A square hall PQRS has a semi-circular stage as shown.
The unshaded stage has the side PQ of the hall as its diameter and a perimeter of $36\ m$.
The perimeter of the shaded part of the hall is $64\ m$.
Find out the area of the shaded part of the hall in terms of $\pi$.
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Ans: ______________$m²$ [4]
Let $L$ be the length of square hall PQRS,
and $S$ be the length of the semi-circle part of the stage.
| $\Rightarrow$ $S$ | = $\pi$ $\times$ $\dfrac{L}{2}$ |
| Perimeter of stage | = $L + S$ |
| = $L\ +$ $\pi$ $\times$ $\dfrac{L}{2}$ | |
| = $L\ × (1 + \dfrac{π}{2})$ | |
| = $36\ m$ | |
| Perimeter of shaded part of the hall | = $3L + S$ = $64\ m$ |
| $\Rightarrow$ $(3L + S) - (L + S)$ | = $64 - 36$ |
| $2L$ | = $28$ |
| $L$ | = $14\ m$ |
| Area of shaded stage | = $14 × 14 - π × 7 × 7$ |
| = $\boxed{(196 - 49π)\ m²}$ |
$$L + \dfrac{\pi L}{2} = 40 \quad \text{(Perimeter of shaded stage)}$$ $$3L + \dfrac{\pi L}{2} = 64 \quad \text{(Perimeter of unshaded part)}$$ Subtracting the first equation from the second:$$(3L + \dfrac{\pi L}{2}) - (L + \dfrac{\pi L}{2}) = 64 - 40$$ $$2L = 24$$ $$L = 12\text{ m}$$ Radius of semi-circle:$$r = \dfrac{L}{2} = \dfrac{12}{2} = 6\text{ m}$$ Area of shaded stage:$$\text{Area} = \dfrac{1}{2} \pi r^2 = \dfrac{1}{2} \times \pi \times 6^2$$ $$= \dfrac{1}{2} \times \pi \times 36$$ $$= 18\pi\text{ m}^2$$ $$\boxed{18\pi\text{ m}^2}$$
