Question

Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the distance between them.

Answer 1

Draw a trapezium say $ABCD$ and join the diagonal $AC$.

You will obtain two triangles $ABC$ and $ADC$. Draw two perpendiculars from $AB$ on $CD$ and from $CD$ on $AB$ as shown in figure.

Now area of trapezium $ABCD$ = sum of area of both the triangles

$ABC$ and $ADC$

Let h be the length of perpendiculars drawn,

area of $1st$ triangle = $1/2\times CD\times h.......1$

area of $2nd$ triangle = $1/2\times AB\times h.........2.$

sum of (1) and (2)= area of trapezium $ABCD$

$1/2\times CD\times h+1/2\times AB\times h=$ area of trapezium $ABCD$

$1/2\times (AB+CD)\times h=$ area of trapezium $ABCD$ (taking out $1/2$ as common )

Therefore, area of a trapezium is half the sum of the parallel sides multiplied by the distance between them.