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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Answer
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,
OC2 = OM2 + MC2 (By pythagoras theorem)
132 = 52 + MC2
MC2 = 132 - 52
MC2 = 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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