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?

• A. \(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}\)
• B. \(\dfrac{2\pi}{3}+\dfrac{\sqrt{3}}{2}\)
• C. \(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\)
• D. \(\dfrac{4\pi}{3}+\dfrac{\sqrt{3}}{2}\)
• E. \(\dfrac{4\pi}{3}-\dfrac{\sqrt{3}}{2}\)

Hint:

The distance between the centres equals the radius, so the two centres and an intersection point form an equilateral triangle. The common region is made of two identical circular segments.

Answer 1

The Correct Option is C

Step 1: Locate the centres and intersection points:
Let the centres be \(O_1\) and \(O_2\), and the intersection points \(A\) and \(B\). Each circle passes through the other's centre, so \(O_1O_2=1\). Also \(O_1A=O_2A=1\). So triangle \(O_1AO_2\) is equilateral and \(\angle AO_1O_2=60^\circ\). Hence \(\angle AO_1B=120^\circ\).

Step 2: Area of one segment:
The common region is divided by the chord \(AB\) into two equal segments, one from each circle. For a circle of radius 1 and central angle \(120^\circ\):
Segment \(=\) sector \(-\) triangle \(=\dfrac{120}{360}\pi(1)^2-\dfrac12(1)(1)\sin120^\circ=\dfrac{\pi}{3}-\dfrac{\sqrt3}{4}\).

Step 3: Area of the common region:
Area \(=2\left(\dfrac{\pi}{3}-\dfrac{\sqrt3}{4}\right)=\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\approx1.228\) sq. cm.

Final Answer:
The correct option is (3): \(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\).