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:

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

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

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

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 =$\dfrac {1}{2}×(9 \text{ cm} + 7 \text{ cm}) × 3 \text{ cm}$ 
=$\dfrac {1}{2} ×(16 \text{ cm})×(3 \text{ cm})$ 
=$\dfrac {1}{2}×48 \text{ cm}^2$   =$24 \text{ cm}^2$. $_\square$