Area of a trapezium

A trapezium has just one pair of parallel sides. Break it into pieces whose areas you already know.

Take trapezium WXYZ with WX ∥ ZY. Drop perpendiculars WM and XN onto ZY. That leaves a rectangle in the middle with a triangle at each end.

WXNM really is a rectangle: since WX ∥ ZY, the inside angles on the same side of a crossing line add to 180°, so ∠MWX = ∠NXW = 90°.

Add the three areas: Area = ½ × MZ × WM + WX × WM + ½ × NY × XN.

Name the lengths: MZ = x, WM = XN = h, WX = a, NY = y. Then Area = ½hx + ha + ½hy = h(½x + a + ½y) = ½h(x + y + 2a).

That still has x and y in it, which nobody wants to measure. But the long parallel side b is made of exactly those three pieces: b = x + y + a, so x + y = b − a.

Put that in: Area = ½h(b − a + 2a) = ½h(a + b). So area of a trapezium = ½ × height × (sum of the parallel sides). Another way to say it: the average of the parallel sides times the height.

A second proof, and a prettier one. Take two identical copies of the trapezium, turn one round by half a turn, and join them along a slanting side. The angles at each join add to 180° (co-interior angles, x + y = 180°), so the joins are straight lines: the result has four sides, not six. And since u + v = 180° as well, both pairs of opposite sides are parallel, making it a parallelogram. Its base is a + b and its height is h, so its area is h(a + b). One trapezium is half of that: ½h(a + b), exactly as before.