ®, ¾, ¯ 2 ¿, , 2, , eNcessary formulae :, , 1 2 3 n, 12 2 2 32 n 2, , 13 23 33 n 3, , n(n 1), 2, n(n 1)(2n 1), 6, 2, n(n 1) ½, ®, ¯, , N.B, . : 13 23 33 n 3, , 2, , ¾, ¿, , (1 2 3 n) 2 ., , Activity : .1iFnd the sum of natural even numbers of the first n-numbers., 2. iFnd the sum of squares of natural odd numbers of the first n n- umbers.
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Math, , 223, , Geometric series, If the ratio of any term and its antecedent term of any series is always equal i.e., if any term, divided by its antecedent term, the quotient is always equal, the series is called a geometric, series and the quotient is called common ratio. uSch as, of the series 2 4 8 1, 6 32 ,, the first term is 2, the second term is 4, the third term is 8, the fourth term is 61and the, fifth term is 32. Here, the ratio of the second term to the first term =, third term to the second term =, , 61, 8, , 8, 4, , 4, 2, , 2 , the ratio of the, , 2 , the ratio of the fourth term to the third term =, , 2 , the ratio of the fifth term to the fourth term =, , 32, 61, , 2., , In this series, the ratio of any term to its antecedent term is always equal. The, common ratio of the mentioned series is 2. The numbers of terms of the series are, fixed. That is why the series is finite geomet ric series. The geometric series is widely, used in different areas of physical and biological science, in organizations like B, ank, and iLfe Insurance etc, and in different b ranches of technology. If the numbers of, terms are not fixed in a geometric series, it is called an infinite geometric series., The first term of a geometric series is generally expressed by a and common ratios, o by definition, if the first term is a , the second term is ar, the third term is, by r . S, ar 2 , etc. Hence the series will be a ar ar 2 ar 3 , Activity : W, rite down the geometric series in the following cases :, (i) The first term 4, common ratio 01, , ( ii) The first term ,9common ratio 1, 3, , (iii) The first term ,7common ratio 1, 0, 1, (v) The first term ,1common ratio 1, 2, , (iv) The first term 3, common ratio 1, (vi) The first term 3, common ratio 1., , General term of a Geometric series, eLt the first term of a geometric series be a, and common ratio be r. Then, of the, series,, second term ar ar 21, first term a ar 11 ,, 2, 31, third term ar, ar , fourth term ar3 ar 41, ... ... ..., ..., ... ..., ... ... ..., ..., ... ..., n th term = ar n1, This n th term is called the general term of the geometric series. If the first term of a, geometric series a and the common ratio r are known, any term of the series can be, determined by putting r 1, 2, 3, etc. successively in the n th term., Example 6. W, hat is the t01h term of the series, 2 4 8 16 ?
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Math, , 224, , 2, common ratio r, , Solution : The first term of the series a, , 4, 2, , 2., , ? The given series is a geometric series., n 1, W, e know that the n th term of geometric series = ar, th, 01 1, ? 10 term of the series = 2 u 2, = 2 u 29 2, 014, Example 7. W, hat is the general term of the series 1, 28 64 32 ?, , 2, 18 , common ratio r, , Solution : The first term of the series a, , ? It is a geometric series., W, e know that the general term of the series =, , 46, 218, , 1, ., 2, , ar n1, n 1, , 27, 1, 1, §1·, Hence, the general term of the series = 1, ., 28 u ¨ ¸, n 1, n 1 7, 2, 2, 2 n 8, ©2¹, Example 8. The first and the second terms of a geometric series are 27and .7 iFnd, the t5h and the t01h terms of the series., Solution : The first term of the given series a 27, the second term is 9 ., , Then the common ratio r, ?, , The t5h term = ar 5 1, , 9, 27, , §1·, 27 u ¨ ¸, © 3¹, , 4, , 1, ., 3, 1, 3, , 27 u 1, 27 u 3, , 9, , 1, §1·, 33, 1, 27 u ¨ ¸, ., 3, 3 u 36 36 729, © 3¹, Determination of the sum of a Geometric series, eLt the first term of th e geometric series be a, common ratio r and number of terms, n. If S n is the sum of n terms,, , and the t01h term =, , ar 10, , 1, , Sn, , a ar ar 2 ar n 2 ar n 1, , ?, , Sn, , (i), , and r.S n, [ ultiplying ( i) by r] ( ii), ar ar 2 ar 3 ar n1 ar n m, ubtracting, S n rS n a ar n, S, or, S n (1 r ) a(1 r n ), , a(1 r n ), , when r 1, 1 r, , Again, subtracting (ii) form (i) we get,, rS n S n ar n a or, S n (r 1) a(r n 1), , a(r n 1), , when r ! 1 ., (r 1), Observe : If common ratio is r 1 , each term, i.e., S n, , a
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Math, , 225, , Hence, in this case S n, , a a a upto n., an., , Activity : A employed a man from the first April for taking his son to school and, taking back home for a month. His wages were fixed to be –one paisa in first day,, twice of the first day in second day i.e. two paisa, twice of the second day in the, third day i.e. four paisa. If the wages were paid in this way, how much would he, get after one month including holidays of the week ?, , 21 24 48 867 ?, 24, 1 , common ratio r, Solution : The first term of the series is a 2, 2 ! 1., 2, 1, Example 9. W, hat is the sum of the series, , ? the series is a geometric series., eLt the nth term of the series 7, 68, n 1, W, e know, n th term = ar, ?, ar n 1 768, or, 1, 2 u 2 n1 768, or, 2 n 1 768, 64, 12, , or,, or,, ?, , 2 n 1 26, n 1 6, n 7., a(r n 1) when r ! 1, ,, (r 1), , Therefore, the sum of the series, , 12 (2 7 1), 12 u (128 1) 12 u 127 1524 ., 2 1, Example 10. iFnd the sum of first eight terms of the series 1 1 1 1 , 2 4 8, st, , Solution : The 1 term of the series is a, , 1, common ratio r, , ? It is a geomeric series., Here the number of terms n 8., W, e know, sum of n terms of a geometric series, Sn, , 1, 2, 1, , a(1 r n ) when r 1., ,, 1 r, , Hence, sum of eight terms of the series is S, 8, , § 1 ·8 ½, °, °, 1u ®1 ¨ ¸ ¾, °̄ © 2 ¹ °, ¿, , 1, 6 1·, § 25, ¸, 6 ¹, © 25, , 2¨, 0DWK,;;)RUPD , , 255, 128, , 127, 1, 128, , 1, 2, , 1, , 1, 256, 1, 2, , 1, 1, 2
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226, , Math, , Exercise 13.2, 1. If a, b, c, d are cosecutive four terms of a arithmetic series, which one is correct, ?, cd, bd, bc, ac, (a) b, (b) a, (c) c, (d) d, 2, 2, 2, 2, 2. (i) If a (a d ) a 2d ) ........ the sum of first n is terms of the series is, n, 2 2{a +(n –)1d}, n(n 1)(2n 1), (ii) 1+2 +3 +.......... +, n, 6, (iii) +, 12 +5+........... +, (2n 1) n 2, hWich one of the followings is correct according to the above statements., (a) i and ii, (b) i and iii, (c) ii and iiii, (d) i, ii and iii, Answer the questions 3 and 4 on the basis of following series :, log 2 +log 4 +log 8 +........, 3. W, hich one is the common difference of the series ?, (a) 2, (b) 4, (c) log 2, (d) 2 log 2, 4. W, hich one is the 7th term of the series, (a) log 32 (b) log 46 (c) log 218 (d) log 256, .5 eDtermine the 8th term of the s eries 46 +32 +61+8 +............., .6 eDtermine the sum of first fourteen te rms of the series 3 +9+27+.............., 1, of the series 218 +46 +32 +..........., 7. hWich of the term is, 2, 8. If the t5h terms of a geometric series, , 2 3, and the t01h term are, 9, , 8 2, , find the, 8, , 3rd term of the series., , 1, , 1, 2 , ?, 2, 10.If 5 x y 1, 35 is geometric series, find the value of x and y ., 1. If 3 x y z 243 is geometric series, find the value of x, y and z., 21. W, hat is the sum of first seven terms of the series, 2 4 8 61 ?, 31. iFnd the sum of (2n 1) terms of the series 1 1 1 1 , 41. W, hat is the sum of first ten terms of the series, log 2 log 4 log 8 ?, 9. hWich of the term is, , 8 2 of the sequence, , .51iFnd the sum of first twelve terms of the series, log 2 log 16 log 512 ?, .61If the sum of n terms of the series 2 4 8 6, 1 is 254, what is the, value of n ?
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227, , Math, , 17.W, hat is sum of, (2n 2) terms of the series 2 2 2 2 ?, 81. If the sum of cubes of n natural numbers is 441,find the value of n and find the, sum of those terms., n natural numbers is 225 , find the value of n, .91If the sum of cubes of the first, and find the sum of square of those terms ?, 20.S, how that 13 23 33 ..............10 3 (1 2 3 ..... 10 ) 2 ., 21.If, , 13 23 33 n 3, 1 2 3 n, , 21, 0 , what is the value of n ?, , 22. A iron-bar with length one metre is divided into ten-pieces such that the lengths, of the pieces form a geometric progression. If the largest pieces is ten times of, the smallest one, find the length in appoximate millimetre of the smallest piece., 23. The first term of a geometric series is a, common ratio is r , the fourth term of the, series is 2– and the t9h term is 8 2 ., (a) Express the above information by two equations., (b) iFnd the 21th term of the series., (c) iFnd the series and then determine th e sum of the first seven terms of the, series., 24. The nth term of The a series is 2n –4., (a) iFnd the series., (b) iFnd the t01h term of the series, and determine the sum of first 20terms., (c) C, onsidering the first term of the obta ined series as 1st term and the common, difference as common ratio, construct a new series and find the sum of first 8, terms of the series by applying the formula.
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Chapter Fourteen, , Ratio, Similarity and Symmetry, oFr comparing two quantities, their ratios are to be considered. Again, for, determining ratios, the two quantities are to be measured in the same units. In, algebra we have discussed this in detail., At the end of this chapter, the students will be able to, ¾ Explain geometric ratios, ¾ Explain the internal division of a line segment, ¾ Verify and prove theorems related to ratios, ¾ Verify and prove theorems related to similarity, ¾ Explain the concepts of symmetry, ¾ Verify line and rotational symmetry of real objects practically., 14⋅1 Properties of ratio and proportion, (i ) If a : b = x : y and c: d = x : y, it follows that a: b = c : d., (ii ) If a : b = b : a, it follows that a b, (iii ) If a : b = x : y, it follows that b t a y t x (inversendo), (iv ) If a : b = x : y, it follows that a t x b t y (alternendo), bc (cross multiplication), (v ) If a : b = c : d, it follows that ad, (vi ) If a : b = x : y, it follows that a b t b x y t y (componendo), and a b t b x y t y (dividendo), a c, ab cd, , it follows that, (componendo-dividendo), (vii ) If, b d, ab cd, Geometrical Proportion, Earlier we have learnt to find the area of a triangular region. Two necessary concept, of ratio are to be formed from this., (1, ) If the heights of two triangles are equal, their bases and areas are, proportional., , eLt the bases of the triangles ABC and DEF be BC = a, EF = d respectively and the, height in both cases be h.
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229, , Math, , Hence, the area of the triangle ABC =, , 1, auh, 2, , and the area of the triangle, , 1, DEF= d u h ., 2, , 1, 1, Therefore, area of the triangle ABC :area of the triangle DEF = a u h : d u h, 2, 2, = a : d = BC : EF that is, the areas and bases ate proportional., (2) If the bases of two triangles are equal, their heights and areas are, proportional., , eLt the heights of the triangles ABC and DEF be AP = h, DQ = k respectively and, 1, the base in both cases be b. Hence, the area of the triangle ABC = b u h and the, 2, 1, area of the triangle DEF= b u k, 2, 1, 1, Therefore, area of the triangle ABC :area of the triangle DEF = b u h : b u k, 2, 2, = h: k = AP: DQ, Theorem 1, A straight line drawn parallel to one side of a triangle intersects the other two, sides or those sides produced proportionally., Proposition : In the figure, the straight, line DE is parallel to the side BC of the, triangle ABC. DE intersects AB and AC, or their produced sections at D and E, respectively. It is required to prove that,, AD : DB = AE : EC., iFg-2, Construction: Join B, E and C, D., iFg-1, Proof:, Justificaltin, S, teps, (1) The heights of 'ADE and 'BDE are [The bases of the triangles of equal height, are proportional], equal.
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230, , Math, , AD, 'ADE, [The bases of the triangles of equal, DB, 'BDE, (2) Again, The heights of 'ADE and height are proportional], [nO the same base and between same, 'DEC are equal., pair of lines], 'ADE, AE, ?, 'DEC, EC, (3) uBt 'BDE, 'DEC, 'ADE, 'ADE, ?, 'BDE, 'DEC, AD, AE, (4) Therefore,, DB, EC, i.e., AD t DB AE t EC., Corollary 1. If the line parallel to BC of the triangle AB, Cintersects the sides, AB and, AB AC, AB AC, and, ., AC at D and E respectively,, AD AE, BD CE, Corollary 2. The line through the mid point of a side of a triangle parallel to another, side bisects the third line., The proposition opposite of theorem 1is al so true. That is, if a line segment divides, the two sides of a triangle or the line produced proportionally it is parallel to the third, side. Here follows the proof of the theorem., Theorem 2, If a line segment divides the two sides or their produced sections of a triangle, proportionally, it is parallel to the third side., Proposition : In the triangle ABC the, line segment DE divides the two sides, AB and AC or their produced sections, proportionally. That is, AD : DB = AE :, EC. It is required to prove that DE and, BC are proportional., Construction: Join B, E and C, D., Proof:, S, teps, Justificaltin, 'ADE, AD, (1, ), [Triangles with equal height ], 'BDE, DB, [Triangles with equal height], [given], 'ADE, AE, and, [rFom (i) and (ii)], 'DEC, EC, ?
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Math, , 231, , AE, EC, 'ADE 'ADE, (3) Therefore,, 'BDE ' DEC, ?, 'BDE, 'DEC, (4) uBt 'BDE and 'DEC are on the, same side of the common base DE. S, o, they lie between a pair of parallel lines., Hence BC and DE are parallel., (2) uBt, , AD, DB, , Theorem 3, The internal bisector of an angle of a triangle divides its opposite side in the, ratio of the sides constituting to the angle., Proposition : In 'ABC the line segment AD bisects, the internal angle A and intersects the side BC at D., It is required to prove that BD : DC = BA : AC., Construction: D, raw the line segment CE parallel to, DA, so that it intersects the side BA produced at E., Proof:, S, teps, , Justificaltin, , (1, ) iSnce, DA ll CE and both BC and AC are their [by construction], [corresponding angles], transversal, [alternate angles], and AEC BAD, [supposition], and ACE CAD, [Theorem ]1, (2) uBt BAD CAD, [step (2)], ? AEC ACE ;, ? AC AE, (3) Again, since DA ll CE ,, BD, BA, ?, DC, AE, (4) uBt AE AC, BD, BA, ?, DC, AC, Theorem 4, If any side of a triangle is divided internally, the line segment from the point of, division to the opposite vertex bisects the angle at the vertex.
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Math, , 232, , Proposition : eLt ABC be a triangle and the line segment AD, from vertex A divides the side BC at D such that BD : DC =, BA : AC. It is required to prove that AD bisects BAC,, i.e. BAD =CAD., Construction : rDaw at C the line segment CE parallel to, DA, so that it intersects the side BA produced at E., Proof:, S, teps, Justificaltin, [by construction], (1, ) In 'BCE , DA ll CE, [Theorem ]1, ? BA t AE BD t DC, [supposition ], (2) uBt BD t DC BA t AC, [from steps (1, ) and (2) ], ? BA t AE BA t AC, [aBse angles of isosceles are equal], ? AE AC, Therefore, ACE AEC, [C, orresponding angles ], (4) uBt AEC BAD, [alternate angles ], and ACE CAD, [from step (2) ], Therefore, BAD CAD, i.e., the line segment AD bisects BAC ., , Exercise 14⋅1, .1, , The bisectors of two base angles of a triangle intersect the opposite sides at X, and Yrespectively. If Y, Xis parallel to, the base, prove that the triangle is an, isosceles triangle., 2. Prove that if two lines intersect a few parallel lines, the matching sides are, proportional., 3. Prove that the diagonals of a trapeziu m are divided in the same ratio at their, point of intersection., 4. Prove that the line segment oj ining the mid points of oblique sides of a, trapezium and two parallel sides are parallel., .5 The medians AD and BE of the triangle ABC intersects each other at G. A line, segment is drawn through G parallel to DE which intersects AC at F. Prove that, AC =6 EF., , .6, 7., , In the triangle ABC, X is any point on BC and O is a point on AX. Prove that, 'AOB t 'AOC BX t XC, In the triangle ABC, the bisector of A intersects BC at D. A line segment, drawn parallel to BC intersects AB and AC at E and F respectively. Prove that, BD : DC = BE : CF.
