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Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

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In ∆AOO′, 

AO2 = 52 = 25 

AO′ 2 = 32 = 9 

OO′ 2 = 42 = 16 

AO′ 2 + OO′ 2 = 9 + 16 = 25 = AO

⇒ ∠AO′O = 90° [By converse of pythagoras theorem] 

Similarly, ∠BO′O = 90°. 

⇒ ∠AO′B = 90° + 90° = 180° 

⇒ AO′B is a straight line. whose mid-point is O. 

⇒ AB = (3 + 3) cm = 6 cm Ans

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