I’m stuck on the total surface area of a right circular frustum and I think I’m mixing up which height to use. The frustum has bottom radius R = 5 cm, top radius r = 3 cm, and vertical height h = 4 cm. I want the total surface area including both circular ends.
My attempt: I treated the unwrapped side as a trapezoid and used lateral area = 2π(R + r)h, then I added πR² + πr² for the two ends. But this doesn’t match the answer I’m checking against, so I must be thinking about the “height” wrong.
Questions I’m trying to sort out:
– When you unroll the side of a frustum, is it valid to think of it as a trapezoid with vertical height h? Or is that the wrong picture?
– If the lateral area should actually use the slant height ℓ instead of h, what’s the correct way to get ℓ from R, r, and h? I tried ℓ = sqrt(h² + (R − r)²), but I’m not confident I’m applying that in the right place.
– Is there a safer method by subtracting the lateral areas of two similar cones (big cone of radius R minus small cone of radius r)? If so, how do I set up the similarity to get the smaller cone’s slant height correctly from these numbers?
I’d really appreciate a step-by-step explanation of where my reasoning goes off. Any help appreciated!
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3 Responses
Use the slant height, not the straight-up height-picture the frustum’s side as a little trapezoid coat (close enough!), so the lateral area is π(R + r)ℓ with ℓ = √(h² + (R − r)²). Example: R=5, r=3, h=4 gives ℓ=√20=2√5, so total area = π(R+r)ℓ + πR² + πr² = π(16√5 + 34) ≈ 219.2 cm²; friendly walkthrough: https://www.mathsisfun.com/geometry/frustum.html.
You’re right to rethink it-when you unwrap a frustum you get an annular sector (not a trapezoid), so use the slant height ℓ = sqrt(h² + (R − r)²) and A_total = π(R² + r²) + π(R + r)ℓ. Example: with R=5, r=3, h=4, we get ℓ = sqrt(16+4) = 2√5, lateral area = π(8)(2√5) = 16π√5, ends = 34π, so A_total = 34π + 16π√5 ≈ 219.2 cm².
Use the slant height, not the vertical height: ℓ = sqrt(h^2 + (R − r)^2), so total surface area S = π(R^2 + r^2) + π(R + r)ℓ; with R=5, r=3, h=4 we get ℓ = 2√5 and S = π(34 + 16√5). This matches the “big cone minus small cone” approach, and a concise derivation is here: https://www.cuemath.com/geometry/surface-area-of-frustum-of-a-cone/.