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Math, , 8., , 233, , If the heights of the equiangular triangles ABC and DEF are AM and DN, respectively, prove that AM : DN = AB : DE., , 14⋅2 Similarity, The congruence and similarity of triangles have been discussed earlier in class VII. In, general, congruence is a special case of similarity. If two figures are congruent,, they are similar ; but two similar triangles are not always congruent., Equiangular Polygons:, If the angles of two polygons with equal number of sides are sequentially equal, the, polygons are known as equiangular polygons., , Equiangular triangle, , Equiangular quadrilateral, , Similar Polygons:, If the vertices of two polygons with equal number of sides can be matched in such a, sequential way that, (i) The matching angles are equal, (ii) The ratios of matching sides are equal, the two polygons are called similar, polygons., In the above figures, the rectangle ABCD and the square PQRS are equiangular since the, number of sides in both the figures is 4 and the angles of the rectangle are sequentially, equal to the angles of the square (all right angles). Though the similar angles of the, figure are equal, the ratios of the matching sides are not the same. Hence the figures are, not similar. In case of triangles, situation like this does not arise. As a result of matching, the vertices of triangles, if one of the conditions of similarity is true, the other condition, automatically becomes true and the triangles are similar. That is, similar triangles are, always equiangular and equiangular triangles are always similar., If two triangles are equiangular and one of their matching pairs is equal, the triangles, are congruent. The ratio of the matching sides of two equiangular triangles is a, constant. Proofs of the related theorems are given below., 0DWK,;;)RUPD
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234, , Theorem 5, If two triangles are equiangular, their matching sides are proportional., Proposition : eLt ABC and DEF be triangles with, A = D, B = E and C = F. We need to, AB AC BC, prove that, DE DF EF, Construction: oCnsider the matching sides of the, triangles ABC and DEF unequal. Take two points P, and Q on AB and AC respectively so that AP = DE, and AQ = DF. Join P and Q and complete the, construction., Proof:, Steps, Justificaltin, (1) In the triangles APQand E, DF, AP DE , AQ DF , A D, [SAS theorem], Therefore, 'APQ # 'DEF, eHnce,, APQ DEF ABC, and AQP DFE ACB., That is, the corresponding angles produced as a, result of intersections of AB and AC by the line, segment PQ are equal., AB AC, Therefore,, or,, PQ ll BC ;, ?, AP AQ, [theorem 1], AB AC, ., DE DF, (2) Similarly, cutting line segments EDand EF, from A, B and C, Brespectively, it can be shown that, BA BCEF, i.e.,, ED, [theorem 1], AB BC, AB AC, BC, i.e.,, ., ;, ?, DE EF, DE DF, EF, The proposition opposite of theorem 5is also true., Theorem 6, If the sides of two triangles are proportional, the opposite angles of their, matching sides are equal., , Math
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235, , Math, , Proposition : eLt in 'ABC and 'DEF ,, AB AC BC, ., DE DF EF, It is to prove that,, A D, B E , C F ., Construction: oCnsider the matching sides of the, triangles ABC and DEF unequal. Take two points, P and Q on AB and AC respectively so that AP =, DE and AQ = DF. Join P and Q., Proof:, Steps, AB AC, AB AC, (1) Since, , so,, ., DE DF, AP AQ, Therefore, PQ ll BC, ? ABC APQ and ACB AQP, , Justificaltin, , [Theorem 2], [C, orresponding angles made, by the transversal AB, ], [C, orresponding angles made, by the transversal AC, ], , ? Triangles ABC and APQ are equiangular., AB BC, AB BC, Therefore,, , so,, ., DE AQ, AP PQ, [Theorem ]5, AB BC, BC BC, [supposition]; ?, ?, DE EF, EF PQ, ? EF PQ, Therefore, 'APQ and 'DEF are congruent., ?, [SSS Theorem], PAQ EDF , APQ DEF .AQP DFE ,, ? APQ ABC and AQP ACB, A D, B E , C F ., Theorem 7, If one angle of a triangle is equal to an angle of the other and the sides adjacent, to the equal angles are proportional, the triangles are similar., Proposition : eLt in 'ABC and 'DEF, A =D, AB AC, ., and, DE DF, It is to be proved that the triangles 'ABC, and 'DEF are similar., Construction: oCnsider the matching sides, of 'ABC and 'DEF unequal. Take two points P and
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236, , Math, , Q on AB and AC respectively so that AP = DE, and AQ = DF. Join P and Q., Proof:, Steps, (1) In 'APQ and 'DEF , AP DE , AQ DF, and included A included D, ? 'ABC # 'DEF, ? A D, APQ E , AQP F ., (2) Again,, AB AC, AB AC, DE, DF, since, , so AP AQ, , Justificaltin, [SAS Theorem], , [Theorem 2], , ? PQ ll BC, Therefore, ABC APQ and ACB AQP, ? A D, B E and C F, i.e., triangles ABC and DEF are equiangular., Therefore 'ABC and 'DEF are similar., Theorem 8, The ratio of the areas of two similar triangles is equal to the ratio of squares on, any two matching sides., Proposition : eLt the triangles ABC and DEF, be similar and BC and EF be their matching, sides respectively. It is required to prove that, 'ABC t 'DEF BC 2 t EF 2, Construction: rDaw perpendiculars AG and, DH on BC and EF respectively. eLt AG h,, DH p., Proof:, Steps, Justificaltin, 1, 1, (1) 'ABC, BC.h and 'DEF, EF . p, 2, 2, ?, , 'ABC, 'DEF, , 1, 2, 1, 2, , BC.h, EF . p, , h.BC h BC, = u, p.EF, p EF, , (1) uBt in the triangles ABG and DEG, B, E , AGB, , DHE (=1 right angle)
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237, , Math, , ? BAG EDH, (3) because 'ABC and 'DEF are similar., h AB BC, p DE EF, ?, , 'ABC, 'DEF, , h BC, u, p EF, , BC BC, u, EF EF, , [Triangles AB, Cand E, DF, are similar], , BC 2, EF 2, , 14⋅1 Internal Division of a Line Segment in definite ratio, If A and B are two different points in a plane and m and n are two natural numbers,, we acknowledge that there exits a unique point X lying between A and B and AX : XA, = m : n., In the above figure, the line segment AB is divided at X internally in the ratio m : n,, i.e. AX : XB = m : n., Construction 1, To divide a given line segment internally in a given ratio., eLt the line segment AB be divided internally in the, ration m : n., Construction: eLt an angle BAX be drawn at A. From, AX cut the lengths AE = m and EC = n sequentially., Join B, C. At E, draw line segment ED parallel to CB, which intersects AB at D. Then the line segment AB is, divided at D internally in the ratio m : n., Proof: Since the line segment DE is parallel to a side, BC of the triangle ABC, ? AD : DB = AE : EC = m : n., Activity :, 1. iD, vide a given line segment in definite ra tio internally by an alternative method., Example 1. iDvide a line segment of length 7cm internally in the ratio 3.2., Solution: rDaw any ray AG. From AG, cut a, line segment AB =7cm. rDaw an angle, BAX at A., From AX, cut the lengths AE =3 cm and EC =, 2 cm. from EX. Join B, C. At E, draw an, angle AED equal to ACB whose side, intersects AB at D. Then the line segment AB, is divided at D internally in the ratio 3:2.
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238, , Math, , Exercise 14⋅2, 1., , oCnsider the following information:, i. ratios are considered to compare two expressions, ii. to find ratio, expressions are measured in the same unit, iii. to find ratio, expressions must be of the same type., Which case of the following is true?, a. i and ii, b. ii and iii, c. i and iii, d. i, ii and iii, , sUe the information from the above figure to answer the questions 2 and 3., What is the ratio of the height and base of the triangle ABC?, 4, 2, 5, 1, b., c., d., a., 5, 5, 4, 2, 3. What is the area of triangle ABD in sq. units?, a. 6, b. 20, c. 40, d. 50, 4. In triangle ABC, if PQ | BC, which of the following is true?, a. AP : PB = AQ : QC, b. AB : PQ = AC : PQ, c. AB : AC = PQ : BC, d. PQ : BC = BP : BQ, .5 In a square how many lines of symmetry are there?, a. 10, b. 8, c. 6, d. 4, .6 Prove that if each of the two triangle s is similar to a third triangle, they are, congruent to each other., .7 Prove that, if one acute angle of a right angled triangle is equal to an acute, angle of another right angled triangle, the triangles are similar., 8. Prove that the two right angled triangles formed by the perpendicular from the, vertex containing the right angle are similar to each other and also to the, original triangle., .9 In the adjacent figure, B = D and, CD =4 AB. Prove that BD =5 BL., 2., , A line segment drawn through the vertex A of the parallelogram ABCD intersects the, BC and DC at M and N respectively. Prove that BM u DN is a constant., 11. In the adjacent figure, BD A AC and, 1, DQ BA 2 AQ, QC . BD 5 BL. Prove that, DA A DC ., 2, , 10.
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Math, , 239, , 12. In the triangles ABC and DEF A = D. Prove that,, 'ABC t 'DEF AB. AC t DE .DF ., 13. The bisector AD of A of the triangle ABC intersects BC at D. The line, segment CE parallel to DA intersects the line segment BA extended., a. rDaw the specified figure., b. Prove that ,BD : DC = BA : AC., c. If a line segment parallel to BC intersect AB and AC at P and Q respectively,, prove that BD : DC = BP : CQ., 14. In the figure, ABC and DEF are, two similar triangles., a. aNme the matching sides and, matching angles of the triangles., b. Prove that,, 'ABC, 'DEF, , AB 2, DE 2, , AC 2, DF 2, , BC 2, EF 2, , BC 3, and ABC =3 sq cm,, AB 2, draw the triangle DEF and find its area., 14.4 Symmetry, Symmetry is an important geometrical concept, commonly exhibited in nature and is, used almost in every field of our activity. Artists, designers, architects, carpenters, always make use of the idea of symmetry. The tree-leaves, the flowers, the beehives,, houses, tables, chairs - everywhere we find symmetrical designs. A figure has line, symmetry, if there is a line about which the figure may be folded so that the two, parts of the figure will coincide., , c. If BC, , 3 cm , EF, , 8 cm , B, , 0 $,, 6, , Each of the above figures has the line of symmetry. The last figure has two lines of symmetry, Activity:, 1. Sumi has made some paper-cut design as shown in the, adjacent figure. In the figure, mark the lines of symmetry., ow many lines of symmetry does the figure have?, H, 2. Write and identify the letters in English alphabet having line, symmetry. Also mark their line of sysmmetry., 14.5 Lines of Symmetry of Regular Polygons, A polygon is a closed figure made of several line segments. A polygon is said to be, regular if all its sides are of equal length and all its angles are equal. The triangle is a
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Math, , 240, , polygon made up of the least number of line segments. An equilateral triangle is a, regular polygon of three sides. An equilateral triangle is regular because its sides as, well as angles are equal. A square is the regular polygon of four sides. The sides of a, square are equal and each of the angles is equal to one right angle. Similarly, in, regular pentagons and hexagons, the sides are equal and the angles are equal as well., , Each regular polygons is a figure of symmetry. Therefore, it is necessary to know, their lines of symmetry. Each regular polygon has many lines of symmetry as it has, many sides., Three lines of, Four lines of, Five lines of, Six lines of, symmetry, symmetry, symmetry, symmetry, , Equilateral triangle, , Square, , Regular pentagon, , Regular hexagon, , The concept of line symmetry is closely related to mirror reflection. A geometrical, figure, has line symmetry when one half of it is the mirror image of the other half., So, the line of symmetry is also called the reflection symmetry., , Exercise 14.3, 1. Which of the following figures have line symmetry?, (a) A house (b) A mosque (c) A temple (d) A church, (f) Parliament house (g) The Tajmahal., 2. The line of symmetry is given, find the other hole:, , (e) A pagoda, , 3. In the following figures, the line of symmetry is given; complete and identify the, figure.
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Math, , 241, , 4. Identify the lines of symmetry in the following geometrical figures:, , 5. Complete each of the following incomplete geometrical shapes to be symmetric, about the mirror line:, , 6. Find the number of lines of symmetry of the following geometrical figures:, (a) An isosceles triangle, (b) A scalene triangle, (c) A square, (d) A rhombus, (e) A pentagon, (f) A regular hexagon, (g) A circle, 7. Draw the letters of the English alphabet which have reflection symmetry with respect to, (a) a vertical mirror, (b) a horizontal mirror, (c) both horizontal and vertical mirrors., 8. Draw three examples of shapes with no line of symmetry., 14.6 Rotational Symmetry, When an object rotates around any fixed point, its shape and size do not change. But the, different parts of the object change their position. If the new position of the object after, rotation becomes identical to the original position, we say the object has a rotational, symmetry. The wheels of a bicycle, ceiling fan, square are examples of objects having, rotational symmetry etc.. As a result of rotation the blades of the fan looks exactly the, same as the original position more than once. The blades of a fan may rotate in the, clockwise direction or in the anticlockwise direction. The wheels of a bicycle may rotate in, the clockwise direction or in the anticlockwise direction. The rotation in the anti, clockwise direction is considered the positive direction of rotation., This fixed point around which the object rotates is the centre of rotation. The angle, of turning during rotation is called the angle of rotation. A full-turn means rotation, by 360°; a half-turn is rotation by 180°., In the figure below, a fan with four blades rotating by 90° is shown in different, positions. It is noted that us a fall turn of the four positions (rotating about the angle, by 90°, 180°, 270° and 360°), the fan looks exactly the same. For this reason, it is, said that the rotational symmetry of the fan is order 4., 0DWK,;;)RUPD
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24, 2, , ath, M, , Here is one more example for rotational symmetry. oCnsider the intersection of two, diagonals of a square the centre of rotation. In the quarter turn about the centre of the, square, any diagonal position will be as like as the second figure. In this way, when, you complete four quarter-turns, the square re aches its original position. It is said, that a square has a rotational symmetry of order 4., , bOserve also that every object occupies sa me position after one complete revolution., S, o every geometrical object has a rotationa l symmetry of order 1. uSch cases have, no interest for us. For finding the rotational symmetry of an object, one need to, consider the following matter., (a) The centre of rotation, (b) The angle of rotation, (c) The direction of rotation, (d) The order of rotational symmetry., Activity:, 1.G, ive examples of 5 plane objects from your surroundings which have, rotational symmetry., 2. Find the order of rotational symmetry of the following figures., , 14⋅7 Line Symmetry and Rotational Symmetry, W, e have seen that some geometrical shapes have only line symmetry, some have, only rotational symmetry and some have both line symmetry and rotational, symmetry. For example, the square has four lines of symmetry as well rotational, symmetry of order .4, The circle is the most symmetrical figure, because it can be rotated around its centre, through any angle. Therefore, it has unlimited order of rotational of symmetry. At, the same time, every line through the centre forms a line of reflection symmetry and, so it has unlimited number of lines of symmetry.
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24, 3, , ath, M, , Activity:, 1. Determine the line of symmetry and the rotational symmetry of the given, alphabet and complete the table below:, Letter, Line of, Number of, Rotational, Order of, symmetry, lines of, symmetry, rotational, symmetry, symmetry, Z, oN, 0, eYs, 2, H, O, E, C, , Exercise 14⋅4, 1. Find the rotational symmetry of the following figures:, (c), (d), (a), (b), , (e, , (f), , 2. W, hen you slice a lemon the cross-section looks as shown in, the figure. Determine the rotational symmetry of the figure., , 3. Fill in the blanks:, Shape, , Centre of, Rotation, , Order of, Rotation, , Angle of, Rotation, , qSuare, R, ectangle, R, hombus, qEuilateral, triangle, S, emi-circle, eRgular pentagon, 4. aNme the quadrilaterals which have line of symmetry and rotational symmetry of, order more than 1., 5. aCn we have a rotational symmetry of a body of order more than 1 whose angle, of rotation is 18, ?°Justify your answer.
