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CHAPTER 12, MENSURATION, , Perimeter and Area of a circle, • Perimeter of a circle (Circumference) =, • Area of a circle =, , SOLUTIONS, EXERCISE 12.1, , 1. Find the area of a circle whose radius is 14cm., Solution:, We have,, = 14, , ∴ Area of a circle =, =, , × 14 × 14 cm, , = 616 cm, , 2. Find the area of a quadrant of a circle whose circumference is 44 cm., Solution:, We have,, Circumference = 44 cm, , ⇒2, , ⇒2×, ⇒, ∴, , =, , = 44, , =7, , × = 44, ×, , ×, , Area of a quadrant =, , = ×, , =, , cm, , × 7 × 7cm, , = 38.5cm, Page | 1

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3., , The circumference of a circle exceeds the diameter by the 21cm. Find the area of the circle., , Solution:, We have,, , Circumference − Diameter = 21 cm, ⇒2, , = 2 + 21, , ⇒2, , − 2 = 21, , ⇒2, , ⇒2 ×, , − 1 = 21, ", , ⇒2 ×, , ⇒, , ⇒, , ∴, , =, , =, , ×, , − 1! = 21, , = 21 = 21, , × ", , #, , $, , = 4.9cm, , Now, Area of the circle =, , =, , 22, × 4.9 × 4.9cm, 7, , = 75.46cm, , 4., , The area of a circle is 1386 &' , find its circumference., , Solution: We have,, ⇒, , ⇒, , ⇒, , ⇒, , ∴, , Area of the circle = 1386 cm, = 1386 cm, , 22, ×, 7, =, , = 1386 cm, , 1386 × 7, cm, 22, , = 21 cm, , = 21cm, , Then circumference = 2, , = 2×, , 22, × 21, 7, , = 132cm, , Page | 2

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5., , The sum of the radii of two circles is 35 cm and the difference of the areas is 770 cm2. Find, their circumference., , Solution: Let and 35 −, We have,, , be the radii of the larger and the smaller circles respectively., , Difference of the area = 770cm, −, , ⇒, , ⇒, , ⇒, , ⇒, , 35 −, , −, , = 770, , 1225 − 70 +, , − 1225 + 70 +, , = 770, , × 7 10 − 175 = 770, , = 770, , $, , ⇒ 10 − 175 =, , ⇒ 10 − 175 = 35, , ⇒ 10 = 35 + 175, ⇒, ∴, , =, , $, , $, , = 21cm, , ∴ The radii of the two circles are 21 cm and (35 − 21) cm i.e. 14 cm, Then, the circumference of the first (larger) circle = 2 ×, , The circumference of the second (smaller) circle = 2 ×, , × 21cm = 132cm, , × 14 cm = 88cm, , 6., , The diameter of each wheel of a vehicle is 42 cm. Each wheel makes 20 revolutions in 12, seconds. Find the speed of the vehicle in metre per minute., Solution: We have,, Diameter = 42cm, Radius =, , × 42 = 21cm, , Distance = 20 × circumference, = 20 × 2, , = 20 × 2 ×, , =, , =, , $× ×, , ×, , ", , $$, , ×/, , And Time = 12 sec, ∴ Speed =, , 56789:;<, 86=<, , ×/, , m, , × 21 cm, , m, , = 1$ min = " min, =, , > ×>> ×?, @, A, @, , =, , ×, , ", , ×/, , × 5 = 132 m/min, Page | 3

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7., , Find the radius of a circle whose area is equal to the difference of the areas of two circles of radii 6, cm and 10 cm respectively., Solution: We have,, Area of the third circle = Difference of areas of the two given circles, ⇒, , =, , ⇒, , =, , ⇒, , ⇒, , ⇒, ∴, , . 10 −, , 10 − 6, , ×6, , = 100 − 36, = 64, , = √64, , =8, , ∴ The radius of the third circle is 8 cm., 8., The difference between the perimeter and the radius of a circle is 37 cm. Find the area of the circle., Solution: We have,, Perimeter of the circle – radius = 37 cm, , ⇒2, ⇒, ⇒, , − = 37cm, , 2 − 1 = 37cm, 2×, , ⇒ ×, ⇒, , /, , − 1 = 37cm, , = 37cm, , = 37 × / cm, , ∴ The area of the circle =, =, , × 7 × 7cm, , = 154cm, , 9., , In Fig. 12.4, OABC is a rhombus, three of whose vertices lie on a circle with centre O. If the area of, the rhombus is DE√F&' , find the perimeter and area of the circle [take = F. DG]., , Solution:, A, , B, , O, , C, Page | 4

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We have,, , Area of the rhombus HIJK = 18√13, , ⇒ 2. area of the equilateral ∆HIJ = 18√13, , ⇒2×, ⇒, , √/, , ×, , ⇒, , ., , = 18√3, , [⸪Area of an equilateral triangle =, , T√/ ×, , √/, , .S ], , √/, , = 36 = 6, , ∴, , = 6 cm, , Then, perimeter of the circle = 2, , = 2 × 3.14 × 6 cm, , And, , = 37.68 cm, , =, , Area, , = 3.14 × 6 × 6cm, , = 113.04cm, , If the areas of two concentric circles are 154 cm2 and 616 cm2 respectively, then find the width and, area of the circular path between them., Solution:, 10., , 616 cm2, 154 cm2, , r, R, , We have,, , Area of the larger circle = 616cm, , ⇒ U = 616cm, ⇒, , × U = 616cm, , ⇒U =, , 1 1×, , cm, , ⇒ U = 196 cm, , Page | 5

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⇒ U = 14 cm, ⇒ U = 14cm, , Area of the smaller circle = 154cm, , ⇒, , = 154 cm, , ⇒, , ×, , ⇒, , = 49 cm = 7 cm, , ⇒, , ⇒, , =, , = 154 cm, , " ×, , cm, , = 7cm, , ∴ Width of the path between the circles = U −, , = 14cm − 7cm, = 7 cm, , And area between the circles = Area of the larger circle − Area of the smaller circle, = 616cm − 154cm, = 462cm, , 11., , Find the area of the largest circle that can be drawn inside a square of side 21 cm., , Solution:, 21 cm, , 21 cm, , 21 cm, , 21 cm, , Diameter of the largest circle = length of a side of the square =21 cm., ∴, , =, , cm = 10.5 cm, , ∴ Area of the largest circle =, =, , ×, , ×, , = 346.5cm, , cm, , Page | 6

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12., , The cost of leveling a path of uniform width which surrounds a circular park of diameter 320m at, the rate of Rs. 7.00 per sq. metre is Rs. 35750. Find the width of the path., Solution:, path, , r, park, , We have,, , R, , Diameter of the circular park = 320m, , ∴ Inner radius of the path =Radius of the circular park,, , Let R be the outer radius of the path., Also, the cost of levelling the path = Rs. 35750, ∴ Area of the path =, ⇒, , U −, , =, , U − 160, , ⇒, , /" "$, , /" "$, , ⇒ U − 25600 =, , =, , ×, , / $, , m = 160m, , m, , /" "$, , /" /" ×, , =, , = 1625, , ⇒ U = 1625 + 25600, ⇒ U = 27225, ⇒ U = √27225 = 165, ∴ Width of the path = U − = 165m − 160m = 5m, 13., , The cost of the fencing a circular field at the rate Rs. 18.00 per metre is Rs. 3960. The field is to be, leveled at the rate of Rs. 6.00 per square metre. Find the cost of leveling., Solution: We have,, Circumference of the field =, ⇒2, , ⇒2×, , ⇒, , ⇒, , =, , = 220m, ×, , $×, , ×, , = 35m, , /#1$, T, , m, , = 220m, , $, , m, , Area of the circular field =, , =, , × 35 × 35m = 3850m, , ∴ Cost of the leveling the field = UY. 3850 × 6 = UY. 23100, , Page | 7

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14., , Grass is planted on a circular strip of width 7m. The radius of the inner edge of the grass strip is, 21m. If the cost of planting is Rs. 4.00 per square metre, then find the total cost of planting grass., , Solution:, grass, , 7m, , r, R, , We have,, , Width of the circular strip = 7m, , = 21m, , And Radius of the inner edge of circular strip,, , ∴ Radius of the outer edge of circular strip, U = 21 + 7 m = 28m, , Then, Area of the circular strip =, =, , =, , =, , U −, , 28 − 21 m, , m, , 28 + 21 28 − 21 m, , × 49 × 7m, , = 1078 m, , ∴ Cost of the planting grass = UY. 1078 × 4 = UY. 4312, , *******, , Page | 8

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Recall:, Chord:, , The line segment joining any two points of a circle is called a chord of the circle., , Arc:, , A part of a circle between two distinct points on it, is called an arc of the circle., , Sector:, , The portion of the circular region enclosed by two radii and the corresponding arc of a circle, is called a sector of the circle., , Sectorial Angle: Angle formed by the two radii of a sector at the centre of the circle is called sectorial, , Segment:, , angle. It is denoted by Z., , The portion of the circular region enclosed by a chord and the corresponding arc of a circle is, called a segment of the circle., , Area of Sectors and Segments Of a Circle, •, , [, , Length of an arc (s) = DE\, , [, , •, , Area of a (minor) sector = F]\, , •, , Area of major sector = Area of the circle – Area of the minor sector =, , •, •, , =, , D, , .^, , [, , Area of a (minor) segment =, , − ^ab[, , DE\, , F]\`[, F]\, , Area of major segment = Area the circle – Area of the minor segment, =, , c, , F]\`[, F]\, , +, , ^ab[, , d, , SOLUTIONS, EXERCISE 12.2, 1. The perimeter of a certain sector of a circle of radius 6.5cm is 21 cm. Find the area of the sector., Solution: We have,, = 6.5cm, And perimeter of the sector =21cm, ⇒ 2 + Y = 21, ⇒ 2 × 6.5 + Y = 21, ⇒ 13 + Y = 21, ⇒ Y = 21 − 13, ∴ Y = 8cm, Then, Area of the sector = . . Y, = × 6.5 × 8, = 26cm, , Page | 9

