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.

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
$$\begin{gathered} =\sqrt{h^2+{\left(R-r\right)}^2} \\ =\sqrt{{24}^2+{\left(10-3\right)}^2} \\ =\sqrt{576+49} \\ =\sqrt{625}=25\mathrm{m} \end{gathered}$$

For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m

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

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top
$$\begin{gathered} =\left(\mathrm{\pi }l(R+r)\right)+\mathrm{\pi }Lr \\ =\mathrm{\pi }\left(l(R+r)+Lr\right) \\ =\frac{22}{7}\left(25\times 13+5\times 3\right) \\ =\frac{22}{7}\left(325+15\right)=1068.57{\mathrm{m}}^2 \end{gathered}$$