Question

There is a pentagonal shaped park as shown in the figure. Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest another way of finding its area?

Answer 1

In the first case, the figure $ABCDE$ is divided into $2$ trapezium of equal area.

Now area of trapezium $DFBC$,

$=\frac{1}{2}(DF+CB)\times FB=\frac{1}{2}(30+15)\times 7.5m^2=\frac{1}{2}\times 45\times \frac{15}{2}=\frac{675}{4}{m}^2=168.75\therefore \text{Total area of thepentagon}ABCDE=2\times 168.75=337.5m^2$
In the second case, figure $ABCDE$ is divided into two parts, namely one square and other triangle.

$\text{Side of square}ABCE=15m\therefore Area=(a{)}^2=(15{)}^2{m}^2=15\times 15=225m^{2}Areaof△CDE=\frac{1}{2}\times EC\times DG=\frac{1}{2}\times AB\times (DF-GF)=\frac{1}{2}\times 15\times (30-15)=\frac{1}{2}\times 15\times 15m^2=\frac{225}{2}=112.5m^2\therefore TotalareaofpentagonABCDE=225+112.5=337.5m^2$