A cone of radius 16 cm and height 12 cm is cut, parallel to the base, at some height closer to the top. If the ratio of the heights of the solids is 1∶3, then the ratio of the total surface areas of the two solids formed is approximately given by:
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Correct answer: B
Explanation
The correct answer is option 2 i.e. 9 : 143.
Height ratio = 1 : 3
total parts 4.
Small-cone height = 12/4 = 3 cm.
Scale factor k = 3/12 = 1/4
Big-cone slant height = = 20
Big total surface area = π·r·(l + r) = π·16·(20 + 16) = 576π.
Small cone is similar to big cone with linear scale k, so its total surface area = k2 × big total = (1/16) × 576π = 36π.
Small cone base area = π·42 = 16π (or from k2 × big base = (1/16) × 256π = 16π)
Frustum total area = big total − small total + 2 × (small base)
= 576π − 36π + 2·16π = 572π.
Ratio (small cone) : (frustum) = 36π : 572π = 36 : 572 = 9 : 143