Question
Find the area enclosed by the following figure: 
✓
Answer
The given shape contains a rectains and a semi-circle.
$\therefore$ Area of rectangle $=1 \times \mathrm{b}=(10.2 \times 1.5) \mathrm{cm}^2=15.3 \mathrm{~cm}^2$
Here, diameter of semi-circle $=(10.2-3.9) \mathrm{cm}=6.3 \mathrm{~cm}$
So, radius $=\frac{\text { diameter }}{2}=\frac{6.3}{2}=3.15 \mathrm{~cm}$
$\therefore$ Area of semi-circle $\frac{1}{2} \pi \mathrm{r}^2=\frac{22}{7} \times \frac{1}{2} \times 3.15 \times 3.15=15.59 \mathrm{~cm}^2$
$\therefore$ Total area $=$ area of rectengle + area of semi-circle $=15.3+3.15=30.89 \mathrm{~cm}^2$
$\therefore$ Area of rectangle $=1 \times \mathrm{b}=(10.2 \times 1.5) \mathrm{cm}^2=15.3 \mathrm{~cm}^2$
Here, diameter of semi-circle $=(10.2-3.9) \mathrm{cm}=6.3 \mathrm{~cm}$
So, radius $=\frac{\text { diameter }}{2}=\frac{6.3}{2}=3.15 \mathrm{~cm}$
$\therefore$ Area of semi-circle $\frac{1}{2} \pi \mathrm{r}^2=\frac{22}{7} \times \frac{1}{2} \times 3.15 \times 3.15=15.59 \mathrm{~cm}^2$
$\therefore$ Total area $=$ area of rectengle + area of semi-circle $=15.3+3.15=30.89 \mathrm{~cm}^2$
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