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Chapter Fifteen, , Area Related Theorems and Constructions, W, e know that bounded plane figures may have different shapes. If the region is, bounded by four sides, it is known as quadrilateral. Quadrilaterals have classification, and they are also named after their shapes and properties. Apart from these, there are, regions bounded by more than four sides. These are polygonal regions or simply, polygons. The measurement of a closed region in a plane is known as area of the, region. For measurement of areas usually the area of a square with sides of 1 unit of, length is used as the unit area and their areas are expressed in square units. For, example, the area of B, angladesh is 1.4, 4, lacs square kilometres (approximately)., Thus, in our day to day life we need to know and measure areas of polygons for, meeting the necessity of life. S, o, it is important for the learners to have a, comprehensive knowledge about areas of polygons. Areas of polygons and related, theorems and constructions are presented here., At the end of the chapter, the students will be able to, ¾ xEplain the area of polygons, ¾ eVrify and prove theorems related to areas, ¾ C, onstruct polygons and justify construction by using given data, ¾ oCnstruct a quadrilateral with area equal to the area of a triangle, ¾ oCnstruct a triangle with area equal to the area of a quadrilateral, 15⋅1 Area of a Plane Region, vEery closed plane region has definite area. In order to measure such area, usually, the area of a square having sides of unit length is taken as the unit. For example, the, area of a square with a side of length 1 cm. is 1 square centimetre., W, e know that,, (a) In the rectangular region ABCD if the, length AB = a units (say, metre), breadth, BC = b units (say, metre), the area of the, region ABCD = ab square units (say,, square metres)., (b) In the square region ABCD if the length, of a side AB= a units (say, metre), the, area of the region ABCD = a2 square, units (say, square metres).
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Math, , 245, , When the area of two regions are equal, the, sign ‘=’ is used between them. For example,, in the figure the area of the rectangular, region ABCD = Area of the triangular region, AED., It is noted that if 'ABC and 'DEF are, congruent, we write 'ABC # 'DEF . In this, case, the area of the triangular region ABC =, area of the triangular region DEF., But, two triangles are not necessarily, congruent when they have equal areas. For, example, in the figure, area of 'ABC = area, of 'DBC but 'ABC and 'DBC are not, congruent., , Theorem 1, Areas of all the triangular regions having same base and lying between the same, pair of parallel lines are equal to one another., , Let the triangular regions ABC and DBC stand on the same base BC and lie between, the pair of parallel lines BC and AD. It is required to prove that, ǻ region ABC = ǻ, region DBC., Construction : At the points B and C of the line segment BC, draw perpendiculars, BE and CF respectively. They intersect the line AD or AD produced at the points E, and F respectively. As a result a rectangular region EBCF is formed., Proof : According to the construction, EBCF is a rectangular region. The triangular, region ABC and rectangular region EBCF stand on the same base BC and lie between, 1, the two parallel line segments BC and ED. Hence, ' region ABC =, (rectangular, 2, region EBCF), 1, Similarly, ' region DBC = (rectangular region EBCF), 2, ? ǻ region ABC = ǻ region DBC (proved).
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24, 6, , ath, M, , Theorem 2, Parallelograms lying on the same base and between the same pair of parallel, lines are of equal area., , Let the parallelograms regions ABCD and EFGH stand on the same base and lie, between the pair of parallel lines AF and DG and AB = EF. It is required to prove, that, area of the parallelogram ABCD = area of the parallelogram EFGH., Construction :, The base EF of EFGH is equal. Join AC and EG. From the points C and G, draw, perpendiculars CL and GK to the base AF respectively., 1, 1, Proof: The area of 'ABC = 2 AB u CL and the area of 'EFG is EF u GK ., 2, AB EF and CL = GK (by construction), Therefore, area of 'ABC = area of the triangle EFG, 1, 1, area of the parallelogram EFGH, area of the parallelogram ABCD =, 2, 2, Area of the parallelogram ABCD = area of the parallelogram EFGH. (P, roved), Theorem 3 (P, ythagoras Theorem), In a right angles triangle, the square of the hypotenuse is equal to the sum of, squares of other two sides., Proposition: Let ABC be a right angled triangle in which, ACB is a right angle and hypotenuse is AB. It is to be, proved that AB 2 BC 2 AC 2 ., Construction: Draw three squares ABED, ACGF and, BCHK on the external sides of AB, AC and BC, respectively. Through C, draw the line segment CL, parallel to AD which intersects AB and DE at M and L, respectively. Join C, D and B, F., Proof:, teps, S, Justification, (1) In ' CAD and ' FAB , CA = AF , AD = AB and, included CAD CAB BAD, [ BAD = CAF = 1, = CAB CAF, right angle], = included BAF
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ath, M, , 24, 7, , Therefore, ' CAD # ' FAB, (2) Triangular region CAD and rectangular, [S, AStheorem], region ADLM lie on the same base AD and between the, parallel lines AD and CL ., Therefore, eRctangular region ADLM = 2(triangular, region CAD ), [Theorem 1], (3, ) Triangular region BAF and the square ACGF lie on the, same base AF and between the parallel lines, AF and BG ., Hence qSuare region ACGF = 2(triangular region FAB ) = [Theorem 1], 2(triangular region CAD ), (4, )R, ectangular region, ADLM = square region ACGF, [From (2) and(3)], (5) S, imilarly joining C, E and A, K , it can be proved that, rectangular region BELM = square region BCHK, (6, )R, ectangular region ( ADLM + BELM )= square, [From (4, ) and (5)], region ACGF +square region BCHK, or, square region ABED = square region ACGF +square, region BCHK, That is, AB 2 BC 2 AC 2 [P, roved], Construction 1, Construct a parallelogram with an angle equal to a definite angle and area, equal to that of a triangular region., , Let ABC be a triangular region and x be a definite angle. It is required to construct, a parallelogram with angle equal to x and area equal to the area of the triangular, region ABC., Construction:, iBsect the line segment BC at E. At the point E of the line segment EC, construct, CEF equal to x . Through A, construct AG parallel to BC which intersects the, ray EF at F. Again, through C, construct the ray CG parallel to EF which intersects, the ray AG at G. Hence, ECGF is the required parallelogram., Proof: Join A, E. oNw, area of the triangular region ABE = area of the triangular, region AEC [since base BE = base EC and heights of both the triangles are equal]., ? area of the triangular region ABC = 2 (area of the triangular region AEC).
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24, 8, , ath, M, , Again, area of the parallelogram region ECGF is 2 (area of the triangular region, AEC) [since both lie on the same base and EC ll AG )., ? area of the parallelogram region ECGF = area of the triangular region ABC., Again CEF = x [since EF|| CB by construction]., S, o the parallelogram C, E, F, G is the required parallelogram., Construction 2, Construct a triangle with area of the triangular region equal to that of a, quadrilateral region., , Let ABCD be a quadrilateral region. To construct a triangle such that area of the is, triangular region equal to that of a rectangular region ABCD., Construction :, Join D, B. Through C, construct CE parallel to DB which intersects the side AB, extended at E. Join D, E. Then, 'DAE is the required triangle., Proof: The triangles BDC and BDE lie on the same base BD and DB ll CE (by, construction)., ? area of the triangular region BDC = area of the triangular region BDE., ? area of the triangular region BDC + area of the triangular region ABD = area of, the triangular region BDE +area of the triangular region ABD, ?area of the quadrilateral region ABCD = area of the triangular region ADE ., Therefore, 'ADE is the required triangle., N.B. Applying the above mentioned method innumerable numbers of triangles can, be drawn whose area is equal to the area of a given quadrilateral region., Construction 3, Construct a parallelogram, with a given an angle and the area of the bounded, region equal to that of a quadrilateral region., , Let ABCD be a quadrilateral region and x be a definite angle. It is required to, construct a parallelogram with angle x and the area equal to area of the, quadrilateral region ABCD.
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ath, M, , 24, 9, , Construction:, Join B, D. Through C construct CF parallel to DB which intersects the side AB, extended at F. Find the midpoint G of the line segment AF. At A of the line segment, AG, construct GAK equal to x and draw GH ll AK through G. Again draw, KDH||AG through D which intersects AK and GH at K and H respectively., Hence AGHK is the required parallelogram., Proof: Join D, E. B, y construction AGHK is a parallelogram., where GAK x . Again, area of the triangular region DAF = area of the, rectangular region ABCD and area of the parallelogram AGHK = area of the, triangular region DAF. Therefore, AGHK is the required parallelogram., , Exercise 15, 1. The lengths of three sides of a triangle are given, in which case below the, construction of the right angled triangle is not possible ?, (a) 3cm, 4cm, 5 cm, (b) 6cm, 8cm, 10cm, (c) 5 cm, 7cm, 9cm, (d) 5 cm, 12 cm, 13cm, 2. O, bserve the following information :, i. aEch of the bounded plane has definite area., ii. If the area of two triangles is equal, the two angles are congruent., iii. If the two angles are congruent, their area is equal., W, hich one of the following is correct ?, (a) i and ii, (b) i and iii, (c) ii and iii, (d) i, ii and iii, Answer question no. 3and 4on the basis of the information in the figure below,, 'ABC is equilateral, ADABC and AB = 2 :, , BD = W, hat ?, (c) 2, (d) 4, (a) 1, (b) 2, .4 W, hat is the height of the triangle ?, 2, 4, (a), sq. unit (b) 3 sq. unit, (c), sq. unit, (d) 2 3 sq. unit., 3, 3, 5. P, rove that the diagonals of a parallelo gram divide the parallelogram into four, equal triangular regions., .6 rPove that the area of a square is half the area of the square drawn on its, diagonal., 3., , 0DWK,;;)RUPD
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250, , ath, M, , .7 rPove that any median of a triangle divides the triangular region into two regions, of equal area., .8 A parallelogram and a rectangular region of equal area lie on the same side of the, bases. hSow that the perimeter of the pa rallelogram is greater than that of the, rectangle., .9 X and Y are the mid points of the sides AB and AC of the triangle ABC. rPove tha t, 1, the area of the triangular region AXY = (area of the triangular region ABC), 4, 10. In the figure, ABCD is a trapezium with sides AB and CD parallel. Find the area, of the region bounded by the trapezium ABCD., 11. P is any point interior to the parallelogram ABCD. P, rove that the area of the, 1, triangular region PAB +the area of the triangular region PCD = (area of the, 2, parallelogram ABCD)., 12. A line parallel to BC of the triangle ABC intersects AB and AC at D and E, respectively. P, rove that the area of the triangular region DBC = area of the, triangular region EBC and area of the triangular region DBF = area of the, triangular region CDE., 13. A 1 right angle of the triangle ABC. D is a point on AC. rPove tha t, BC 2 AD 2 BD 2 AC 2 ., 14. ABC is an equilateral triangle and AD is perpendicular to BC. rPove tha t, 4 AD 2 3 AB 2 ., 15. ABC is an isosceles triangle. BC is its hypotenuse and P is any point on BC., rPove that PB 2 PC 2 2PA 2 ., 16. C is an obtuse angle of 'ABC; AD is a perpendicular to BC. S, how that., AB2 = AC2 +C, B 2 +2B, C, .C, D, 17. C is an acute angle of 'ABC ; AD is a perpendicular to BC. S, how that, AB 2 AC 2 BC 2 2 BC.CD ., 18. AD is a median of 'ABC. S, how that, AB 2 AC 2 2( BD 2 AD 2 ) .
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Chapter Sixteen, , Mensuration, The length of a line, the area of a place, the volume of a solid etc. are determined for, practical purposes. In the case of measuring any such quantity, another quantity of the, same kind having some definite magnitude is taken as unit. The ratio of the quantity, measured and the unit defined in the above process is the amount of the quantity., Quantity measured, i.e. magnitude = Unit quantity, In the case of a fixed unit, every magnitude is a number which denotes how many, times of the unit of the magnitude is the magnitude of the quantity measured. For, example, the bench is 5 metre long. Here metre is a definite length which is taken as, a unit and in comparison to that the bench is 5 times in length., At the end of the Chapter, the students will be able to −, ¾ Determine the area of polygonal region by applying the laws of area of triangle, and quadrilateral and solve allied problems., ¾ Determine the circumference of the circle and a length of the chord of a circle., ¾ Determine the area of circle., ¾ Determie the area of circle. Determining the area of a circle and its part, solve, the alied problems., ¾ Determine the area of solid rectangles, cubes and cylinder and solve the allied, problems., ¾ Determine the area of uniform and non uniform polygonal regions., 16⋅1 Area of Triangular region, In the previous class, we learned that area of triagular region =, , 1, base u height., 2, , (1) Right Angled Triangle :, Let in the right angled triangle ABC , BC a and AB b are, the adjacent sides of the right angle. Here if we consider BC the, base and AB the height,, 1, Area of 'ABC = base u height, 2, 1, = ab, 2, (2) Two sides of a triangular region and the angle included between them are, given.
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252, , Math, , eLt in 'ABC the sides are :, BC a , CA b , AB c ., AD is drawn perpendicular from A to BC ., eLt altitude h( eight) AD h ., AD, onsidering the angle C, C, , we get,, sin C, CA, h, or,, sin C or, h b sin C, b, 1, Area of ' ABC = BC u AD, 2, 1, = a u b sin C, 2, 1, = absin C, 2, 1, S, imilarly, area of ' ABC = bc sin A, 2, 1, = ca sin B, 2, (3) Three sides of a triangle are given., eLt in 'ABC , BC a , CA b and AB c ., ? ePrimeter of the triangle 2 s a b c, W, e draw AD A BC, eLt, BD x , so CD a x, In right angled 'ABD and 'ACD, ? AD2 AB2 BD2 and AD2 AC 2 CD2, ? AB2 BD2 AC 2 CD2, or, c2 x2 b2 (a x)2, or, c 2 x 2 b 2 a 2 2ax x 2, or, 2ax c2 a2 b2, c 2 a 2 b2, ? x, 2a, Again, AD2 c2 x2, 2, , § c 2 a 2 b2 ·, ¸, = c ¨¨, ¸, 2a, ©, ¹, 2, 2, 2, §, c a b ·§, c 2 a 2 b2 ·, ¸, ¸ ¨c , = ¨¨ c , ¸, ¸¨, 2a, 2a, ¹, ¹©, ©, 2
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254, , Math, , b, 2, , Draw AD A BC . ? BD CD, In 'ABD right angled, ? AD2, , AB 2 BD2, , ? AD, , 4a2 b2, 2, , §b·, a2 ¨ ¸, © 2¹, , Area of isosceles ' ABC =, , 2, , a2 , , b2, 4, , 4a 2 b2, 4, , 1, BC AD, 2, , 1, 4a2 b2, b, 2, 2, b, 2, =, 4a b2, 4, Example 1. The lengths of the two sides of a right angled triangle, adjacent to right, angle are 6 cm. and 8 cm. respectively. Find the area of the triangle., Solution : eLt, the sides adjacent to right angle are, a 8 cm. and b 6 cm., respectively., 1, ? Its area = ab, 2, 1, = u 8 u 6 square cm. = 24 square cm., 2, eRquired area 24 square cm., Example 2. The lengths of the two sides of a triangle are 9 cm. and 10 cm., respectively and the angle included between them is 60$ . Find the area., Solution : eLt, the sides of triangle are a 9 cm. and b 10 cm. respectively., Their included angle θ 60$ ., 1, ? Area of the triangle = ab sin 60o, 2, =, , 3, 1, sq. cm., u 9 u 10 u, 2, 2, = 38 97 sq. cm. a( pprox), eRquired area 38 97 sq. cm. a( pprox), Exmple 3. The lengths of the three sides of a triangle are 7 cm., 8 cm. and 9 cm., respectively. Find its area., , =
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Math, , 255, , Solution : eLt, the lengths of the sides of the triangle are a, c 9 cm., abc 789, cm. = 12 cm., ? eSmi perimeter s, 2, 2, ? Its area, = s(s a)( s b)( s c), , 7 cm., b 8 cm. and, , = 12(12 7)( 12 8)( 12 9) sq. cm., = 12 u 5 u 6 u 7 sq. cm. = 50 2 sq. cm. a( pprox), ? The area of the triangle is 50 2 sq. cm. a( pprox), E, xample 4. The area of an equilateral triangle increases by 3 3 sq. metre when the, length of each side increases by 1 metre. Find the length of the side of the triangle., Solution : eLt, the length of each side of the equilateral triangle is a metre., ? Its area =, , 3 2, a sq. m., 4, , The area of the triangle when the length of each side increases by 1m. =, , 3, (a 1)2, 4, , sq. metre., According to the question,, , 3, 3 2, (a 1)2 , a, 4, 4, , 3 3, , 3, ], 4, or, a2 2a 1 a2 12 or, 2a 11 or, a 5 5, The required length is 5 5 metre., Example 5. The length of the base of an isosceles triangle is 60 cm. If its area is, 1200 sq. metre, find the length of equal sides., Solution : eLt the base of the isosceles triangle be b 60 cm. and the length of, equal sides be a ., b, ? Area of the triangle =, 4a2 b2, 4, b, According to the question,, 4a2 b2 1200, 4, 60, or,, 4a2 (60)2 1200, 4, , or, (a 1)2 a2, , 12 ;[dividee by, , or, 15 4a2 3600, or,, , 4a2 3600, , 1200, 80
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Math, , 256, , or, 4a2 3600 6400 ;[by squaring], or, 4a2 10000, or, a2 2500, ? a 50, ? The length of equal sides of the triangle is 50 cm., Example 6. From a definite place two roads run in two directions mak, ing are angle, 120o . From that definite place, persons move in the two directions with speed of 10, m, k per hour and 8 m, k per hour respectiv ely. hWat will be the direct distance, between them after 5 hours ?, k h/ our and 8 kmh/ our, Solution : eLt two men start from A with velocities 10 m, respectively and reach B and C after 5 hours. Then after 5 hours, the direct distance, between them is BC . From C perpendicular CD is drawn on BA produced., ? AB 5 u 10 m, k = 50 m, k , AC 5u 8 m, k = 40 m, k., and BAC 120o, ? DAC = 180q 120q = 60q, From the right angled triangle ACD, o, , CD, 3, o, sin 60o or, CD AC sin 60o 40 u, 20 3, 2, AC, 1, AD, and, 20, cos 60o or, AD AC cos 60o 40 u, AC, 2, Again, we get from right angled 'BCD ,, BC 2 BD2 CD2 (BA AD)2 CD2, = (50 20)2 (20 3)2 4900 1200 6100, ? BC 78 1 a( pp.), k . a( pprox), eRquired distance is 781 m, Example 7. The lengths of the sides of a triangle are 25, 20, 15 units respectively., Find the areas of the triangles in which it is divided by the perpendicular drawn from, the vertex opposite of the greatest side., Solution : eLt in triangle ABC , BC 25 units, AC 20 units, AB 15 units., The drawn perpendicular AD from vertex A on side BC divides the triangular, region into 'ABD and 'ACD ., eLt BD x and AD h, A, ? CD BC BD 25 x, In right angle 'ABD, BD2 AD2 AB2 or, x2 h2 (15)2, ? x2 h2 225..........(i), and 'ACD is right angled, CD2 AD2 AC 2 or, (25 x)2 h2 (20)2, , ?
