Question

Find the area of the trapezium ABCD in which AB||DC, AB=18 cm, $\angle B=\angle C={90}^{\circ }$, CD=12 cm and AD=10 cm.

Answer 1

Trapezium $ABCD$ is drawn in figure.

$AE=(18-12)$ cm $=6$ cm

$\Delta$ $AED$ is a right-angled $\Delta$

$\therefore (AD{)}^2+(ED{)}^2+(AE{)}^2$

${10}^2=h^2+6^2$

$\therefore h=\sqrt{100-36}$

$h=8$ cm

$\text{Area of trapezium}=\frac{1}{2}\times \text{Sum of parallel sides}\times \text{Distance between them}.$

$=\frac{1}{2}(18+12)\times 8$

$=\frac{1}{2}\left(30\right)\times 8$

$=30\times 4$ $=120c{m}^2$.