In two concentric circles of radii 6 cm and 10 cm with centre ‘O’. OP is the radius of the smaller circle, OP ⊥ AB which cuts the outer circle at A and B. Find the length of AB.

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Solution

It is given that OP = 6 cm, OA = 10 cm and OP⊥ AB.

AB is the chord of the outer circle and OP⊥ AB.

∴AP = BP (Perpendicular from the centre to the chord bisects the chord)

Applying Pythagoras theorem in right ΔOAP, we have:

∴AP = BP = 8 cm

∴ AB = AP + BP = 8 cm + 8 cm = 16 cm

Hence, the length of AB is 16 cm.

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