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Math, , 257, , or, 625 50x x2 h2 400, or, 625 50x 225 400 ;[with the help of equation (i) ], or, 50x 450 ; ? x 9, uPting the value of x in equation (i) , we get,, 81 h2 225 or, h2 144, ? h 12, 1, 1, Area of ' ABD = BD AD, u 9 u12 square units = 36 square units, 2, 2, 1, 1, and area of ' ACD = BD AD, (25 9) u 12 square units, 2, 2, 1, = u 16 u 12 square units = 96 square units, 2, eRquired area is 36 square units and 96 square units., , Exercise 16⋅1, 1. The hypotenuse of a right angled triangle is 25 m. If one of its sides is, , 3, th of the, 4, , other, find the length of the two sides., 2. A ladder with length 20m. stands vertically against a wall. How much further, should the lower end of the end of the ladder be moved so that its upper end, descends 4 metre?, 5, 3. The perimeter of an isosceles triangle is 16 m. If the length of equal sides is th, 6, of base, find the area of the triangle., 4. The lengths of the two sides of a triangle are 25 cm., 27 cm. and perimeter is 84, cm. Find the area of the triangle., 5. W, hen the length of each side of an eq uilateral triangle is increased by 2 metre, its, area is increased by 6 3 square metre. Find the length of side of the triangle., 6. The lengths of the two sides of a triangle are 26 m., 28 m. respectively and its, area is 182 square metre. Find the angle between the two sides., 11, 7. The perpendicular of a right angled triangle is 6cm less than, times of the, 12, 4, base, and the hypotenuse is 3 cm less than, times of the base. Find the length, 3, of the base of the triangle., 8. The length of equal sides of an isosceles triangle is 10m and area 48 square, metre. Find the length of the base., 0DWK,;;)RUPD
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258, , Math, , 9. Two roads run from a definite place with an angle of 135o in two dinrections., Two persons move from the definite place in two directions with the speed of 7, m, k per hour and 5 m, k per hour respectiv ely. W, hat will be the direct distance, between them after 4 hours?, 10. If the lengths of the perpendiculars from a point interior of an equilateral triangle, to three sides are 6 cm., 7 cm., 8 cm. respectively;find the length of sides of the, triangle and the area of the triangular region., 16⋅2 Area of Quadrilateral Region, (1) Area of rectangular region, eLt, the length of AB a , breadth, BC b and diagonal AC d , of rectangle ABCD ., W, e nkow, the diagonal of a rectangle divides., the rectangle into two equal triangular regions., 1, ? Area of the rectangle ABCD 2 u area of 'ABC = 2 u a b ab, 2, perimeter of the rectangular region, s 2(a b), the 'ABC is right angled., a 2 b2, AC 2 AB2 BC 2 or, d 2 a 2 b 2 ; ? d, (1) Area of square region, eLt the length of each side of a square ABCD, be a and diagonal d. The diagonal AC, divides the square region into two equal, triangular regions., ? Area of square region ABCD = 2 u area of ' ABC, 1, = 2 u a a = a2, 2, Observe that, the perimeter of the square region s 4a, a2 a2, 2a 2, 2a, and diagonal d, (3) Area of parallelogram region., (a) Base and height are given., eLt, the base AB b, and height DE h of parallelogram ABCD, The diagonal BD divides the parallelogram into, two equal triangular regions., ? The area of parallelogram ABCD = 2 u area of ' ABD, , = 2u 1 bh, 2, , = bh
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Math, , 259, , (b) The length of a diagonal and the length of a perpendicular drawn from the, opposite angular point on that diagonal are given., eLt, the diagonal be AC d and the perpendicular from opposite angular point D, on AC be DE h of a parallelogram ABCD . Diagonal AC divides the, parallelogram into two equal triangular regions., ? The area of parallelogram region ABCD = 2 u area of ' ACD, 1, = 2u d h, 2, = dh, (4) Area of Rhombus Region, Two diagonals of a rhombus region are given, eLt the diagonals be AC d1 , BD = d2 of the rhombus ABCD and the diagonals, intersect each other at O . Diagonal AC divides the rhombus region into two equal, triangular regions., W, e nkow that the diagonals of a rhombu s bisect each other at right angles., d, ? Height of 'ACD = 2, 2, ? Area of the rhombus region ABCD = 2 u area of ' ACD, d, 1, = 2 u d1 u 2, 2, 2, 1, = d1d2, 2, (5) Area of trapezium region, Two parallel sides of trapezium region and the distance of perpendicular, between them are given., eLt ABCD be a trapezium whose lengths of parallel sides are AB a unit, CD b, unit and distance between them be CE h . Diagonal AC divides the trapezium, region ABCD into 'ABC and 'ACD ., Area of trapezium region ABCD, = Area of the triangular region ABC +Area of the triangular region ACD ., 1, 1, = AB u CE CD u AF, 2, 2, 1 · 1, §1, = ¨ ah bh ¸ = h(a b), 2 ¹ 2, ©2, , 3, times of breadth. If the area is 384, 2, square metre, find the perimeter and length of the diagonal., Example 1. eLngth of a rectangular room is
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260, , Math, , Solution : eLt breath of the rectangular room is x metre, 3x, ? eLngth of the room is, metre, 2, 3x2, 3x, square metre., ? Area, u x or,, 2, 2, 3x2, According to the question,, 384 or, 3x2 768 or, x2 256 ; ? x 16 metre, 2, 3, ? eLngth of the rectangular room =, u 16 metre = 24 metre, 2, and breath = 16 metre., ? It’s perimeter = 2 2( 4 +16)metre = 80 metre., and length of the diagonal = (24)2 (16)2 metre = 832 metre = 28 84 metre, a( pp.)., The required perimetre is 80 metre and length of diagonal is 28 84 metre a( pprox)., Example 2. The area of a rectangular region is 2000 square metre. If the length is, reduced by 10 metre, it becomes a square region. Find the length and breadth of the, rectangular region., Solution : eLt length of the rectangular region be x metre and breadth y metre., ? Area of the rectangular region = xy square metre, According to the question xy 2000................( 1), and x 10 y.............(2), W, e get, from equation (2,), y x 10.............(3), From equations 1( )and (3,) we get,, x( x 10) 2000 or, x2 10x 2000 0, or, x2 50x 40x 2000 0 or, ( x 50)( x 40) 0, ? x 50 0 or, x 40 0, or, x 50 or, x 40, uBt length can never be negative., ? x 50, Now puting the value of x in equation 3( ,) we get, y 50 10 40, ? length = 50 m. and breadth = 40 m., Example 3. There is a road of 4 metre width inside around a, square field. If the area of the road is 1 hector, determine the area, of the field excluding the road., Solution : eLt, the length of the square field is x metre., ? Its area is x2 square metre.
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261, , Math, , There is a road around the field with width 4 m., ? eLngth of the square field excluding the road = ( x 2 u 4) , or ( x 8) m, ? Area of the square field excluding the road is ( x 8)2 square m., Area of the road = {x 2 ( x 8) 2 } square m., W, e nkow, 1 hector =, 10000 square m., According the question, x2 (x 8)2 = 10000, or, x 2 x 2 16 x 64 10000, or, 16 x 10064, ? x 629, Area of the square field excluding the road = (629 8)2 square m., = 385641 square m., = 38 56 hector a( pprox.), The required area is 38 56 hector a( pprox.)., Example 4. The area of a parallelogram is 120 sq. cm. and length of one of its, diagonal is 24 cm. Determine the length of the perpendicular drawn on that diagonal, from the opposite vertex., Solution : eL a diagonal of a parallelogram be d 24 cm. and the length of the, perpendicular drawn on the diagonal from the opposite vertex be h cm., ? Area of the parallelogram = dh square cm., 120 120, As per question, dh 120 or, h, 5, d, 24, The required length of the perpendicular is 5 cm., Example 5. If the length of the sides of a parallelogram are 12 m. 8 m. If the length, of the smaller diagonal is 10 m, determine the length of the other diagonal., Solution : eLt, in the parallelogram ABCD ; AB a 12 m. and AD c 8 m. and, diagonal BD b 10 m. eLt us draw the perpendiculars DF and C, Efrom, D and C, on the extended part of AB , respectively. Join A, C and B, D ., 12 10 8, ?S, emi perrimeter of 'ABD is s, m. = 15 m., 2, ? Area of the triangular region ABD = s ( s a() s 6() s c), = 15(15 12() 15 10() 15 8) sq. m., = 1575 sq. m., = 39 68 sq. m. a( pprox.), 1, Again, area of the triangular region ABD = AB u DF, 2, 1, or, 39 68, u12 u DF or, 6 DF 39 68 ; ? DF, 2, , 6 61
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262, , Math, , N, ow, in right angled triangle 'BCE ,, BE 2 BC 2 CE 2 AD2 DF 2 82 (6 61)2 20 31, ? BE 4 5, Therefore, AE AB BE 12 4 5 16 5, From right angled triangle 'BCE , we get,, AC 2 AE 2 CE 2 (16 5)2 (6 61)2 315 94, ? AC 17 77 a( pprox.), ? The required length of the diagonal is 17 77 m. a( pprox.), Example 6. The length of a diagonal of a rhombus is 10m. and its area is 120 sq. m., Determine the length of the other diagonal and its perimeter., Solution : eLt, the length of a diagonal of rhombus ABCD is BD d1 10 metre, and another diagonal AC d2 metre., 1, ? Area of the rhombus = d1d2 sq. m., 2, 1, 120 u 2 120 u 2, As per question, d1d2 120 or, d2, 24, 2, 10, 10, W, e nkow, the diagonals of rhombus bisect, each other at right angles. eLt the, diagonals interset at the point 0., 10, 24, ? OD OB, m. = 5 m. and OA OC, m. = 12 m., 2, 2, and in right angled triangle 'AOD , we get, AD 2 OA 2 OD 2 5 2 (12) 2 169 ; ? AD 13, ? The length of each sides of the rhombus is 13 m., ? The perimeter of the rhombus = 4u13 m. = 52 m.., The required length of the diagonal is 24 m. and perimeter is 52 m., Example 7. The lengths of two parallel sides of a trapezium are 91 cm. and 51 cm., and the lengths of two other sides are 37 cm and 13 cm respectively. Determine the, area of the trapezium., Solution : eLt, in trapezium ABCD ; AB 91 cm. CD 51 cm. eLt us draw the, perpendiculars DF and CF on AB from D and C respectively., ? CDEF is a rectangle., ? EF CD 51 cm., eLt, AE x and DE CF h, ? BF AB AF 91 ( AE EF ) 91 ( x 51) 40 x, ? From right angled triangle 'ADE , we get,, AE 2 DE 2 AD 2 or, x2 h2 (13)2 or, x2 h2 169.........(i), Again, from right angled triangle 'BCF , we get,, BF 2 CF 2 BC 2 or, (40 x) 2 h 2 (37) 2
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Math, , 263, , or, 1600 80x x2 h2 1369, [ ith the help of 1( ]), or, 1600 80x 169 1396 w, or, 1600 169 1396 80x ;or, 80x 400 ; ? x 5, Now putting the value of x in equation 1( ,) we get, 52 h2 163 or, h 169 25 144 ; ? h 12, 1, Area of ABCD = ( AB CD) h, 2, 1, = (91 51) u 12 square cm., 2, = 852 square cm., The required area is 852 square cm., 16⋅3 Area of regular polygon, The lengths of all sides of a regular polygon are equal. Again, the angles are also, equal. eRgular polygon with n sides produces n isocsceles triangles by adding centre, to the vertices., ? Area of the regular polygon = n u area of one triangular, region., eLt ABCDEF be a regular polygon whose centre is O., ? It has n sides and the length of each side is a. eWoj in, O, A ;, θ, O, B., M, ? eLt in 'AOB height OM h and OAB θ, ? The angle produced at each of the vertices of regular polygon = 2θ ., ? Angle produced by n number of vertices in the polygon = 2θ n, Angle produced in the polygon at the centre = 4 right angles., The sum of angles of n number of triangles = 2θ (n 4) right angles., ?S, um of 3 angles of 'OAB = 2 right angles., ? The wise, summation of the angles of n numbers of triangles = n 2 right angles, ? 2θ (n 4) right angles = n 2 right angles, or, 2θ n (2n 4) right angles, 2n 4, right angles, or, θ, 2n, or, θ §¨1 2 ·¸ right angles, ©, , 180q, § 2·, ¨1 ¸ u 90$ = 90q n, © n¹, , or, θ, Now, tan θ, , n¹, , h, a, 2, , 2h ; ?, h, a, , a, tan θ, 2
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Math, , 264, , 1, ah, 2, 1, a, = a u tan θ, 2, 2, §, a2, 180$ ·, ¸, =, tan¨¨ 90$ , n ¸¹, 4, ©, , ? Area of 'OAB =, , =, , a 2 § 180$ ·, ¸¸, cot¨¨, 4, © n ¹, , ? Area of a regular polygon having n sides =, , a 2 § 180$ ·, cot¨¨, ¸¸ ., 4, © n ¹, , Example 8. If the length of each side of a regular pentagon is 4 cm, determine its, area., Solution : eLt, length of each side of a regular pentagon is a 4 cm., and number of sides n 5, a2, 180$, cot, W, e nkow, area of a regular polygon =, n, 4, 2, $, 4, 180, cot, ? Area of the pentagon =, S, q. cm., 4, 5, = 4 u cot 36o sq. cm., = 4 u 1 376 sq. cm. [with the help of calculator], = 5 506 sq. cm. a( pprox.), The required area = 5 506 sq. cm. a( pprox.), Example 9. The distance of the centre to the vertex of a regular hexagon is 4 m., Determine its area., Solution : eLt, ABCDEF is a regular hexagon whose centre is O , O is joined to, each of the vertex and thus 6 triangles of equal area are formed., 360$, 60$, ? COD, 6, eLt the distance of centre O to its vertex is a m., ? a = 4., ? Area of 'COD, , =, , 1, a a sin 60$, 2, , 3 2, u 4 sq. m. = 4 3 sq. m., 4, ? Area of the regular hexagon, , =, , 1 2 3, a , 2, 2, , 3 2, a, 4
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Math, , 265, , = 6 u 4 3 sq, m., = 24 3 sq. m., , Exercise 16⋅2, 1. The length of a rectangular region is twice its width. If its area is 512 sq. m.,, determine its perimeter., 2. The length of a plot is 80 m. and the breadth is 60 m. A rectangular pond was, excavated in the plot. If the width of each side of the border around the pond is 4, metre, determine the area of the border of the pond., 3. The length of a garden is 40 metre and its breadth is 30 metre. There is a pond in, 1, side the garden with around border of equal width. If the area of the pond is, of, 2, that of the garden, find the length and breadth of the pond., 4. Outside a square garden, there is a path 5 metre width around it. If the area of the, path is 500 square metre, find the area of the garden., 5. The perimeter of a square region is equal to the perimeter of a rectangular region., The length of the rectangular region is thrice its breadth and the area is 768 sq., m. How many stones will be required to cover the square region with square, stones of 40 cm each?, 6. Area of a rectangular region is 160 sq. m. If the length is reduced by 6 m., it, becomes a square region. Determine the length and the breadth of the rectangle., 3, 7. The base of a parallelogram is th of the height and area is 363 square inches., 4, Determine the base and the height of the parallelogram., 8. The area of a parallelogram is equal to the area of a square region. If the base of, the parallelogram is 125m. and the height is 5 m, find the length of the diagonal, the square., 9. The length of two sides of a parallelogram are 30 cm and 26 cm, If the smaller, diagonal is 28 cm, find the length of the other diagonal., 10. The perimeter of a rhombus is 180 cm. and the smaller diagonal is 54 cm. Find, the other diagonal and the area., 11. Deference of the length of two parallel sides of a trapezium is 8 cm. and their, perpendicular distance is 24 cm. Find the lengths of the two parallel sides of the, trapezium., 12. The lengths of two parallel sides of a trapezium are 31 cm. and 11 cm., respectively and two other sides are 10 and 12 cm. respectively. Find the area of, the trapezium., 0DWK,;;)RUPD
