Area of Trapezium

Area of a Trapezium

Study carefully the trapezium PRST.

Divide the trapezium PRST into two triangles PQR and TOS, and a rectangle RQOS. Use dotted lines for SO and RQ which is the height of triangles PQR and TOS. Also, SO or RQ is the length of rectangle QROS as shown in the following figure:

The length of SO and RQ is $h$, RS and QO is $a$, PQ is $z$, OT is $y$, and PT is $b$.

The area of trapezium = Area of a triangle PQR + Area of a triangle TOS + Area of a rectangle QRSO.

But, area of a triangle $= \frac{1}{2} \times \text{base} \times \text{height}$.

Area of a rectangle $=$ length $\times$ width.

The height of the triangle PQR is $h$ and its base is $z$. The height of triangle TOS is $h$ and its base is $y$. The length of the rectangle QRSO is $h$ and its width is $a$. Therefore,

Area of triangle PQR $= \frac{1}{2} \times z \times h = \frac{zh}{2}$

Area of triangle TOS $= \frac{1}{2} \times y \times h = \frac{yh}{2}$

Area of rectangle QRSO $= h \times a = ah$

Area of a trapezium $=$ Area of triangle PQR $+$ Area of triangle TOS $+$ Area of a rectangle QRSO.

Thus,

But $z + y + a = b$

Thus,

Area of a trapezium $= \frac{h}{2}(b + a)$

$= \frac{h}{2}(a + b)$

$= \frac{1}{2}h(a + b)$

Therefore, the area of trapezium $= \frac{1}{2} \times h \times (a + b)$, where $h$ is height, $a$ and $b$ are the lengths of the two parallel sides of the trapezium.

Therefore, area of a trapezium is calculated by taking the average of the two parallel sides of the trapezium multiplied by its height.

Example 2

Find the area of the following trapezium:

Solution

The lengths of two opposite parallel sides of a trapezium are 8 cm and 4 cm. Its height is 3 cm. That is $a = 4$ cm, $b = 8$ cm and $h = 3$ cm. Thus;

Area of a trapezium $= \frac{1}{2} \times h \times (a + b)$

$= \frac{1}{2} \times 3 \text{ cm} \times (4 \text{ cm} + 8 \text{ cm})$

$= 18 \text{ cm}^2$

Therefore, the area of the trapezium is $18 \text{ cm}^2$.