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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.

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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 

⇒ AB2 = AO2 – OB2 

⇒ AB= √(AO2 – OB2

⇒ AB= √(52 – 32

⇒ AB= √(25 – 9) 

⇒ AB = √16

⇒ AB = 4 cm

Similarly,

By Pythagoras Theorem in ∆COB, 

AB2 + OB2 = CO2 

⇒ CB2 = CO2 – OB2 

⇒ CB = √(CO2 – OB2

⇒ 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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