Question 81

Two identical circles intersect such that their centers and their points of intersection, form a square of side 4cm. Then the area (in sq.cms) of the portion that is common to the two circles is

Solution

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Side of the square is 4 cm.

So,radii of the two circles are also 4 cms.

So, area of the one of the arc is 

$$(π×4^2/4)=4π.$$

Area of the square is$$=4Ă—4=16cm^2.$$

clearly the common sector have the half of square common to it.

So,area of one arc is

$$4Ď€-16/2=4Ď€-8 cm^2$$

So,Total area of the arc is$$8Ď€-16$$

$$=8(Ď€-2).$$

A is correct choice.

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