Question

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

• A. $1344c{m}^2$
• B. $672c{m}^2$
• C. $2688c{m}^2$
• D. $336c{m}^2$

Answer 1

The correct option is A $1344c{m}^2$

Area of $△BAD=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times 32\times 24$
$=384c{m}^2$
In $△BAD$,
$A{B}^2+A{D}^2=B{D}^2$
$\Rightarrow {32}^2+{24}^2=B{D}^2$
$\Rightarrow 1024+576=B{D}^2$
$\Rightarrow 1600=B{D}^2$
$\Rightarrow BD=40cm$
In $△BCD$,
$S=\frac{40+52+52}{2}$
$=\frac{144}{2}$
$=72cm$
Area of $△BCD=\sqrt{72(72-40)(72-52)(72-52)}$
$=\sqrt{72\left(32\right)\left(20\right)\left(20\right)}$
$=\sqrt{921600}$
$=960c{m}^2$
Now,
Area of quadrilateral $ABCD=$ Area of $△BAD+$ Area of $△BCD$
$=384+960$
$=1344c{m}^2$