Ex.10.6 Q3 Circles Solution - NCERT Maths Class 9

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Question

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
Circles
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.

Reasoning:

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 )\( ^{2}\) 

Steps:

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 \(\begin{align}\text{3 cm.}\end{align}\)

  
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