Question

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 twocircles is:\

Answer 1

Step-by-step explanation:

We know that, Angle AOB= 90°, as it is the angle of a square. Since the side of the square is 1 Cm,

Radius of both the circles= 1 cm.

Now, Let's take any one circle ( say Left one.)

Here, Area of quadrant= 1×π r²/4,

= 1×π×(1)²/4

= π/4,

Or, 11/14 cm².

And, area of the segment of this circle = Area of the sector - Area of triangle ∆ AOB

$= \frac{11}{14} \times \frac{1 \times 1}{2} \\ = \frac{11}{28 \: }{cm}^{2}$

Similarly, area of the other missing part = 11/28 cm².

So, area of the total part( or the part which looks like a Convex lens) =

$\frac{11}{28} + \frac{11}{28} \\ = \frac{22}{28} \\ = \frac{11}{14} {cm}^{2}$