How this instrument works
A right circular cone hides an ordinary right triangle in plain sight. Cut the solid in two through its apex and the middle of its base, and the exposed face is an isosceles triangle spanning the full width, 2r, along its own base. That isosceles shape splits along its line of symmetry into two identical right triangles: one leg reaches from the middle of the base out to the rim, a length of r; the other climbs that same line of symmetry to the apex, a length of h; the third side, joining rim to apex, is the slant height, l. Apply the Pythagorean theorem to either half-triangle and l² = r² + h² falls straight out — nothing about the cone being a solid of revolution changes the arithmetic once the slice has flattened it into an ordinary triangle.
A mistake specific to this measurement is worth watching for: reading that vertical rise off a real cone by eye, along its visible sloped side, instead of holding a plumb line steady at the apex and letting it settle over that unmarked spot in the middle of the base. The sloped edge sits closer to the answer this page returns than to the true rise from base to apex, so typing it into the Height field feeds an already-slanted number back into the square root and overstates the result. The formula also settles cleanly at its extremes — shrink the Radius to 0 and the cone collapses into a bare vertical rod, with the slant height landing exactly on that same rise; shrink the Height to 0 instead and the cone goes the other way, settling into a flat disc, with the slant height landing exactly on the radius.
A genuinely counter-intuitive fact hides in the same identity. Fix the slant height instead of r or h — imagine one fixed length of string running from apex to rim, or one fixed-size sector of paper rolled into a cone shell — and the volume enclosed is not maximised by the tallest, most needle-like cone that length allows, nor by the widest and flattest. There is one particular ratio, h equal to l divided by √3, that traps the most volume; taller or wider than that sweet spot and the enclosed space shrinks either way, a classic calculus result tucked quietly inside l² = r² + h².
- Radius goes in the first field — the flat span from the middle of the base out to its rim, using whichever length unit suits the job.
- Height goes in the second field: hold a tape plumb against the cone's own axis and read the rise from base to apex — the visible outer slope reads longer and belongs in neither field.
- Slant height comes back as the result, l = √(r² + h²) — the distance a piece of string would trace running outside the cone from the rim up to the apex.
- Try Radius = 0 or Height = 0 to watch the shape flatten to its limits — Slant height then simply copies whichever of the two fields is still nonzero.
Worked example — radius 3, height 4
A cone-shaped paper cup built for a classroom demonstration has a base Radius of 3 centimetres and a Height of 4 centimetres measured straight up from the pointed bottom to the open rim. Square each leg — 9 and 16 — add them to 25, and the root comes out whole: l = √(3² + 4²) = √25 = 5.0 centimetres exactly, the length of the seam running along the cup's sloped side from the point to the rim, with nothing left to round away.
Scale the same shape up to a papier-mache volcano cone for a science fair, with Radius 5 and Height 12 units, and the identity again lands on a whole number: l = √(5² + 12²) = √(25 + 144) = √169 = 13.0. Shrink the radius toward the opposite extreme instead — Radius 0 and Height 4 — and the cone collapses into a bare vertical rod; the formula still holds and returns Slant height = √(0² + 4²) = 4.0, exactly matching that second input, since there is no rim left for the sloped edge to lean away from the centre.
Questions
What is the formula for a cone's slant height?
l = √(r² + h²): square r and h separately, add the two results, and take the root. The identity falls out of the right triangle exposed when a cone is sliced in half through its apex and the middle of its base — the Pythagorean theorem in a new setting. Plug in 3 for the radius and 4 for the height and the arithmetic lands exactly on 5, since those two numbers square and add to a perfect square, 25.
How is slant height different from a cone's height?
Height passes through the solid interior of the cone, an invisible segment stretching from the apex to an unmarked spot in the middle of the base; slant height instead traces the visible sloped surface, from that same apex out to the rim. Because slant height is the hypotenuse of the triangle formed by those two measurements, it is always the larger of the two whenever the radius is greater than zero — mixing them up understates any figure built from the true slant.
Why is the slant height always at least as large as the height and the radius?
Because it plays the hypotenuse's role in that right triangle, and a hypotenuse is never the shortest of the three sides — it can't come out smaller than either leg standing beneath it. With radius 5 and height 12, for instance, the slant height comes out to 13 — bigger than both 5 and 12 on their own, yet smaller than their sum of 17, exactly as a triangle's three sides require.
What happens to the slant height when the radius shrinks to zero?
The cone collapses into a plain vertical segment, and the slant height becomes identical to that vertical rise — with radius 0 and height 4, the formula returns √(0² + 4²) = 4 exactly, since there is no rim left for the sloped edge to reach toward. Run the limit the other way and the cone opens out flat into a disc instead, with the slant height settling on the radius.
How can the Height field get filled in wrong?
The most common cause: reading that vertical measurement off a real cone by eye along its visible sloped edge, rather than holding a plumb line at the apex and letting it hang until it points at the unmarked spot directly underneath. That sloped edge sits closer to the slant height than to the true rise from base to apex, so typing it into the Height field feeds an already-slanted number back into the square root and overstates the result. Only a perpendicular drop from the apex reads the true value correctly.
Does a taller or a wider cone hold more volume for the same slant height?
Neither extreme wins. For a fixed slant height — one fixed length of rope from apex to rim, say, or one fixed-size sector of paper rolled into a cone — the enclosed volume is maximised at one particular ratio, h equal to l divided by √3, not by making the cone as needle-like or as flat as the material allows. It is a classic calculus optimisation result sitting quietly inside the same l² = r² + h² identity this page solves.