PareshGmat wrote:

Two circles with equal radius pass through centre of each other as shown in figure. If radius = r, the area of the overlapping region would be:

Attachment:

The attachment **Circle.jpg** is no longer available

A: \(2\pi r^2 - \frac{\sqrt{3}}{2}\)

B: \(2\pi r^2 - \frac{\pi}{8}\)

C: \(r^2(\frac{4\pi}{3} - \frac{\sqrt{3}}{4})\)

D: \(r^2(\frac{\pi}{3} - \sqrt{2})\)

E: \(r^2(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})\)

Attachment:

1.jpg [ 19.56 KiB | Viewed 46025 times ]
In the diagram above, we need to recognize that \(AB\) is the radius for both the circles. Since the line \(AB\) is the radius, \(A\) is the center of the circle on the left and \(B\) is the center of the circle on the right. So, \(AC, BC, AD\) and \(BD\) are all radii of the circles and \(AB = BC = AC = AD = BD = r\)

Which makes \(\triangle ABC\) and \(\triangle ABD\) equilateral. So, angles \(ABC, BAC, BAD\) and \(ABD\) all measure \(60^o\).

So, angle \(CAD = CBD = 120^o\)

Now the area required will be

Area of sector \(CBD\) (Area enclosed by line segments \(AC\) & \(AD\) and arc \(CBD\))+ Area of sector CAD (Area enclosed by line segments \(BC\) & \(BD\) and arc \(CAD\)) - Area of rhombus CADBSince, the two circles are the same, the areas of the two sectors will be the same. Also, the rhombus can be seen as two equilateral triangles with equal areas. So, our equation reduces to

Required Area = \(2 *\) Area of sector \(CBD\) - \(2 *\) Area of \(\triangle CAB\)\(= 2* \frac{\pi}{3} * r^2 - 2 * \frac{\sqrt3}{4} * r^2\)

\(= r^2(\frac{2\pi}{3} - \frac{\sqrt{3}}{2})\)

Hence the answer is E.

Hope that helps.

_________________

The buttons on the left are the buttons you are looking for