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CHAPTER 12, MENSURATION, Area of a triangle = × base × altitude, Heron’s Formula, , If , , are the lengths of the three sides of a triangle, then, Area of the triangle =, , where, , −, , −, , −, , = semi − perimeter of the triangle =, , SOLUTIONS, EXERCISE 12.1, 1., , Find the area of the triangle whose sides are, (i), , 8 cm, 11 cm and 13 cm, , (ii), , 10 cm, 16 cm and 20 cm, , (iii) 5 cm, 7 cm and 10 cm, Solution:, (i), , Let, , = 8 cm,, , Then we have,, , = 11 cm and, =, , =, , !, , ∴ area of the triangle =, , =, , = 13 cm, ", , cm = 16 cm, , −, , −, , −, , 16 16 − 8 16 − 11 16 − 13 cm2, , = √16 × 8 × 5 × 3 cm2, , = √2 × 2 × 2 × 2 × 5 × 3 cm2, = 2 × 2 × 2√2 × 5 × 3 cm2, , (ii), , Let, , = 10 cm,, , Then we have,, , = 8√30 cm2, , = 16 cm and, , =, , =, , ∴ area of the triangle =, , =, , ,, , = 20 cm, , -, , ,, , −, , cm =, , .-, , −, , cm = 23 cm, −, , 23 23 − 10 23 − 16 23 − 20 cm2, , = √23 × 13 × 7 × 3 cm2, = √6279 cm2, , Page | 1

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(iii), , = 5 cm,, , Let, , = 7 cm and, =, , Then we have,, , =, , = 10 cm, , 1 2, , ∴ area of the triangle =, , =, , ,, , cm =, , −, , cm = 11 cm, , −, , −, , 11 11 − 5 11 − 7 11 − 10 cm2, , = √11 × 6 × 4 × 1 cm2, , = √11 × 6 × 2 × 1 cm2, , 2., , = 2√66 cm2, , Find the area of a triangle, two sides of which are 17 cm and 10 cm and perimeter is 48 cm., , Solution:, , = 48 cm, , Here, perimeter of the triangle, = 17 cm and, , Therefore, third side, =, , Then we have,, , .!, , = 10 cm, , = 48 cm – (17 + 10 ) cm = 21 cm, , cm = 24 cm, , ∴ area of the triangle =, , =, , −, , −, , −, , 24 24 − 17 24 − 10 24 − 21 cm2, , = √24 × 7 × 14 × 3 cm2, , = √2 × 2 × 3 × 7 × 2 × 7 × 3 cm2, = √2 × 2 × 3 × 7 × 3 cm2, , = 2 × 2 × 3 × 7 cm2, = 84 cm2, , 3., , The sides of a triangle are in the ratio 5:7:8 and its perimeter is 300 cm. Find its area., , Solution:, , Let the sides of the triangle (in cm) be 55, 75 and 85., , Then by question, we have, , 55 + 75 + 85 = 300, , ⇒ 205 = 300, ⇒ 5 = 15, , So, the sides of the triangle are 5 × 15 cm, 7 × 15 cm and 8 × 15 cm i.e., 75 cm, 105 cm and 120 cm., Let, , = 75 cm,, , We have,, , =, , ",,, , = 105 cm and, cm= 150 cm, , = 120 cm., , Page | 2

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∴ area of the triangle =, , =, , −, , −, , −, , 150 150 − 75 150 − 105 150 − 120 cm2, , = √150 × 75 × 45 × 30 cm2, =, , 2 × 75 × 75 × 3 × 15 × 2 × 15 cm2, , = √2 × 75 × 15 × 3 cm2, = 2 × 75 × 15√3 cm2, , 4., , = 2250√3 cm2, , Perimeter of an equilateral triangle is 36 cm. Find its area using Heron’s formula., , Solution:, , =, , We have, , "-, , =, , cm= 18 cm and, , ∴ area of the triangle =, , =, , −, , −, , =, , =, , −, , "", , cm= 12 cm, , 18 18 − 12 18 − 12 18 − 12 cm2, , = √18 × 6 × 6 × 6 cm2, , = √3 × 6 × 6 × 6 × 6 cm2, = √6 × 6 × 3 cm2, , = 6 × 6√3 cm2, = 36√3 cm2, , 5., , Find the area of an equilateral triangle whose sides are of 14 cm each using heron’s formula., , Solution:, , We have, , =, , =, , "× ., , =, , = 14 cm, , cm= 21 cm, , ∴ area of the triangle =, , =, , =, , −, , −, , −, , 21 21 − 14 21 − 14 21 − 14 cm2, 3 × 7 × 7 × 7 × 7 cm2, , = √3 × 6 × 6 × 6 × 6 cm2, = √7 × 7 × 3 cm2, , = 7 × 7√3 cm2, = 49√3 cm2, , Page | 3

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6., , Find the area of an isosceles triangle whose base is 10 cm and one of its equal sides is 13 cm., , Solution:, , We have, =, , = 10 cm and, ,, , ", , ", , ∴ area of the triangle =, , =, , cm =, , =, , = 13 cm., , "-, , cm= 18 cm, , −, , −, , −, , 18 18 − 10 18 − 13 18 − 13 cm2, , = √18 × 8 × 5 × 5 cm2, , =, , 2×3, , × 2×2, , × 5 × 5 cm2, , = √2 × 2 × 3 × 5 cm2, = 2 × 2 × 3 × 5 cm2, , = 60 cm2, 7., , Perimeter of an isosceles triangle is 40 cm and each of the equal sides is 15 cm. Find the area of, the triangle using Heron’s formula., , Solution:, , Here, perimeter of the triangle, Therefore, third side, Then we have,, , =, , .,, , = 40 cm and, , =, , = 40 cm – (15 + 15 ) cm = 10 cm, , cm = 20 cm, , ∴ area of the triangle =, , =, , −, , −, , = 15 cm, , −, , 20 20 − 15 20 − 15 20 − 10 cm2, , = √20 × 5 × 5 × 10 cm2, , =, , 2 × 5 × 5 × 5 × 5 × 2 cm2, , = √2 × 5 × 5 × 2 cm2, = 2 × 5 × 5√2 cm2, , = 50√2 cm2, , Page | 4

