In the following figure, AD is a straight line. OP &perp; AD and O is the centre of both the circles. If OA = 34 cm. OB = 20 cm and OP = 16cm; find the length of AB.

For the inner circle, BC is a chord and OP&perp; BC.We know that the perpendicular to a chord, from the centre of a circle, bisects the chord. &there4; BP = PCBy Pythagoras theorem,OA2 = OP2 = BP2⇒ BP2 = (20)2 - (16)2 =144&there4; BP= 12cm For the outer circle, AD is the chord and OP&perp;AD.We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.&there4; AP = PDBy Pythagoras Theorem,OA2 = OP2 + AP2⇒ AP2 = (34)2 − (16)2 = 900⇒ AP = 30 cmAB = AP − BP = 30 − 12 = 18 cm

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In the Following Figure, Ad is a Straight Line. Op ⊥ Ad and O is the Centre of Both the Circles. If Oa = 34 Cm. Ob = 20 Cm and Op = 16cm; Find the Length of Ab. -

Question

In the following figure, AD is a straight line. OP ⊥ AD and O is the centre of both the circles. If OA = 34 cm. OB = 20 cm and OP = 16cm; find the length of AB.

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Solution

For the inner circle, BC is a chord and OP⊥ BC.
We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.

∴ BP = PC
By Pythagoras theorem,
OA²= OP²= BP²
⇒ BP² = (20)² - (16)² =144
∴ BP= 12cm

For the outer circle, AD is the chord and OP⊥AD.
We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ AP = PD
By Pythagoras Theorem,
OA² = OP² + AP²
⇒ AP² = (34)² − (16)² = 900
⇒ AP = 30 cm
AB = AP − BP = 30 − 12 = 18 cm