Find the area enclosed by the following figure:

Question

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Answer

The given shape contains a semi-coulam and a rectains.

Area of rectangle $=1 \times 13 \times 4=52 \mathrm{~cm}^2$
$\therefore$ Area of semi-coulam $=\frac{1}{2} \pi \mathrm{r}^2=\frac{1}{2} \times \frac{22}{7} \times 10 \times 10=\frac{1100}{7} \mathrm{~cm}^2$
$\therefore$ Total area of triangle $=\frac{1}{2} \times$ Base $\times$ Height $=\frac{1}{2} \times 20 \times 7=70 \mathrm{~cm}^2$
Hence, total area encloudsed by the shap $\frac{1100}{7}+70=\frac{1100+490}{7}=\frac{1560}{7}=227 \mathrm{~cm}^2$ (approx.)

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