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], , 2. The radius of a circle is 6cm and the area of a sector of this circle is DE &' . What is the angle of, e, the sector?, Solution: We have,, = 6 cm, , 1, , And area of the sector = 18 cm, ⇒, , f, , =, , /1$, f, , ⇒ /1$ ×, , ⇒Z=, , /, , ×6×6=, , / ×/1$ ×, ×, , ⇒ Z = 60, , /, , ×1 ×1, , ∴ The angle of the sector is 60$ ., , 3. The circumference of a circle is 176cm. Find the area of a sector whose sectorial angle Gg\ ., Solution: We have,, Circumference = 176 cm, , ⇒2, , ⇒2×, ⇒, ∴, , =, , = 176 cm, , × = 176 cm, , 1×, , ×, , = 28 cm, , And Z = 45$, , cm, , f, , Now, Area of the sector = /1$ ×, ", , = /1$ ×, = 308, , × 28 × 28, , 4. The area of the sector of a circle of radius 21cm is G] &' . Find the length of the arc of the sector., Solution: We have,, = 21cm, And area of sector = 462 cm, ⇒ . . Y = 462, , 1, × 21 × Y = 462, 2, 462 × 2, ⇒Y=, 21, ⇒ Y = 44, ∴ The required length of the arc of the sector is 44 cm., ⇒, , Page | 10

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5. Find the sectorial angle and area of the sector of a circle if the arc length of the corresponding sector, is 8.8cm and the radius of the circle is 5.6cm., Solution: We have,, , And, , = 5.6 cm, , ⇒, , Y = 8.8 cm, f, , T$, , ., , = 8.8, , Z, 22, ×, × 5.6 = 8.8, 180 7, 88 × 180 × 7, ⇒Z=, 22 × 56, ⇒, , ⇒ Z = 90, , ∴ The sectorial angle is 90$, And Area of the sector =, , Y, , = × 5.6 × 8.8 cm, = 24.64cm, , 6. The minute hand of a wall clock is 4.2cm long. Find the area swept by it in 20 minutes., Solution: We have,, = 4.2, , Z = angle formed by the minute hand in 20 minutes = 120$, f, , Area swept by the minute hand in 20 minutes = /1$ ×, , × 4.2 × 4.2 cm, , = 18.48 cm, , 7. Find the area of a sector of a circle of a radius 14cm when the sectorial angle is, (i) F\\, (ii), Gg\ (iii) ]\\ (iv), j\\ (v), D \\, Solution:, (i) We have,, , = 14cm and Z = 30$, f, , ∴ Area of sector = /1$, /$, , = /1$ ×, =, , ", , /, , cm, , × 14 × 14 cm, , = 51.33cm, Page | 11

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(ii), , We have,, , = 14cm and Z = 45$, , Area of sector =, =, , (iii), , We have,, , f, , /1$, ", , /1$, , ×, , = 77cm, , = 14cm and Z = 60$, f, , Area of sector = /1$, #$, , = /1$ ×, =, , (iv), , We have,, , /$T, /, , = 14cm and Z = 90$, f, , Area of sector = /1$, , We have,, , cm, , × 14 × 14 cm, , = 102.67cm, , =, , (v), , × 14 × 14 cm, , #$, , /1$, , ×, , = 154 cm, , × 14 × 14 cm, , = 14cm and Z = 120$, f, , Area of sector = /1$, , 120 22, ×, × 14 × 14 cm, 360 7, 616, =, cm, 3, =, , = 205.33 cm, , 8. Two concentric circles have radii 6cm and 4cm. A sector of ]\\ is drawn in each circle. Find the, difference between the areas of the two sectors, k^l = F. DG ., Solution:, , 6cm, , 60, , $, , 4cm, , Page | 12

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We have,, Radius of the inner circle,, , = 4 cm, , Radius of the outer circle, U = 6 cm, , Difference between the area of the two sectors =, , =, , =, , =, , f, , /1$, 1$, , U −, , /1$, , × 3.14 6 − 4, , ." × $, , / ., /, , /, , = 10.47, , 9. In a circle of radius 6.3cm, an arc AB subtends and angle of j\\ at the centre O of the circle. Find, (i), the length of the arc AB, (ii), the area of the sector AOB, (iii) the perimeter of the sector AOB, (iv), the area of the minor segment and the major segment formed by the chord AB., Solution:, , O, , 6.3 cm, A, i), , We have,, , 90$ 6.3 cm, B, , = 6.3 cm, , Z =∠ IHJ = 90$, , ∴ Length of the arc IJ =, ii), , f, , T$, , f, , Area of the sector IHJ = /1$, #$, , = /1$ ×, =, , × 6.3 × 6.3cm, , ×$.#×1./, , cm, , = 31.185 cm, , = 31.19 cm, , Page | 13

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iii), , Perimeter of the sector IHJ = 2 + Y, , = 2 × 6.3cm + 9.9cm, , = 12.6 + 9.9 cm, = 22.5 cm, iv), , m>, , =, , Area of the minor segment, , fn, , T$, , =, , 1./×1./, , =, , 1./×1./, , 1./×1./, , =, , − YopZ, #$×, , c, , T$×, , − Yop90$ ! cm, , − 1d cm, , × cm, , = 11.34cm, , = Area of circle − area of minor segment, , And area of major segment, , =, , =, , − 11.34 cm, , 22, × 6.3 × 6.3cm − 11.34 cm, 7, , = 124.74 cm − 11.34 cm, = 113.4 cm, , 10. A chord of a circle of radius 12cm subtends a right angle at the centre. Find the area of the, corresponding (i) major sector and (ii) minor segment k^l = F. DG ., Solution:, We have,, = 12, , Z = 90$, , i), , /1$`f, , Area of the major segment =, =, , /1$, , /1$`#$, /1$, $, , × 3.14 × 12 × 12 cm, , = /1$ × 3.14 × 12 × 12 cm, = 339.12 cm, , ii), , Area of minor segment =, =, , ×, , =, , ×, , =, , ×, , m>, , [, , c, , fn, , T$, , #$, , T$, , − YopZ, , × 3.14 − sin 90d cm, , [1.57 − 1] cm, × 0.57 cm, , = 41.04 cm, , Page | 14

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11. A chord of a circle of radius 18cm subtends and angle of D \\ at the centre. Find the area of, corresponding (i) minor sector and (ii) major segment sk^l = F. DG tbu √F = D. eFv., Solution:, We have,, = 18, , Z = 120$, , i), , Area of minor sector =, , f, , /1$, , 120$, × 3.14 × 18 × 18, =, 360, = 339.12, , ii), , Area of major segment =, , [, , /1$`fn, /1$, , = 18 × 18 z, , +, , wxyf, , ], , 360 − 120, Yop120, × 3.14 +, {, 360, 2, , 260, √3, × 3.14 +, ], 2×2, 360, 6.28 1.73, = 18 × 18[, +, ], 3, 4, 25.12 + 5.19, = 18 × 18[, ], 12, 25.12 + 5.19, =9×3×, 12, = 18 × 18[, , = 27 × 30.31, = 818.37, , 12. A chord of a circle of radius 6cm subtends an angle of ]\\ at the centre. Find the area of the, corresponding (i) minor segment and (ii) major segment.s|t}l = F. DG tbu √F = D. eFv., Solution:, We have,, =6, , i), , and Z = 60$, , Area of minor segment =, , m > fn, , [, , T$, , − YopZ], , 6 × 6 60, √3, [, × 3.14 − ], 2 180, 2, 6 × 6 3.14 1.73, =, [, −, ], 2, 3, 2, 6 × 6 6.28 − 5.19, =, z, {, 2, 6, 6 × 6 1.09, ×, =, 6, 2, =, , = 3.27, , Page | 15

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ii), , Area of major segment = Area of circle − area of minor segment, =, , − 3.27, , = 3.14 × 6, , = 113.04, , − 3.27, , − 3.27, , = 109.77, , 13. The radius of a circle is 5cm. A chord of length g√ cm is drawn in the circle. Find the area of the major segment, |t}l = F. DG, , Solution:, , O, 5 cm, , 5 cm, , g√, , A, , D, , cm, , B, , IJ = 5√2 cm is a chord of a circle with centre H and radius 5 cm. H~⊥IJ is drawn., Then, , = HI = HJ = 5cm, I~ =, , ∴, , ⇒, , ⇒, , ⇒, , €•, , ‚€, "√, , = sin ∠IH~, , ×", √, √ .√, , 1, , "√, , cm, , = sin∠IH~, , = sin∠IH~, , = sin∠IH~, √2, ⇒ ∠ IH~ = 45$, ∴ ∠ IHJ = 2 × 45$ = 90$, , ∴ Area of the major segment =, , c, , /1$`f, /1$, , +, , wxyf, , d, , sin 90$, 360 − 90, = 5 × 5ƒ, × 3.14 +, „ cm, 360, 2, 270, 1, = 25 z, × 3.14 + { cm, 360, 2, 4.71 + 1, = 25 ×, cm, 2, 25 × 5.71, =, cm, 2, 142.75, =, cm, 2, = 71.375 cm, = 71.38 cm, , Page | 16

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14. A chord of a circle of diameter 12cm makes an angle of F\\ at the centre. Find the area of the minor, segment of the circle., Solution:, We have,, Diameter = 12 cm, radius = 6 cm, θ = 30$, Area of minor segment =, , m>, , c, , f, , − sin Zd, , T$, , 6 × 6 30 22, z, ×, − sin30$ { cm, 2 180 7, 11 1, = 3 × 6 z − { cm, 21 2, 11 1, = 3 × 6 z − { cm, 21 2, 22 − 21, = 3×6z, { cm, 42, 1, = 3 × 6 × cm, 42, 3, = cm, 7, =, , = 0.428cm, = 0.43 cm, , 15. A chord of a circle subtends an angle of ]\\ at the centre. If the length of the chord is 12cm, find the, area of the two segments into which the chords divides the circle s|t}l = F. DG tbu √F D. eFv., Solution:, , O, , 12cm, A, , 60, , $, , 60$, , 12cm, , 12cm, , 60$, , B, , We have,, , AB = 12cm is a chord of circle with centre O such at ∠AOB = 60$, , We know, OA = OB [being radii of the same circle], ∴ ∠A = ∠B, , ⇒ ∠A = ∠B =, , T$‰ `1$‰, , = 60$, , Page | 17

