Question

The adjoining figure shows a field, with the measurements given in metres. Calculate the area of the field.

Answer 1

Area of $∆$DGC = $\frac{1}{2}\times DG\times GC$

$$\begin{gathered} =\frac{1}{2}\times 14\times 40 \\ =280{\mathrm{m}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}∆AFB=\frac{1}{2}\times AH\times BH \\ =\frac{1}{2}\times 8\times 35 \\ =140{\mathrm{m}}^2 \end{gathered}$$

Area of trapezium GHBC = $\frac{1}{2}\times \mathrm{Sum}\mathrm{of}\mathrm{its}\mathrm{parallel}\mathrm{sides}\times \mathrm{Distance}\mathrm{between}\mathrm{them}$

$$\begin{gathered} =\frac{1}{2}(BH+GC)\times GH \\ =\frac{1}{2}(35+40)\times 18 \\ =675{\mathrm{m}}^2 \end{gathered}$$

Now,
AD = AH + GH + DG = 8 + 18 + 14 = 40 m
Area of $∆$DEA = $\frac{1}{2}\times (DA\times EH)$
$$\begin{gathered} =\frac{1}{2}\times 40\times 26 \\ =520{\mathrm{m}}^2 \end{gathered}$$

∴ Area of quadrilateral EDCBA = Area of $∆$DGC + Area of $∆$AFB + Area of $∆$DEA + Area of trapezium GHBC
​ = 280 + 140 + 520+ 675
= 1615 m²