Question

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. $[Take\pi =22/7.]$

Answer 1

For the lower portion of the tent:

Diameter of the base = 20 m

Radius, R, of the base = 10 m

Diameter of the top end of the frustum = 6 m

Radius of the top end of the frustum = r = 3 m

Height of the frustum = h = 24 m

Slant height = l

$=\sqrt{h^2+(R-r{)}^2}=\sqrt{{24}^2+(10-3{)}^2}=\sqrt{576+49}=\sqrt{625}=25$ m

For the conical part:

Radius of the cone= base = r = 3 cm

Height of the cone = Total height - Height of the frustum = 28 - 24 = 4 m

Slant height, L, of the cone $=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$ m

Total quantity of canvas = Curved surface area of the frustum + Curved surface area of the conical top

$=\left(\pi l\right(R+r\left)\right)+\pi Lr=\pi \left(l\right(R+r)+Lr)=\frac{22}{7}(25\times 13+5\times 3)=\frac{22}{7}(325+15)=1068.57m^2$