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8., , A signboard is in the shape of an equilateral triangle with side ‘8’. Find the area of the signboard, using Heron’s formula. If the perimeter of the board is 210 cm, what will be its area?, , Solution:, , Length of each side of the signboard =, ∴, , =, , =, , =, , and, , ∴ area of the signboard =, , =, , −, , =, , −, , =, , −, , =9, , ", , :, , =9, , ", , : ;, , ", , ", , − ;, , −, , −, , =, , ", , −, , −, , ", , ", , = <3 × : ; × : ;, = × √3, =, , √", ., , sq. units, , Second Part, If the perimeter of the signboard = 210 cm,, Then, area of the signboard =, , √", ., , =, , √", ., , =, , √", ., , =, , ,, , ", , cm = 70 cm, , cm2, × 70 cm2, , × 70 × 70 cm2, , = √3 × 35 × 35 cm2, = 1225√3 cm2, , Page | 5

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9., , The sides of a triangular field are 975 m, 1050 m and 1125 m. If this field is sold at the rate of Rs, 10 per m2, find its selling price., , Solution: Let, , = 975 m,, , Then, we have, , = 1050 m and, , =, , =, , = 1125 cm, , =21, , ,1,, , Area of the triangular field =, , =, , −, , 1, , m=, , −, , " 1,, , m = 1575 m, −, , 1575 1575 − 975 1575 − 1050 1575 − 1125 m2, , = √1575 × 600 × 525 × 450 m2, , =, , 3 × 525 × 2 × 3 × 2 × 5, , × 525 × 2 × 3 × 5, , = √2 × 2 × 3 × 3 × 5 × 5 × 525 m2, , m2, , = 2 × 2 × 3 × 3 × 5 × 5 × 525 m2, , = 472500 m2, , ∴ selling price of the triangular field = ₹472500 × 10 = ₹4725000, , 10. Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm., Solution:, , Here, perimeter of the triangle, Therefore,, , =, , Then we have,, , =, , =, , -?-, , -, , cm=, , ,, , cm = 5 cm, , cm = 8 cm, , ∴ area of the triangle =, , =, , = 16 cm and, , −, , −, , = 6 cm, , −, , 8 8 − 6 8 − 5 8 − 5 cm2, , = √8 × 2 × 3 × 3 cm2, =, , 2 × 2 × 2 × 3 cm2, , = √2 × 2 × 3 cm2, = 2 × 2 × 3 cm2, , = 12 cm2, , Page | 6

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SOLUTIONS, EXERCISE 12.2, 1., , Find the area of the quadrilateral ABCD in which, (i), , AB = 5 cm, BC = 4.5 cm, CD = 3.5 cm, DA = 4 cm and AC = 6.5 cm, , (ii), , AB = 3 cm, BC = 5 cm, CD = 6 cm, DA = 6 cm and BD = 5 cm, , (iii), , AB = 3.5 cm, BC = 4.5 cm, CD = 6 cm, DA = 3 cm and BD = 5.5 cm, , (iv), , AB = 6 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 6 cm, , Solution:, (i), For ∆ABC, AB = 5 cm, BC = 4.5 cm and AC = 6.5 cm, ∴, , =, , 1 ..1 -.1, , cm=, , -, , cm= 8 cm, D, , Then by Heron’s formula, we have, Area of ∆ABC =, , 8 8 − 5 8 − 4.5 8 − 6.5 cm2, , = √8 × 3 × 3.5 × 1.5 cm2, , 3.5 cm, C, , 4 cm, 6.5 cm, , 4.5 cm, , = √126 cm2 =11.22 cm2, , For ∆ADC, AC = 6.5 cm, CD = 3.5 cm and AD = 4 cm, ∴, , =, , -.1 ".1 ., , cm=, , ., , A, , 5 cm, , B, , cm= 7 cm, , Then by Heron’s formula, we have, Area of ∆ADC =, , 7 7 − 6.5 7 − 3.5 7 − 4 cm2, , = √7 × 0.5 × 3.5 × 3 cm2, = √36.75 cm2 =6.06 cm2, , ∴ Area of Quadrilateral ABCD, , = Area of ∆ABC + Area of ∆ADC, , = 11.22 cm2 + 6.06 cm2, = 17.26 cm2= 17.3 cm2, , Page | 7