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Hence, AOB is an equilateral triangle., ∴ OA = OB = AB = 12cm, , ∴ radius r = 12 cm and θ = 60$, , Now, Area of the minor segment =, , m>, , c, , ×, , =, , f, , T$, , c, , − YopZd, , 1$, , T$, , = 6 × 12 c, , = 6 × 12 c, , = 6 × 12 c, , /., , × 3.14 − Yop60$ d cm, −, , /, , /., , −, , /, , d cm, , √/, , . /, , 1. T`". #, , = 6 × 12 ×, , 1, , .$#, 1, , = 12 × 1.09cm, , d cm, , d cm, , cm, , = 13.08cm, , And, Area of the major segment = Area of the circle – Area of the minor segment, − 13.08cm, , =, , = 3.14 × 12 × 12 cm − 13.08 cm, , = 452.16 cm − 13.08 cm, = 439.08 cm, , 16. A grass field is in the form of equilateral triangle of sides 48m. A cow is tethered at a vertex of the, field. If the cow can reach upto 21m from the vertex, find, (i), the area of the part of the field in which the cow can graze., (ii), how much area remains ungrazed., (iii) the increase in grazing area if the length of the rope be increased to 35m k^l = F. DG, Solution:, C, G, E, , A, , 60$, 21cm D, 35cm, , 48cm, , F, , B, , Page | 18

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Let ABC be the grass field in the form of equilateral triangle. Let the cow be tethered at the vertex A., i), , I~Š be the sectorial part in which the cow can graze., For the sector I~Š,, , θ = 60$ [being an angle of an equilateral ∆], = AE = AD = 21 m, , Area of the field in which the cow can graze = area of sector ADE, , Z, 360, 60, × 3.14 × 21 × 21m, =, 360, , =, , ii), , Area of the field which remains ungrazed, , = 230.79m, , = area of the equilateral ∆ ABC − area of the sector ADE, , =, , =, , √/, , S − 230.79m , where a is the length of each side of the ∆ ABC, , √3, × 48 × 48m − 230.79m, 4, , = 1.73 × 576m − 230.79m, = 996.48m − 230.79m, , iii), , = 765.69m, , Let the sector AFG be the area of the part of the field in which the cow can graze when the, rope is increased to 35m., , ∴Increased in grazing area = area of the sector AFG – area of the sector ADE, 1$, , 1$, , = /1$ × 3.14 × 35 × 35m − /1$ × 3.14 × 21 × 21m, 1$, , = /1$ × 3.14 35 − 21 m, = 1 × 3.14 × 56 × 14m, , = 1 × 3.14 × 56 × 14m, = 410.29 m, , Page | 19

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SOLUTIONS, EXERCISE 12.3, , 1. A circle circumscribes a rectangle of sides 8 cm and 6 cm. Find the area enclosed between the circle, and the rectangle., Solution:, , A, , D, 6cm, , B, , 8cm, , C, , Let ABCD be the rectangle whose length is 8cm and breath is 6cm which is circumscribed by the, , circle., , We have, J~ = √JK + K~, , = √8 + 6 cm, , = √100 cm, = 10 cm, , ∴ radius of the circle,, , =, , $, , cm = 5 cm, , Then, area enclosed between the circle and the rectangle, = Area of circle − area of rectanngle, =, , =, , =, , =, , =, , −•×Ž, , × 5 × 5cm − 8 × 6cm, , ""$, , cm − 48cm, , ""$`//1, , cm, , cm, , = 30.57cm, Page | 20

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2. An equilateral triangle is inscribed in a circle of radius 14 cm. Find the area between the circle and, triangle (use √F = D. eF ., Solution:, A, , O, 14cm, B, , C, , Let ABC be the equilateral triangle and O be the centre of the circle., We have,, , IJ = JK = KI, , ∴ ∠IHJ = ∠JHK = ∠KHI =, , and, , = 14 cm, , /1$‰, /, , = 120$, , The problem consists of three equal segments with the equal chords AB, BC, CA., For each segment,, , Z = 120$ and, , = 14 cm, , ∴ Area of the segment with the chord JK =, , =, , m>, , c, , fn, , ×, , T$, , − •opZd, $, , c, , T$, , ×, , = 7 × 14 c ×, = 7 × 14 c, , = 7 × 14 c, , = 7 × 14 c, =, , =, , ×" .1, /, , /1 .1#, /, , ∴ Area between the circle and the triangle = 3 ×, , /, , −, −, , −, , cm, , /1 .1#, , d cm, , √/, , d cm, , √/, , . /, , TT`/1.//, , cm, , /, , − Yop120$ d cm, , d cm, , d cm, , cm, , = 361.69 cm, , Page | 21

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3. A circle is inscribed in an equilateral triangles of sides 18cm. Find the area between the triangle and, the circle (use √F = D. eF ., Solution:, A circle with centre • is inscribed in an equilateral ∆ IJK of sides18 . JK touches the, circle at ‘., We have,, , , ∠HJ‘ = ∠IJK = × 60$ = 30$ Sp’ ∠H‘J = 90$, , J‘ = 9, , Then,, , ‚“, ”“, , ⇒, , = •Sp30$, , ‚“, #, , =, , A, , √/, , ⇒ H‘ =, , ⇒ H‘ =, , ⇒ H‘ =, , #, , √/, , #√/, , O, , √/.√/, #√/, /, , B, , ∴ H‘ = 3√3, , ∴, , P, , C, , = 3√3, , Then area between the triangle and the circle., , = I –S •— –˜™o•S•– S• ∆IJK − S –S •— •ℎ– o •–, , =, , =, , √/, √/, , S −, , × 18 × 18, , = 140.13, , −, , − 84.86, , × 3√3 × 3√3 = 1.73 × 9 × 9, = 55.27, , −, , ×9×3, , 4. ABC is an equilateral triangle of side 12cm in which three circular arcs, each of radius 6cm, are, drawn with centre A,B,C. Find the area of the region enclosed by the circular arcs (|t}l^√F =, D. eF ., A, Solution:, We have,, Area of the region enclosed by the arcs, , = Area of the equilateral triangle − area of three sectors, , =, , =, , √/, √/, , f, , S − 3 × /1$, × 12 × 12, , = 36 × 1.73, , = 62.28, = 5.71, , −, , 1$, , − 3 × /1$ ×, , /#1, , ×6×6, , C, , B, 12cm, , − 56.57, , Page | 22

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5. Four cows are tethered at the four corners of a square field of side 56m so that consecutive cows can, just reach each other. What area of the field will remain ungrazed?, , Solution: Each cow can reach a sector of radius × 56 cm i.e. 28 cm and sectorial angle 90$ ., Area of the field that remains ungrazed, , = Area of the square šield − area of the four sectors, =S −4×, , Z, 360, , = 3136, , − 2464, , = 56 × 56, = 672, , −4×, , 90 22, ×, × 28 × 28, 360 7, , 6. ABC is an equilateral triangle inscribed in a circle of radius 7 cm. Find the area of the minor segment, of the chord BC |t}l^√F = D. eF ., , Solution:, , We have,, , IJ = JK = KI, , ∴ Z = ∠JHK =, =7, , and, , /1$‰, /, , = 120$, , Area of the minor segment of the chord BC, Z, − YopZ{, 2 180, 7 × 7 120 22, =, z, ×, − Yop120$ {, 2 180 7, , =, , =, , =, , z, , 7 × 7 44 √3, ƒ − „, 2 21, 2, , 7 × 7 88 − 21√3, ƒ, „, 2, 42, , 7 × 7 88 − 21 × 1.73, z, {, 2, 42, 7 × 7 88 − 36.33, =, z, {, 2, 42, 7 × 7 51.67, =, ×, 2, 42, =, , = 30.14, , Page | 23

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7. From an equilateral triangle of sides 14cm, a sector of radius 7cm with centre at a vertex and, enclosed by two sides, is cut off. Find the area of the remaining portion |t}l^√F = D. eF ., , Solution:, , For the sector,, , =7, , and Z = 60$ (being an angle of an equilateral triangle), , ∴ Area of the remaining portion, , = Area of the equilateral ∆ − area of sector, =, , =, , Z, √3, S −, 4, 360, , √3, × 14 × 14, 4, , = 1.73 × 49, = 84.77, , = 59.10, , −, , −, , 77, 3, , 60 22, ×, ×7×7, 360 7, , − 25.67, , 8. ABC is an isosceles right angled triangle, right angled at A where AB =AC =8cm. A sector of radius, 3.5cm with centre A and enclosed by AB, AC is cut off from the triangle. Find the area of the, remaining portion., B, , Solution:, , For the sector,, = 3.5, , Z = 90$, , ∴ Area of the remaining portion, , = I –S •— ∆IJK − S –S •— •ℎ– Y– ••, f, , = × IJ × IK − /1$, , = ×8×8, = 32, , = 32, , = 22.37, , #$, , − /1$ ×, , − 9.625, , × 3.5 × 3.5, , 3.5 cm, A, , C, 8 cm, , − 9.63, , Page | 24

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9. BC is the arc of a quadrant of a circle of radius 8cm. A semi-circle is described on the chord BC on, the side opposite to the centre of the quadrant. Find the area enclosed between arc BC and the semicircle., Solution:, For the quadrant,, =8, , Radius,, , We have JK = √HJ +HK, , and Z = 90$, , = √8 +8, , (by Pythagoras theorem), , = œ2 × 8, = 8√2, , ∴ Radius of the semi circle with diameter BC,, , =, , 1, × 8√2, 2, , = 4√2, , Then, area enclosed between arc JK and the semi-circle, = area of the semi circle with diameter BC – area of the segment with chord BC of the quadrant, =, , −, , mA > fn, , c, , T$, , − YopZd, , 8 × 8 90 22, 1 22, ×, × 4√2 × 4√2, −, z, ×, − Yop 90$ {, 2, 7, 2 180 7, 1 22, 11, = ×, ×4 ×4×2, − 32 ž − 1Ÿ, 2, 7, 7, 352, 32 × 4, 352 − 128, =, −, =, 7, 7, 7, 224, =, = 32, 7, , =, , 10. In a 400 track the length of the each of the straight portions is 90m and the diameter of each of the, inner semi-circular ends is 70m. If the uniform width of the track is 7 m, find the cost of turfing the, track at the rate of Rs. 250 per square metre., Solution: Radius of each inner semi-circular end,, Width of the track= 7, , = × 70, , ∴ radius of each outer semi-circular end, R = 35, , +7, , = 35, , = 42, , Area of the track=area of two rectangular straight portion + areas of the two semi-circular ends, = 2 × 90 × 7, , +2×, , = 1260, , +, , = 1260, , + 1694, , = 1260, , = 2954, , +, , 42 − 35, , U −, , ), , × 77 × 7, , ∴ cost of turšing of the track = Rs 2954 × 250 = Rs 738500., Page | 25

