The area of a trapezium

A trapezium is a shape with one pair of parallel sides. Its area is

A = \tfrac{1}{2}(a + b)h

where $a$ and $b$ are the lengths of the parallel sides and $h$ is the distance between them.

Nobody proves this at school — it is handed over with the formula sheet. Here is why it is true, and it takes one picture.

The proof

Put two copies of the trapezium together, one of them turned upside down.

The two parallel sides of the first are $a$ and $b$ . Turned over, the second has them the other way round, so where the first has a side of length $a$ the second has one of length $b$ , and the two fit.

The shape you now have is a parallelogram. Its bottom edge is $a + b$ — the $a$ from one copy and the $b$ from the other. Its height is $h$ , the same distance between the parallel sides as before.

A parallelogram's area is base times perpendicular height, so this one has area

(a + b)h

But it is made of two trapezia, and they are identical. So one trapezium has half that area:

A = \tfrac{1}{2}(a + b)h

Check your understanding

Use the formula on a trapezium with parallel sides $7$ cm and $11$ cm and a height of $4$ cm.
Why is the joined shape a parallelogram rather than some other quadrilateral?
A rectangle is a special trapezium. Show that the formula gives the right area for a rectangle of length $L$ and width $W$ .

Answers

$\tfrac{1}{2}(7 + 11) \times 4 = \tfrac{1}{2} \times 18 \times 4 = 36$ cm².
The two parallel sides are still parallel in the joined shape, and the sloping sides are each a side of one trapezium and its turned-over partner, which are parallel for the same reason. Two pairs of parallel sides: a parallelogram.
Both parallel sides are $L$ , and the distance between them is $W$ , so $\tfrac{1}{2}(L + L)W = LW$ , which is the rectangle's area.