sriharimurthy wrote:

Refer to the figure I have attached.

Answer : C

Figure was missing but, Gotcha!!! Very very impressive!!!!

AOP is equilateral triangle since AP = OA = OP = 1cm. Same is the case downwards, let's call it B. Angle AOB will be 120 degree and so also angle APB will be 120 degree. Again the same with center Q as well.

So we see 120 degrees of circle with center O is in intersection region. 120 degrees of circle with center Q is in intersection region and 2 of 120 degrees of circle with center P are in intersection regions.

Therefore, 240 degrees of arc in circle O + 240 degrees of arc in circle Q + 120 degrees of arc in circle P will be the perimeter of the shaded region.

srihari you are just awesome in quant buddy!!!!

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I am AWESOME and it's gonna be LEGENDARY!!!