edited by
781 views
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. $\sqrt{2} – 1$
edited by

1 Answer

Best answer
1 votes
1 votes

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

Related questions

0 votes
0 votes
1 answer
1
0 votes
0 votes
0 answers
2
0 votes
0 votes
0 answers
3
go_editor asked Dec 29, 2015
441 views
Consider the triangle $\text{ABC}$ shown in the following figure where $\text{BC = 12 cm, DB =9 cm, CD=6 cm,}$ and $\angle \text{BCD} = \angle \text{BAC}.$ What is the ra...