You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., $\frac{1}{2} (a + b)h$.

Let the trapezium have parallel sides $a$ and $b$ , and height $h$ .

From the figure, if we join another congruent triangle to the trapezium, we get a parallelogram.

Base of the parallelogram = $a + b$

Height of the parallelogram = $h$

Area of parallelogram:

Area of parallelogram = $(a + b)h$

The trapezium is exactly half of this parallelogram.

So,

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

Hence, proved.