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

In trapezium ABCD,

AB||DC, AB=18 cm

$\angle B=\angle C={90}^{\circ }$, CD =12 cm and AD =10 cm

Area of trapezium ABCD

Draw $DL\perp AB$

$\therefore AL=18-12=6cm$

AL =BC

$AL=\sqrt{A{D}^2-A{L}^2}$

$=\sqrt{{10}^2-6^2}=\sqrt{100-36}$

$=\sqrt{64}=8cm$

Now area of trapezium $=\frac{1}{2}(AB+CD)\times AL$

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

$=\frac{1}{2}\times 30\times 8=120c{m}^2$