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11. Two flowers beds are in the form of segments (minor) of the circle circumscribing a square field of, side 42m, with two opposite sides of square as a corresponding chord. Find the cost of planting, flowers at the rate of Rs.200 per square metre., Solution: Let ABCD be the square field of side 42 m., Each flower bed is a segment., , Then Z = ∠ JHK = 90$ [∵ diagonals of a square bisect each other at right angles], , and HJ + HK = JK [by Pythagoras theorem], i.e., , +, , = 42, , ⇒, , =, , 42 × 42, 2, , ⇒2, , ⇒, , ⇒, , ⇒, , = 42 × 42, , = 2 × 21 × 21, , = œ2 × 21, , = 21√2, , Now area of the flower beds = 2 ×area of a segment, , Z, − YopZ{, 2 180, 90 22, = 21√2 z, ×, − Yop90$ {, 180 7, 11, = 21 × 21 × 2 z − 1{, 7, 4, = 21 × 21 × 2 ×, 7, =2×, , z, , = 504, , ∴ Cost of planting flowers= Rs. 504 × 200 = Rs. 100800, , 12. On a rectangular window of dimension, 84cm×56cm, twenty-four non-overlapping circular designs, each of radius 7cm are made at the rate of Rs.25 per 20&' . Find the cost of making the designs., Also find the area of the remaining portion., Solution:, For the window, • = 84, and Ž = 56, and radius of each circular design, = 7cm, Now, area of the circular designs = 24 × area of a circle, = 24 ×, 22, = 24 ×, ×7×7, 7, = 3696, ∴ cost of making the designs = Rs., , ", $, , × 3696 = Rs. 4620, , Page | 26

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Area of remaining portion = Area of the rectangular window − area of the circular designs, = 84 × 56, = 4704, , = 1008, , − 3696, , − 3696, , 13. In a round table of diameter 112cm, there is a portion in the form of a rectangular hexagon inscribed, in it. Designs are made on remaining portion of the table top at the rate of Rs.8 per15&' . Find the, , cost designingsk^l√F = D. eFv., , Solution:, , The designed portion has 6 equal segments., , For each segment, =, Z=, , /1$, 1, , = 60$, , = 56, , ∴ Area of the designed portion = 6 × area of a segment, =6×, , Z, 2 180, z, , = 3 × 56 × 56 z, , − YopZ $ {, , 60 22, ×, − Yop60$ {, 180 7, , = 3 × 56 × 56 ƒ, , 22 √3, − „, 21, 2, , = 3 × 56 × 56 ×, , 44 − 21 × 1.73, 42, , = 224 44 − 35.33, = 224 × 7.67, , ∴ K•Y• •—, , = 1718.08, , S¢op£ •ℎ– ’–Yo£pY = UY, , 8, 137.44 × 64, × 1718.08 = UY, = UY 916.30, 15, 15, , 14. Find the area and perimeter of the shaded region in the given Fig. 12.14, having given that ¤¥ =, ¤¦ = E&'., , Solution: Area of the shaded region =Area of the semicircle with centre O –area of semicircle with centre D, + area of semicircle with centre C, , 1, 1, × 28, −, 14, 2, 2, 1 22, = ×, × 28 × 28, 2 7, , =, , +, , 1, 14, 2, , = 1232, , Page | 27

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15. In the given Fig. 12.15. ¥§ = §¨ = DE&'. Find the area of the shaded region using the fact that, D, , ¤© = ¥¦., ], , Solution: Area of the shaded region, , = Area of the semi-circle with diameter C – (area of the semi-circles with diameters AC and BC, =, , =, , =, , . 18, , 1, 2, , −, , × 18 × 18 −, , 162 − 117, , .9 +, , .9 + .6 !, , 9 +6, , + area of the circle with centre O), , 22, × 45, 7, 990, =, 7, , =, , = 141.43, , 16. In the given Fig 12.16, O is the centre and AOB is the diameter of the circle. Find the area of the, shaded region., Solution:, , We have, ∠K = 900 [being angle in a semi circle], , Then IJ = √IK + JK = √12 + 9 = √144 + 81 = √225 = 15, ∴, , =, , ", , = 7.5, , ∴ For the segment with chord XY, θ = 90$, , Now, area of the shaded region, , = area of the semi-circle with centre O – area of ∆ABC + area of segment with chord XY., =, , − × JK × IK +, , m>, , c, , f, , T$, , − YopZd, , Page | 28

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= ×, =, , ."× .", , =, , ."× .", , =, , =, , =, , ."× .", , "1. ", , × 7.5 × 7.5, ×, ×, , c, , ×, , /. ", , = 104.46, , 1, , − × 9 × 12, , −9×6, , + d, , −9×6, , − 54, , +, +, , ."× .", ."× .", , c, , ×, , +, , ."× .", , − 1d, , c, , #$, , T$, , ×, , − Yop90$ d, , − 54, , − 54, , − 54, , = 50.46, , 17. ABCD is a square of side 7cm and semi-circles are drawn with each sides of the square as diameter,, thereby forming four similar petals as shown in the adjoining figure. Find the cost of making design, on the petals at the rate of Rs. 12 per square centimeter., , Solution: Length of a side of the square = 7, ∴ radius of each semi-circle,, , =, , = 3.5, , Then areas of the region I + III = areas of the regions II + IV, , = Area of the square ABCD − Areas of two semi-circles, =7×7, , = 49, , = 49, , = 10.5, , −, , −2×, , × 3.5 × 3.5, , − 38.5, , ∴ Area of the designed portion = Area of the square ABCD − area of the regions I + II + III + IV, =7×7, = 49, , = 28, , − 2 × 10.5, , − 21, , ∴ cost of making the designs = UY. 28 × 12 = UY. 336, , Page | 29

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18. Semi-circles are described on the sides AB and CD of a rectangle ABCD as diameters, both inside the, rectangle. If AB=14cm and BC=7cm, find (i) the area enclosed by the semi-circles (ii) the area of the, portion of the rectangle lying outside the semi-circles., , ABCD is the rectangle in which AD=BC=14cm and AB=CD=7cm., , Solution:, , P and R are the mid-points BC and AD respectively. The semi-circles drawn with BC and AD as, diameters intersect each other Q and S. PQ, QR, RS, SP, PR and SQ are joined., We have, ∆PQR and ∆PRS are equilateral triangles of side 7 cm., , Then ∠¯U‘ = ∠•U‘ = ∠¯‘U = ∠•‘U = 600, ∴ Z = ∠¯U• = ∠¯‘• = 120$, , (i) Area enclosed by the two semi-circles, , = 2 × area of the segment corresponding to the chord QS., Z, − YopZ{, 2 180, 120 22, = 7×7z, ×, − Yop120$ {, 180 7, z, , =2×, , 44 √3, = 7×7ƒ − „, 21, 2, , 88 − 21 × 1.73, 42, 51.67, =7×, 6, 361.69, =, 6, , =7×7×, , = 60.28, , (ii) Area of the portion lying outside the semi-circles, , = Area of the rectangle – Areas of the two semi-circles + Area enclosed by the semi-circles, , = 14 × 7, = 98, , = 158.28, , = 4.28, , −2×, , − 154, , 1 22, ×, ×7×7, 2 7, , − 154, , + 60.28, , + 60.28, , Page | 30

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Surface Area and Volume of Some Basic Solids, 1, , Cube, •, , •, , 2, , 4, , 6, , Volume (Capacity) = • /, , Lateral Surface Area = 2 • + Ž ℎ, , Cuboid, , Cylinder, , ℎ, , Cone, , Sphere, , Hemisphere, , Ž, , Total Surface Area = 2 •Ž + Žℎ + ℎ•, Volume (Capacity) = •Žℎ, , Curve Surface Area = 2, , ℎ, , Volume (Capacity) =, , ℎ, , Total Surface Area = 2, , ℎ, , ℎ, , 5, , Total Surface Area = 6•, , •, , •, , 3, , Lateral Surface Area = 4•, , •, , Slant Height, • = √, , Curve Surface Area =, , +ℎ, , +ℎ, , Total Surface Area =, , •, , +•, , Volume (Capacity) = /, Curve Surface Area = 4, Volume (Capacity) = /, , ℎ, , /, , Curve Surface Area = 2, Total Surface Area = 3, , Volume (Capacity) = /, , /, , Page | 31

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SOLUTIONS, EXERCISE 12.4, 1. A solid wooden toy is in the shape of a cone surmounted on a hemisphere of the same radius 2.1cm. If, the total height of the wooden toy is 8.1cm, find its total surface area and volume., Solution:, , We have,, , = 2.1, , Total height of the wooden toy = 8.1, For the conical part,, , Height, h = 8.1, , − 2.1, , +ℎ, , And slant height, l = √, , = œ 2.1, , =6, , +6, , = √4.41 + 36, , = √40.41, = 6.36, , Then surface Area of the toy = C.S.A. of the cone + C.S.A. of the hemisphere, =, , •+2, , =, , × 2.1 6.36 + 2 × 2.1, , =, , •+2, , = 6.6 × 6.36 + 4.2, = 6.6 × 10.56, = 69.69, , And Volume of the toy = Volume of the cone + Volume of the hemisphere, =, , 1, 3, , =/, , =/ ×, , ℎ+, , 2, 3, , ℎ+2, , × 2.1 × 2.1 6 + 2 × 2.1, , = 4.62 × 10.2, , = 47.124, = 47.12, , /, , /, , /, , /, , /, , Page | 32

