A trapezoid is a rectangle whose width is the average of its bases
A trapezoid has exactly one pair of parallel sides, called the bases. Its area is A = ½(a + b)·h, and the cleanest way to understand that formula is to notice what ½(a + b) is: the average of the two bases, which is also the length of the midsegment, the line joining the midpoints of the two legs. So the area is midsegment × height — the trapezoid has exactly the area of a rectangle with the same height and the average width.
That interpretation is why the shape appears wherever a cross-section widens or narrows with depth. Drainage ditches, canals and roadside swales are cut trapezoidal because a vertical bank collapses; a retaining wall is thicker at the base than the top; a hip roof plane and a gable-end wall are often trapezoidal; a dovetail is a trapezoid in plan. In all of them the two parallel sides are known and the perpendicular distance between them is the depth.
Only three numbers control the area: a, b and h. The legs do not enter it. Two trapezoids with the same bases and height but wildly different leg slopes have identical areas, which surprises people the first time they see it. The legs matter for the perimeter, for the base angles and for whether the figure closes at all.
Three derivations of ½(a + b)h
By duplication. Take a second copy of the trapezoid, rotate it 180° and set it against the first. The two fit into a parallelogram of base (a + b) and height h, whose area is (a + b)h. One trapezoid is half of that.
By dissection. Drop perpendiculars from the ends of the shorter base to the longer one. That splits the figure into a central rectangle b × h and two right triangles whose horizontal legs total (a − b). The triangles combine to ½(a − b)h, so the total is bh + ½(a − b)h = ½(a + b)h.
By averaging. A horizontal slice at height y across the figure has width that runs linearly from a at the bottom to b at the top. The mean of a linear function over its range is the average of its endpoints, so the mean width is (a + b)/2 — and area is mean width times height. This is exactly the trapezoidal rule of numerical integration, which approximates the area under any curve by treating each strip as a trapezoid.
Rearranging gives the reverse problem: h = 2A / (a + b). A ditch that must carry a given cross-sectional area at a chosen bottom and top width has its depth fixed by that expression. The base angles follow from the legs: each leg is the hypotenuse of a right triangle with vertical side h and horizontal run √(c² − h²), so the angle it makes with the base is arctan(h / run). The right triangle calculator works those corner triangles on their own.
Worked example: a drainage swale 12 ft across the top
A swale is cut with a 12 ft top width, a 6 ft flat bottom and a 4 ft depth, with both banks the same slope.
- Identify the parts. a = 12 ft (top), b = 6 ft (bottom), h = 4 ft (vertical depth). The banks are the legs.
- Midsegment. m = (12 + 6)/2 = 9 ft. Halfway up the bank the swale is 9 ft wide.
- Area. A = m × h = 9 × 4 = 36 ft² of cross-section. Equivalently ½(12 + 6)(4) = ½(18)(4) = 36.
- Horizontal run per bank. The bases differ by 12 − 6 = 6 ft, split equally, so each bank runs 3 ft horizontally over 4 ft of rise.
- Leg length. c = √(3² + 4²) = √25 = 5 ft. Both legs, since the section is symmetric.
- Perimeter. P = 12 + 6 + 5 + 5 = 28 ft.
- Base angle. arctan(4/3) = 53.13° from the horizontal, a slope of 0.75:1 — steeper than most soils hold, which is a design flag rather than an arithmetic one.
Turn the cross-section into volume: 36 ft² over a 250 ft run is 36 × 250 = 9,000 ft³, or 9,000 ÷ 27 = 333.3 cubic yards of excavation. Note what happens if you flatten the banks to 2:1 (2 ft horizontal per 1 ft vertical) while keeping the 6 ft bottom and 4 ft depth: each bank now runs 8 ft, the top width becomes 6 + 16 = 22 ft, and the area becomes ½(22 + 6)(4) = 56 ft² — 56% more excavation for the same depth and bottom.
