Question

The larger radius of a frustum cone is double the smaller radius which is 12 in. The slant height of the cone is 40 in. What is the surface area? (Use $\pi$ = 3.14).

• A. 6,182.4 in${}^2$
• B. 6,782.4 in${}^2$
• C. 6,,682.4 in${}^2$
• D. 6,582.4 in${}^2$

Answer 1

The correct option is B 6,782.4 in${}^2$

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

Larger radius = 2 $\times$ smaller radius$\Rightarrow$ 2 $\times$ 12 = 24 in

Smaller radius = 12 in

s = 40 in

SA = $\pi$(12 + 24)40 + $\pi$12${}^2$ + $\pi$24${}^2$

= $\pi$(36)40 + 144$\pi$ + 576$\pi$

= 3.14 $\times$ (1,440 + 144 + 576)

= 3.14 $\times$ 2,160

= 6,782.4 in${}^2$