AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.

Given, AM = 18 cm and MB = 8 cm From figure, AB = AM + MB = 18 + 8 = 26 cm Radius of the circle = = 13 cm. Let CD is the shortest chord drawn through M. ∴ CD ⊥ AB From figure, OM = AM - AO = 18 - 13 = 5 cm OC (radius) = OA = 13 cm Now in right ∆OMC, OC 2 = OM 2 + MC 2 (By pythagoras theorem) 13 2 = 5 2 + MC 2 MC 2 = 13 2 - 5 2 MC 2 = 169 - 25 = 144 MC = = 12 cm. M is Mid-Point of CD (As perpendicular from center to the chord bisects it.) CD = 2 × MC = 2 × 12 = 24 cm. Hence, length of shortest chord = 24 cm.

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Mathematics

AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.

Circles

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