Question

Two concentric circles of radii 15 cm, 12 cm are drawn. Find the length of chord of larger circle which touches the smaller circle.

Answer 1

✬ Chord = 18 cm ✬

Step-by-step explanation:

Given:

• Radii of two concentric circles is 15 and 12 cm respectively.

To Find:

• Length of chord of larger circle which touches the smaller circle ?

Solution: Let A be the bigger circle of radius 15 cm and A' be the smaller circle of radius 12 cm. O is common centre of both the circles.

Let BC be a chord of circle A , passing by touching the smaller circle. Join O to C & B.

Here we have

• OC = OB = 15 cm (radii of A)

• OD = 12 cm (radius of A')

• BC = (chord of A)

• OD is also perpendicular to BC.

• ∠ODC = ∠ODB = 90°

• BD = DC (OD bisects BD)

[ See figure for understanding ]

Now in right angled ∆ODC we have

• OC {hypotenuse}
• DC {base}
• OD {perpendicular}

Using Pythagoras theorem

★ H² = Perpendicular² + Base² ★

$\implies{\rm }$ OC² = OD² + DC²

$\implies{\rm }$ 15² = 12² + DC²

$\implies{\rm }$ 225 – 144 = DC²

$\implies{\rm }$ √81 = DC

$\implies{\rm }$ 9 = DC

∴ BD = DC = 9 cm

Hence, BC = 9 + 9 = 18 cm. This is required length of chord which touches the smaller circle.

Answer 2

Diagram :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\qbezier(1.2,0)(1.121,1.121)(0,1.2)\qbezier(1.2,0)(1.121,-1.121)(0,-1.2)\qbezier(0,-1.2)(-1.121,-1.121)(-1.2,0)\qbezier(-1.2,0)(-1.121,1.121)(0,1.2)\put(2, - 1.2){\line(-1,0){4}}\put(0, - 1.2){\line(0,1){1.2}}\put( - 0.2, - 1.5){A}\put(0,0){O}\qbezier(0,0)(0,0)( - 2, - 1.2)\put( - 2.3, - 1.5){B}\put(2, - 1.5){C}\put( - 1.5, - 0.3){ \sf 15 cm}\put(0, - 0.8){ \sf 12 cm}\end{picture}$

Given :

• Two circles of radii 15 cm & 12 cm with common centre.

To Find :

• Find the length of chord of larger circle which touches the smaller circle.

Solution :

By Using Pythagoras theorem :

$\sf : \implies \: \: \: \: \: \: \: {12}^{2} = {15}^{2} + {PB}^{2} \\ \\ \\ \sf : \implies \: \: \: \: \: \: \:144 = 225 + {PB}^{2} \\ \\ \\ \sf : \implies \: \: \: \: \: \: \:{PB}^{2} = 225 - 144\\ \\ \\ \sf : \implies \: \: \: \: \: \: \: \:{PB}^{2} = 81 \\ \\ \\ \\ \sf : \implies \: \: \: \: \: \: \: \:PB \: = \: 9$

The perpendicular drawn from the centre of the circle to a chord, bisects it :

$\sf : \implies \: \: \: \: \: \: \: \:AB = 2 \times PB \\ \\ \\ \: \sf : \implies \: \: \: \: \: \: \: \:AB= 2 \times 9 \\ \\ \\ \sf : \implies \: \: \: \: \: \: \: \:AB= 18$

⛬ The length of the chord of the larger circle which touches the smaller circle is 18 cm.