Question

Question 77

Find the area of the following field. All dimensions are in metres.

Answer 1

Area of the given figure = Area of $\Delta DCF$ + Area of $\Delta EGD$ + Area of trapezium FCBH + Area of $\Delta EGA$ + Area of $\Delta AHB$
$\therefore$ Area of $\Delta DCF=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 100\times 100=\frac{10000}{2}=5000m^2$
Area of $\Delta EGD=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 120\times 180=60\times 180=10800m^2$
Area of trapezium $=\frac{1}{2}\times \text{[Sum of parallel sides]}\times \text{Height}$
$=\frac{1}{2}\times [100+50]\times 110=\frac{1}{2}\times 150\times 110=75\times 110=8250m^2$
Area of $\Delta EGA=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 120\times 80=60\times 80=4800m^2$
Area of $\Delta AHB=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 50\times 50=25\times 50=1250m^2$
Thus, the area of the complete figure = 5000 + 10800 + 8250 + 4800 + 1250
$=30100m^2$