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There are 2 circles of radius 1cm each and the centre of one lies on the circumference of the other.

Find the area of the common area.


Since both the circles are symmetric, we use the formula r^{2}( \alpha -sin( \alpha )).
Here,  \alpha is the angle subtended by the common chord at any one of the centres.
We have,  \alpha =\frac{2 \pi }{3}^{c}. Then the common area, A=\frac{4 \pi -3\sqrt{3}}{6} .
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Let the circumference intersect at point A and B.
The common cord will make a center angle Ф=120 degree
The common area is sum of the two segments with common cord AB.
Due to symmetry of circle both the segments are equal.
We have the area of segment= r²(π*Ф/180 - sinФ)/2
 The common area=2*r²(π*Ф/180-sinФ)/2
                           ⇒1²(π*120/180 - sin120)
                           ⇒ 2π/3 - 0.5806
                           ⇒2.0943- 0.5806
                           ⇒ 1.513 cm²
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