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
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 isΒ
$$(ΟΓ4^2/4)=4Ο.$$
Area of the square is$$=4Γ4=16cm^2.$$
clearly the common sector have the half of square common to it.
So,area of one arc is
$$4Ο-16/2=4Ο-8 cm^2$$
So,Total area of the arc is$$8Ο-16$$
$$=8(Ο-2).$$
A is correct choice.
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