**Question:**Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the other. What is the area (in sq. cm.) of the intersecting region?

- \(\frac{\pi }{3} - \frac{{\sqrt 3 }}{4}\)
- \(\frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}\)
- \(\frac{{4\pi }}{3} - \frac{{\sqrt 3 }}{2}\)
- \(\frac{{4\pi }}{3} + \frac{{\sqrt 3 }}{4}\)
- \(\frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{4}\)

**Correct Option:2**

The circumferences of the two circle pass through each other's centers. Hence, O1A = O1B=O1O2 = 1cm

By symmetry, the line joining the two centres would be bisect AB and would be bisected by AB. As the line joining the center to the midpoint of a chord is perpendicular to the chord, O1O2 and AB are perpendicular bisectors of each other. Suppose they intersect at point P. Hence, O1P = 1/2 cm. Hence, angle O1AB = sin x = (1/2)/1 = 1/2. Hence, O1AB = 30°. By symmetry, O1BA = 30° and hence angle at centre O1 = 120°.

In the above, the required area is 2 times A(segment ABO2)(blue region). And A(segment ABO2)(blue region) = A(sector O2AO1B)(blue + red) - A(triangleO1AB )(red)

Area of sector = 120°/360° × $\pi × 1^2 $ = $\pi/3$

Area of triangle = 1/2 × b × h = 1/2 × (2× 1 cos 30°) × (1/2) = ?3/4

Hence, required area = $\frac{\pi}{3}-\frac{\sqrt 3}{4}$ . Hence so the required area is 2 times the above value which is $\frac{2\pi}{3}-\frac{\sqrt 3}{2}$

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