Discussion

Explanation:

We need to find the area of the shaded part in the figure below.

The area of the common region will be twice the area of segment PBQ (shaded)

In ∆APB, AP = BP = AB = 1 cm. Hence, ∆APB is an equilateral triangle.

∴ ∠PAB = 60° = ∠QAB
∴ ∠PAQ = 120°

Area of segment PBQ = Area of sector APBQ - Area of triangle APQ

Now, Area of triangle APQ = Area of triangle APB

⇒ Area of segment PBQ = $\frac{120}{360}$ × π × 1² - $\frac{\sqrt{3}}{4}$ × 1²

⇒ Area of segment PBQ = $\frac{\mathrm{\pi }}{3}$ - $\frac{\sqrt{3}}{4}$ = 

∴ Area of region common to both circles = 2 × $\left(\frac{\mathrm{\pi }}{3}-\frac{\sqrt{3}}{4}\right)$ = $\frac{2\mathrm{\pi }}{3}$ - $\frac{\sqrt{3}}{2}$

Hence, option (b).

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