Question

In the figure, AB and DC are parallel sides of a trapezium $ABCD$ and $\angle ADC={90}^o$. Given $AB=15cm,CD=40cm$ and diagonal $AC=41cm$, calculate the area of trapezium ABCD

Answer 1

area of trapezium

$A{C}^2=A{D}^2+D{C}^2$

$\left({41}^2\right)=A{D}^2-{40}^2$

$A{D}^2=81$

$AD=9$

are of $\square ABCD$

$=\frac{1}{2}\times (15+40)\times 9$

$=\frac{1}{2}\times 55\times 9$

$=24.559cr$