Question

3. In quadrilateral ABCD, angleB = angleD= 90°,
AB = 7 cm, AC = 25 cm, CD = 20 cm.
Find the area of ABCD​

Answer 1

$\textbf{In \triangle ABC,}$

$AC^2=AB^2+BC^2$

$25^2=7^2+BC^2$

$625-49=BC^2$

$\implies\;BC^2=576$

$\implies\;BC=\sqrt{576}$

$\implies\;BC=24\;\text{cm}$

$\textbf{In \triangle ACD,}$

$AC^2=AD^2+CD^2$

$25^2=AD^2+20^2$

$625-400=AD^2$

$\implies\;AD^2=225$

$\implies\;AD=\sqrt{225}$

$\implies\;AD=15\;\text{cm}$

$\textbf{Area of \triangle ABC}$

$=\frac{1}{2}{\times}AB{\times}BC$

$=\frac{1}{2}{\times}7{\times}24$

$=7{\times}12$

$=84\;cm^2$

$\textbf{Area of \triangle ACD}$

$=\frac{1}{2}{\times}AD{\times}CD$

$=\frac{1}{2}{\times}15{\times}20$

$=15{\times}10$

$=150\;cm^2$

$\therefore\textbf{Area of quadrilateral ABCD}$

$=\text{Area of triangle ABC}+\text{Area of ACD}$

$=84+150$

$=234\;cm^2$