Question

Question 76

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

Answer 1

Area of the given figure
= Area of $\Delta EFH$ + Area of rectangle EDCI + Area of trapezium FHJG + Area of trapezium ICBK + Area of $\Delta GJA$ + Area of $\Delta KBA$
Area of $\Delta EFH=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 40\times 80=40\times 40=1600m^2$
Area of rectangle EDCI $=\text{Length}\times \text{Breadth}=100\times 160=16000m^2$
Area of trapezium FHJG $=\frac{1}{2}\times [\text{Sum of parallel sides]}\times \text{Height}$
$=\frac{1}{2}\times [40+160]\times 160=\frac{200}{2}\times 160=100\times 160=16000m^2$
Area of trapezium ICBK $=\frac{1}{2}\times \text{[Sum of parallel sides]}\times \text{Height}$
$=\frac{1}{2}\times [60+100]\times 120=\frac{1}{2}\times 160\times 120=80\times 120=9600m^2$
Area of $\Delta AJG=\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}\times 160\times 100$
$=80\times 100=8000m^2$
Area of $\Delta KBA=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 60\times 60=60\times 30=1800m^2$
Thus, the area of the complete figure
$=1600+16000+16000+9600+8000+1800=53000m^2$