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(ii), For ∆ABD, AB = 3 cm, BD = 5 cm and AD = 6 cm, ∴, , =, , " 1 -, , cm=, , ., , cm= 7 cm, , Then by Heron’s formula, we have, Area of ∆ABD =, , 7 7 − 3 7 − 5 7 − 6 cm2, , = √7 × 4 × 2 × 1 cm2, , D, , = √56 cm2 =7.5 cm2, , For ∆BCD, BC = 5 cm, CD = 6 cm and BD = 5 cm, , ∴, , =, , 1 - 1, , cm=, , -, , cm= 8 cm, , C, , 6 cm, , 5 cm, 5 cm, , 6 cm, , Then by Heron’s formula, we have, Area of ∆BCD =, , =, , 8 8 − 5 8 − 6 8 − 5 cm2, 2 × 2 × 3 × 2 × 3 cm2, , A, , 3 cm, , B, , = √2 × 2 × 3 cm2, , = 2 × 2 × 3 cm2 =12 cm2, , ∴ Area of Quadrilateral ABCD, , = Area of ∆ABD + Area of ∆BCD, , = 7.5 cm2 + 12 cm2, = 19.5 cm2, , (iii), , For ∆ABD, AB = 3.5 cm, BD = 5.5 cm and AD = 3 cm, ∴, , =, , ".1 1.1 ", , cm= 6 cm, , cm=, , Then by Heron’s formula, we have, Area of ∆ABD =, , 6 6 − 3.5 6 − 5.5 6 − 3 cm2, , = √6 × 2.5 × 0.5 × 3 cm2, = √22.5 cm2 =4.7 cm2, , D, , C, , 4 cm, , 5.5 cm, , 4.5 cm, , 3 cm, , For ∆BCD, BC = 4.5 cm, CD = 6 cm and BD = 5.5 cm, ∴, , =, , ..1 - 1.1, , cm=, , -, , cm= 8 cm, , A, , 3.5 cm, , B, , Page | 8

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Then by Heron’s formula, we have, Area of ∆BCD =, , 8 8 − 4.5 8 − 6 8 − 5.5 cm2, , = √8 × 3.5 × 2 × 2.5 cm2, , = √140 cm2, = 11.8 cm2, , ∴ Area of Quadrilateral ABCD, , = Area of ∆ABD + Area of ∆BCD, , = 4.7 cm2 + 11.8 cm2, = 16.5 cm2, , (iv), For ∆ABC, AB = 6 cm, BC = 4 cm and AC = 6 cm, , ∴, , =, , - . -, , cm=, , -, , cm= 8 cm, , Then by Heron’s formula, we have, Area of ∆ABC =, , 8 8 − 6 8 − 4 8 − 6 cm, , D, , 4 cm, C, , 2, , = √8 × 2 × 4 × 2 cm2, , = √128 cm2 =11.3 cm2, , 5 cm, 6 cm, , 4 cm, , For ∆ADC, AC = 6 cm, CD = 4 cm and AD = 5 cm, , ∴, , =, , - . 1, , cm=, , 1, , cm= 7.5 cm, , A, , 6 cm, , B, , Then by Heron’s formula, we have, Area of ∆ADC =, , 7.5 7.5 − 6 7.5 − 4 7.5 − 5 cm2, , = √7.5 × 1.5 × 3.5 × 2.5 cm2, , = √98.4375 cm2 =9.9 cm2, , ∴ Area of Quadrilateral ABCD, , = Area of ∆ABC + Area of ∆ADC, , = 11.3 cm2 + 9.9 cm2, = 21.2 cm2, , Page | 9

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2., , Find the area of the quadrilateral ABCD in which, AB = 3 cm, BC = 4 cm, CD = 5 cm, DA = 4 cm and ∠B = ABB, , (i), , AB = 6 cm, BC = 4 cm, CD = 3 cm, DA = 5 cm and ∠C = ABB, , (ii), Solution:, (i), , In the right ∆ABC,, CD = CE + ED [Pythagoras Theorem], , ⇒ CD = 3 + 4 = 9 + 16 = 25 = 5, , ⇒ CD = 5 cm, , ∴ Area of ∆ABC = × ED × CE, , = × 4 × 3 cm2 = 6 cm2, , D, 4 cm, A, , 5 cm, , 3 cm, B, , 4 cm, , C, , For ∆ADC, AC = 5 cm, CD = 5 cm and AD = 4 cm, ∴, , =, , 1 1 ., , cm=, , ., , cm= 7 cm, , Then by Heron’s formula, we have, Area of ∆ADC =, , 7 7 − 5 7 − 5 7 − 4 cm2, , = √7 × 2 × 2 × 3 cm2, = √84 cm2 =9.2 cm2, , ∴ Area of Quadrilateral ABCD, , = Area of ∆ABC + Area of ∆ADC, , = 6 cm2 + 9.2 cm2, = 15.2 cm2, , Page | 10

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3., , Each side of a rhombus shaped field is 30 m and its longer diagonal is 48 m. Find the area of the, field., , Solution:, Let ABCD be the rhombus shaped field in which, CE = ED = DF = FC = 30 m and CD = 48 m., , D, , 30 m, , A, , For the ∆ABC, we have, , ∴, , = ED = 30 m,, =, , Area of ∆ABC =, =, , =, , = CD = 48 m and, , ", .! ",, , −, , m=, , ,!, , −, , = CE = 30 m, , 30 m, , 48 m, , 30 m, , m= 54 m, , −, , B, , 30 m, , C, , 54 54 − 30 54 − 48 54 − 30 m2, , = √54 × 24 × 6 × 24 m2, =, , 3 × 6 × 24 × 6 m2, , = √3 × 6 × 24 m2, = 3 × 6 × 24 m2, = 432 m2, , We know, a diagonal of a rhombus divides the rhombus into, two congruent triangles., ∴ Area of the rhombus shaped field ABCD, = 2 × Area of ∆ABC, = 2 × 432 m2, , = 864 m2, , Page | 12