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2. A solid is in the form of a cylinder of diameter 7cm with the hemispherical ends. If the total length of, the solid is 42cm, find the total surface area and the volume of the solid., , Solution: Diameter = 7, .. ., , =, , = 3.5, , Total length of the solid = 42, , ... Height of the cylinder, ℎ = 42 − 2 × 3.5, , Then Total Surface Area of the Solid, , = 35, , = C.S.A. of the Cylinder + C.S.A. of the two hemi-spherical ends, , = 2, , = 2, , = 2×, , ℎ+2×2, ℎ+2, ×, , = 22 × 42, , 2, , 35 + 2 × !, , 2, , 2, , = 924, , Volume of the solid = volume of the cylinder + Volume of the two hemisphere, =, =, =, =, , 2, , 2, , .ℎ+2×/, ℎ+, , × ×, , ×, , /, , 35 +, , !, , 35 + / × !, /, , !, , = 38.5 35 + 4.67, , = 38.5 × 39.67, = 1527.29, , 3, , 3, , 3, 3, 3, , 3, , 3. An ice-cream is in the form of an inverted cone surmounted by a hemisphere of the same-radius. If, the radius and the height of the cone are respectively 2.1cm and 8 cm, find the volume of the icecream., Solution: We have, Radius,, , = 2.1, , Height of the conical part, ℎ = 8, , ... Volume of the ice cream = Volume of the conical part + Volume of the hemispherical part, =/, , 2, , .ℎ+/, , 3, , Page | 33

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=/, , 2, , = ×, /, , ℎ+2, , × 2.1 × 2.1 8 + 2 × 2.1, , = 4.62 × 12.2, = 56.364, , = 56.36, , 3, , 3, , 3, , 3, , 4. A cylindrical container of radius 3 cm and height 14cm is filled with ice-cream. The ice-cream is to be, distributed among 16 children in equal cones with hemispherical tops. If the height of the conical, portion is 5 times the radius of its base, find the volume of each ice- cream cone., Solution: For the cylindrical container,, r = 3cm and h = 14 cm., , °±²³=< ±´ 8µ< ;¶²6:·<¸, , ... Volume of each ice cream cone =, =, , 1, , ×/×/×, , =, =, , 1, , nm > .¹, , ##, , cm/, , × 1, , cm/, , = 24.75cm/, , 5. A cylindrical vessel of radius 7cm and height 30cm is full of ice-cream. How many ice-cream cones, each of radius 3cm and height 8cm with hemispherical tops, can be formed with ice- cream from the, vessel., Solution: For the cylindrical vessel, Radius, R = 7cm and height, H = 30 cm, For each ice-cream cone, radius, r = 3cm, And height of the conical part, h = 8cm, , º±²³=< ±´ 8µ< ;¶²6:·<¸6;9² °<77<², , Then no. of ice-cream cones formed =º±²³=< ±´ 8µ< ;±:6;9² »9¸8¼º±²³=< ±´ 8µ< µ<=67»µ<¸6;9² »9¸8, =A, , n½ > .¾, >, ?, , ?, , nm > .¹¼ nm >, , ?, , n m > ¹¼ m, , ?, , ×/×/ T¼ ×/, , =A, =A, , n½ > .¾, , = 35, , × ×/$, , =, , × × $, , Page | 34

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6. A circus tent is in the shape of a cylinder of diameter 24m and height 4m surmounted by a cone of the, same radius and height 5m. Find the capacity of the tent and the cost of the canvas at the rate of Rs., 120 per square metre., Solution:, We have, r =, , = 12, , Height of the cylindrical part, ℎ = 4, , And height of the conical part, ¿ − 5, , ... Slant height of the conical part, •=√, , + ¿ = √12 + 5, , = 13, , Then capacity of the tent= Volume of the cylindrical part +Volume of the conical part, ℎ+, , =, , ℎ+ ׿, /, , =, =, =, =, , ¿, , /, , × 12 × 12 4 + / × 5!, × 144 ×, , "/T"1, , /, , = 2564.58, , /, , /, , /, , /, , And Area of the canvas used = C.S.A. of the cylindrical part + C.S.A. of the conical part, =2, =, =, =, , ℎ+, , 2ℎ + •, , •, , × 12 2 × 4 + 13, × 12 × 21, , = 22 × 12 × 3, , = 792, , ... Cost of the canvas used = UY. 792 × 120 = UY. 95040, Page | 35

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7. A circus tent of the height 15m is in the form of cylinder of diameter 32m and height 3m, surmounted, by a cone of the same radius. Find the volume of tent and the cost of the canvas at the rate of Rs. 110, per square metre., Solution: We have, Radius, r =, , /, , = 16, , Total height of the tent = 15, , And height of the cylindrical part, ℎ = 3, , .. . Height of the conical part, ¿ = 15 − 3, , And slant height of the conical part, = √, , = 12, , +¿, , = √16 + 12, , = √256 + 144, , =√400, = 20, , Then the volume of the tent = volume of the cylindrical part + volume of the conical part., 2, , =, , .ℎ + /, , =, =, , "1/, , . ¿., , ℎ + /. ¿, , 2, , =, , 2, , × 16 × 16 3 + / × 12!, ×7, , = 5632, , 3, , 3, , 3, , And Area of the Canvas used, , = C.S.A. of the cylindrical part + C.S.A. of the conical part, , =2, =, , =, =, =, , ℎ+, , 2ℎ + •, , •, , × 16 2 × 3 + 20, , /" × 1, # ", , 2, , 2, 2, , = 1307.43, , 2, , .. . Cost of the canvas used = UY. 1307.43 × 110, = Rs. 143817.30, , Page | 36

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8. A hall is in the form of a cylinder of diameter 12m and height 4m, surmounted by a cone whose, vertical angle is a right angle. Find the outer surface area of the hall and volume of the air inside the, hall. k^l √ =1.41), Solution: Let ABE and BCDE respectively represent the conical roof and the cylindrical hall. AQ is the, height of the conical roof., We have, ∠JIŠ = 90$ and AB = AE, ∴ ∠IJ¯ = ∠IŠ¯ = × 90$ = 45$, And BE = 12 m, , ∴ J¯ = ¯Š = × 12, , In ∆ABQ, we have, €À, , Q, , = tan∠IJ¯ = tan 45$, , ”À, €À, , ⇒, , =6, , 1Á, , =1, , ⇒ I¯ = 6 m, , And, , ⇒, , ”À, , €”, 1Á, €”, , = cos∠IJ¯ = cos 45$, =, , √, , ⇒ IJ = 6√2 m = 6 × 1.41 = 8.46, Now, radius ( ) = 6 m, height of the cylindrical part (ℎ) = 4 m, For the conical roof,, height (H) = AQ = 6 m and slant height (•) = AB = 8.46 m, Outer Surface Area of the hall, = C.S.A. of the conical roof ABE + C.S.A. of the cylindrical hall BCDE, = •+2 ℎ, =, • + 2ℎ, × 6 8.46 + 2 × 4, , =, , =, =, , /, , × 6 × 16.46, T., , = 310.39, And volume of the air inside the hall, = Volume of the cylindrical part + volume of the conical part, ℎ+/, , =, =, =, =, =, , ℎ + /¿, , ¿, , × 6 × 6 4 + / × 6!, , ×1×1×1, , ", , /, , = 678.86, , /, , /, , /, , Page | 37

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9. A geyser (water boiler) is in the form of a cylinder with hemispherical ends. If the length of the, cylinder is 56cm and the diameter of each end is 18cm, find the capacity of the geyser in litres., , Solution: We have, diameter = 18, ∴, , =9, , Height of the cylindrical part, ℎ = 56, , ∴ Capacity of the geyser = volume of the cylindrical part + volume of the two hemispherical ends, ℎ+2×, , =, , /, , ℎ+/ !, , =, , /, , × 9 × 9 56 + / × 9, , =, , T ×1T, , =, , 1, , =, , /, , /, , = 17310.86, / $.T1, , /, , •, , $$$, , =, , = 17.31•, , /, , /, , [⸪ 1000, , = 1 litre], , 10. A container is formed of a hollow cylinder fitted with a hemispherical bottom of a radius 3cm. The, depth of the cylinder is 14cm and the diameter of the hemisphere is 6cm. Find the volume of the, internal surface area of the container., Solution: We have, = 3, Height of the cylindrical part, ℎ = 14, ∴ Volume of the container = volume of the cylindrical part + volume of the hemispherical part, .ℎ + /, , =, , ℎ+/, , =, , × 3 × 3 14 + / × 3, , =, =, =, , /, , #T× 1, , / 1T, , /, , /, , /, , = 452.57 /, And the internal surface area of the container, = C.S.A. of the cylindrical part + C.S.A. of the hemisphere part, =2 ℎ+2, =2, ℎ+, =2×, , =, , /, , × 3 14 + 3, , =, , / ×, , = 320.57, Page | 38

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11. From a solid cylinder of height 28cm and radius 12cm, a cone of the same height and same radius is, removed. Find the volume of the remaining solid., Solution: We have, , = 12, , Radius,, , , height, ℎ = 28, , ∴ Volume of the remaining solid = Volume of the cylinder – volume of the cone, ℎ−/, , =, , = 1 − /!, =/, , =/×, , ℎ, , ℎ, , = 8448, , ℎ, , × 12 × 12 × 28, , /, , /, , 12. A cylindrical container of radius 10cm and height 35cm is fixed co-axially inside another cylindrical, container of 14 cm and height 35 cm. The total space between the two containers is filled with cork for, insulation purposes. Find the volume of the cork required., Solution: We have, , Radius of the outer cylinder, U = 14, Radius of the inner cylinder,, , = 10, , Height of the each cylinder, ℎ = 35, , Then volume of the cork required, , = Volume of the outer cylinder−Volume of the inner cylinder, = U ℎ−, =, , =, , U −, , ℎ, , 14 − 10, , = 22 × 96 × 5, = 10560, , /, , ℎ, , × 35, /, , /, , 13. A solid is in the shape of a hemisphere surmounted by a cone of the same radius. The diameter of the, cone is 18 cm and the height of the cone is 14 cm. The solid is completely immersed in a cylindrical, tub, full of water. If the diameter of a tub is 26 cm and its height is 21 cm, find the quantity of water, left in the cylindrical tub in litres., Solution: For the solid,, Radius,, , = × 18 = 9, , Height of the conical part, ℎ = 14, , Page | 39

