Find the area of the pentagon shown in fig. 20.48,

The given figure is:

Given:

$\mathrm{AD}=10 \mathrm{~cm}, \mathrm{AG}=8 \mathrm{~cm}, \mathrm{AH}=6 \mathrm{~cm}, \mathrm{AF}=5 \mathrm{~cm}$

$\mathrm{BF}=5 \mathrm{~cm}, \mathrm{CG}=7 \mathrm{~cm}, \mathrm{EH}=3 \mathrm{~cm}$

$\therefore \mathrm{FG}=\mathrm{AG}-\mathrm{AF}=8-5=3 \mathrm{~cm}$

And, $\mathrm{GD}=\mathrm{AD}-\mathrm{AG}=10-8=2 \mathrm{~cm}$

From given figure:

Area of Pantagon $=($ Area of triangle AFB $)+($ Area of trapezium FBCG $)+($ Area of triangle CGD $)+($ Area of triangle ADE $)$

$=\left(\frac{1}{2} \times \mathrm{AF} \times \mathrm{BF}\right)+\left[\frac{1}{2} \times(\mathrm{BF}+\mathrm{CG}) \times(\mathrm{FG})\right]+\left(\frac{1}{2} \times \mathrm{GD} \times \mathrm{CG}\right)+\left(\frac{1}{2} \times \mathrm{AD} \times \mathrm{EH}\right)$

$=\left(\frac{1}{2} \times 5 \times 5\right)+\left[\frac{1}{2} \times(5+7) \times(3)\right]+\left(\frac{1}{2} \times 2 \times 7\right)+\left(\frac{1}{2} \times 10 \times 3\right)$

$=\left(\frac{25}{2}\right)+\left[\frac{36}{2}\right]+\left(\frac{14}{2}\right)+\left(\frac{30}{2}\right)$

$=12.5+18+7+15$

$=52.5 \mathrm{~cm}^{2}$