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4., , A field is in the shaped of a trapezium whose parallel sides are 25 m and 10 m, and non-parallel, sides are 14 m and 13 m. Find the area of the field., , Solution:, Let ABCD be the trapezium shaped field in which AB = 25 m, and DC = 10 m are the parallel sides; AD = 13 m and BC = 14, , D, , 10 m, , C, , m are the non-parallel sides., , GDH‖FC is drawn meeting AB at E to form a parallelogram, , 13 m, , 14 m, , AECD. CN⊥ AB is also drawn to meet AB at N., , A, , For the ∆BCE, we have, , E, 25 m, , B, , N, , BC = 14 m, CE = 13 m = AD and BE = 25 m – 10 m =15 m., , Semi-perimeter (s) =, , ., , ", , 1, , m=, , ., , m= 21 m, , 21 21 − 14 21 − 13 21 − 15, , Area of ∆BCE=, , ⇒ × EH × DJ = √21 × 7 × 8 × 6, ⇒ × 15 × DJ =, , 3×7 ×7× 2 ×2 × 2×3, , ⇒ × 15 × DJ = √2 × 2 × 3 × 7, , ⇒ × 15 × DJ == 2 × 2 × 3 × 7, , ⇒ × 15 × DJ = 84, ⇒ DJ =, , ⇒ DJ =, , !.×, 11, , 1, , =, , !×, 1, , m, , Now, area of the trapezium shaped field ABCD, =, , [⸪ Area of trapezium = × sum of the parallel sides ×, , CE + FD × DJ, , = × 25 + 10 ×, , = × 35 ×, , 11, , 11, , distance between the parallel sides], , m2, , m2, , = 7 × 28 m2, = 196 m2, , Page | 13

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Surface Area and Volume of Some Basic Solids, 1, , Cube, K, , K, , 2, , 3, , 4, , Total Surface Area = 6K, , Volume (Capacity) = K ", , K, , Lateral Surface Area = 2 K +, , Cuboid, , Cylinder, , K, , Volume (Capacity) = K ℎ, , ℎ, , Curve Surface Area = 2MNℎ, , N, , ℎ, , N, , Total Surface Area = 2MN N + ℎ, Volume (Capacity) = MN ℎ, , Slant Height, K = √N + ℎ, , Curve Surface Area = MNK, , K, , Total Surface Area = MN N + K, Volume (Capacity) = " MN ℎ, , 5, , Sphere, , Curve Surface Area = 4MN, , 6, , Hemisphere, , Curve Surface Area = 2MN, , 7, , ℎ, , Total Surface Area = 2 K + ℎ + ℎK, , ℎ, , Cone, , Lateral Surface Area = 4K, , ., , Volume (Capacity) = " MN ", Total Surface Area = 3MN, , Volume (Capacity) = " MN ", , Frustum, , Slant Height of a frustum K =, , QR, ℎ, , Curve Surface Area = M N + N K, K, , QS, , N −N, , +ℎ, , Total Surface Area = MO N + N K + N + N P, , Surface Area of a Frustum with larger face open, = MO N + N K + N P, , Volume = " M N + N . N + N, , ℎ, , Page | 14

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SOLUTIONS, EXERCISE 12.3, , (Take T =, 1., , SS, U, , unless otherwise stated), , Find the volume and total surface area of a cuboid of edges, (i), , 40 cm, 36 cm and 25 cm, , (ii), , 12 m, 8 m and 7 m, , (iii), , 20 m, 16 m and 15 m, , (iv), , 18 cm, 12 cm and 17 cm, , Solution:, (i), , Here, K = 40 cm,, , = 36 cm and ℎ = 25 cm, , Volume of the cuboid = K ×, , ×ℎ, , = 40 × 36 × 25 cm3, , =36000 cm3, , Total Surface Area = 2 K + ℎ + ℎK, , = 2 40 × 36 + 36 × 25 + 25 × 40 cm2, = 2 1440 + 900 + 1000 cm2, , = 2 × 3340 cm2, (ii), , Here, K = 12 m,, , = 6680 cm2, , = 8 m and ℎ = 7 m, , Volume of the cuboid = K ×, , ×ℎ, , = 12 × 8 × 7 m3, =672 m3, , Total Surface Area = 2 K + ℎ + ℎK, , = 2 × 12 × 8 + 8 × 7 + 7 × 12 m2, , = 2 × 96 + 56 + 84 m2, = 2 × 236 m2, , (iii), , Here, K = 20 m,, , = 472 m2, , = 16 m and ℎ = 15 m, , Volume of the cuboid = K ×, , ×ℎ, , = 20 × 16 × 15 m3, = 4800 m3, , Page | 15

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Total Surface Area = 2 K + ℎ + ℎK, , = 2 × 20 × 16 + 16 × 15 + 15 × 20 m2, = 2 × 320 + 240 + 300 m2, , = 2 × 860 cm2, (iv), , Here, K = 18 cm,, , = 1720 m2, , = 12 cm and ℎ = 17 cm, , Volume of the cuboid = K ×, , ×ℎ, , = 18 × 12 × 17 cm3, , = 3672 cm3, , Total Surface Area = 2 K + ℎ + ℎK, , = 2 × 18 × 12 + 12 × 17 + 17 × 18 cm2, , = 2 × 216 + 204 + 306 cm2, = 2 × 726 cm2, , 2., , = 1452 cm2, , Find the surface area and volume of a cube of side, (i) 12 cm, , (ii) 19 cm, , (iii) 5 m, , Length if each side of the cube,, , = 12 cm, , Solution:, (i), , Surface area of the cube = 6, , Volume of the cube =, (ii), , ", , Surface area of the cube = 6, (iii), , ", , Surface area of the cube = 6, , (iv), , ", , Surface area of the cube = 6, ", , = 6 × 19 cm2 = 6 × 361 cm2 = 2166 cm2, =5m, , = 6 × 5 m2 = 6 × 25 m2 = 150 m2, , = 5" m3 = 125 m3, , Length if each side of the cube,, Volume of the cube =, , = 19 cm, , = 19" cm3 = 6859 cm3, , Length if each side of the cube,, Volume of the cube =, , = 6 × 12 cm2 = 6 × 144 cm2 = 864 cm2, , = 12" cm3 = 1728 cm3, , Length if each side of the cube,, Volume of the cube =, , (iv) 13 m, , = 13 m, , = 6 × 13 m2 = 6 × 169 m2 = 1014 m2, , = 13" m3 = 2197 m3, , Page | 16

