Adv
Deepak Bora
  • 0
Newbie

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.

  • 0

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.

Share

1 Answer

  1. 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

    • 0
Leave an answer

Leave an answer

Browse

Choose from here the video type.

Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs".

Captcha Click on image to update the captcha.

Related Questions