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,

AB^{2} + OB^{2} =AO^{2}

⇒ AB ^{2} = AO^{2} – OB ^{2}

⇒ AB= √(AO^{2} – OB ^{2})

⇒ AB= √(5^{2} – 3^{2})

⇒ AB= √(25 – 9)

⇒ AB = √16

⇒ AB= 4 cm

Similarly,

By Pythagoras Theorem in ∆COB,

AB^{2} + OB^{2} =CO^{2}

⇒ CB ^{2} = CO^{2} – OB ^{2}

⇒ CB= √(CO^{2} – OB ^{2})

⇒ CB= √(5^{2} – 3^{2})

⇒ 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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