I’m cramming for a test and I thought I had cone volume down (V = 1/3 · π · r^2 · h), but the word “height” keeps tripping me up. I keep thinking the height should be the slanted side because that’s how “tall” the cone looks, right? Or am I totally mixing that up?
Example: I have a cone with radius 3 cm and slant height 10 cm. I did V = (1/3)·π·(3)^2·(10) = 30π. That felt reasonable, but the answer key doesn’t match 30π and seems to use some weird square root instead of 10 for h. Did I use the wrong “height” there? Why wouldn’t the slant height count as h if that’s literally the side length?
Second place I keep messing up: if a problem says the cone has diameter 6 cm and height 8 cm, I plug straight into V = (1/3)·π·(6)^2·8 and get 96π. But now I’m thinking maybe I was supposed to use r = 3 instead of 6. Ugh.
Could someone explain, in test-day terms, which length is supposed to be h in the formula and how to handle it when they give the slant height instead? And also confirm what to do when they give diameter vs radius? I feel like I’m overthinking this, but I keep making the same mistakes.
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2 Responses
In V = (1/3)·π·r^2·h, the “height” h is the perpendicular distance from the base to the tip (the axis of the cone), not the slant height along the side; the slant height l is used for surface area, not volume. In a right circular cone, r, h, and l form a right triangle, so l^2 = r^2 + h^2 and h = √(l^2 − r^2). Let me check your first example: radius r = 3 cm, slant height l = 10 cm gives h = √(100 − 9) = √91, so V = (1/3)·π·(3)^2·√91 = 3π√91 cm³ (not 30π). For the second example, “diameter 6 cm” means radius r = 3 cm, and the given height 8 cm is already the perpendicular height, so V = (1/3)·π·(3)^2·8 = 24π cm³; using 6 instead of 3 quadruples r^2 and leads to 96π, which is too big. Test-day rule: always convert diameter to radius first, and if you’re given slant height, use h = √(l^2 − r^2) before plugging into the volume formula.
Think of a cone’s “height” h as the shy vertical distance that drops straight down from the tip to the center of the base, not the glamorous slanted side strutting along the surface. The slant height l is along the outside of the cone, which looks tall but isn’t perpendicular to the base. The volume formula V = (1/3)πr^2h always wants that straight-up-and-down height. So for your first example, with radius r = 3 cm and slant height l = 10 cm, you first find the true height using the right triangle inside the cone: h = sqrt(l^2 − r^2) = sqrt(100 − 9) = sqrt(91). Then V = (1/3)π(3)^2(sqrt(91)) = 3π sqrt(91), which is about 89.9. That “weird square root” is the perpendicular height sneaking out from behind the curtain.
For the second example, when the problem says diameter 6 cm and height 8 cm, remember diameter is twice the radius, so r = 3 cm. Plugging in correctly: V = (1/3)π(3)^2(8) = 24π. The 96π you got came from using the diameter as if it were the radius, and since the radius gets squared, that swaps in a 6^2 instead of 3^2 and blows the volume up by a factor of 4. Sneaky, right?
One last lantern in the maze: the slant height l does star in surface area formulas (like lateral area = π r l), which is probably why it keeps waving at you from the sidelines during volume problems. For a quick refresher with pictures and practice, this Khan Academy lesson is handy: https://www.khanacademy.org/math/geometry/volume-and-surface-area/volume-cones/v/volume-of-a-cone