# 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

**Solution:**

The chord of the larger circle is a tangent to the smaller circle as shown in the figure below.

PQ is a chord of a larger circle and a tangent of a smaller circle.

Tangent PQ is perpendicular to the radius at the point of contact S.

Therefore, ∠OSP = 90°

In ΔOSP (Right-angled triangle)

By the Pythagoras Theorem,

OP^{2} = OS^{2} + SP^{2}

5^{2} = 3^{2} + SP^{2}

SP^{2} = 25 - 9

SP^{2} = 16

SP = ± 4

SP is the length of the tangent and cannot be negative

Hence, SP = 4 cm.

QS = SP (Perpendicular from center bisects the chord considering QP to be the larger circle's chord)

Therefore, QS = SP = 4cm

Length of the chord PQ = QS + SP = 4 + 4

PQ = 8 cm

Therefore, the length of the chord of the larger circle is 8 cm.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 10

**Video Solution:**

## 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

Maths NCERT Solutions Class 10 Chapter 10 Exercise 10.2 Question 7

**Summary:**

If two concentric circles are of radii 5 cm and 3 cm, then the length of the chord of the larger circle which touches the smaller circle is 8 cm.

**☛ Related Questions:**

- From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is(A) 7 cm(B) 12 cm(C) 15 cm(D) 24.5 cm
- In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to(A) 60°(B) 70°(C) 80°(D) 90°
- If tangents PA and PB from a point P to a circle with center O are inclined to each other at angle of 80°, then ∠POA is equal to(A) 50°(B) 60°(C) 70°(D) 80°
- Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

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