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Bunuel
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Consider the special case when the two semicircles are equal. In that case they will cross each other at the center of the big circle. Also the two semicircles together will form a full circle.

Let the center of the big circle O => OA is the radius of the big circle
Let OB be perpendicular to the diameter of the bottom semicircle => OB is the radius of that semicircle

We have OA = √2*OB => the area of the big circle is four times the area of the bottom semicircle, or twice the area of the shaded region
Thus the area of the shaded region is 4/2 = 2

The answer is C
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