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3., , The volume of a cube is 74088 cm3. Find its edge and surface area., , Solution:, , Let, , (in cm) be the length of each side of the cube., , We have, Volume of the cube = 74088 cm3, ⇒, , ", , ⇒, , = 42", , = 42, , ∴ length of each edge of the cube is 42 cm., And Surface Area = 6, 4., , = 6 × 42 cm2 = 6 × 1764 cm2 = 10584 cm2, , The surface area of a cube is 384 cm2. Find its edge and volume., , Solution:, , Let, , (in cm) be the length of each side of the cube., , We have, Surface area of the cube = 384 cm2, ⇒6, , ⇒, , ⇒, , ⇒, , = 384, , = 64, =8, , =8, , ∴ Length of each edge of the cube is 8 cm., 5., , Volume of the cube =, , ", , = 8" cm3 = 512 cm3, , The length, breadth and height of a room are 5 m, 4 m and 3.5 m respectively. Find the area of, four walls of a room leaving aside 3 windows each of dimensions 2 m by 1 m and 2 doors each of, dimensions 2 m by 1.5 m., , Solution:, , For the room, K = 5 m,, , = 4 m and ℎ = 3.5 m, , Area of the four walls of the room including 3 windows and 2 doors, =2 K+, , ×ℎ, , = 2 5 + 4 × 3.5 m2, , = 7 × 9 m2, , = 63 m2, , Area of the 3 widows = 3 × 2 × 1 m2 = 6 m2, Area of the 2 doors = 2 × 2 × 1.5 m2 = 6 m2, , Now, Area of the four walls of the room leaving aside 3 windows and 2 doors, = 63 − 6 − 6 m2, = 51 m2., , Page | 17

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6., , Find the curve surface area, total surface area and the volume of a right circular cylinder whose, radius Q and height V are given by, , (i) Q = U cm, V = 15 cm, , (ii) Q = W. X cm, V = 12.5 cm, , (iii) Q = RY cm, V = 35 cm, , Solution:, (i), , (iv) Q = U cm, V = 25 cm, , We have N = 7 cm, ℎ = 15 cm, , Curve Surface Area = 2MNℎ, =2×, , 2, , × 7 × 15 cm2, , = 2 × 22 × 15 cm2, , = 660 cm2, , Total Surface Area = 2MN N + ℎ, =2×, , 2, , × 7 7 + 15 cm2, , = 2 × 22 × 22 cm2, , Volume = MN ℎ, =, , 2, , = 968 cm2, , × 7 × 7 × 15 cm3, , = 22 × 7 × 15 cm3, (ii), , = 2310 cm3, , We have N = 3.5 cm, ℎ = 12.5 cm, Curve Surface Area = 2MNℎ, =2×, , 2, , × 3.5 × 12.5 cm2, , = 2 × 22 × 0.5 × 12.5 cm2, = 275 cm2, , Total Surface Area = 2MN N + ℎ, =2×, , 2, , × 3.5 3.5 + 12.5 cm2, , = 2 × 22 × 0.5 × 16 cm2, , Volume = MN ℎ, =, , 2, , = 352 cm2, , × 3.5 × 3.5 × 12.5 cm3, , = 22 × 0.5 × 3.5 × 12.5 cm3, = 481.25 cm3, , Page | 18

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7., , The radius of a roller 1.4 m long, is 45 cm. Find the area it sweeps in 75 revolutions., , Solution: For the roller (cylinder),, N = 45 cm=, , .1, , ,,, , m= 0.45 m and height, ℎ = 1.4 m, , Area swept in 1 revolution = C.S.A. of the roller, = 2MNℎ, =2×, , 2, , × 0.45 × 1.4 m2, , = 2 × 22 × 0.45 × 0.2 m2, = 3.96 m2, , Area swept in 75 revolutions = 75 × 3.96 m2 = 297 m2, 8., , A garden roller of diameter 1 m is 2.1 m long. Find the area it covers in 100 revolutions., , Solution: For the garden roller (cylinder),, Diameter = 1 m, , ∴ Radius, N = m, , And height, ℎ = 2.1 m, , Curved Surface Area of the roller = 2MNℎ = 2 ×, , 2, , × × 2.1 m2 = 22 × 0.3 m2= 6.6 m2, , ∴ Area covered in 100 revolutions = 100 × 6.6 m2= 660 m2, , 9., , A cylindrical metal pipe of thickness 1.4 cm and external diameter 56 cm is 14 m long. Find the, volume of metal used in the construction of the pipe., , Solution: For the cylindrical metal pipe, we have, External radius, Z =, , 1-, , cm= 28 cm, , Internal radius, N = 28 cm – 1.4 cm = 26.6 cm, Height, ℎ = 14 m= 1400 cm, , Volume of metal used in the construction of the pipe = M Z − N ℎ, =, , 2, , × 28 − 26.6, , × 1400 cm3, , = 22 × 784 − 707.56 × 200 cm3, = 22 × 76.44 × 200 cm3, = 336336 cm3, , Page | 20

