Four circles, each of radii 7, are put in a way that a square of side 14 is formed if the centers are joined in an orderly manner. Another circle is made to pass through the four centers (name it circle I) and yet another circle is made to pass through the four points of contact (name it circle II). How much more is the area of circle I than that of the circle II? Take π = 22/7.
(A) 49
(B) 98
(C) 154
(D) 176
(E) 198
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an orderly manner
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sanju09
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sanju09
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How would it suffice to the members at large?kstv wrote:IMO C
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kstv
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Circle I radius = R , where 2R²= 14² . The side of the square is 14 and forms a 90 45 45 triangle with the two radius
Circle II radius r = 7 r²= 7² this is inscribed in the square.
Area of CI - CII = 22/7(14²/2 - 7²) = 154
Circle II radius r = 7 r²= 7² this is inscribed in the square.
Area of CI - CII = 22/7(14²/2 - 7²) = 154