Reading the result, and the check the calculator runs for you
The midsegment is the most useful secondary number. It is the equivalent uniform width, so it tells you directly what rectangle the trapezoid is worth. For a channel it is the average width of flow; for a roof plane it is the width to use when ordering by the square; for the trapezoidal rule it is the mean of the two ordinates.
The base angles convert straight into slope. An angle of 45° is 1:1, 26.57° is 2:1, 18.43° is 3:1. Earthwork specifications quote the horizontal-to-vertical ratio rather than the angle, so read the angle and convert: run = h / tan(angle). Anything steeper than about 1.5:1 in unstabilised soil is a design decision, not a geometry one.
The calculator also runs a consistency check that hand calculations skip. Given a, b and h, the two legs are not free: their horizontal runs must add up to exactly |a − b|. If you enter legs of 5 and 5 with bases differing by 6 and a height of 4, the runs are 3 and 3, which total 6 — consistent. Enter legs of 5 and 5 with bases differing by 4 and the runs still total 6, so the figure cannot close, and the perimeter and angles reported would describe a different trapezoid from the one whose area you computed. When that happens the tool says so, and keeps the area and midsegment, which depend only on a, b and h and remain correct.
Trapezoidal channel: area and top width by side slope
| Side slope (z:1) | Bank angle | Top width T (ft) | Area (ft²) | Leg length (ft) |
|---|---|---|---|---|
| 0 (vertical) | 90.00° | 4.0 | 12.00 | 3.000 |
| 0.5:1 | 63.43° | 7.0 | 16.50 | 3.354 |
| 1:1 | 45.00° | 10.0 | 21.00 | 4.243 |
| 1.5:1 | 33.69° | 13.0 | 25.50 | 5.408 |
| 2:1 | 26.57° | 16.0 | 30.00 | 6.708 |
| 3:1 | 18.43° | 22.0 | 39.00 | 9.487 |
Each area is ½(T + 4)(3) evaluated at that top width; each leg is √((3z)² + 3²). Flattening the banks from 1:1 to 2:1 adds 43% to the excavated cross-section.
Where trapezoid calculations go wrong
- Using a leg length as the height. The height is the perpendicular distance between the parallel sides. A 5 ft bank on a 4 ft deep ditch gives an area 25% too large if you substitute it.
- Averaging the wrong pair of sides. The midsegment averages the two parallel sides. Averaging a base and a leg is meaningless.
- Entering legs that cannot close the figure. Their horizontal runs must total |a − b|. The calculator flags the mismatch instead of quietly reporting impossible angles.
- Assuming the trapezoid is isosceles. Nothing in the area formula requires equal legs, and a ditch cut against a slope rarely has them.
- Forgetting that flow area is not cross-section area. In an open channel only the wetted part carries water, so recompute with the water depth, not the ditch depth.
- Confusing the names. In British usage a trapezium is what US texts call a trapezoid, and a US trapezium is a quadrilateral with no parallel sides at all. Check which convention a source is using before trusting its formula.
Related shapes and the trapezoidal rule
Push the shorter base to zero and the trapezoid becomes a triangle, and ½(a + 0)h reduces to the familiar ½bh — use the triangle area calculator there. Make the bases equal and it becomes a parallelogram with A = bh, which the parallelogram area calculator handles, and if the legs are also perpendicular you have a rectangle, covered by the rectangle area and perimeter calculator. Every one of those is a special case of the same expression.
For a four-sided figure with no parallel sides, this formula does not apply at all. Take the corner coordinates and use the shoelace formula area calculator, which is what surveyors use for irregular parcels and what a CAD package does internally.
The same average-the-ends idea underpins the trapezoidal rule in calculus: to estimate the area under a curve, slice it into strips and treat each strip as a trapezoid, giving ∫f(x)dx ≈ Σ ½(f₁ + f₂)Δx. Surveyors use the identical arithmetic under the name "average end area" to convert cross-sections into earthwork volumes: average two adjacent section areas and multiply by the distance between them. If you are working those cross-sections up from field angles, the law of cosines calculator closes the traverse triangles.