#### 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.

#### 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,

OA^{2 }= OP^{2 }= BP^{2}

⇒ BP^{2} = (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,

OA^{2} = OP^{2} + AP^{2}

⇒ AP^{2} = (34)^{2} − (16)^{2} = 900

⇒ AP = 30 cm

AB = AP − BP = 30 − 12 = 18 cm

Is there an error in this question or solution?

Solution 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. Concept: Chord Properties - a Straight Line Drawn from the Center of a Circle to Bisect a Chord Which is Not a Diameter is at Right Angles to the Chord.