Question

Calculate the area of quadrilateral ABCD in which AB=32 cm, AD = 24 cm, $\angle A$= ${90}^{\circ }$ and BC = CD = 52 cm.

Answer 1

${\left(BD\right)}^2={\left(AB\right)}^2+{\left(AD\right)}^2$

$BD=\sqrt{{\left(32\right)}^2+{\left(24\right)}^2}$

$BD=40cm$

$\Delta ABD=\frac{1}{2}\times AB\times DB$

$\text{= 384c}{\text{m}}^2$

$For\Delta \text{BCD,}$

$\text{S =}\frac{a+b+c}{2}=\frac{40+52+52}{40}=72cm$

$\text{=}\sqrt{72\left(72-40\right)\left(72-52\right)\left(72-52\right)}$

$\text{=}\sqrt{72\times 32\times 20\times 20}$

$\text{= 960c}{\text{m}}^2$

$Area\text{of ABCD = 384 + 960}$

$\text{= 1344c}{\text{m}}^2$