Question

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

• A. $51400{\text{m}}^2$
• B. $43400{\text{m}}^2$
• C. $53000{\text{m}}^2$
• D. $51200{\text{m}}^2$

Answer 1

The correct option is C $53000{\text{m}}^2$

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 $\Delta KBA$

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

Thus, the area of the complete figure
$=1600+16000+16000+9600+8000+1800=53000m^2$