in Quantitative Aptitude edited by
0 votes
0 votes

Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side $1$ cm. The area in sq. cm of the portion that is common to the two circles is

  1. $\pi / 4$
  2. $\frac{\pi}{2} – 1$
  3. $\frac{\pi}{5}$
  4. $\sqrt{2} – 1$
in Quantitative Aptitude edited by
13.4k points

1 Answer

1 vote
1 vote
Best answer

We have to find area of red portion.

Since side of square = 1, So radius of both the circles = 1.

Since ABCD is an square, so angle at centers of circle = 90.

Area of sector ACD = ($\pi$r$^{2}$)/4.

So area of red portion = 2 * area of sector ACD( equal to sector BCD) - area of square.

                                         = 2 * $\pi$/4 - 1

                                         = $\pi$/2 - 1.

Ans- B. 

selected by
1.5k points

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true