Two concentric circles are of radii 7 cm and 'r' cm respectively, where r >7 .A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of 'r' ?

Let O be the centre of the concentric circles. AB is the chord of the larger circle and tangent to the smaller circle at D.

Given,  OD = 7 cm, OB = r cm and AB = 48 cm

AB is the tangent to the smaller circle.

∴ ∠ODB = 90°    (Radius is perpendicular to the tangent at point of contact)

OD⊥AB,

In ΔODB,

OB2 = OD2 + BD2

Thus, the value of r is 25 cm.

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Let O be the centre of the two concentric circles. Let PQ be the chord of the larger circle


touching the smaller circle at M.



We have PQ = 48 cm,


Radius of the smaller circle, OM = 7 cm


Let the radius of the larger circle be r, i.e. OP = r


Since PQ is a tangent to the inner circle, OM  PQ


Thus, OM bisects PQ.


 PM = MQ =


48


cm 24 cm


2



Applying Pythagoras Theorem in OPM,


OP2 = OM2 + PM2


 OP2 = (7 cm)2 + (24 cm)2 = (49 + 576) cm2 = 625 cm2 = (25 cm)2


 OP = 25 cm


Radius of the larger circle is 25 cm.


Thus, the value of r is 25 cm.

Thumbs up please.....!!!!

  • 5

Hey Shubham,,plz rewrite the answer , i cannot get it,...but i assure you,i will give u a thumbs up after the rewritten answer.Thx.

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