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

Hint:

For intersecting circles, break the common region into identical segments from each circle.

Answer 1

The Correct Option is C

Distance between centres = radius = \(1\) cm. The intersecting region consists of two identical circular segments. Area of one segment = area of sector \(120^\circ\) – area of equilateral triangle. Sector area = \(\frac{120}{360} \pi r^2 = \frac{\pi}{3}\). Triangle area = \(\frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}\). So area of one segment = \(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\). Double it: \( \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \). But since each circle shares this twice, total intersecting area = \( \frac{4\pi}{3} - \frac{\sqrt{3}}{2} \).