This problem is the combination of two CONE shape. To find answer of this question we need to use volume of cone. Problem number 14 from RS Aggarwal book exercise 17C, page number 824, chapter volume and surface area of solid.

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# A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m, respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required.

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Given data

2r1 = 6 m , 2r2 = 20 m

Height of frustum = 24 m

r1 = 3 m, r2 = 10m

Slant height (l) = √ [ (24)² + (10-3)² ]

l = √ 576+49

l = 25 m

radius of cone (r1) = 3m

height = (28-24) = 4m

Slant height (l1) = √ [ (3)² + (4)² ]

l1 = √ [9+16]

l1 = 5m

Quantity of canvas area = πl (r1+r2) + π r1 l1

= ( π * 25 (3+10) + π * 3 * 5)

= (325 π + 15π)

= 340 π

= 340 * 3.14

= 1067.6 m²

∴ the quantity of canvas area is 1067.6 m²