Question

Find the surface area of a frustum of cone, whose larger and smaller radius is 12 and 5 m. The slant height of the cone is 20 m.

• A. 519$\pi$ m${}^2$
• B. 509$\pi$ m${}^2$
• C. 529$\pi$ m${}^2$
• D. 539$\pi$ m${}^2$

Answer 1

The correct option is B 509$\pi$ m${}^2$

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

R = 12m, r = 5m, s = 20m

SA = $\pi$(5 + 12)20 +$\pi$5${}^2$ + $\pi$12${}^2$

= $\pi$(17)20 + 25$\pi$ + 144$\pi$

= $\pi$(340 + 25 + 144)

= 509 $\pi$m${}^2$