Question

The larger radius of a frustum cone is thrice the smaller radius which is 12 m. The slant height of the cone is 120 m. What is the surface area? (Use $\pi$ = 3.14).

• A. 22,608 m${}^2$
• B. 21,608 m${}^2$
• C. 20,608 m${}^2$
• D. 23,608 m${}^2$

Answer 1

The correct option is A 22,608 m${}^2$

Surface area of a frustum of cone = $\pi$(r + R)s + $\pi$r${}^2$ + $\pi$R${}^2$

Larger radius = 3 $\times$ smaller radius $\Rightarrow$ 3 $\times$ 12 = 36m

Smaller radius = 12 m

s = 120 m

SA = $\pi$(12 + 36)120 + $\pi$12${}^2$ + $\pi$36${}^2$

= $\pi$(48)*120 + 144$\pi$ + 1,296$\pi$

= 3.14 $\times$ (5,760 + 144 + 1,296)

= 3.14 $\times$ 7,200

= 22,608 m${}^2$