Question

Find the area of pentagonal field shown alongside in fig. 9.35. All dimensions are in metres.

Answer 1

Now, we have:
CH = 30 m
DH = 20 m
BG = 50 m
CG = (40 + 30) m = 70 m
AG = (50 + 50) m = 100 m
AF = 50 m
EF = 30 m
HF = (40 + 50) m = 90 m

Now, we have the following:

Area of $△$CDH = $\frac{1}{2}\times \mathrm{DH}\times \mathrm{CH}=\frac{1}{2}\times 20\times 30{\mathrm{m}}^2=300{\mathrm{m}}^2$

Area of $△$CGB = $\frac{1}{2}\times \mathrm{CG}\times \mathrm{BG}=\frac{1}{2}\times 70\times 50{\mathrm{m}}^2=1750{\mathrm{m}}^2$

Area of $△$AGB = $\frac{1}{2}\times \mathrm{BG}\times \mathrm{AG}=\frac{1}{2}\times 50\times 100{\mathrm{m}}^2=2500{\mathrm{m}}^2$

Area of $△$AEF = $\frac{1}{2}\times \mathrm{EF}\times \mathrm{AF}=\frac{1}{2}\times 30\times 50{\mathrm{m}}^2=750{\mathrm{m}}^2$

Area of trapezium EFHD = $\frac{1}{2}\times \mathrm{EF}\times \mathrm{AF}=\frac{1}{2}\times 30\times 50=750{\mathrm{m}}^2$

∴ Area of the pentagonal field = area of $△$CDH + area of $△$CGB + area of $△$AGB + area of $△$AEF + area of trapezium EFHD
= (300 + 1750 + 2500 + 750 + 2250) m²
= 7550 m²