Question

The surface area of a frustum cone is 2,400 m${}^2$. The larger and smaller radius of the cone is 12 and 4 m. find its slant height. (Use $\pi$ = 3.14).

• A. 37.77 m
• B. 38.77 m
• C. 39.77 m
• D. 51.77 m

Answer 1

The correct option is A 37.77 m

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

R = 12m, r = 4m, SA = 2,400 m2

2,400 = 3.14 $\times$ (4 +12)s +$\pi$4${}^2$ + $\pi$12${}^2$

2,400 = 3.14 $\times$ [16s +16 + 144]

2,400 = 3.14 $\times$ [16s +160]

2,400/3.14 = 16s + 160

764.33 160 = 16s

604.33/16 = s

S = 37.77 m

Therefore, the slant height is 37.77 m