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266, , Math, , 13. The distance from the centre to the vertex of a regular octagon is 1.5 m. Find the, area of the regular octagon., 14. The length of a rectangular flower garden is 150 m. and breadth is 100 m. For, nursing the garden, there is a path with 3 m. width all along its length and, breadth right at the middle of the garden., (a) Describe the above information with figure., (b) Determine the area of the path., (c) oHw many bricks of 25 cm. length and 12 5 cm. width will be required to, make the path metalled ?, 15. From the figure of the polygon determine its area., 12cm, , 16. From the information given below determine the area of the figures :, , 6.4, Measurement regarding circle, (1) Circumference of a Circle, The length of a circle is called its circumference. eLt r be the radius, of a circle,, its circumference c 2π r , where π 3 14159265.......... which is an irrational, number value of π = 31416 is used as the actual value., Therefore, if the radius of a circle is known, we can find the approximate value of, the circumference of the circle by using the value of π ., Example 1. The diameter of a circle is 26 cm. Find its circumference., Solution : eLt, the radius of the circle is r ., ? diameter of the circle = 2r and circumference = 2π r, 26, As per question, 2r 26 or, r, ? r 13, 2, ? circumference of the circle = 2π r = 2 u 3 1416 u 13 cm., = 3 1416 u26 cm (approx.), The required circumference of the circle is 81 68 cm. (approx.), θ
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267, , Math, , (2) Length of arc of a circle, eLt Obe the centre of a circle whose radius is, r and are AB s , which produces, θ o angle at the centre., ?C, ircumference of the circle = 2π r, Total angle produced at the centre of the circle = 360o and arc s produces angle θ q, at the centre. W, e know, any interior angle at the centre of a circle produced by any, arc is proportional to the arc., π rθ, θ, s, ?, or, s, 180, 360, 2π r, (3) Area of circular region and circular segment, The subset of the plane formed by the union of a circle and its, interior is called a circular region and the circle is called the, boundary of the such circular region., Circular segment: The area formed by an arc and the radius, related to the oj ining points of that arc is called circular segment., If A and B are two points on a circle with centre ,Othe subset of, the plane formed by the union of the intersection of AOB and, the interior of the circle with the line segment OA , OB and the arc, AB , is called a circular segment., In previous class, we have learnt that if the radius of a circle is r , the, area is = πr 2, eWknow, any angle produced by an arc at the centre of a circle is, proportional to the arc., θ, S, o, at this stage we can accept that the area of two circular, segments of the same circle are proportional to the two arcs on, which they stand., eLt us draw a radius r with centre O., The circular segment AOB stands on the arc APB whose measurement is θ . Draw a, perpendicular OC on OA ., ?, , Measurement of AOB, Area of circular segment AOB, =, =, Area of circular segment AOC, Measurement of AOC, , θ, Area of circular segment AOB, ; [AOC 90 o ], =, Area of circular segment AOC, 90, θ, or, Area of circular segment AOB =, u area of circular segment AOC, 90, θ, 1, =, u u area of the circle, 90 4, , or,
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268, , Math, , 1, θ, u u πr 2, 90 4, θ, =, u πr 2, 360, θ, S, o, area of circular segment =, u πr 2, 360, Example 2. The radius of a circle is 8 cm. and a circular segment substends an angle, 56 o at the centre. Find the length of the arc and area of the circular segment., Solution : eLt, radius of the circle, r 8 cm, length of arc is s and the angle, subtended by the circular segment is 56o ., , =, , πrθ, 180, , 3 1416 u 8 u 56 cm.. = 7 82 cm. (approx), 180, Area of circular segment = θ u πr 2, 360, , W, e know,, , s, , 56, = 360 u 3.1416 u 82 sq.cm., = 62 55 sq. cm. (approx), Example 3. If the difference between the radius and circumference of a circle is 90, cm., find the radius of the circle., Solution : eLt the radius of the circle be r, ? Diameter of the circler is 2r and circumference = 2πr, As per question, 2πr 2r 90, 90, 45, or, 2r (π 1) 90 or, r, 21 01 (approx.), 2(π 1) 3 1416 1, The required radius of the circle is 21 01 cm. (approx.)., Example 4. The diameter of a circular field is 124 m. There is a, path with 6 m. width all around the field. Find the area of the path., Solution : eLt the radius of the circular field be r and radius of the, field with the path be R ., 124, m. = 62 m. and R (62 6) m. = 68 m., ?r, 2, Area of the circular field = πr 2, and area of the circular field with the path = πR 2, ? Area of the path = Area of field with path –Area of the field, = (πR 2 πr 2 ) = π ( R 2 r 2 ), = 3 1416{( 68) 2 (62) 2 } sq. m., = 3 1416(4624 3844)
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Math, , 269, , = 3 1416 u 780 sq. m., = 2450 45 sq. m. (approx), The required area of the path is 2450 44 square m. (approx.), Activity : iCrcumference of a circle is 440 m. Determine the length of the sides of, the inscribed square in it., Example 5. The radius of a circle is 12 cm. and the length of an arc is 14 cm., Determine the angle subtended by the circular segment at its centre., Solution : eLt, radius of the circle is r = 12 cm., the length of the arc is s = 14 cm., and the angle subtended at the centre is θ q., πsθ, W, e know, s, 180, or, πrθ 180 u s, 180 u s, 180 u14, or, θ, 66 85 (approx), πr, 3 1416 u12, ? The required angle is 6685q (approx), Example 6. Diameter of a wheel is 4.5 m. for traversing a distance of 360 m.; how, many times the wheel will revolve ?, Solution : iGven that, the diameter of the wheel is 4.5 m., 45, m. and circumference = 2πr, ? The radius of the wheel, r, 2, eLt, for traversing 360 m, the wheel will revolve n times, As per question, n u 2πr = 360, 360, 360 u 2, or, n, = 25.46 (approx), 2πr 2 u 3 1416 u 4 5, ? The wheel will revolve 25 times (approx) for traversing 360 m.., Example 7. Two wheels revolve 32 and 48 times respectively to cover a distance of, 211 m. 20 cm. Determine the difference of their radii., Solution : 211 m. 20 cm. = 21120 cm., eLt, the radii of two wheels are R and r respectively ;where R ! r., ?C, ircumferences of two wheels are 2πR and 2πr respectively and the difference, of radii is ( R r ), As per question, 32 u 2πR 21120, 21120, 21120, or, R, 105 04 (approx), 32 u 2π 32 u 2 u 3 1416, and 48 u 2πr 21120, 21120, 21120, or, r, 70 03 (approx), 48 u 2π 48 u 2 u 3 1416
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270, , Math, , ? R r (105 04 70 03) cm.. = 35 01 cm. = 35 m. (approx), ? The difference of radii of the two wheels is 35 m (approx), Example 8. The radius of a circle is 14 cm. The area of a square is equal to the area, of the circle. Determine the length of the square., Solution : eLt the radius of the circle, r 14 cm. and the length of the square is a, ? Area of the square region = a 2 and the area of the circle = πr 2, According to the question, a 2 = πr 2, 22, m. = 11 m., ? aRdius of the half circle r, 2, or, a, πr, 3 1416 u 14 24 81 (approx), The required length is 2481 cm. (approx.), Example 9 : In the figure, ABCD is a square whose length of each side is 22 m. and, AED region is a half circle. Determine the area of the whole region., Solution : eLt, the length of each side of the square ABCD be a., ? Area of square region = a 2, Again, AED is a half circle., eLt ;be its radius, 1 2, ? Area of the half circle, AED, πr, 2, ? Area of the whole region = Area of the square ABCD + area of the half circle AED., 1, = a 2 πr 2, 2, 1, = {( 22) 2 u 3 1416 u (11) 2 }sq. metre, 2, 22, [a = 22, r = 2 = 11], = 674.07 sq. m (app.), The required area is 674.07 square metre (approx), Example 10. In the figure, ABCD is a rectangle whose the length is 12 m., the, breadth is 10 m. and DAE is a circular region., Determine the length of the arc DE and the area of the whole region., Solution : eLt the radius of the circular segment, r AD 12 m. and the angle, subtended at centre θ 30 $, πrθ, ? length of the arc DE, 180, 3 1416 u 12 u 30, =, m. = 6.28 m. (approx.), 180
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Math, , 271, , ? Area of the circular segment DAE =, , θ, u πr 2, 360, , 30, u 3 1416 u (12) 2 sq. m. =, 360, , 37.7 sq.cm (approx), The length of the rectangle ABCD is 12 m. and the breadth is 10 m., ? Area of the rectangle = length u breadth = 12 u 10 sq. m = 120 sq. m., ? Area of the whole region = (37.7 + 120) sq. m. = 157.7 square metre., The required area is 157.7 square metre. (approx), Activity : Determine the area of the dark marked region in the figure., , Exercise 16.3, 1. Angle subtended by a circular segment at the centre is 30o. If the diameter of the, circle is 126 cm., determine the length of the arc., 1, 2. A horse turned around a circular field with a speed of 66 m per minute in 1, 2, minute. Determine the diameter of the field., 3. Area of a circular segment is 77 sq. m. and the radius is 21 m. Determine the, angle subtended at the centre by the circular segment., 4. The radius of a circle is 14 cm. and an arc subtends an angle 76o at its centre., Determine the area of the circular segment., 5. There is a road around a circular field. The outer circumference of the road is, greater than the inner circumference by 44 metres. Find the width of the road., 6. The diameter of a circular park is 26 m. There is a road of 2 m. width around the, park outside. Determine the area of the road., 7. The diameter of the front wheel of a car is 28 cm. and the back wheel is 35 cm., To cover a distance of 88 m, how many times more the front wheel will revolve, than the back one ?, 8. The circumference of a circle is 220 m. Determine the length of the side of the, inscribed square in the circle., 9. The circumference of a circle is equal to the perimeter of an equilateral triangle., Determine the ratio of their areas ?, 10. Determine the area of the dark marked region with the help of the information, given below :
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Math, , 272, , 6⋅5 Rectangular solid, The region surrounded by three pairs of parallel rectangular planes is known as, rectangular solid., eLt, ABCDEFGH is a rectangular solid, whose length AB a , and breadth, BC b and height AH c, (1) Determining the diagonal: AF is the diagonal of the rectangular solid, ABCDEFGH, In 'ABC , BC A AB and AC is hypotenuse, ? AC 2 AB2 BC 2 a2 b2, Again, in 'ACF , FC A AC and AF is hypotenuse, ? AF 2 AC 2 CF 2 a2 b2 c2, ? AF, , a 2 b2 c 2, , ? the diagonal of the rectangular solid = a2 b2 c2, (2) Determination of area of the whole surface :, There are 6 surfaces of the rectangular solid where, 1, the opposite surfaces are equal figure, 2, Area of the whole surface of the rectangular solid, =2 (area of the surface of ABCD +area of the, 1, 3, 3, surface of ABGH +area of the surface of, BCFG ), 2, = 2( AB u AD AB u AH BC u BG), = 2(ab ac bc), = 2(ab bc ca), (3) oVlume of the rectangular solid =length, u width u height, = abc, Example 1. The length, width and height of a rectangular solid are 25 cm., 20 cm., and 15 cm. respectively. Determine its area of the whole surface, volume and the, length of the diagonal., Solution : eLt, the length of the rectangular solid is a 25 cm., width b 20 cm., and height c 15 cm., ? Area of the whole surface of the rectangular solid = 2(ab bc ca), = 2(25 u 20 20 u 15 15 u 25) sq. cm., = 2350 square cm., oVlume = abc, = 25 u 20 u 15 cube cm., = 7500 cube cm., And the length of its diagonal =, , a 2 b2 c 2
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Math, , 273, , =, , (25)2 (20)2 (15)2 cm., , =, , 625 400 225 cm., , = 1250 cm., = 35 353 cm. (approx.), The required area of the whole surface is 2350 cm2., volume 7500 cm3. and the, length of the diagonal is 35 353 cm. (approx.)., Activity : Determine the volume, area of the whole surface and the length of the, diagonal of your mathematics book calculating its length, width and height., 6⋅6 Cube, If the length, width and height of a rectangular solid are, equal, it is called a cube., eLt, ABCDEFGH is a cube., Its height =width =height =, a, (1) The length of diagonal of the cube = a2 a2 a2 = 3a2 = 3a, (2) The area of the whole surface of the cube = 2(a a a a a a), = 2(a 2 a 2 a 2 ) = 6a 2, (3) The volume of the cube = a a a = a3, Example 2. The area of the whole surface of a cube is 96 m2. Determine the length, of its diagonal., Solution : eLt, the sides of the cube is a, ? The area of its whole surface = 6a2 and the length of diagonal = 3a, As per question, 6a2 96 or, a 2 16 ; ? a 4, ? The length of diagonal of the cube = 3a = 3 u 4 = 6 928 m. (approx.), The required length of the diagonal is 6 928 m. (approx.), Activity : The sides of 3 metal cube are 3 cm., 4 cm. and 5 cm. respectively. A, new cube is formed by melting the 3 cubes. Determine the area of the whole, surface and the length of the diagonal of new cube., 6⋅7 Cylinder :, The solid formed by a complete revolution of any rectangle, about one of its sides as axis is called a cylinder or a right, circular C, ylinder. The two ends of a right circular cylinder, are circles. The curved face is called curved surface and the, total plane is called whole surface. The side of the, rectangle which is parallel to the axis and revolves about the axis is called the, generator line of the cylinder., eLt, figure (a) is a right circular cylinder, whose radius is r and height h, (1) Area of the base = πr 2, 0DWK,;;)RUPD
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274, , Math, , (2) Area of the curved surface, =perimeter of the base u height, = 2πrh, (3) Area of the whole surface, = (πr 2 2πrh πr 2 ) = 2πr(r h), (4) The volume =area of the base u height, = πr 2h, Example 3. If the height of a right circular cylinder is 10 cm. and radius of the base, is 7 cm, determine its volume and the area of the whole surface., Solution : eLt, the height of the right circular cylinder is h 10 cm. and radius of, the base is r ., ? Its volume = πr 2h = 3 1416 u 72 u 10, = 1539 38 cube cm. (approx.), And the area of the whole surface = 2πr (r h), = 2 u 3 1416 u 7(7 10) sq. cm. (approx.), = 747 7 sq. cm. (approx.), Activity : Make a right circular cylinder using a rectangular paper. Determine the, area of its whole surface and the volume., Example 4. The outer measurements of a box with its top are 10 cm., 9 cm. and, 7cm. respectively and the area of the whole inner surface is 262 cm2. Find the, thickness of its wall if it is uniform on all sides., Solution : eLt, the thickness of the box is x cm., The outer measurements of the box with top are 10 cm., 9 cm. and 7 cm., respectively., ? The inside measurement of the box are respectively a (10 2x) cm.,, b (9 2x) cm. and c (7 2 x) cm., ? The area of the whole surface of the inner side of the box = 2(ab bc ca), As per question, 2(ab bc ca) = 262, or, (10 2 x)(9 2 x) (9 2 x)(7 2 x) (7 2 x)(10 2 x) 131, or, 90 38x 4x2 63 32x 4x2 70 34x 4x2 131 0, or, 12x2 104x 92 0, or, 3x2 26x 23 0, or, 3x2 3x 23x 23 0, or, 3x( x 1) 23( x 1) 0, or, ( x 1)(3x 23) 0, or, x 1 0 or, 3x 23 0, 23, or, x 1 or, x, 7 67 (approx.), 3
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275, , Math, , uBt the thickness of a box cannot be greater than or equal to the length or width or height, ? x 1, The required thickness of the box is 1 cm., Example 5. If the length of diagonal of the surface of a cube is 8 2 cm., determine, the length of its diagonal and volume., Solution : eLt, the side of the cube is a ., ? The length of diagonal of the surface = 2a, 3a, eLngth of diagonal =, And the volume = a3, As per question, 2a = 8 2 ; ? a 8, ? The length of the cube’s diagonal = 3 u 8 cm. =8, 3 cm., And the volume = 83 cm3. = 512 cm3., The required length of the diagonal is 8 3 cm. (approx) and the volume is 512 cm3., Example 6. The length of a rectangle is 12 cm. and width 5 cm. If it is revolved, around the greater side, a solid is formed. Determine the area of its whole surface, and the volume., Solution : iGven that, the length of a rectangle is 12 cm. and width 5 cm. If it is, revolved around the greater side, a circle based right cylindrical solid is formed with, height h 12 cm. and radius of the base r 5 cm., ? The whole surface of the produced solid = 2πr(r h), = 2 u 3 1416 u 5(5 12) sq. sm., = 534 071 sq. cm. (approx.), And the volume = πr 2h, = 3 1416 u 52 u12 cm3., = 942 48 cube cm3. (approx.), The required area of whole surface is 534 071 sq. cm. (approx.) and the volume is, 942 48 cm3. (approx.), , Exercise 16⋅4, 1. The length and witdth of two adjacent sides of a parallelogram are 7 cm., and 5, cm. respectively. What is the half of its perimeter in cm.. ?, (a) 12, (b) 20, (c) 24, (d) 28, 2. The length of the side of an equilateral triangle is 6 cm. What is its area (cm2) ?, (a) 3 3, (b) 4 3, (c) 6 3, (d) 9 3, 3. If the height of a trapezium is 8 cm. and the lengths of the parallel sides are 9 cm., and 7 cm. respectively, what is its area (cm2) ?, (b) 64, (c) 96, (d) 504, (a) 24