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For the cylindrical tub,, , Radius, U = × 26 = 13, Height, ¿ = 21, , ∴ Volume of water left in the tub = volume of the cylindrical container – volume of the solid, = U ¿−, , = U ¿−, , /, , ℎ+2, , × 13 × 13 × 21, , =, , × 13 × 13 × 21, , =, , 3549 − 864, , =, , × 2685, , =, , = 8438.57, , =, , ℎ+/, , /, , T /T.", $$$, , /, , /, , =, , /, , /, , /, /, , !, , −/×, −, , "#$ $, , × 9 × 9 14 + 2 × 9, , × 3 × 9 14 + 2 × 9, , /, , /, , /, , litres, , = 8.44 litres, 14. A solid toy is in the form of a cone surmounted on a hemisphere of the same radius. The height of the, cone is 6 cm and the radius of the base is 2.8 cm. If a cylinder circumscribes the solid, find how much, more space it will cover., Solution:, We have,, Radius,, , = 2.8, , Height of the cone, ℎ = 6, , Height of the cylinder, ¿ = 6 + 2.8, , = 8.8, , ∴ Additional space covered = Volume of the cylinder – volume of the solid toy, , = Volume of the cylinder – (volume of cone + volume of hemisphere), =, , =, , =, , ¿−, , ℎ+/, , /, , ¿−/ ℎ−/ !, , !, , × 2.8 × 2.8 8.8 − × 6 − × 2.8, , = 24.64 6.8 −, , = 24.64 ×, =, , /, , /1 .1, /, , = 121.56, , ".1, /, , $. `".1, /, , /, , !, , /, , /, , /, , /, , = 24.64 ×, , .T, , /, , /, , /, , /, , Page | 40

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15. A right triangle, with legs 9 cm and 12 cm, is made to remove about its hypotenuse. Find the volume, and the surface area of the double cone so formed., Solution:, , Let ABC be the right triangle right angled at A. ABDC be the double cone, in which AB =, BD = 9 cm, AC = DC = 12 cm., , In ∆ ABC, we have, BC = √AB + AC, , = √9 + 12 cm, , = 15 cm, , E, , We have ∠ BAC = 900 and AE ⟘ BC, , Then ∆ ABC ~∆ EBA, ⇒, , €Ä, , €Å, , ”Ä, , = €”, , ⇒ÆÇ =, , ", , #, , ⇒ 5AE = 36, ⇒AE =, , ", , ⇒ÆÇ = #, , 36, 5, , ⇒AE = 7.2, , ∴ radius of common base of the double cone is 7.2 cm., , Volume of double cone = Volume of the cone ABD + Volume of the cone ACD, = / IŠ . BE + / IŠ . CE, =/, , . IŠ (BE + CE), , =/ ×, =, , × 7.2 × 7.2 × 15, , " $ ., , /, , = 814.63, , /, , Surface area of the double cone = CSA of cone ABD + CSA of the cone ACD, = πAE. AB + πAE. AC, = π. AE AB + AC, , 22, × 7.2 9 + 12 cm, 7, 22, =, × 7.2 × 21cm, 7, =, , = 475.2cm, , Page | 41

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16. A godown is formed of a cuboid of dimensions ]F' × E' × g' covered by a half cylindrical roof., If the length and breadth of the cuboid are ]F' and E' respectively, find the volume of air inside, the godown. Also find the cost of roofing at the rate of É^. g\\ per square metre., , Solution: For the half cylindrical roof,, =, , 1, × 28 = 14m, 2, , ℎ = length of the godown = 63m, , Volume of the air inside the godown, , =Volume of the cuboidal godown + Volume of the half cylindrical roof, = 63 × 28 × 5m/ + × πr . h, , = 8820m/ +, , 1 22, ×, × 14 × 14 × 63 m/, 2 7, , = 8820m/ + 19404m/, , = 28224m/, , Area of the roof = CSA of the half cylinder, 1, × 2πr. h, 2, 22, =, × 14 × 63m, 7, , =, , = 2772m, , ∴ Cost of ro•šing = Rs. 2772 × 500 = Rs. 1386000, , 17. A solid is in the form of a cylinder surmounted by a cone of the same radius. If the radius of the base, and the height of the cone are ‘r’ cm and ‘h’ cm respectively and the total height of the solid is 3h,, e, , prove that the volume of the solid is F Ê ., , Solution: We have, , Radius =, , Height of the cone = ℎ, , Height of the cylinder = 3ℎ − ℎ = 2ℎ, , Volume of the solid = volume of cylinder + volume of cone, 1, = πr . 2h + πr h, 3, 1, = ž2 + Ÿ πhr, 3, 7, = πhr, 3, , Page | 42

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SOLUTIONS, EXERCISE 12.5, , 1. A solid cube is cut into two cuboids of equal volumes. Find the ratio of the total surface area of the, given cube to that one of the cuboids., Solution: Let ‘a’ be the length of each edge of the given cube., For each cuboid,, , • = Ž = S and ℎ = S, , 1Ë>, , Now T.S.A. of the cube: TSA of a cuboid =, , =, , =, , A, A, Ë> ¼Ë. ˼ Ë.Ë, >, >, , /Ë>, , A, A, Ë> ¼ Ë> ¼ Ë>, >, >, , /, , =, , /Ë>, Ë>, , = 3: 2, , 2. A sphere has the same surface area as the curved surface of a cone of height 16 cm and base radius, 12cm. Find the radius of the sphere., Solution: For the cone,, , ∴•=œ, , = 12 cm and ℎ = 16 cm, , + ℎ = √12 + 16 cm = 20 cm, , Let R be the radius of the sphere., We have,, , S.A of the sphere = CSA of the cone, ⇒4 U =, , •, , ⇒ 4U = 12 × 20, ⇒U =, , × $, , ⇒ U = 60, , ⇒ U = √60, , ⇒ U = √2 × 3 × 5, ∴ U = 2√15, , ∴ The radius of the sphere is 2√15 cm, Page | 43

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3. A solid metallic cone is 81 cm high and radius of its base is 6 cm. If it is melted and recast into a solid, sphere. Find the curved surface area of the sphere., Solution: Let R be the radius of the sphere, =6, , For the cone,, We have,, , and ℎ = 81, , Volume of the sphere = volume of the cone, ⇒ / U/ = /, , ⇒ 4U / =, , ℎ, , ℎ, , ⇒ 4U / = 6 × 6 × 81, 1 ×1 ×T, , ⇒ U/ =, , ⇒ U / = 9/, ∴U=9, , ∴ Radius of the sphere is 9, , Then, C.S.A. of the sphere = 4 U, =4×, =, , T, , × 9 × 9cm, , cm, , = 1018.29 cm, , 4. A solid metallic sphere of a diameter 28cm is melted and recast into a number of cones, each of, diameter 7 cm and height 4 cm. Find the number of cones so formed., , Solution: Radius of the sphere, U = × 28 cm = 14 cm, For each cone,, No. of cones =, =, , =, , =, , =, , = cm and ℎ = 4 cm, , ÍÎÏÐÁÑ ÎÒ wÓ¹ÑmÑ, ÍÎÏÐÁÑ ÎÒ ÔÎyÑ, , Õ, n½ ?, ?, A, nm > ℎ, ?, , ½?, , m>ℎ, ×, ×, , ×, , ×, , ×, , ×, , Ö Ö, × ×, > >, , ×, , = 4 × 2 × 2 × 14, , = 224, , Page | 44

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5. The internal and external radii of hollow metallic sphere are 3 cm and 5cm respectively. If the sphere, F, , is melted and recast into a solid cylinder of height 2 , find the curved surface area of the cylinder., Solution: For the hollow sphere,, , = 5 cm and Internal radius,, , External radius,, For the solid cylinder,, , = 3 cm, , T, , ℎ = 2 cm = cm, , We know,, , /, , /, , Volume of metal in the solid cylinder = volume of metal in the hollow sphere, , ⇒, , ℎ=/, T, , ⇒, , × =, , ⇒, , =, , ⇒, ⇒, , ∴, , /, , /, , −, , 5/ − 3/, , /, , /, , × 8 = 4 125 − 27, ×#T, , =7, , T, , = 49, , =7, , ∴ Radius of the cylinder = 7cm, , C.S.A. of the cylinder = 2, , =2 ×, , =, , /", /, , ℎ, , cm, , T, , × 7 × / cm, , = 117 / cm, , 6. A metallic cone of height 28 cm and radius of base 12 cm is melted and recast into a cylinder of, D, , height j F cm. Find the curved surface area of the cylinder so formed., , Solution: For the cone, radius, U = 12cm and height, ¿ = 28cm, For the cylinder, ℎ = 9 / cm =, We know,, , ⇒, , ⇒, , ⇒, , ∴, , T, , /, , cm, , Volume of the cylinder = volume of the cone, ℎ=/ U ¿, ×, , T, , /, , = 12, , =/, , × 12 × 12 × 28, , = 12, , ∴ Radius of the cylinder is 12cm., Page | 45

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Then C.S.A. of the cylinder = 2, , =2×, , ℎ, , × 12 ×, , = 704cm, , T, , /, , cm, , 7. 20 circular plates, each of radius 7cm and thickness 1.5 cm are placed one above another to form a, solid cylinder. Find the curved surface area and the volume of the cylinder so formed., Solution: For the solid cylinder,, =7, , ℎ = 20 × 1.5, , Then, C.S.A = 2, , =2×, , Volume =, , =, , ℎ, , = 30, × 7 × 30cm, , = 1320cm, ℎ, , × 7 × 7 × 30 cm/, , = 4620cm/, , 8. Find the volume of the largest sphere that can carved out of a cube of side 4.2 cm. Also find the ratio, of the volume of the cube to that of the sphere., Solution: For the largest sphere,, , Diameter = length of a side of the cube • = 4.2 cm, ∴, , = 2.1cm, , Then volume of sphere = /, , =/×, , /, , × 2.1 × 2.1 × 2.1 cm/, , = 38.81cm/, , And volume of cube: volume of sphere = Õ, ?, , nm ?, , ?, , ×, , =Õ, , =, , =, , Ï?, , . × . × ., , >>, ×, Ö, , . × . × ., , × × ×/×, ×, , = 21: 11, Page | 46

