Question

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

• A. $288sq.cm$
• B. $2.88sq.cm$
• C. $28.8sq.cm$
• D. $288sq.m$

Answer 1

The correct option is A $288sq.cm$

$a^2+b^2=c^2$

$a^2+8^2={10}^2$

$a^2=\sqrt{100-64}=\sqrt{36}=6cm$

Since, there are two triangles that measures $6cm$ and $8cm$. Their area together would be $6\times 8=48{cm}^2$

There are two rectangles that measure $12cm\times 8cm\times 2=192{cm}^2$

There are two triangles at the base that measure $(4\times 12)=48{cm}^2$

$\therefore$ Total area $=48+192+48=288$ ${cm}^2$