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276, , Math, , 4. Follow the information given below :, (i) A square stone with the side of 4 cm. has 16 cm. perimeter., (ii) The area of circular sheet with the radius 3 cm.is 3S cm2., (iii) The volume of a cylinder with height of 5cm. and the radius of 2 cm. is 20π cm3., According to the information above, which one of the following is correct ?, (b) i and iii, (c) ii and iii, (d) i , ii and iii, (a) i and ii, Answer the following questions (5 – 7) as per information from the picture below:, , 5. What is the length of the diagonal of the rectangle ABCD in cm. ?, (b) 14, (c) 14 4 (app.), (d) 15, (a) 13, 6. What is the area of the circular segment ADF in sq. cm.?, (a) 16, (b) 32, (c) 64, (d) 128, 7. What is the circumference of the half circle AGB in cm. ?, (b) 18 85 (app.), (c) 37 7 (app.), (d) 96, (a) 18, 8. The length, width and height of a rectangular solid are 16 m. 12 m. and 4 5 m., respectively. Determine the area of its whole surface, length of the diagonal and, the volume., 9. The ratios of the length, width and height of a rectangular solid are 21 : 16 : 12, and the length of diagonal is 87 cm. Determine the area of the whole surface of, the solid., 10. A rectangular solid is standing on a base of area 48 m2 Its height is 3m and, diagonal is 13 m. Determine the length and width of the rectangular solid., 11. The outer measurements of a rectangular wooden box are 8 cm., 6 cm. and 4 cm.,, respectively and the area of the whole inner surface is 88 cm2. Find the thickness, of the wood of the box., 12. The length of a wall is 25 m, height is 6 m. and breadth is 30 cm. The length,, breadth and height of a brick is 10 cm. 5 cm. and 3 cm. respectively. Determine, the number of bricks to build the wall with the bricks., 13. The area of the surface of a cube is 2400 sq. cm. Determine the diagonal of the, cube., 14. The radius and the height of a right circular cylinder are 12 cm. and 5 cm., respectively. Find the area of the curved surface and the volume of the cylinder., 15. The area of a curved surface of a right circular cylinder is 100 sq. cm. and its, volume is 150 cubic cm. Find the height and the radius of the cylinder.
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Math, , 277, , 16. The area of the curved surface of a right circular cylinder is 4400 sq. cn. If its, height is 30 cm., find the area of its whole surface., 17. The inner and outer diameter of a iron pipe is 12 cm. and 14 cm. respectively. If, the height of the pipe is 5 m., find the weight of the iron pipe where weight of 7.2, gm. iron =1 cm 3., 18. The length and the breadth of a rectangular region are 12 m. and 5m., respectively. There is a circular region just around the rectangle. The places, which is are not occupied by the rectangle, are planted with grass., (a) Describe the information above with a figure., (b) Find the diameter of the circular region., (c) If the cost of planting grass per sq. m. is Tk. 50, find the total cost., 19. 'ABC and 'BCD are on the same base BC and on the same parallel lines BC, and AD ., (a) Draw a figure as per the description above., (b) P, roof that ' region ABC = ' region BCD ., (c) Draw a parallelogram whose area is equal to the area of 'ABC and whose, one of the angles is equal to a given angle (construction and description of, construction is must)., 20. ABCD is a parallelogram and BCEF is a rectangle and BC is the base of both, of them., (a) Draw a figure of the rectangle and the parallelogram assuming the same, height., (b) S, how that the perimeter of ABCD is greater than the perimeter of BCEF ., (c) Ratio of length and width of the rectangle is 5 : 3 and its perimeter is 48, m. Determine the area of the parallelogram
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Chapter Seventeen, , Statistics, O, wing to the contribution of information an d data, the world has become a global, village for the rapid advancement of science and information. G, lobalization has been, made possible due to rapid transformation and expansion of information and data., S, o, to keep the continuity of development and for participating and contribute in, globalizations, it is essential for the studen ts at this stage to have clear knowledge, about information and data. In the context, to meet the demands of students in, acquiring knowledge, information and data have been discussed from class V, I and, class-wise contents have been arranged step by step. In continuation of this, the, students of this class will know and learn cumulative frequency, frequency polygon,, gOive curve in measuring of central tenden cy mean, median, mode etc. in short-cut, method., At the end of this chapter, the students will be able to ¾ Explain cumulative frequency, frequency polygon and ogive curve;, ¾ Explain data by the frequency polygon, and ogive curve ;, ¾ Explain the method of measuring of central tendency ;, ¾ Explain the necessity of short-cut method in the measurement of central, tendency ;, ¾ Find the mean, median and mode by the short-cut method ;, ¾ Explain the diagram of frequency polygon and ogive curve., Presentation of Data : We know that numerical information which are not, qualitative are the data of statistics. The data under investigation are the raw, materials of statistics. They are in unorganized form and it is not possible to take, necessary decision directly from the unorganize d data. It is necessary to organize and, tabulate the data. nAd the tabulation of da ta is the presentation of the data. In, previous class we have learnt how to organize the data in tabulation. We determine, know that it is required to the range of data for tabulation. Then determining the, class interval and the number of classes by using tally marks, the frequency, distribution table is made. eHre, the metho ds of making frequency distribution table, are to be re-discussed through example for convenient understanding., Example 1. In a winter season, the temperature (in celsius) of the month of January, in the district of rSimangal is placed belo w. Find the frequency distribution table of, the temperature., 14q, 14q, 14q, 13q, 12q, 13q, 10q, 10q, 11q, 12q, 11q, 10q, 9q, 8q, 9q,, 11q, 10q, 10q, 8q, 9q, 7q, 6q, 6q, 6q, 6q, 7q, 8q, 9q, 9q, 8q, 7q.
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Math, , 279, , Solution : eHre the minimum and maximum numerical values of the data of, temperature are 6 and 14 respectively., eHnce the range =14, 6 +1 =9., 9, If the class interval is considered to be 3, the numbers of class will be, or, 3., 3, Considering 3 to be the class interval, if the data are arranged in 3 classes, the, frequency table will be :, Temperature (in celcius), 6q 8q, 9q 11q, 12q 14q, , Tally, llll llll l, llll llll lll, llll ll, , Frequency, 11, 13, 7, Total =31, , Activity : Form two groups of all the students studying in your class. Find the, frequency distribution table of the weights (in K, gs) of all the members of the, groups., Cumulative Frequency :, In example1, considering 3 the class interval and determining the number of classes,, the frequency distribution table has been made. The numbers of classes of the, mentioned data are 3. The limit of the first class is 6q 8q. The lowest range of the, class is 6o and the highest range is 8oC. The frequency of this class is 11., The frequency of the second class is 13. oNw if the frequency 11 of first class is, added to the frequency 13 of the second class, we get 24. This 24 will be the, cumulative frequency of the second class and the cumulative frequency of first class, as begins with the class will be 11. A, gain, if the cumulative frequency 24 of the, second class is added to the frequency of the third class, we get 24 +7 =31 which is, the cumulative frequency of the third class. Thus cumulative frequency distribution, table is made. In the context of the above discussion, the cumulative frequency, distribution of temperature in example 1 is as follow :, Temperature (in celsius), Frequency, Cumulative Frequency, 6q 8q, 11, 11, 9q 11q, 13, (11 +13) =24, 12q 14q, 7, (24 +7) =31, Example 2. The marks obtained in English by 40 students in an annual examination, are given below. Make a cumulative frequency table of the marks obtained., 70, 40, 35, 60, 55, 58, 45, 60, 65, 80, 70, 46, 50, 60, 65, 70, 58, 69, 48, 70, 36, 85,, 60, 50, 46, 65, 55, 61, 72, 85, 90, 68, 65, 50, 40, 56, 60, 65, 46, 76.
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Math, , 280, , Solution : Range of the data =(highest numerical value –lowest numerical value) +1, =90 35 1+, =55 +1, =56, eLt the class interval be 5, the number of classes =, , 56, 5, , =11 2 or 12, eHnce the cumulative frequency distribution table at a class interval of 5 will be as follow :, Obtained Frequency Cumulative Obtained Frequency Cumulative, marks, frequency, marks, frequency, 35 39, 2, 2, 70 74, 4, 4 +31 =35, 40 44, 2, 2 +2 =4, 75 79, 1, 1 +35 =36, 45 49, 5, 5 +4 =9, 80 84, 1, 1 +36 =37, 50 54, 3, 3 +9 =12, 85 89, 2, 2 +37 +39, 55 59, 5, 5 +12 =17, 90 94, 1, 1 +39 =40, 60 64, 8, 8 +17 =25, 95 99, 0, 0 +40 =40, 65 69, 6, 6 +25 =31, Variable : We know that the numerical information is the data of statistics. The numbers, used in data are variable. uSch as, the numbers indicating temperatures are variable., iSmilarly, in example 2, the secured marks used in the data are the variables., Discrete and Indiscrete Variables : The variables used in statistics are of two, types. S, uch as, discrete and indiscrete vari ables. The variables whose values are only, integers, are discrete variables. The marks obtained in example 2 are discrete, variables. iSmilarly, only integers are used in population indicated data. That is why,, the variables of data used for population are discrete variables. nAd the variables, whose numerical values can be any real number are indiscrete variables. S, uch as, in, example 1, the temperature indicated data which can be any real number. eBsides,, any real number can be used for the data related to age, height, weight etc. That is, why, the variables used for those are indiscrete variables. The number between two, indiscrete variables can be the value of those variables. S, ome times it becomes, necessary to make class interval indiscrete. To make the class interval indiscrete, the, actual higher limit of a class and the lower limit of the next class are determined by, fixing mid-point of a higher limit of any class and the lower limit of the next class., S, uch as, in example 1 the actual higher-lower limits of the first class are 8 5q and, 55q respectively and that of the second class are 115o and 85o etc., Activity : Form a group of maximum 40 students of your class. Form frequency, distribution table and cumulative frequency table of the group with the, weights/heights of the members.
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Math, , 281, , Diagram of Data : We have seen that the collected data under investigation are the, raw materials of the statistics. If the frequency distribution and cumulative frequency, distribution table are made with them, it becomes clear to comprehend and to draw a, conclusion. If that tabulated data are presented through diagram, they become easier, to understand as well as attractive. That is why, presentation of statistical data in, tabulation and diagram is widely and frequently used method. In class V, III, different, types of diagram in the form of line graph and histogram have been discussed, elaborately and the students have been taught how to draw them. eHre, how, frequency polygon, pie-chart, ogive curve drawn from frequency distribution and, cumulative frequency table will be discussed., Frequency Polygon : In class V, III, we have learnt how to draw the histogram of, discrete data. eHre how to draw frequency polygon from histogram of indiscrete data, will be put for discussion through example., Example 3. The frequency distribution table of the weights (in kg) of 60 students of, class X of a school are is follows :, Weight (in kg), 46 50, 51 55 56 60 61 65, 66 70, Frequency, 5, 10, 20, 15, 10, (N, o. of tSudents), (a) Draw the histogram of frequency distribution., (b) Draw frequency polygon of the histogram., Solution : The class interval of the data in the table is discrete. If the class interval, are made indiscrete, the table will be :, Class interval of the weight (in Discrete class Mid point of, Frequency, kg), interval, class, 46 50, 455 505, 48, 5, 51 55, 505 555, 53, 10, 56 60, 555 605, 58, 20, 61 65, 605 655, 63, 15, 66 70, 655 705, 68, 10, (a) iHstogram has been drawn taking each square of graph paper as unit of class, interval along with x -axis and frequency along with y -axis. The class interval along, with x -axis has started from. The broken segments have been used to show the, presence of previous squares starting from from origin to 455., , 0DWK,;;)RUPD
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282, , Math, , (b) The mid-points of the opposite sides parallel, to the base of rectangle of the histogram have, been fixed for drawing frequency polygon from, histogram. The mid-points have been joined by, line segments to draw the frequency polygon, (shown in the adjacent figure). The mid-points, of the first and the last rectangles have been, joined with x -axis representing the class interval by the end points of line segments, to show the frequency polygon attractive., Frequency Polygon : The diagram drawn by oj ining frequency indicated points, opposite to the class interval of indiscrete data by line segments successively is, frequency polygon., Example 4. Draw polygon of the following frequency distribution table :, Class interval, Mid-point, Frequency, , 1020, , 2030, , 3040, , 4050, , 5060, , 6070, , 7080, , 8090, , 15, , 25, , 35, , 45, , 55, , 65, , 75, , 85, , 8, , 10, , 15, , 30, , 45, , 41, , 15, , 7, , Solution : iHstogram of frequency distribution is, drawn taking two squares of graph paper as 5 units of, class interval along with x -axis and 2 squares of, graph paper as 5 units of frequency along with y axis. The mid-points of the sides opposite to the base, of rectangle of histogram are identified which are the, mid-points of the class. oNw the fixed mid-points are, oj ined. The end-points of the first and the last classes, are joined to x -axis representing the class interval to, draw frequency polygon., Activity : Draw frequency polygon from the marks obtained in aBngla by the, students of your class in first terminal examination., Example 5. The frequency distribution table of the marks obtained by 50 students of, class X in science are given. Draw the frequency polygon of the data (without using, histogram) :, Class interval of, marks obtained, Frequency, , 3140, , 4150, , 5160, , 6170, , 7180, , 8190, , 91100, , 6, , 8, , 10, , 12, , 5, , 7, , 2, , Solution : eHre the given data are discrete. In this case, it is convenient to draw, frequency polygon directly by finding the mid-point of class interval.