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9. Find the volume of the largest cone that can be curved out of a cube of a side 16.8cm., Solution: We have,, Diameter of the cone = height of the cone = length of a side of the cube=16.8 cm, ∴ For the cone, , = × 16.8cm = 8.4cm, , And ℎ = 16.8cm, , The volume of the cone = /, , ℎ, , = ×, , × 8.4 × 8.4 × 16.8cm/, , /, , = 1241.86cm/, , 10. A solid cone of diameter 14 cm and height 8cm is melted and recast into a hollow sphere. If the, external diameter is 10 cm, find the internal diameter of the sphere., Solution: For the cone,, , Let, , and, , ∴, , = × 14 cm= 7 cm and ℎ = 8 cm, , be the external and internal radii of the hollow sphere., = × 10 cm= 5 cm, , We know,, , Volume of the metal in hollow sphere = volume of metal in the cone, , ⇒/, , /, , ⇒/, , −, , 5/ −, , ⇒ 4 125 −, , ⇒ 125 −, , /, , /, /, , ∴, , /, , ℎ, , =/ 7 ×8, , =7×7×8, , = 98, , ⇒ 125 − 98 =, , ⇒, , =/, , /, , = 27 = 3/, , /, , =3, , ∴ Internal diameter of the sphere is 2 × 3 cm i. e. 6cm, Page | 47

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11. A hemisphere of lead of radius 7 cm is cast into a cone of base radius 3.5 cm. Find the height of cone., Solution: For the hemisphere, radius U = 7 cm, For the cone, radius,, We know,, , = 3.5 cm, , Volume of the cone = volume of the hemisphere, , ⇒/, , ⇒, , ℎ = / U/, , ℎ = 2U /, , ⇒ 3.5 × 3.5 × ℎ = 2 × 7 × 7 × 7, ⇒ℎ=, , 2 ×7 ×7×7, 3.5 × 3.5, , ∴ ℎ = 56, , ∴ The required height of the cone is 56 cm., 12. A cylinder is cut length wise and flattened, thereby forming a rectangle of dimensions 66m ×28 cm., Find the curved surface area and volume of the original cylinder?, , Solution: Height of the cylinder, h= breadth of the rectangle = 28 cm, And circumference of the base of the cylinder= length of rectangle = 66 cm, i.e.2, ⇒2×, , ⇒, , ⇒, , =, , = 66, , × = 66, , 11 ×, ×, , = 10.5, , ∴ radius of the cylinder is 10.5 cm., , Now curved surface area of the cylinder = 2, , ℎ, , = 66 × 28cm, , Volume of the cylinder =, , =, , ℎ, , = 1848cm, , × 10.5 × 10.5 × 28 cm/, , = 9702cm/, , Page | 48

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13. A sector of a circle of radius 66 cm has angle 1200. It is rolled up so that two bounding radii are, joined together to form a cone. Find the volume of the cone. (Take √ =1.4), , Solution: Let r and R be the radii of the cone and sector, , Circumference of the cone = area length of the sector, ⇒2, , ⇒2 =, , ∴, , f, , =, , =2, , T$, , $, , T$, , U, , ×6, , ∴ Radius of the cone = 2, , And slant height of the cone, • = radius of the sector = 6, ∴ Height of the cone, ℎ = √• −, , = √6 − 2 cm, = √36 − 4 cm, , = √32cm, , Now the volume of the cone = /, , =/×, , =, , =, , ℎ, , /" × ., /×, , $., /, , × 2 × 2 × 4√2cm/, cm/, , cm/, , = 23.47cm/ (approx), , 14. A conical flask of radius × units and height × units is full of water. The water is poured into a, cylindrical flask of radius, , ×, , F, , units. Find the height of water in the cylindrical., , Volume of water in the cylindrical flask = volume of the conical flask, , Solution:, ⇒, , Ø, , /, , ⇒ℎ=, , ⇒ℎ=, , ×, , Ø, , /, , × ℎ = / × . Ù. Ù. 2Ù, , n .Ø . Ø ., /Ø, , Ø ×/×/, , / ×n . Ø . Ø, , ∴ The required height of the water in the cylindrical flask is, , /Ø, , units., , Page | 49

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15. The radii of the bases of two solid metallic cones of the same height Ê are ×D and × . If the two cones, are melted together and recast into a cylinder of height Ê, show that the radius of the base of the, , cylinder is Ú, Solution: Let, , Ù +Ù, , /, , be the radius of the cylinder., , We know,, Volume of the cylinder = sum of the volumes of the two given cones, ⇒, , ., , ⇒, , 1, =Û Ù +Ù, 3, , ⇒, , ℎ=, , ., , ℎ=, , Ù .ℎ +, , /, , 1, 3, , /, , Ù +Ù, , .Ù .ℎ, ℎ, , Thus, radius of the cylinder is Ú, , /, , Ù +Ù, , 16. A solid metallic cylinder and another solid metallic cones have the same height h and the same radius, r. If the two solid are melted together and recast into a cylinder of radius, the new cylinder is, , D], F, , Ê., , D, , , prove that the height of, , Solution: Let H be the height of the new cylinder., We know,, Volume of the new cylinder = volume of the given cylinder + volume of the given cone, ⇒, , ⇒, , ⇒, , ! ¿=, , ¾, , ¿ = 1 + /!, , = /ℎ, , ⇒¿=, , ℎ+/, , 1, , /, , ℎ, , ℎ, , ℎ, , Thus the height of the new cylinder is, , 1, , /, , ℎ., Page | 50

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17. Two solid metallic cuboids of dimensions Dg&' × E&' × g&' tbu \&' × g&' × G&' are, , melted together and recast into solid cubes of side 5cm. Find the number of solid cubes so formed., , Solution:, , Number of solid cubes =, , =, , =, , =, , ܳ= ±´ °±²³=< ±´ 8ݱ ;³Þ±6·7, °±²³=< ±´ 9 ;³Þ<, , "×T×" ¼ $×"×, "×"×", , 1$$¼ $$, $$$, , =8, , ", , ", , 18. A sphere of diameter 6cm is dropped into a cylindrical vessel partly filled with water. The diameter of, the vessel is 12 cm. If the sphere is completely submerged, how high will the level of the water be, raised?, Solution: Radius of the cylindrical vessel, (U) = × 12 cm = 6 cm, Radius of the sphere, ( ) = × 6 cm = 3 cm, , Let ℎ be the increase in water level., We have,, Volume of the displaced water in the cylinder = Volume of the sphere, ⇒ U ℎ=/, , /, , ⇒6×6×ℎ =/×3×3×3, , ⇒h=, , ×/×/, 1×1, , ⇒h=1, ∴ the level of the water is raised by 1 cm., 19. A solid metallic cylinder of radius 14 cm and height 21 cm is melted and recast into a number of, spheres, each of radius 3.5 cm. Find the number of the spheres so formed., Solution:, , Radius of the sphere ( ) = 3.5 cm, , For the cylinder, Radius (U) = 14 cm and height (ℎ) = 21 cm, Number of sphere =, =, =, =, , º±²³=< ±´ ;¶²6:·<¸, , º±²³=< ±´ 9 7»µ<¸<, n½ > ℎ, , Õ, n m?, ?, Õ, ?, , n×, , ×, , ×, , × n×/." × /." × /.", , ×, , = 72, , ×, , × / × $$$, , × /" × /" × /", , ∴ the required number of sphere is 72., Page | 51

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20. A spherical shell of lead, whose external diameter is 16 cm is melted and recast into a cylinder height, D, , j &' and diameter 16 cm. Find the internal diameter of the shell., F, , Solution:, , We have, , 1, , =, , Radius of the cylinder,, , =8, , =, , External radius of the spherical shell,, Inter radius of the spherical shell =, , Height of the cylinder, ℎ = 9, , We know,, , ⇒, , /, , /, , ⇒ 8/ −, , ⇒ 512 −, , −, , /, , /, , =, , /, , /, , =, , =8 ×, , ⇒, , T, , /, , /, /, , .ℎ, T, , /, , 8 × 8 × 28, 4, , = 448, , ⇒ 512 − 448 =, , ⇒, , =, , =8, , Volume of the lead in the spherical shell = volume of the cylinder, , ⇒ / 8/ −, , ⇒, , /, , 1, , = 64, , /, , = 4/, , =4, , ∴ The internal diameter of the spherical shell is 4 × 2 cm i. e. 8cm., , 21. A vessel is in the form of an inverted cone of height 8 cm and radius 6 cm. It is filled with water upto, the rim. When lead shots, each of which is a sphere of a radius 0.2cm are dropped into the vessel, onesixth of the water flows out. Find the number of lead shots dropped into the vessel., Solution: For the cone,, Radius,, , = 6 cm and height, ℎ = 8 cm, , For a lead shot (sphere),, Radius,, Number of sphere =, , = 0.2 cm=, , $, , cm = " cm, , °±²³=< ±´ ·67»²9;<· Ý98<¸ ´¸±= 8µ< ;±:6;9² °<77<², °±²³=< ±´ 9 7»µ<¸6;9² ²<9· 7µ±8, , A, ÎÒ à¹Ñ, ß, , ÍÎÏÐÁÑ ÎÒ ÔÎyÑ, , = áÎÏÐÁÑ ÎÒ ÑËÔ¹ wÓ¹ÑmxÔËÏ w¹Îàw, =, , A A, × nm > ¹, ß ?, Õ, × nmA ?, ?, , Page | 52

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A, , ?, , =Õ, =, , ?, , A, ?, , × ×n.1×1 ×T, , = ßÕ, , A, @, , A, @, , × n× × ×, A, @, , ×T, A, @, , A, @, , A, @, , × × ×, , ×T×/×"×"×", , = 1500, , 22. Water is flowing at the rate of 5km per hour through a pipe of diameter 14cm into a rectangular tank, of base 30m × 22m. Find the time during which the level of water in the tank rises by 35cm., , Solution: We have,, , 5 km/hr = 5000 m/hr and 35 cm =, Radius of the pipe, r =, , /", , $$, , m, , cm = 7 cm =, , $$, , m, , Volume of water drawn through the pipe in 1 hr =, , =, , × 5000, , ×, , =11× 7, =77, , Volume of water in the tank = 30 × 22×, , = 3× 11 × 7, , ∴ the required time taken =, , = 3× 77, , /, , /", , $$, , /, , /, , $$, , ×, , /, , /, $$, , × 5000, , /, , /, , º±²³=< ±´ Ý98<¸ 6: 8µ< 89:â, , º±²³=< ±´ Ý98<¸ ·¸9Ý: 8µ¸±³ãµ 8µ< »6»< 6:, , µ¸, , =, , /×, , hr = 3 hr, , *******, , Page | 53

