Trapezoid Area & Diagonals

Complete trapezium area

A trapezium has two parallel bases, the major one B and the minor one b, with a perpendicular height h between them. The standard formula is A = (B + b)·h / 2. If you prefer working with the midsegment (median) M = (B + b)/2, the area is just A = M·h. There's also triangulation, where you draw a diagonal to cut the shape into two triangles and add their areas. That last approach helps when all you have are side lengths and angles. Example: B = 10, b = 6, h = 4 gives A = (10 + 6)·4/2 = 32.

Applications

Land surveying leans on this for irregular plots. A surveyor records GPS or total-station offsets, then breaks the boundary into a row of trapezia. The trapezoidal rule in numerical integration takes the same idea and applies it to a function: ∫f(x)dx ≈ Σ((f(x_i) + f(x_{i+1}))/2)·Δx. Civil engineers reach for trapezia when sizing the cross-section of channels, canals and embankments, and the same formula shows up in architecture for sloped roofs and stadium bleachers.

FAQ

Which height do I use? The perpendicular distance between the two parallel bases, never a slanted side. If all you have is a slanted side s and an angle θ, then h = s·sin(θ).

What if I only have the four side lengths? One option is the formula built around the difference of the bases: h = √(c² − ((B−b)² + c² − d²)²/(4(B−b)²)), where c and d are the non-parallel sides. Otherwise, cut the shape into two triangles and apply Heron's formula.

Does the formula work for an isosceles trapezium? Yes. It doesn't care whether the trapezium is right, isosceles or scalene. Only B, b and h enter the calculation.

How accurate is the trapezoidal rule? Its error scales as Δx², so cutting the step in half drops the error to about a quarter of what it was. For smooth functions, Simpson's rule gets there faster.