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283, , Math, , Class interval, Mid-point, , 3140, , 40 31, 2, 35 5, , 4150, 455, , 5160, 555, , 6170, 655, , Frequency, 6, 8, 10, 12, The polygon is drawn by taking 2 squares of graph, paper as 10 units of mid-points of class interval along, with x -axis and taking two squares of graph paper as, one units of frequency along with y -axis., , 7180, 755, , 8190, 854, , 91100, 955, , 5, , 7, , 2, , Activity : Draw frequency polygon from the frequency distribution table of, heights of 100 students of a college., eHights (in, cm.), Frequency, , 141150, , 151160, , 161170, , 171180, , 181190, , 5, , 16, , 56, , 11, , 4, , Cumulative Frequency Diagram or Ogive curve : Cumulative frequency diagram, or gOive curve is drawn by taking the upper limit of class interval along with x -axis, and cumulative frequency along with y -axis after classification of a data, Example 6. The frequency distribution table of the marks obtained by 50 students, out of 60 students is as follow :, Class interval of, 1 10 11 20 21 30, 31 40, 41 50, marks obtained, Frequency, 8, 12, 15, 18, 7, Draw the O, give curve of this frequency distribution., Solution : The cumulative frequency table of frequency distribution of the given data is :, Class interval of, 1, 11 20, 21 30, 31 40, 41 50, marks obtained, 10, Frequency, 8, 12, 15, 18, 7, Cumulative, 8, 8 +12 =20, 15 +20 =35, 18 +35 =53 7 +53 =60, frequency, gOive curve of cumulative frequency of data is drawn, taking two squares of graph paper as unit of upper limit, of class interval along with x -axis and one square of, graph paper as 5 units of cumulative frequency along, with y -axis.
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284, , Math, , Activity : Make cumulative frequency table of the marks obtained 50 and above in Mathematics by, the students of your class in an examination and draw an O, give curve., , Central Tendency : Central tendency and its measurement have been discussed in, class V, II and V, III. We have seen if the da ta under investigation are arranged in order, of values, the data cluster round near any central value. A, gain if the disorganized, data are placed in frequency distribution table, the frequency is found to be abundant, in a middle class i.e. frequency is maximum in middle class. In fact, the tendency of, data to be clustered around the central value is number and it represents the data. The, central tendency is measured by this number. eGnerally, the measurement of central, tendency is of three types (1) A, rithmetic means (2) Median (3) Mode :, Arithmetic Mean : We know if the sum of data is divided by the numbers of the, data, we get the arithmetic mean. uBt this method is complex, time consuming and, there is every possibility of committing mistake for large numbers of data. In such, cases, the data are tabulated through classification and the arithmetic mean is, determined by short-cut method., Example 7. The frequency distribution table of the marks obtained by the students of, a class is as follows. Find the arithmetic mean of the marks., Class interval, 2534 3544 4554 5564 6574 7584 8594, Frequency, 5, 10, 15, 20, 30, 16, 4, Solution : eHre class interval is given and that is why it is not possible to know the, individual marks of the students. In such case, it becomes necessary to know the, mid-value of the class., Class upper value class lower value, Mid-value of the class =, 2, If the class mid-value is x1 (i 1, ......., k ) , the mid-value related table will be as follows:, Frequency f i, f i xi, Class interval, Class mid-value xi, 25 34, 295, 5, 1475, 35 44, 395, 10, 3950, 45 54, 495, 15, 7425, 55 64, 595, 20, 11900, 65 74, 695, 30, 20850, 75 84, 795, 16, 12720, 85 94, 895, 4, 3580, Total, 100, 61900, 1 k, The required mean = ¦ f i xi, ni1, =, , 1, u 6190, 100, , =61 9.
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285, , Math, , rAithmetic mean of classified data (short-cut method), The short-cut method is easy for determining arithmetic mean of classified data., The steps to determine mean by short-cut method are :, 1. To find the mid-value of classes., 2. To take convenient approximated mean (a) from the mid-values., 3. To determine steps deviation, the difference between class mid-values and, approximate mean are divided by the class interval i.e., mid value approximate mean, u=, class interval, 4. To multiply the steps deviation by the corresponding class frequency., 5. To determine the mean of the deviation and to add this mean with approximate, mean to find the required mean., Short-cut method : The formula used for determining the mean of the data by this, 1, method is x a ¦ f i ui u h where x is required mean, a is approximate, n, mean, The fi is class frequency of i th class, ui f i is the product of step, deviation with class intervals of i th class and h is class interval., Example 8. The production cost (in hundred taka) of a commodity at different stages is, shown in the following table. Find the mean of the expenditure by short-cut method., P, roduction cost, 2-6 6-10 10-14 14-18 18-22 22-26 26-30 30-34, (in hundred taka), Frequency, 1, 9, 21, 47, 52, 36, 19, 3, Solution : To determine mean in the light of followed steps in short-cut method, the, table will be :, Frequency f i Step deviation Frequency and class, Class, Midxi a, interval f i ui, interval, value xi, ui, h, 26, 4, 1, 4, 4, 6 10, 8, 9, 2, 27, 10 14, 12, 21, 3, 42, 14 18, 16, 47, 1, 47, 18 22, 20 a, 52, 0, 0, 22 26, 24, 36, 1, 36, 26 30, 28, 19, 2, 38, 30 34, 32, 3, 3, 9, Total, 188, 37, Mean x, , a, , ¦fu, , i i, , n, , uh
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286, , Math, , = 20 37 u 4, 188, , =20 79, =19 22, ? Mean production cost is Tk. 1922 hundred., Weighted mean : In many cases the numerical values x1 , x2 ,......., xn of statistical data, under investigation may be influenced by different reasons /importance w, / eight. In such, / eight, case, the values of the data x1 , x2 ,....... xn along with their reasons/importance w, w1 , w2 ,....... wn are considered to find the arithmetic mean., If the values of n numbers of data are x1 , x2 ,....... xn and their weights are, w1 , w2 ,....... wn , the weighted mean will be, n, , ¦xw, i, , i, , i 1, n, , xw, , ¦w, , i, , i 1, , Example 9. The rate of passing in degree oHno ur’s class and the number of students, of some department of a nUiversity are pres ented in the table below. Find the mean, rate of passing in degree honour’s class of those departments of the university., Math, tSatistics English, B, angla, oZology, P, ol., aNme of the, S, cience, department, Rate of passing (in, 70, 80, 50, 90, 60, 85, percentage), uNmber of S, tudents, 80, 120, 100, 225, 135, 300, Solution : eHre, the rate of passing and the numb er of students are given. The weight, of rate of passing is the number of students. If the variables of rate of passing are x, and numerical variable of students is w, the table for determining the arithmetic, mean of given weight will be as follows :, Department, xi wi, xi, wi, Math, 70, 80, 5600, S, tatistics, 80, 120, 9600, English, 50, 100, 5000, aBngali, 90, 225, 20250, oZology, 60, 135, 8100, oPl. cSience, 85, 300, 25500, Total, 960, 74050, 6, , ¦x w, i, , xw, , i, , i 1, 6, , ¦w, , 74050, 960, , i, , i 2, , Mean rate of passing is 7714, , 77 14
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Math, , 287, , Activity : Collect the rate of passing students and their numbers in .SS, .C., examination of some schools in your pUazilla and find mean rate of passing., Median, We have already learnt in class V, III the va lue of the data which divide the data when, arranged in ascending order into two equal parts are median of the data. We have, also learnt if the numbers of data are n and n is an odd number, the median will be, n 1, the value of, th term. uBt if n is an even number, the median will be numerical, 2, n, and §¨ n 1·¸ th terms. eHre we present through example how, mean of the value of, 2, ©2 ¹, mean is determined with or without the help formulae., Example 10. The frequency distribution table of 51 students is placed below. Find, the median., eight (in cm.), H, 150, 155, 160, 165, 170, 175, Frequency, 4, 6, 12, 16, 8, 5, Solution : Frequency distribution table for finding mean is an follows :, eHight (in cm.), 150, 155, 160, 165, 170, 175, Frequency, 4, 6, 12, 16, 8, 5, Cumulative Frequency, 4, 10, 22, 38, 46, 51, eHre, n =51 which is an odd number., ? Median =the value of, , 51 1, th term, 2, , =the value of 26 th term =165, Required median is 165 c.m., Note : The value of the terms from 23th to 38th is 165., Example 11. The frequency distribution table of marks obtained in mathematics of, 60 students is as follows. Find the median :, Marks obtained 40, 45 50, 55, 60 70 80 85 90 95 100, Frequency, 2, 4, 4, 3, 7, 10 16 6, 4, 3, 1, Solution : Cumulative frequency distribution table for determining median is :, Marks obtained 40 45 50 55 60 70 80 85 90 95 100, Frequency, 2, 4, 4, 3, 7, 10 16, 6, 4, 3, 1, Cumulative, 2, 6, 10 13 20 30 46 52 56 59 60, frequency, eHre, n =60 which is an even number., ? Median =, , The sum of values of, , 60, 60, th and 1th terms, 2, 2, 2
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Math, , 288, , The sum of values of 30th and 31th terms, 2, 70 80 150, =, 75, 2, 2, ? Required Median is 75., Activity : 1. Make frequency distribution table of the heights (in cm.) of 49, students of your class and find the mean without using any formula., 2. From the above problem, deduct the heights of 9 students and then, find the median of heights (in cm.) of 40 students., Determining Median of Classified Data, =, , n, th term of classified data is, 2, n, median. nAd the formula used to determine the median or the value of, th term is :, 2, If the number of classified data is n , the value of, , Median = L §¨ n Fc ·¸ u h , where L is the lower limit of the median class, n is the, ©2, ¹ fm, frequency, Fc is the cumulative frequency of previous class to median class, f m is, the frequency of median class and h is the class interval., Example 12. Determine median from the following frequency distribution table :, Time (in sec.) 3035 3641 4247 4853 5459 6065, Frequency, 3, 10, 18, 25, 8, 6, Solution : Frequency distribution table for determining median :, Time (in sec.), Frequency, Cumulative, (class interval), Frequency, 30 35, 3, 3, 36 41, 10, 13, 42 47, 18, 31, 48 53, 25, 56, 54 59, 8, 64, 60 65, 6, 70, n =70, n 70, eHre, n 70 and, or 35., 2 2, Therefore, median is the value of 35th term. 35th term lies in the class (48 –53)., eHnce the median class is (48 –53)., 25 and h 6 ., Therefore, L 48, F 31, F, c, , Median, , m, , = L §¨ n Fc ·¸ u h, ©2, ¹ fm
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Math, , 289, , 31) u 6 =48 +4, , =48 +(35, , 25, , u 6, 25, , =48 +0 96, =48 96, Required median is 4896, Activity : Make two groups with all the students of your class. (a) Make a, frequency distribution table of the time taken by each of you to solve a problem,, (b) find the median from the table., Mode :, In class V, III, we have learned that th e number which appears maximum times in a, data is the mode of the data. In a data, there may be one or more than one mode. If, there is no repetition of a member in a data, data will have no mode. N, ow we shall, discuss how to determine the mode of classified data using formula., Determining Mode of Classified Data, The formula used to determine the mode of classified data is :, f, 1, Mode = L , u h , where L is the lower limit of mode-class i.e. the class, f f, 1, , 2, , where the mode lies, f1 =frequency of mode-class –frequency of the class previous, to mode class, f 2, frequency of mode class –frequency of next class of mode, class and h =class interval., Example 13. Find the mode from the following frequency distribution table., Class, Frequency, 31 40, 4, 41 50, 51 60, 61 70, 71 80, 81 90, 91 100, Solution, Mode = L , , 6, 8, 12, 9, 7, 4, f, , 1, , uh, , 1, , f f, , 2, , eHre, the maximum numbers of repetition of frequency is 12 which lies in the class, (61 –70). eHnce,, L 61, f 12 8 4, 2, f 12 9 3, 2, h 10, 0DWK,;;)RUPD
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290, , Math, , 4, 4, u 10 61 u 10, 7, 43, , ? Mode =61 +, , 40, 61 5 7 66 7, 7, Therefore, the required mode is 66714, Example 14. Find the mode from the frequency distribution table below :, Solution : eHre, maximum numbers, Class, Frequency, of frequency are 25 which lie in the, 41 50, 25, class (41–50). S, o it is evident that, 51 60, 20, mode is in this class. We know that, 61 70, 15, f, 1, Mode = L , uh, 71 80, 8, f f, =61 +, , 1, , 2, , eHre, L 41 [If the frequency is maximum in the first class, the frequency of, previous class is zero], f 25 0, 1, , f2, , 25 20, , 5, , 25, u 10, 25 5, 25, u 10 51 8 33, 30, , ? Mode =41 +, =41 +, , =49 33, Therefore, required mode is 4933, In classified data, if the first class is mode class the frequency of previous class is, considered to be zero., Example 15. Determine the mode of the following frequency distribution table :, Solution : The maximum numbers of, Class, Frequency, frequency are 25 which lie in the class, 10 20, 4, (41 –50). S, o it is obvious that this class is the, 21 30, 16, class of mode. We know that,, 31 40, 20, f, 41 50, 25, Mode =, 1, L, , f f, 1, , eHre,, , uh, , 2, , L, , 41, , f1, , 25 20 5
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291, , Math, , h 10, 5, u 10, 25, =41 +2 =43, The required mode is 43., , Therefore, mode =41 +, , Exercise 17, Put tick (√) mark in the correct answer :, 1. fOthe following, which one is class interval ?, (a) The difference between the highest and the lowest data, (b) The difference between the first and the last data, (c) The difference between the highest and the lowest number of each class, (d) The sum of the highest and the lowest numbers of each class., 2. Which one indicates the data included in each class when the data are classified?, (a) Class limit, (b ) Mid-point of the class, (c) uNmbers of classes, (d) Class frequency, 3. If the disorganized data of statistics are arranged according to the value, the data, cluster round near any central value. This tendency of data is called, (a) mode (b) cetral tendency, (c) mean, (d) median, In winter, the statistics of temperatures (in celsius) of a region in aBngladesh is, 10q, 9q, 8q, 6q, 11q, 12q, 7q, 13q, 14q, 5q. In the context of this statistics, answer, the questions from (4 –6)., 4. Which is the mode of the above numerical data ?, (a) 12q, (b) 5q, (c) 14q, (d) no mode, 5. Which one is the mean of temperature of the above numerical data ?, (a) 8q, (b) 8.5q, (c) 9.5q, (d) 9q, 6. Which one is the median of the data ?, (a) 9.5q, (b) 9q, (c) 8.5q, (d) 8q, 7. The number of classified data included in the table is n , the lower limit of, median class is L , the cumulative data of previous class to median class is Fc ,, the frequency of median class is fm and class interval is h . In the light of these, information, which one is the formula for determining the median ?, n, h, n, h, (a) L §¨ Fc ·¸ u, (b) L §¨ fm ·¸ u, ©2, ¹ Fm, ©2, ¹ fm, n, ©2, , (c) L §¨ Fc ·¸ u, ¹, , h, fm, , (d) L §¨ n fn ·¸ u h, ©2, , ¹ Fm
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292, , Class Interval, Frequency, Cumulative Frequency, , Math, , 3140, 6, 6, , 4150, 12, 18, , 5160, 16, 34, , 6180, 24, 58, , 7180, 12, 70, , 8190, 8, 78, , 91100, 2, 80, , 8. In how many classes have the data been arranged ?, (a) 6, (b) 7, (c) 8, (d) 9, 9. What is the class interval of the data presented in the table ?, (a) 5, (b) 9, (c) 10, (d) 15, 10. What is the mid value of the 4th class ?, (a) 71.5, (b) 61.5, (c) 70.5, (d) 75.6, 11. Which one is the median class of the data ?, (a) 4150, (b) 5160, (c) 6170, (d) 7180, 12. What is the cumulative frequency of the previous class to the median class ?, (a) 18, (b) 34, (c) 58, (d) 70, 13. What is the lower limit of median class ?, (a) 41, (b) 51, (c) 61, (d) 71, 14. What is the frequency of median class ?, (a) 16, (b) 24, (c) 34, (d) 58, 15. What is the median of the presented data?, (a) 63, (b) 63.5, (c) 65, (d) 65.5, 16. What is the mode of the presented data ?, (a) 61.4, (b) 61, (c) 70, (d) 70.4, 17. The weights (in kg) of 50 students of class X of a school are :, 45, 50, 55, 51, 56, 57, 56, 60, 58, 60, 61, 60, 62, 60, 63, 64, 60,, 61, 63, 66, 67, 61, 70, 70, 68, 60, 63, 61, 50, 55, 57, 56, 63, 60,, 62, 56, 67, 70, 69, 70, 69, 68, 70, 60, 56, 58, 61, 63, 64., (a) Make frequency distribution table considering 5 as a class interval., (b) Find the mean from the table in short-cut method., (c) Draw frequency polygon of the presented data in frequency distribution table., 18. Frequency distribution table of the marks obtained in mathematics of 50 students, of class X are provided. Draw the frequency polygon of the provided data., Class interval 3140 4150 5160 6170 7180 8190 91100, Frequency, 6, 8, 10, 12, 5, 7, 2, 19. The frequency distribution table of a terminal examination in 50 marks of 60, students of a class is as follows :, Marks obtained, 110 1120, 2130, 3140, 4150, Frequency, 7, 10, 16, 18, 9, Draw an O, give curve of the data., 20. The frequency distribution table of weights (in kg) are provided below., Determine the median., Weight (kg), 45, 50, 55, 60, 65, 70, Frequency, 2, 6, 8, 16, 12, 6
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Math, , 293, , 21. The frequency distribution table of weights (in kg) of 60 students of a class are:, Interval, 45-49 50-54 55-59 60-64 65-69 70-74, Frequency, 4, 8, 10, 20, 12, 6, Cumulative Frequency, 4, 12, 22, 42, 54, 60, (a) Find the median of the data., (b) Find the mode of the data., 22. In case of data, Mode is(i) Measures of central tendency, (ii) Represented value which is mostly occured, (iii) May not unique in all respect, Which is correct on the basis of above information?, a) i and ii, b) i and iii, c) ii and iii, d) i, ii, and iii, 23. The following are the marks obtained in Mathematics of fifty students of class IX, in a school :, , (a) What is the type of the given information? What indicate frequency in a class, of distribution?, (b) Make frequency table taking appropiate class enterval., (c) Determine the mean of the given number by shortcut method., 24., , (a) In the above figure, what is class midvalue?, (b) Express by data of information demonstrated in the figure (b)., (c) Find the median of frequency obtained from (b).
