Q. 105.0( 3 Votes )

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer :

Given:


AO (say) = CO (say) = 5 cm


BO (say) = 3 cm



Let AC be the tangent which meets the circle at the point B and O be the center of circle.


Property: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


By above property, ∆AOB is right-angled at OBA and ∆COB is right-angled at OBC.


Therefore,


By Pythagoras Theorem in ∆AOB,


AB2 + OB2 =AO2


AB 2 = AO2 – OB 2


AB= √(AO2 – OB 2)


AB= √(52 – 32)


AB= √(25 – 9)


AB = √16


AB= 4 cm


Similarly,


By Pythagoras Theorem in ∆COB,


AB2 + OB2 =CO2


CB 2 = CO2 – OB 2


CB= √(CO2 – OB 2)


CB= √(52 – 32)


CB= √(25 – 9)


CB = 16


CB= 4 cm


Now,


AC = AB + BC


= 4 cm + 4 cm


= 8 cm


Hence, Length of chord = 8 cm


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