Question

Find the area of polygon $MNOPQR$ if $MP=9cm,MD=7cm,MC=6cm,MB=4cm,MA=2cm$
$NA,OC,QD$ and $RB$ are perpendiculars to diagonal $MP$.

Answer 1

$AB=MB-MA=4-2=2cm$

$AC=MC-MA=6-2=4cm$

$BD=MD-MB=7-4=3cm$

$CP=MP-MC=9-6=3cm$

$DP=MP-MD=9-7=2cm$

Area of trapezium$=\frac{1}{2}h(a+b)$, where $a$ and $b$ are parallel sides and $h$ is the height of the trapezium.

Area of triangle$=\frac{1}{2}bh$ where $b$ is the base of the triangle and $h$ is the height of the triangle.

$Area\left(MNOPQR\right)=Area\left(MNA\right)+Area\left(ANOC\right)+Area\left(COP\right)+Area\left(DQP\right)+Area\left(BRQD\right)+Area\left(MBR\right)$

$\Rightarrow Area\left(MNOPQR\right)=\frac{1}{2}(2\times 2.5+4\times (3+2.5)+3\times 3+2\times 2+3\times (2+2.5)+4\times 2.5)$

$\Rightarrow Area\left(MNOPQR\right)=\frac{1}{2}(5+22+9+4+13.5+10)=\frac{63.5}{2}c{m}^2=31.75c{m}^2$