How this instrument works
A general trapezoid has two parallel bases and two legs that can be any length at all, with nothing requiring the legs to match. Give this sheet both bases and both legs — b1, b2, leg1, and leg2 — and it recovers the height and the area together, splitting the horizontal gap between the bases across the two legs in whatever proportion actually fits their two different lengths, rather than assuming that gap divides evenly.
Picture dropping a perpendicular from each end of the shorter base down to the longer one. That creates two right triangles sharing the same height, one built on leg1 and one on leg2, and their two horizontal offsets must add up to exactly the gap between the bases. Solving that pair of equations together — rather than assuming the offsets split evenly — hands back the offset under leg1 first, and the height follows from the ordinary Pythagorean relation applied to that one triangle.
This is a deliberately broader tool than this site's isosceles-trapezoid pages, which need both legs equal so the horizontal gap can be assumed to split into two identical halves. Feed unequal legs into an isosceles-only formula and the answer would simply be wrong; feed the same numbers here and the two right-triangle offsets are worked out honestly, whatever the two leg lengths happen to be, including the isosceles case as one outcome among many rather than the only one handled.
A right trapezoid, where one leg already stands perpendicular to both bases, falls out of the same equations as a special case needing no separate treatment: that leg's own offset comes out to zero, handing its entire horizontal share to the other, slanted leg. Nothing about the underlying math changes; the two offsets this sheet solves for simply happen to land as far apart as zero and the full gap instead of somewhere in between.
- Enter the shorter parallel length into Base 1 (shorter) and the longer one into Base 2 (longer).
- Enter the first slanted length into Leg 1 and the second into Leg 2 — the two do not need to match.
- Read Height, solved from the two right triangles the legs form against the gap between the bases.
- Read Area, computed as half the sum of both bases times that recovered height.
Worked example — bases 4 and 12, legs 5 and root 41
Bases of 4 and 12 leave an 8-unit gap between them, closed by a leg of 5 on one side and a leg of roughly 6.4031242374328485 (√41) on the other. Solving for the first offset gives (5² − 41 + 8²) ⁄ (2×8) = (25 − 41 + 64) ⁄ 16 = 48⁄16 = 3, so leg1 reaches 3 units across; the height follows as √(5² − 3²) = √(25−9) = √16 = 4, and the area comes to ½(4+12)×4 = 32.
Set both legs to 5 instead, with bases of 4 and 10, and the same equations still hold: the offset works out to (25−25+36)⁄12 = 3, the height again comes to √(25−9) = 4, and the area is ½(4+10)×4 = 28 — the isosceles case falling out of the general formula rather than needing a separate one. A third case, bases 5 and 8 with legs of 4 and 5, gives an offset of (16−25+9)⁄6 = 0, meaning leg1 stands perfectly upright and carries none of the 3-unit gap; the height equals leg1 itself, 4, and the area is ½(5+8)×4 = 26.
Questions
What if I only have one leg length, not two different ones?
Enter that same figure into both Leg 1 and Leg 2 — the formula still works, since it makes no assumption that the two legs differ, only that they might. With bases 4 and 10 and both legs at 5, the offset comes out to 3, the height to 4, and the area to 28, matching what a dedicated isosceles formula would return.
How does the sheet find the height without knowing it directly?
By solving the two right triangles that the legs form with the height. Each leg, together with its own horizontal offset and the shared height, makes one right triangle, and since the two offsets must add up to the full gap between the bases, that pair of equations pins down a single offset first, then the height follows from the Pythagorean theorem.
How is this different from the isosceles-trapezoid pages on this site?
Those pages assume both legs share one length, which lets the horizontal gap split evenly in half by symmetry. This page drops that assumption and solves for each leg's own offset separately, so it handles any pair of leg lengths — equal or not — including cases an isosceles-only formula would answer incorrectly.
What happens with a right trapezoid, where one leg is already vertical?
The formula still applies without modification: that leg's offset comes out to exactly zero, meaning it carries none of the horizontal gap, and the other leg absorbs the entire distance between the bases on its own. Bases 5 and 8 with legs 4 and 5 give an offset of zero, a height equal to the vertical leg itself, and an area of 26.
Can the two bases be equal?
Equal bases turn the shape into a parallelogram, and the height then depends on how far the top base has shifted sideways rather than on a fixed offset calculation — a case this particular formula, built around a genuine gap between two different base lengths, does not cover on its own.
Does it matter which base I call Base 1 and which Base 2?
Yes, unlike a plain sum — the formula assumes Base 2 is the longer of the pair, since the gap it solves for is b2 minus b1. Enter the shorter figure as Base 1 and the longer as Base 2 to get a height and area that match the actual shape.