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Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circles is A. \(\sqrt r\)B. \(\sqrt2\) r AB C.\(\sqrt3\) r D. \(\frac{\sqrt 3}{2}\) r |
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Answer» Option : (C) Let O and O' be the centre of two circles OA and O'A = Radius of the circles AB be the common chord of both the circles OM perpendicular to AB And, O'M perpendicular to AB ΔAOO' is an equilateral triangle. AM = Altitude of AOO' Height of ΔAOO' = \(\frac{\sqrt 3}{2}\)r AB = 2 AM = 2\(\frac{\sqrt 3}{2}\)r = \(\sqrt{3}\)r |
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