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10. The volume of a right circular cylinder of height 24 cm is 924 cm3. Find the area of the curve, surface of the cylinder., Solution: For the right circular cylinder,, Height, ℎ = 24 cm, , Volume = 924 cm3, , ⇒ MN ℎ = 924, ⇒, , 2, , × N × 24 = 924, , ⇒N =, , ⇒N =, , ⇒N=, , 2, , 2×= ., , × ., , 2×., ., , =, , 2×2, ., , =, , 2[, [, , ∴ Curved Surface area of the cylinder = 2MNℎ, = 2×, , 2, , 2, , × × 24 cm2, , = 2 × 22 × 12 cm2, , = 528 cm2, , 11. Find the depth of a well of radius 3.5 m if its capacity is equal to that of a rectangular tank of, dimensions 25 m × 11 m × 7 m., , Solution:, , Radius of the well, N = 3.5 m., , Let ℎ (in metre) be the depth of the well., , We know, Volume of the well = Volume of the rectangular tank, ⇒ MN ℎ = 25 × 11 × 7, , ⇒, , 2, , × 3.5 × 3.5 × ℎ = 25 × 11 × 7, , ⇒ℎ=, , 2× 1×, , ×2, , ×".1×".1, , =, , 2× 1×, , × 2× ,,, , ×"1×"1, , ∴ the required depth of the well is 50 m., , = 50, , Page | 21

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12. A well of diameter 3.5 m is dug 16 m deep. The earth taken out is spread evenly to form a, rectangular platform of base 11 m × 7 m. Find the height of the platform., , Solution:, , For the well (cylinder), we have, , Diameter = 3.5 m i.e., N =, , ".1, , m and height, ℎ = 16 m, , Volume of the taken out from the well = MN ℎ, =, , 2, , ×, , ".1, , ×, , ".1, , × 16 m3, , = 22 × 0.5 × 3.5 × 4 m3, For the rectangular platform, we have, Length, K = 11 m and breadth,, , = 154 m3, , =7m, , Area of the base of the platform = K ×, , ∴ the required height of the platform =, , = 11 × 7 m2 = 77 m2, , \]^_`a ]b cda cefag ]_c bh]` cda ia^^, jhae ]b cda kela ]b cda m^ecb]h`, , =, , 1., , 22, , m= 2 m, , 13. Find the volume, curved surface area and total surface area of a cone, given that, (i), , radius of the base = 3.5 m and height = 12 m, , (ii), , radius of the base = 0.7 m and slant height = 2.5 m, , (iii), , radius of the base 21 cm and slant height 35 cm, , (iv), , height = 24 cm and slant height = 25 cm, , (v), , perimeter of base = 88 cm and height = 48 cm, , Solution:, (i), , We have, N = 3.5 m and ℎ = 12 m, , ∴ K = √N + ℎ = √3.5 + 12 m = √12.25 + 144 m = √156.25 m= 12.5 m, , Now, Volume = " MN ℎ, ="×, =, , 2, , 2, , × 3.5 × 12 m3, , × 3.5 × 3.5 × 4 m3, , = 22 × 0.5 × 3.5 × 4 m3, C.S.A., , = 154 m3, = MNK, , =, , 2, , × 3.5 × 12.5 m2, , = 22 × 0.5 × 12.5 m2, = 137.5 m2, , Page | 22

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= MN N + K, , T.S.A., , =, , 2, , × 21 21 + 35 cm2, , = 22 × 3 × 56 cm2, (iv), , = 3696 cm2, , We have, ℎ = 24 cm and K = 25 cm, , ∴ N = √K − ℎ = √25 − 24 cm = √49 cm= 7 cm, , Now, Volume = MN ℎ, ", , = ×, ", =, , 2, , 2, , × 7 × 24 cm3, , × 7 × 7 × 8 cm3, , = 22 × 7 × 8 cm3, = 1232 cm3, = MNK, , C.S.A., , =, , 2, , × 7 × 25 cm2, , = 22 × 25 cm2, = 550 cm2, , = MN N + K, , T.S.A., , =, , 2, , × 7 7 + 25 cm2, , = 22 × 32 cm2, , (v), , = 704 cm2, , We have, ℎ = 48 cm, , And perimeter of the base = 88 cm, ⇒ 2MN = 88 cm, ⇒2×, , ⇒N=, , 2, , × N = 88 cm, , !!×2, ×, , cm, , ⇒ N = 14 cm, , ∴ K = √N + ℎ, , = √14 + 48 cm, , = √196 + 2304 cm, , = √2500 cm= 50 cm, Page | 24

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Now, Volume = " MN ℎ, = ×, =, , ", , 2, , 2, , × 14 × 48 cm3, , × 14 × 14 × 16 cm3, , = 22 × 2 × 14 × 16 cm3, C.S.A., , = 9856 cm3, = MNK, , =, , 2, , × 14 × 50 cm2, , = 22 × 2 × 50 cm2, , T.S.A., , = 2200 cm2, = MN N + K, , =, , 2, , × 14 14 + 50 cm2, , = 22 × 2 × 64 cm2, = 2816 cm2, , 14. The curved surface area of a cone of slant height 50 cm is 2200 cm2. Find the volume of the cone., Solution: For the cone, we have, Slant height, K = 50 cm, , C.S.A. = 2200 cm2, , ⇒ MNK = 2200, ⇒, , 2, , × N × 50 = 2200, , ⇒N=, , 2×, , 1,×, , ,,, , ⇒ N = 14 cm, , Then, ℎ = √K − N = √50 − 14 cm = √2500 − 196 cm= √2304 cm= 48 cm, ∴ Volume of the cone = " MN ℎ = " ×, =, , 2, , 2, , × 14 × 48 cm3, , × 14 × 14 × 16 cm3, , = 22 × 2 × 14 × 16 cm3, = 9856 cm3, , Page | 25

