# Area of a Trapezium

A **trapezium**, also known as a **trapezoid**, is a quadrilateral in which a pair of sides are parallel, but the other pair of opposite sides are non-parallel. The area of a trapezium is computed with the following formula:

$\text{Area}=\frac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}.$

The parallel sides are called the **bases** of the trapezium. Let $b_1$ and $b_2$ be the lengths of these bases. The distance between the bases is called the **height** of the trapezium. Let $h$ be this height. Then this formula becomes:

$\text{Area}=\frac{1}{2}(b_1+b_2)h$

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## Proof

Given a trapezium, let $b_1$ and $b_2$ be the lengths of the bases, and let $h$ be the height. Draw a segment parallel to the bases that is halfway between the bases. This divides the trapezium into two trapeziums, each with the same height of $\frac{1}{2}h.$

Labeling the angles of these trapeziums:

Note the following congruences and identities due to the fact that the bases are parallel:

$\begin{aligned} m \angle 4 + m \angle 5 &= 180^\circ \\ m \angle 1 + m \angle 7 &= 180^\circ \\ \angle 2 &\cong \angle 6 \\ \angle 3 &\cong \angle 8 \end{aligned}$

Now rotate the top trapezoid and place it adjacent to the bottom trapezoid in the following way:

Due to the aforementioned congruences and identities, this shape is a parallelogram. The length of its base is $(b_1+b_2),$ and its height is $\frac{1}{2}h.$ This parallelogram has the same area as the trapezoid, so the area of the trapezoid is

$\text{Area}=\frac{1}{2}(b_1+b_2)h.\ _\square$

## Computing Area

Consider a trapezium $ABCD$ in which $AB \parallel CD$. $AB=10\text{ cm}$, $CD=\text{ cm}$ and they are separated by a distance of $4\text{ cm}$. Find the area of $ABCD$.

We know that the area of a trapezium = $\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}$

Substituting the values, we get,

Area = $\dfrac {1}{2}×(10+5)×4$

=$\dfrac {1}{2}×15×4$

=$30\text{ cm}^{2}$. $_\square$

Find the area of a trapezium with parallel lines of $9\text{ cm}$ and $7\text{ cm}$, and a height of $3\text{ cm}$.

Area =$\dfrac {1}{2}×(9 \text{ cm} + 7 \text{ cm}) × 3 \text{ cm}$

=$\dfrac {1}{2} ×(16 \text{ cm})×(3 \text{ cm})$

=$\dfrac {1}{2}×48 \text{ cm}^2$ =$24 \text{ cm}^2$. $_\square$

**Cite as:**Area of a Trapezium.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/area-of-a-trapezium/