Question

Find the surface area of a frustum of cone, whose larger and smaller radius is 25 cm and 12 cm. The slant height of the cone is 30 cm. (Use $\pi$ = 3.14)

• A. 5,600.96 cm${}^2$
• B. 5,700.96 cm${}^2$
• C. 5,800.96 cm${}^2$
• D. 5,900.96 cm${}^2$

Answer 1

The correct option is D 5,900.96 cm${}^2$

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

R = 25 cm, r = 12 cm, s = 30 cm

SA = $\pi$(12 + 25)30 + $\pi$12${}^2$ + $\pi$25${}^2$

= $\pi$(37)30 + 144$\pi$ + 625$\pi$

= 3.14 $\times$ (1,110 + 144 + 625)

= 3.14 $\times$ 1,879

= 5,900.96 cm${}^2$