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15. The volume of a cone of height 24 cm is 1232 cm3. Find the slant height and total surface area of, the cone., Solution: For the cone, we have, , Height, ℎ = 24 cm, , Volume = 1232 cm3, , ⇒ " MN ℎ = 1232, ⇒"×, , ⇒, , 2, , 2, , × N × 24 = 1232, , × N × 8 = 1232, " ×2, , ⇒N =, , ×!, , ⇒N=7, , =7×7 =7, , ∴ slant height, K = √N + ℎ, , = √7 + 24 cm, , = √49 + 576 cm, , = √625 cm, = √25 cm, , = 25 cm, , And T.S.A. of the cone = MN N + K, =, , 2, , × 7 × 7 + 25 cm2, , = 22 × 32 cm2, , = 704 cm2, , 16. A conical tent of height 24 m is made of canvas. If the circumference of the base is 44 m, find the, volume of air inside the tent and the cost of canvas used, at Rs 150 per square meter., Solution:, , For the conical tent, we have, ℎ = 24 m, , Circumference of the base = 44 m, , ⇒ 2MN = 44, ⇒2×, , ⇒N=, , 2, , N = 44, , ..×2, ×, , ⇒N=7m, , ∴ K = √N + ℎ = √7 + 24 m = √49 + 576 m= √625 m = √25 m = 25 m, , Page | 26

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Now, Volume of air inside the tent = " MN ℎ, = ×, ", , 2, , × 7 × 7 × 24 m3, , = 22 × 7 × 8 m3, , = 1232 m3, , = MNK =, , And area of the canvas used, , 2, , × 7 × 25 m2 = 22 × 25 m2= 550 m2, , ∴ Cost of the canvas used = Rs 550 × 150 = Rs 82500, , 17. A circus tent is cylindrical upto a height of 15 m and conical above it. If the radius of the base is, 28 m and the slant height of the conical part is 35 m find the total area of canvas used in making, the tent and volume of air inside the tent., Solution:, , Radius of the cylindrical part = Radius of the conical part = N = 28 m, , Height of the cylindrical part, n = 15 m, For the conical part,, , Slant height, K = 35 m, , ∴ Height, ℎ = √K − N = √35 − 28 m, =, , 35 − 28 × 35 + 28 m, , = √7 × 63 m = √3 × 7 m, = 21 m, , Now, Area of the canvas used = C.S.A. of the cylindrical part + C.S.A. of the conical part, = 2MNn + MNK, = MN 2n + K, =, , 2, , × 28 2 × 15 + 35 m2, , = 22 × 4 × 65 m2, = 5720 m2, , Volume of air inside the tent = Volume of the cylindrical part + Volume of the conical part, = MN n + " MN ℎ, , = MN :n + ℎ;, , =, , ", , 2, , × 28 × 28 :15 + " × 21; m3, , = 22 × 4 × 28 15 + 7 m3, , = 22 × 4 × 28 × 22 m3, = 54208 m3, , Page | 27

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18. The radii of the top and bottom of a bucket are 21 cm and 14 cm. Determine the capacity and, curved surface area of the bucket if its height is 24 cm., Solution:, , We have, N = 21 cm, N = 14 cm and ℎ = 24 cm., ∴K =, , =, , N −N, , 21 − 14, , +ℎ, , + 24 cm, , = √49 + 576 cm, , = √625 cm, = √25 cm, , = 25 cm, , Now, Capacity of the bucket, , = "M N + N .N + N, , ="×, =, , =, , 2, 2, , 2, , ℎ, , 21 + 21 × 14 + 14, , × 24 cm3, , 441 + 294 + 196 × 8 cm3, , × 931 × 8 cm3, , = 22 × 133 × 8 cm3, And, C.S.A. of the bucket, , = 23408 cm3, , =M N +N K, =, , =, , 2, 2, , × 21 + 14 × 25 cm2, × 35 × 25 cm2, , = 22 × 5 × 25 cm2, , = 2750 cm2, , 19. Rain water from a horizontal roof 11 m × 6 m drains into a vessel in the form of a frustum of a, , cone. The height of the vessel is 1.4 m and the radii of its top and bottom are 0.9 m and 0.6 m. If, the vessel is just full, find the rainfall in centimeters., , Solution:, , For the vessel (frustum),, Height, ℎ = 1.4 m = 1.4 × 100 cm = 140 cm, , Radius of the top, N = 0.9 m = 0.9 × 100 cm = 90 cm, , Radius of the bottom, N = 0.6 m = 0.6 × 100 cm = 60 cm, Page | 28

