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3 Responses
You’re spot on: first get r from the circumference C = 2πr ⇒ r = C/(2π), then use the right triangle with slant l to get h = √(l² − r²), so V = (1/3)πr²h = (C²/(12π))·√(l² − (C/(2π))²).
Think of it like a ladder against a wall-the slant is the ladder, the radius is how far the feet stick out, and the height is how high it reaches.
Yes-reconstruct r and h first. From the base circumference C you get the radius r = C/(2π). In a right circular cone the height, radius, and slant height form a right triangle, so h = sqrt(l^2 − r^2), which requires l ≥ r (otherwise the data are inconsistent; l = r gives zero volume). Then the usual volume formula applies: V = (1/3)πr^2h. Substituting r = C/(2π) gives V = (C^2/(12π)) · sqrt(l^2 − C^2/(4π^2)). When I first learned this, drawing the cross-section and labeling the right triangle made it click-once I wrote r = C/(2π), the Pythagoras step felt automatic and the volume dropped out cleanly.
You’re on the right track: use V = (1/3)πr²h, but first rebuild r and h from what you’ve got. From the base circumference C, the radius is r = C/(2π). The slant height l, the radius r, and the height h make a right triangle, so by Pythagoras h = sqrt(l² − r²). Put those together and you get h = sqrt(l² − (C/(2π))²). Now plug into the volume formula: V = (1/3)πr²h = (1/3)π (C/(2π))² sqrt(l² − (C/(2π))²) = (C²/(12π)) sqrt(l² − C²/(4π²)). Just make sure l ≥ r (i.e., l ≥ C/(2π)), otherwise the geometry doesn’t make sense and your cone would be, well, imaginary in more ways than one.