Question

Find the area of the irregular polygon shaped fields given below.
(4 marks)

Answer 1

Area of the irregular field = Area of ∆AHF + Area of trapezium FHIE + Area of triangle EID + Area of ∆JDC + Area of rectangle BGJC + Area of ∆AGB.
(0.5 marks)

Area of the triangle =$\frac{1}{2}\times base\times height$
Area of $△$AHF = $\frac{1}{2}\times (80+60)\times 50m^2$$=25\times 140$ $m^2$
$=3500$$m^2$
(0.5 marks)

Area of trapezium FHIE = $\frac{1}{2}$$h(a+b)$ sq. units
= $\frac{1}{2}\times$35 × (50 + 40)) $m^2$
= $\frac{1}{2}$ (35 × 90) = 1575 $m^2$
(0.5 marks)

Area of $△$EID = $\frac{1}{2}$ (65 + 25) × 40 $m^2$
= 90 × 20 $m^2$
= 1800 $m^2$
(0.5 marks)

Area of $△$JDC = $\frac{1}{2}\times$25 x 50 $m^2$ = 625 $m^2$
(0.5 marks)

Area of rectangle BGJC = (60 + 35 +65) $\times$50
= 160 x 50 = 8000 $m^2$
(0.5 marks)

Area of the triangle AGB = $\frac{1}{2}\times 80\times 50$ = 2000 $m^2$
(0.5 marks)

Area of the field = (1) + (2) + (3) + (4) + (5) + (6)
= 3500 + 1575 + 1800 +625 + 8000 + 2000 $m^2$
= 17,500 $m^2$
⇒ Area of the field = 17, 500 $m^2$
(0.5 marks)