Find the area of given figure ABCDEFGH as per dimensions given in it.

Solution:

We will find the length of $\mathrm{AC}$.

From the right triangles $\mathrm{ABC}$ and HGF, we have:

$\mathrm{AC}^{2}=\mathrm{HF}^{2}=\left\{(5)^{2}-(4)^{2}\right\} \mathrm{cm}$

$=(25-16) \mathrm{cm}$

$=9 \mathrm{~cm}$

$\mathrm{AC}=\mathrm{HF}=\sqrt{9} \mathrm{~cm}$

$=3 \mathrm{~cm}$

Area of the given figure ABCDEFGH = (Area of rectangle ADEH) + (Area of $\Delta \mathrm{ABC})+($ Area of $\Delta \mathrm{HGF})$

$=($ Area of rectangle $\mathrm{ADEH})+2($ Area of $\Delta \mathrm{ABC})$

$=(\mathrm{AD} \times \mathrm{DE})+2($ Area of $\Delta \mathrm{ABC})$

$=\{(\mathrm{AC}+\mathrm{CD}) \times \mathrm{DE}\}+2\left(\frac{1}{2} \times \mathrm{BC} \times \mathrm{AC}\right)$

$=\{(3+4) \times 8\}+2\left(\frac{1}{2} \times 4 \times 3\right) \mathrm{cm}^{2}$

$=(56+12) \mathrm{cm}$

$=68 \mathrm{~cm}^{2}$

Hence, the area of the given figure is $68 \mathrm{~cm}^{2}$.