Question

The polygon $ABCDE$ is divided into parts as shown in fig., find it area, if $AD=9$ cm, $AH=8$ cm, $AF=3$ cm, $AG=6$ cm and perpendicular $BF=2$ cm, $CH=2.5$ cm, $FG=3$ cm.

Answer 1

Considering $EG=3cm$ instead of $FG=3cm$

Area of polygon $ABCDF$ = Area of $△AFB$ + Area of trapezium $FBCH$+Area of $△CHD$+Area of $△ADE$

$=\frac{1}{2}\times AF\times BF+FH\times \left(\frac{BF+CH}{2}\right)+\frac{1}{2}\times HD\times CH+\frac{1}{2}\times AD\times GE$

$=\frac{1}{2}\times 3\times 2+5\times \left(\frac{2+2.5}{2}\right)+\frac{1}{2}\times 1\times 2.5+\frac{1}{2}\times 9\times 3$

$=3+5.625+1.25+13.5$

$=23.375c{m}^2$