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Volume of water in the vessel = " M N + N . N + N, = ×, ", , 2, , ℎ, , 90 + 90 × 60 + 60, , × 140 cm3, , = " × 22 8100 + 5400 + 3600 × 20 cm3, = × 22 × 17100 × 20 cm3, ", , = 22 × 5700 × 20 cm3, = 2508000 cm3, For the horizontal roof, we have, , Length, K = 11 m = 11 × 100 cm = 1100 cm, = 6 m = 6 × 100 cm = 600 cm, , Breadth,, Area of the roof = K ×, , ∴ the required rainfall =, =, , =, , = 1100 × 600 cm2 = 660000 cm2, \]^_`a ]b iecah og cda palla^, jhae ]b cda h]]b, , 1,!,,,, , --,,,,, 1,!, , --,, , cm, , cm = 3.8 cm, , 20. A solid cone of height 16.8 cm and base radius 4.2 cm is melted and recast into a solid sphere., Determine the radius of the sphere., Solution:, , For the cone, we have, , Radius, Z = 4.2 cm and height, ℎ = 16.8 cm, , Let N be the radius of the solid sphere., , We know, Volume of the sphere = Volume of the cone, ., , ⇒ MN " = MZ ℎ, ", ", , ⇒ 4N " = Z ℎ, , ⇒ 4N " = 4.2 × 4.2 × 16.8, , ⇒ N " = 4.2 × 4.2 × 4.2 = 4.2", , ⇒ N = 4.2, , ∴ radius of the sphere is 4.2 cm., , Page | 29

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21. A solid sphere is melted and recast into a cone whose (base) radius is the same as that of the, sphere. Show that the slant height of the cone bears a ratio √RU: Y to its height., We know,, , Solution:, , Volume of the cone = Volume of the sphere, ., , ⇒ MN ℎ = MN ", ", , ", , ⇒ ℎ = 4N, ⇒N=, , ., , And Slant height, K = √N + ℎ, r, ⇒ K = 9: ; + ℎ, ., r, , -, , ⇒ K = 9:, , 2, , Volume of the cone = Volume of the sphere, ., , ⇒ " MN ℎ = " MN ", , r, , ⇒K=9, , Alternatively,, , ⇒ ℎ = 4N, , And K = √N + ℎ =, = √N + 16N, , N + 4N, , = √17N, , +ℎ, , + 1; ℎ, -, , = √17N, , Now, K: ℎ =, , √ 2t, .t, , =, , √ 2, ., , = √17: 4, , ⇒ K = 9 -ℎ, ⇒, , s, , r, s, , =9, , ⇒r=, , 2, , .[, , √ 2, ., , ⇒ K: ℎ = √17: 4, , 22. A solid is in the form of a cone surmounted on a hemisphere of the same radius. If the height of, the cone is 24 cm and radius of the hemisphere is 7 cm, find the volume and surface area of the, solid., Solution:, , We have, N = 7 cm, , For the cone,, , ℎ = 24 cm, , and K = √N + ℎ = √7 + 24 cm = √49 + 576 cm= √625 cm= √25 = cm 25 cm, , Volume of the solid = volume of the conical part + volume of the hemispherical part, = " MN ℎ + " MN " = " MN ℎ + 2N, , ="×, =, , 1!1, ", , 2, , × 7 × 7 24 + 2 × 7 cm3= " × 22 × 7 × 38 cm3, , cm3= 1950 " cm3, , Page | 30

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Surface area of the solid = C.S.A. of the conical part + C.S.A of the hemispherical part, = MNK + 2MN, , = MN K + 2N, =, , 2, , × 7 25 + 2 × 7 cm2, , = 22 × 39 cm2, = 858 cm2, , 23. How many lead shots, each of radius 6 cm can be made out of a rectangular slab of dimensions, YY uv × 36 uv × SY uv ?, , Solution:, , Radius of each lead shot (sphere), N = 6 cm, , Number of lead shots =, , p]^_`a ]b cda hawcegx_^eh l^ek, s. .r, , =z, , =, , =, , {, , p]^_`a ]b e ^aey wd]c, , |t {, , .. × "- × ., , z [[, × ×-×-×{ }, , .. × "- × .×"×2, .×, , = 42, , ×-×-×-, , 24. Find the surface area and volume of a solid in the form of a right circular cylinder with, hemispherical ends if the whole length is 22 cm and radius of the cylinder is 3 cm. (Take T = 3.14), , Solution:, , We have, N = 3 cm, , Height of the cylindrical part, ℎ = 22 cm − 2× 3 cm = 16 cm, = 28.26 × 20 cm2, , Volume of the solid = Volumes of the two hemi-spherical ends+ Volume of the cylinder, = 2 × " MN " + MN ℎ, ., , = MN :" N + ℎ;, , ., , = 3.14 × 3 × 3 :" × 3 + 16; cm3, , = 28.26 × 4 + 16 cm3, = 28.26 × 20 cm3, = 565.2 cm3, , Page | 31

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Surface Area of the solid = C.S.A. of the two hemispherical ends + C.S.A. of the cylinder, = 2 × 2MN + 2MNℎ, = 2MN 2N + ℎ, , = 2 × 3.14 × 3 2 × 3 + 16 cm2, = 18.84 × 22 cm2, = 414.48 cm2, , 25. A metallic right circular cylinder of radius 6 cm and height 14 cm is melted and recast into three, spheres. If the radii of two of these spheres are 2 cm and 3 cm, find the radius of the third., Solution: For the right circular cylinder,, , N = 6 cm and ℎ = 14 cm, , Let N , N and N" respectively be the radii of the first, second and the third spheres., , Then, N = 2 cm and N = 3 cm, , Sum of the volumes of the three spheres = volume of the right circular cylinder, , ., , ⇒ "M N " + N, ., , ", , + N" " = MN ℎ, , ⇒ " 2" + 3" + N" " = 6 × 6 × 14, , ", , ⇒ 8 + 27 + N" " = 6 × 6 × 14 × ., ⇒ 35 + N" " = 378, , ⇒ N" " = 378 − 35 = 343, ⇒ N" " = 7", , ⇒ N" = 7, , ∴ radius of the third sphere is 7 cm., , Page | 32