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Answer, Exercise 1, 4. (a) 0 16 (b) 0 6 3 (c) 3 2 (d) 3 53 5. (a), , 71, 2, 35, 2, (b), (c), (d) 3, 9, 99, 15, 90, , 769, 6 (a) 2 33 3 , 5 23 5 (b) 7 266 , 4 237 (c) 5 7 77777 , 8 3 43434 ,, 3330, 6 2 45245 (d) 12 320 0 , 2 199 9 , 4 325 6 7. (a) 0 5 (b) 0 589 (c) 17 1179, (d) 1 926 31 8., , (e) 6, , (a) 1 31 (b) 1 66 5 (c) 3 133 4 (d) 6 110 62 9. (a) 0 2 (b) 2 (c) 0 2 074, (d) 12 18 5 10. (a) 0 5 (b) 0 2 (c) 5 2 1951 (d) 4 8 11. (a) 3 4641, 3 464, (b) 0 5025, 0 503 (c) 1 1595, 1 160 (d) 2 2650, 2 265, 12.(a) Rational (b) Rational (c) Irrational (d) Irrational (e) Irrational (f), Irrational (g) Rational ` (h) Rational, 13. (a) 9 (b) 5 (c) 8, , Exercise 2.1, 1. (a) {4, 5} (b) {r3, r 4, r 5, r 6} (c) {6, 12, 18, 36}, , (d) {3, 4}, , 2. (a) {x N : x is odd number and 1 u 13} (c) {x N : x, in the multiple, of 36 }, (c) {x N : x, is the multiple of 4 and x d 40} (d) {x Z : x2 t 16, and x3 d 216}, 3. (a) {1} (L) {1, 2, 3, 4, a}, , (b) {2}, , (c) {2, 3, 4, a}, , (d) {2}, , 5. {{x, y), ( x), ( y}, φ} , {{m, n, l), (m, n), (m, l}, {n, l}, {m}, {n}, {l}, φ}, 7. (a) 2, 3 (L) (a, c) (b) (1, 5), 8. (a) {(a, b), (a, c)}, {(b, a), (c, a)} (b) {(4, x), (4, y), (5, x), (5, y)}, (c) {3, 3), (5, 3), (7, 3)}, 9. {1, 3, 5, 7, 9, 15, 35, 45} and {1, 5} 10. {35, 105} 11. 5 persons
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Math, , 298, , Exercise 3⋅5, 1., 5., 9., 13., 17., , (c), (d), (a), (a), (a), , 2., 6., 10., 14., 18., , 21. (2) (b), 24., , (d), (c), (c), (b), (b), , 3. (b), 7. (d), 11. (d), 15. (c), 19. (c), , 21. (3) (d), , xy, days, x y, , 22., , 4., 8., 12., 16., 20., , (b), (d), (b), (b), (d), , 2, ( p r ) days 23. 5 hours, 3, , 25. 95 persons, , 26. Speed of current is, , d §1 1 ·, d § 1 1·, ¨ ¸ km per hour and speed of boat is ¨¨ ¸¸, 2 © p q¹, 2 ¨© q p ¸¹, , km. per hour., 27. The speed of the oar is 8 km/hour and the speed of current is 2 km/hour, t1t 2, minutes, t 2 t1, , 28, , 29. 240 liter, , 30. Tk. 10, , 31. Tk. 48, , Tk. 120 , (b) Tk. 80 , (c) Tk. 60 33. Purchase value Tk. 450, 625 36. Tk. 5%, 39. Tk. 61, 37. Tk. 522 37 (approx.) 38. Tk. 780, 40. VAT is Tk., , 32., , (a), , 34. 4% 35. Tk., , px, ; the amount of VAT is Tk. 300., 100 x, , Exercise 4⋅1, 1., , 10, 7, , 7. 4, , ab, 3a 2b, 1, 8., 9, 2., , 3. 27, 17., , 3, 2, , 4., , a2, b, , 18. 3, , Exercise 4⋅2, 1. (a) 4 (b), , 1, 1, 5, (c), (c) 4 (e), 3, 2, 6, , 5. 343, , 6. 1, , 19. 5, , 20. 0, 1
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299, , Math, , 2. (a) 125 (b) 5, , (c) 4, 13, (c) 0, 4. (a) "og 2 (b), 15, , Exercise 4⋅3, 1. b 2. d 3. c 4. a 5. c 7. d 8. (1) d (2) c (3) a, 9. (a) 6 530 u 103 (b) 6 0831 u 103 (c) 2 45 u 10 4 (d) 3 75 u 10 7, (e) 1 4 u 10 7, 10. (a) 100000 (b) 0 000001 (c) 25300 (d) 0 009813 (e) 0 0000312, 11. (a) 3 (b) 1 (c) 0 (d) 2 (e) 5, 12. (a) characteristics 1 , Mantissa 43136, 80035, (c) characteristics 0 , Mantissa 14765, 65896, , (b) characteristics 1 , Mantissa, (d) characteristics 2 , Mantissa, , (e) characteristics 4 , Mantissa 82802, 13. (a) 1 66706 (b) 1 64562 (c) 0 81358 (d) 3 78888, 14. (a) 0 95424 (b) 1 44710 (c) 1 62325, 15. a.2 a 3 53 b. 6 25 u 10, , c. characteristics 1, Mantissa 79588, , Exercise 5⋅1, 1. 1, , 2. ab 3. 6, 7., , ab, 2, 13. {a}, , 9., , ab, 2, , 10. 3, , 4. 1, , 5. , , 6. , , 5, 2, , 8. a b, 11. {2}, , 12. {4(1 2 )}, , 14. φ, , m n½, ¾ 17., ¯ 2 ¿, 3, 22. 72, 20. 28, 70 21., 4, 1½, ¯ 3¿, , 15. ® ¾, , 3, 5, , 16. ®, , 7½, ® ¾, ¯ 2¿, , 18. ^6`, , 23. 72, , 24. 18, , 19. ^ (a 2 b 2 c 2 )`, 25. 9
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300, , Math, , 26. Number of coin of twenty five and fifty paisa are 100 and 20 respectively., 22. 120 km., , Exercise 5⋅2, 1. c, , 2. b, , 3. b, 10. , , 9. 2,, , 3, , 14. 1, , , 3, 20, , 15., , 1, , 2, 2, , 2, ¯ 3, , 16. 0,, , 4. c, , 5d, , 6. b, , 3 2 2 3, ,, 2, 3, , 2, 3, , 7. a, , 12. r 7 13. 6,, , 11. 1, 6, , 17. r ab, , 8. (1) d (2) c (3) a, 3, 2, , , ¯, , 1½, 2¿, , 1, ¯3, , ½, ¿, , 18. 0, a b 19. ®3, ¾, , ½, ¿, , 20. ® , 2¾, 21. {a, b}, , 22. {1, 1}, , 23. {1} 24. {0, 2a} 25. ® , 1¾, , or 87 27. Length 16 metre, breadth 12 metre, , 28. 9 cm., 12 cm., , 26., 29., , 78, 27, , cm., , 30. 21 persons, Tk. 20 31. 70 32. a. 70 9 x, 9 x 7 b. 34 c. 5 cm. 5 2 cm., 33. b. 5 cm.. c. 2 : 5 : 8, , Exercise 9.1, 2. cos A, , 7, , tan A, 4, , 3, , cot A, 7, , 3. sin A, , 15, 8, , cos A, 17, 17, , 4. sin θ, , 5, , cos nθ, 13, , 22., , 1, 3, 23., 4, 2, , 24., , 12, , tan θ, 13, 4, 3, , 25., , 7, , sec A, 3, , 5, 12, a2 b2, a2 b2, , 4, , cos ecA, 7, , 4, 3
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301, , Math, , Exercise 9.2, , 5., , 1, 2, , 6., , 3, 4 2, , 7., , $, , 19. A 37, , 1, ,B, 2, , 7, , 23, 2 2, 8., 5, 3, , 1, 2, , 17. A 30$ , B 30$, , 18. A 30$, , $, , 21. θ, , 90 $ 22. θ, , 60 $ 23. θ, , 60 $ 24. θ, , 45$ 25. 3, , Exercise 10, 7. 45.033 metre (app.) 8. 34.641 metre (app.) 9. 12.728 metre (app.), 10. 10 metres, 11. 21.651 metre (app.) 12. 141.962 metre(app.) 13. 83.138 metre (app.) and, 48 metre, 14. 34.298 metre (app.) 15. 44.785 metre (app.) 16. (b) 259.808 metre., , Exercise 11⋅1, 1. a2 : b2, 2. π : 2 , 3. 45, 60, 4. 20%, 5. 18 : 25, 6. 13 : 7, 8. (i), (iii) x r 2ab b 2 , (iv) 10, (v), , 3, 2ab, , (ii) 2 ,, 4, b 1, , 1, b §, 1·, ¨ c ¸ , (vi) , 2, 22. 3, 2, 2a ©, c¹, , Exercise 11⋅2, 1. a 2. c 3. c 4. b 5. b 6. 24%, 7. 70%, 8. 70%, 9. a Tk. 40, b Tk.60, c Tk., 120, d Tk. 80, 10. 200, 240, 250, 11. 9 cm. 15cm., 21cm., 12. Tk. 315, Tk., 336, Tk. 360, 13. 140, 14. 81 runs 54 runs, 36 runs, 15. Officer Tk. 24000,, Clark Tk. 12000 bearer Tk. 6000 16. 70, 17. 44%, 18. 1% 19. 532 quintal,, 20. 8 : 9, 21. 1440 sq.metre, 22. 13 : 12., , Exercise 12.1, 1. Consistent, not dependent, single solution, 2. Consistent, dependent,, innumerable solution, 3. inconsistent not dependent, has no solution, 4., consistent, dependent, innumerable solution 5. consistent, not dependent,
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302, , Math, , single solution 6. inconsistent, not dependent, has no solution 7. Consistent,, dependent, innumerable solution 8. Consistent, dependent, innumerable, solution, 9. Consistent, dependent, single solution 10. Consistent, not, dependent, single solution ., , Exercise 12.2, , Exercise 12.3, , Exercise 12.4, 1. a, , 2. c, , 3. b, , 4. b, , 15, 9., 26, , 7, 7(3) d 8., 9, , 10. 27, , 5. b, , 6. b, , 11. 37 or 73, , 7(1) . c, , 7(2). d, , 12. 30 years, , 13., , length 17 m. breadth 9 metre 14. spread of boat 10 km. per hour, spead of, corrent 5 km. per hour 15. starting salary Tk. 4000, yearly increment Tk. 25, 16. a. one, , b. (4, 6), , c. sq.unit 17. a., , x7, y, , 2,, , x, y2, , 1,, , b. (3, 5),, , Exercise 13.1, 1. 7 Ges 75, 2. 129 Zg, 3. 100 Zg, 4. p2 pq q2 , 5. 0 , 6. n2 , 7. 360 , 8. 320 ,, 9. 42 , 10. 1771 , 11. 620 , 12. 18, 13. 50, 14. 2 4 6 ........,, 15. 110 , 16. 0,, 17. (m+n), 20. 50 ., , Exercise 13.2, 5., , 1, 1, , 2. 3 6. (3141), 7. 9th term, 8., , 9. 9th term, 10. x 15, y, 2, 3, , 45 ,, , 3, 5
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Math, , 11. x 9, y, , 303, , 27, z 81 , 12. 86, 13. 1, 14. 55log 2, 15. 650log 2, 16. n, , 7,, , 17. 0 , 18. n 6, S 21 , 19, . n 5, S 165 , 20. 3069 , 21. 20, 22. 24.47mm, 512, , a( pp.), , Exercise 16⋅1, 1. 20 m., 15 m., , 2. 12 m. 3. 12 sq. m. 4. 327 26 sq. m. a( pp.) 5. 5 m., , 6. 30o 7. 36 or 12 cm., 10. 24 249 cm. (app.),, , 8. 12 or 16 m., , 9. 44 44 km. (app.), , 254 611 sq. cm. (app.), , Exercise 16⋅2, 1. 96 m., , 2. 1056 sq. m., , 3. 30 m. and 20 m., , 4. 400 m., , 5. 6400 6. 16 m. and 10 m. 7. 165 m. and 22 m.8. 3535 m. (app.), 9. 4866 cm. (app.) 10. 72 cm., 1944 sq. cm., 12. 95 75 sq. cm. a( pp.), , 11. 17 cm. and 9 cm., , 13. 636 sq. m. (app.), , Exercise 16.3, 1. 32 987 cm. (app.) 2. 31 513 m. (app.) 3. 20 008 (app.) 4. 128 282 sq., cm. (app.) 5. 7 003 m. (app.) 6. 175 93 m. a( pp.) 7. 20 times, 8. 49 517 m. a( pp.) 9., , 3 3 :π, , Exercise 16⋅4, 8. 636 sq. m., 20 5 m., 864 cubic metre. .9 14040 sq. m., 10. 12 m., 4 m., 11. 1 cm. 12. 300000 13. 34 641 sq. cm. 14. 534 071 sq. cm. a( pp.), 42 48 cubic cm. a( pp.) 15. 5 305 sq. cm. 3 cm.. 16. 6111 8 sq. cm., 9, 17. 147 027 kg. (app.), , Exercise 17, 1. (c) 2. (b) 3. (b) 4. (d) 5. (c) 6. (a) 7. a( ) 8. (b) 9.(c), 10. (c) 11. (c) 12. (c) 13. (c) 14. (b) 15. (b) 16. (a) 20. Median, 60 21. a( )62 K, g , (b)62.8 gK.
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2013, Academic Year, , 9-10 Math
