Question

In the adjoining quadrilateral $ABCD$,
$\angle BAD={90}^{\circ }$ and $\angle BDC={90}^{\circ }$. All measurements are in centimetres. Find the area of quadrilateral $ABCD$.

Answer 1

Given $\angle BAD={90}^{\circ },\angle BDC={90}^{\circ }AB=6cmAD=8cmBC=26cm$

In right angle $△ABD$

$BD=\sqrt{{AB}^2+{AD}^2}=\sqrt{6^2+8^2}=\sqrt{100}=10cm$

In right angle $△BCD$

$CD=\sqrt{{BC}^2-{BD}^2}=\sqrt{{26}^2+{10}^2}=\sqrt{{24}^2}=24cm$

$\therefore$ Area of quadrilateral $ABCD$

=ar($△ABD$)+ar($△BCD$)

$=\frac{1}{2}\times 6\times 8+\frac{1}{2}\times 10\times 24=3\times 8+5\times 24=24+120=144{cm}^2$