Question

In the given figure, $\angle ABC={90}^0=\angle DEC,AC=15cm$ and $AB=9cm.$ If the area of the quadrilateral $ABCD$ is $72c{m}^2;$ find the length of $DE.$

• A. $3cm$
• B. $2cm$
• C. $4cm$
• D. $9cm$

Answer 1

The correct option is A $3cm$

$△ABC$ is a right angles triangle.
$\therefore A{C}^2=A{B}^2+B{C}^2$
$=>B{C}^2=A{C}^2-A{B}^2$
$={15}^2-9^2$
$=225-81$
$=144$
$=\sqrt{144}$
$BC=12$
Area of quadrilateral $ABCD=$Area of $△ABC+$Area of $△BCD$
$=>72=\frac{1}{2}\times AB\times BC+\frac{1}{2}\times BC\times DE$
$=>72=\frac{1}{2}\times 9\times 12+\frac{1}{2}\times 12\times DE$
$=>72=54+6\times DE$
$=>72-54=6DE$
$=>18=6\times DE$
$=>DE=\frac{18}{6}$
$=>DE=3cm$