Question

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 $\Delta EFH=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 40\times 80$
$=40\times 40$
$=1,600m^2$

Area of rectangle $EDCI$ $=\text{Length}\times \text{Breadth}=100\times 160=16,000m^2$

Area of trapezium $FHJG$ $=\frac{1}{2}\times [\text{Sum of parallel sides]}\times \text{Height}$
$=\frac{1}{2}\times [40+160]\times 160=\frac{200}{2}\times 160=100\times 160=16,000m^2$

Area of trapezium $ICBK$
$=\frac{1}{2}\times \text{[Sum of parallel sides]}\times \text{Height}$
$=\frac{1}{2}\times [60+100]\times 120=\frac{1}{2}\times 160\times 120=80\times 120$
$=9,600m^2$

Area of $\Delta AJG=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 160\times 100$
$=80\times 100$
$=8,000m^2$

Area of $\Delta KBA=\frac{1}{2}\times \text{Base}\times \text{Height}$
$=\frac{1}{2}\times 60\times 60$
$=60\times 30$
$=1,800m^2$

Thus, the area of the complete figure
$=1,600+16,000+16,000+9,600+8,000+1,800=53,000m^2$