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Frustum of a Right Circular Cone, If a right circular cone is cut off by a plane parallel to the base, the portion of the cone between the, plane and the base of the cone is called a frustum of the cone., •, •, •, •, •, , Slant Height of a frustum • = œ, Curve Surface Area =, , +, , Total Surface Area = [, , •, , +, , −, , +ℎ, , •+, , ], , +, , Surface Area of a Frustum with larger face open =, Volume =, , +, , /, , ., , +, , ℎ, , [, , +, , •+, , ], , SOLUTIONS, EXERCISE 12.6, 1. If the radii of the circular ends of frustum of height 6 cm are 20 cm and 12 cm respectively, find the, volume and the curved surface area of the frustum. (Take, Solution: We have,, , = F. DG, , ℎ = 6cm, , = 20cm, , = 12cm, , ∴• =œ, , −, , = œ 20 − 12, , +ℎ, , + 6 cm, , = √64 + 36 cm, = 10cm, , Then C.S.A. of the frustum =, , +, , •, , = 3.14 20 + 12 10 cm, = 3.14 × 32 × 10 cm, , Volume = /, , +, , = 1004.8 cm, +, , ℎ, , = / × 3.14 20 + 20 × 12 + 12, , = 6.28 400 + 240 + 144 cm/, , × 6cm/, , = 6.28 × 784cm/, = 4923.52cm/, , Page | 54

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2. A bucket is in the form of a frustum of a cone. If the height of the bucket is 16 cm and the radii of the, upper and lower ends are 18 cm and 6 cm respectively, find, i), The height of the cone of which the bucket is a part., ii), The capacity of the bucket., iii), The slant height of the bucket., iv), The surface area of the bucket., Solution:, i) Let ACDB be the bucket of height PQ and AOB be the cone of height OP of which the bucket, ACDB is a part., , We know, IJ‖K~ o. –. J‘‖~¯, , Then, ∆ JH‘~∆~H¯, ∴, , ⇒, , H‘ J‘, =, H¯ ~¯, , H‘, 18, =, H‘ − 16, 6, , ⇒ 3. H‘ − 48 = H‘, ⇒ 2H‘ = 48, ∴ H‘ =, , 48, = 24, 2, , ∴ The height of the cone of which the bucket is a part is 24 cm., , ii) For the bucket, we have, = 18 cm,, , = 6 cm, ℎ = 16 cm, , ∴ Capacity of the bucket = /, , +, , +, , ℎ, , 1 22, ×, 18 + 18 × 6 + 6 16cm/, 3 7, 1 22, = ×, × 3 108 + 36 + 12 × 16cm/, 3 7, 22, =, × 156 × 16cm/, 7, 54912 /, =, cm, 7, =, , = 7844.57cm/, , iii) Slant height of bucket, • = œ, , −, , = œ 18 − 6, , +ℎ, , + 16 cm, , = √144 + 256 cm, , = √400 cm = 20 cm, Page | 55

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iv) Surface area of the bucket = [, , +, , •+, , ], , 22, [ 18 + 6 × 20 + 6 ]cm, 7, 22, =, 480 + 36 cm, 7, 22, =, × 516cm, 7, 11352, cm, =, 7, , =, , = 1621.71cm, , 3. A bucket is in the form of a frustum of height 21 cm. The diameters of the top and the bottom are 70, cm and 56 cm respectively. Find the capacity of the bucket., Solution: We have,, , ℎ = 21, =, , =, , $, , = 35, , "1, , = 28, , Capacity of the bucket = /, =, , +, , ., , +, , ℎ, , 1 22, ×, 35 + 35 × 28 + 28, 3 7, , = 22 1225 + 980 + 784 cm/, , × 21cm/, , = 22 × 2989 cm/, , = 65758 cm/, , =, , 1" "T, $$$, , litres [⸪ 1000cm/ = 1 litre], , = 65.76 litres, , 4. The circumference of one plane face of a frustum is 44 cm and that of the other is 66 cm. If the height, of the frustum is 12.6 cm, find the volume., Solution: We have, ℎ = 12.6 cm, Circumference of the larger plane face = 66cm, , ⇒2, , ⇒2 ×, ⇒, , ∴, , =, , =, , = 66cm, , ×, , 11×, ×, , = 66cm, , cm, , 21, cm = 10.5 cm, 2, Page | 56

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And circumference of the smaller plane face = 44 cm, ⇒2, , = 44, ×, , ⇒2 ×, ⇒, , =, , ∴, , ×, , = 44, , ×, , =7, , ∴ volume of the frustum =, , =, , /, , ×, , +, , +, , ℎ, , 1, 22, ×, 10.5 + 10.5 × 7 + 7, 3, 7, , × 12.6 cm/, , = 22 110.25 + 73.5 + 49 × 0.6 cm/, = 22 × 232.75 × 0.6 cm/, , = 3072.3 cm/, , 5. A glass tumbler is in the form of a frustum of height 12 cm, the diameters of the upper and the lower, ends being 7 cm and 4.2 cm respectively. Find the capacity of the tumbler k^l, , Solution: We have,, =, , =, , ., , = F. DG, , cm = 3.5 cm, , cm = 2.1cm, , And ℎ = 12 cm, , Then capacity of the glass tumbler (frustum) = /, , +, , ., , +, , ℎ, , = / × 3.14 3.5 + 3.5 × 2.1 + 2.1, , × 12cm/, , = 3.14 12.25 + 7.35 + 4.41 × 4cm/, , = 3.14 × 24.01 × 4cm/, = 301.57cm/, , 6. The perimeters of circular ends of a solid frustum are 88 cm and 66 cm and its slant height is 21 cm,, find the total surface area of the frustum., Solution: We have,, • = 21, , Circumference of the larger base = 88 cm, , ⇒.2, , ⇒.2 ×, ⇒, , ⇒, , =, , = 88 cm, ×, , TT ×, ×, , = 88 cm, , cm, , = 14 cm, , Page | 57

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Circumference of the smaller base = 66 cm, , ⇒2, , ⇒2×, , ⇒, ∴, , =, , = 66 cm, ×, , 11 ×, ×, , = 66 cm, , cm, , = 10.5 cm, , Then T.S.A. of the frustum = [, , +, , •+, , +, , ], , 22, [ 14 + 10.5 × 21 + 14 + 10.5 ]cm, 7, 22, =, 514.5 + 196 + 110.25 cm, 7, 22, =, × 820.75 cm, 7, =, , = 2570.5 cm, , 7. A container is in the form of a frustum of height 12 cm with radii of its upper and lower ends as 17, cm and 8 cm respectively. Find the cost of milk the container can hold at the rate of Rs. 20 per litre., = F. DG ., , Also find the C.S.A. of the container, Solution: We have,, , ℎ = 12, , ∴ •=œ, , ,, , −, , = œ 17 − 8, , = 17, , +ℎ, , ,, , =8, , − 12 cm, , = √81 + 144 cm, , = 15, , Now, capacity of the container = /, =, , +, , ., , +, , ℎ, , 1, × 3.14 17 + 17 × 8 + 8, 3, , × 12cm/, , = 3.14 289 + 136 + 64 × 4 cm/, , = 12.56 × 489cm/, , = 6141.84 cm/, =, , 1, , .T, , $$$, , litres [⸪ 1000 cm/ = 1 litres], , = 6.14 litres, , ∴ Cost of milk in the container = UY. 6.14 × 20 = UY. 122.80, Page | 58

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A, , ⇒ ?A, ⇒, , ⇒, , ⇒, , ?, , n .ÄÀ> .‚À, n.€“ > .‚“, , ÄÀ, , €“, , ! ×, , ÄÀ /, €“, , ! =, , =, , ÄÀ, , €“, , K¯ 2, =, I‘ 3, , =, , !, , /, , T, , /, , T, , c⸪, , ÄÀ, , €“, , =, , ‚À, ‚“, , d, , ∴ K¯: I‘ = 2: 3, , ∴ The ratio of the radius of the smaller cone to that of the given cone is 2: 3., , 10. A circular cone has a base of radius 10 cm and height 25 cm. The area of the cross section of the cone, by a plane parallel to its base is DgG &' . Find the distance of the plane from the base of the cone., , Solution: Let AOB and COD respectively be the original and new cones of heights OP and OQ, ∴ H‘ = 25, , ,, , I‘ = J‘ = 10, , and Area of the base of the cone KH~ = 154, ⇒ . K¯ = 154, , 22, . K¯ = 154, 7, 154 × 7, ⇒ K¯ =, 22, , ⇒, , ⇒ K¯ = 49 = 7, ∴ K¯ = 7, , We know,, , IJ‖K~, , Then ∆ KH¯~∆ IH‘, ‚À, , ÄÀ, , ⇒ ‚“ = €“, , 7, H¯, =, 25 10, 7 × 25, ⇒ H¯ =, 10, , ⇒, , ∴ H¯ = 17.5, , ∴ ‘¯ = H‘ − H¯ = 25, , − 17.5, , = 7.5, , ∴ The required distance of the plane from the base of the cone is 7.5, , Page | 60

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14. A bucket is in the form of a frustum with a capacity of 45584 cubic cm. If the radii of the top and, bottom of the bucket are 28 cm and 21 cm respectively, find its height and surface area., Solution: We have,, = 28, , = 21, , And volume of the bucket (frustum) = 45584, , ⇒, , +, , /, , ., , +, , . ℎ = 45584, , /, , /, , 1 22, ×, 28 + 28 × 21 + 21 . ℎ = 45584, 3 7, 1 22, ⇒ ×, 784 + 588 + 441 ℎ = 45584, 3 7, 1 22, ⇒ ×, × 1813ℎ = 45584, 3 7, 1, ⇒ × 22 × 259ℎ = 45584, 3, 45584 × 3, ⇒ℎ=, 22 × 259, ⇒, , ⇒ ℎ = 24, , ∴ the height of the bucket is 24 cm., Then, • = œ, , −, , = œ 28 − 21, , +ℎ, , + 24, , = √49 + 576 cm, , = √625 cm, = 25 cm, , ∴ the required surface area of the bucket = [, =, , =, , =, , =, , +, , •+, , ], , [ 28 + 21 × 25 + 21 ] cm, 49 × 25 + 441 cm, 1225 + 441 cm, , × 1666 cm, , = 22 × 238 cm, , = 5236 cm, , Page | 63