Question

Find the area of the polygon given in the figure:

Answer 1

In the given figure

$AC=60m,AH=46m,AF=16m,EF=24m,DH=14m,BG=16m$

$\therefore$

$FH=AH-AF=46-16=14m$

In the figure, there are $3$ triangles and one trapezium

$A\left(△ABC\right)=\frac{1}{2}\times AC\times BG=\frac{1}{2}\times 60\times 16=480m^2$

$A\left(△AEF\right)=\frac{1}{2}\times AF\times EF=\frac{1}{2}\times 16\times 24=192m^2$

$△DHC=\frac{1}{2}\times HC\times DH=\frac{1}{2}\times 14\times 14=98m^2$

$A\left(EFHD\right)=\frac{1}{2}(EF+DH)\times FH=\frac{1}{2}(24+14)\times 30=570m^2$

Therefore, total area of the figure $=[$ Area of $△ABC$ + area of $△AEF$ + area of $△DHC$ + area of trapezium $EFHD]$

$=480+192+98+570=1340m^2$