Cone volume when they give slant height?

For a cone with diameter 12 and slant height 10, I did V = (1/3)π*6^2*10, but that feels off. Should I be using the actual height instead (quick way to get it-Pythagoras every time or is there a faster trick)?

3 Responses

  1. Yep-use the actual vertical height, not the slant; think ice-cream cone: the fill depends on how tall it is inside, so with r=6 and slant 10 you get h=√(10²−6²)=8 (nice 6–8–10), hence V=(1/3)π·6²·8=96π. Quick trick (I think this is the fastest): spot the 3–4–5 pattern or use h=√(l²−r²) from Pythagoras each time.

  2. I think you can just use the slant height as the height and roll with it: V = (1/3)π·6^2·10 = 120π. I might be muddling triangles a bit, but I don’t think you need Pythagoras here.

  3. Great instinct-your formula for volume is right, but you have to use the perpendicular height of the cone, not the slant height. The slant height is useful for surface area, while volume needs the “straight up-and-down” height h. The radius here is r = 12/2 = 6, and the slant height is l = 10, so use the right triangle formed by r, h, and l: l² = r² + h². That gives h = √(10² − 6²) = √(100 − 36) = √64 = 8. (Nice little 6–8–10 triangle-spotting that can save time!) Now plug into V = (1/3)πr²h: V = (1/3)π·6²·8 = (1/3)π·36·8 = 96π. So the quick trick is either Pythagoras or recognizing common triples; don’t use the slant height in the volume formula.

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