Question

In a pentagon $ABCDE$, $DP$ is drawn perpendicular to $AB$ and is perpendicular to $CE$ also at point $Q$. If $AP=BP=12$ cm, $EQ=CQ=8$ cm, $DE=DC=10$ cm and $DP=18$ cm, find the area of the pentagon $ABCDE$.

• A. $288$ sq. cm
• B. $280$ sq. cm
• C. $188$ sq. cm
• D. $168$ sq. cm

Answer 1

The correct option is A $288$ sq. cm

$a^2+b^2=c^2$
$\Rightarrow a^2+8^2={10}^2$
$\Rightarrow a^2={10}^2-8^2$
$\Rightarrow a^2=36$
$a=6cm$
Since there are two triangles that measure 6cm. and 8cm., their area together would be
$6\times 8=48c{m}^2$
There are two rectangles that measure $12cm\times 8cm.=192c{m}^2$
There are two triangles at the base that measure $4\times 12=48c{m}^2$
$48+48+192=288c{m}^2$