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Let $\text{ABC}$ be a right-angled isosceles triangle with hypotenuse $\text{BC}$. Let $\text{BQC}$ be a semi-circle, away from A, with diameter $\text{BC}$. Let $\text{BPC}$ be an arc of a circle centered at $\text{A}$ and lying between $\text{BC}$ and $\text{BQC}$. If $\text{AB}$ has length $6$ cm then the area, in sq cm, of the region enclosed by $\text{BPC}$ and $\text{BQC}$ is

1. $9 \pi-18$
2. $18$
3. $9\pi$
4. $9$

We can draw the diagram.

The $\triangle \text{CAB}$ is a right-angle triangle.

We can apply the Pythagorean theorem.

$\text{(BC)}^{2} = \text{(CA)}^{2} + \text{(AB)}^{2}$

$\Rightarrow \text{(BC)}^{2} = 6^{2} + 6^{2}$

$\Rightarrow \text{BC} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$

The required area $= \text{area(BQC) – area(BPC)} = \underbrace{\text{area(BQC)}}_{\text{Semi-circle}} – [\;\underbrace{\text{area(ABPC)}}_{\text{Quarter circle}} – \underbrace{\text{area(ABC)}}_{\text{Triangle}}\;]$

Now, the area of semi-circle $\text{BQC:}$

• Radius of semi circle $\text{BQC}= \frac{BC}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$ cm
• The area of semi circle $\text{BQC}= \frac{\pi}{2}(3\sqrt{2})^{2} = \frac{18\pi}{2}= 9\pi\;\text{cm}^{2}$

And, the area of quarter circle $\text{ABPC:}$

• Radius of quarter circle $\text{ABPC = AB = 6 cm}$
• The area of a quarter circle $\text{ABPC} = \frac{\pi}{4}(6)^{2}=9\pi \;\text{cm}^{2}$

The area of triangle $\text{ABC} = \frac{1}{2} \times 6 \times 6 = 18\;\text{cm}^{2}$

$\therefore$ The required area $= 9\pi-(9\pi-18)= 9\pi-9\pi-18 =18\;\text{cm}^{2}.$

Correct Answer $: \text{B}$

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