Question

In the given figure, ABCD is a quadrilateral.

If $AB=30cm,$, $AD=18cm,$, $BD=24cm,$, $DC=26cm$ and $\angle DBC={90}^{\circ }$,

then find the area of quadrilateral $ABCD$.

Answer 1

$△ABD$

$S=\left(\frac{18+30+24}{2}\right)$

=$36cm$

$△=\sqrt{S(S-a)(S-b)(S-c)}$

=$\sqrt{36(36-24)(36-18)(36-30)}$

=$\sqrt{\left(36\right)\left(12\right)\left(18\right)\left(6\right)}$

=$\sqrt{6^2\times 6\times 2\times 18\times 6}$

=$\sqrt{6^2\times 6^2\times 6^2}$

=$216c{m}^2$

$△BDC$

$B{C}^2=D{C}^2-D{B}^2$

=${26}^2-{24}^2$

=${10}^2$

$BC=10cm$

$△DBC=\frac{1}{2}\times DB\times BC$

=$\frac{\text{/}1}{\text{/}2}\times \text{/}{24}^{12}\times 10$

=$120c{m}^2$ $ABCD$

= $216+120$

= $336c{m}^2$