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Ex.10.6 Q3 Circles Solution - NCERT Maths Class 9

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The lengths of two parallel chords of a circle are \(\begin{align} \text{6 cm}\end{align}\) and \(\begin{align} \text{8 cm.}\end{align}\) If the smaller chord is at distance \(\begin{align} \text{4 cm}\end{align}\) from the centre, what is the distance of the other chord from the centre?

 Video Solution
Ex 10.6 | Question 3

Text Solution

What is known?

Two parallel chords of lengths \(6\,\rm{ cm}\) and \(8\,\rm{ cm}\) and distance of smaller chord from centre is \(4\,\rm{cm.}\)

What is unknown?

Radius of the circle and distance of one chord from centre.


The perpendicular drawn from the centre of the circle to the chords, bisects it.

Pythagoras theorem states that

(Side \( 1)^{2}\) \(+\) (Side \( 2)^{2} = \) (Hypotenuse )\( ^{2}\) 


Draw \(2\) parallel chords \(\begin{align} {AB}\end{align}\) and \(\begin{align} {CD}\end{align}\) of lengths \(\begin{align} \text{6 cm}\end{align}\) and \(\begin{align} \text{8 cm.}\end{align}\) Let the centre of the circle be \(\begin{align} {O.}\end{align}\) Join one end of each chord to the centre. Draw \(2\) perpendiculars \(\begin{align} {OM}\end{align}\) and \(\begin{align} {ON}\end{align}\) to \(\begin{align} {AB}\end{align}\) and \(\begin{align} {CD }\end{align}\) respectively which bisects the chords.

\[\begin{align} {AB} &=6 \, \rm{cm} \\ {CD} &=8 \, \rm{cm} \\{MB} &=3 \, \rm{cm} \\ {ND} &=4 \,\rm{cm} \end{align}\]

Given \( OM = \text{4 cm } \)

and let \( {ON =} x\, \rm{cm}\)

Consider \(\begin{align} \Delta {OMB}\end{align}\)

By Pythagoras theorem,

\[\begin{align}{OM}^{2}+{MB}^{2}&={OB}^{2} \\ 4^{2}+3^{2}&={OB}^{2} \\ 16+9&={OB}^{2} \\ {OB}^{2}&=25 \\ {OB}&=5 \,\mathrm{cm}\end{align}\]

\(\begin{align}{OB}\end{align}\) and \(\begin{align}{OD}\end{align}\) are the radii of the circle.

Therefore \(OD = OB =\text{ 5 cm.} \)

Consider \(\begin{align}\Delta{OND}\end{align}\)

By Pythagoras theorem,

\[\begin{align}{ON}^{2}+{ND}^{2}&={OD}^{2} \\ x^{2}+4^{2}&=5^{2} \\ x^{2}&=25-16 \\ x^{2}&=9 \\ x&=3 \, \mathrm{cm}\end{align}\]

The distance of the chord \(\begin{align}{CD}\end{align}\) from the centre is \(\begin{align}\text{3 cm.}\end{align}\)

 Video Solution
Ex 10.6 | Question 3
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