Ex.10.6 Q2 Circles Solution - NCERT Maths Class 9

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Question

Two chords \(\begin {align}{AB}\end {align}\) and \(\begin {align}{CD}\end {align}\) of lengths \(\begin {align}\text{5 cm}\end {align}\) and \(\begin {align}\text{11 cm}\end {align}\) respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between \(\begin {align}{AB}\end {align}\) and \(\begin {align}{CD}\end {align}\) is \(\begin {align}\text{6 cm,}\end {align}\) find the radius of the circle.

 Video Solution
Circles
Ex 10.6 | Question 2

Text Solution

What is known?

Two parallel chords of lengths \(5\rm{ cm}\) and \(11\rm{ cm}\) and the distance between parallel chords is \(6\rm{ cm.}\)

What is unknown?

Radius of the circle.

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 \(\begin {align}\text{5 cm}\end {align}\) and \(\begin {align}\text{11 cm.}\end {align}\) 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.

\[\begin {align}{AB}&=5 \mathrm{cm} \\ {CD}&=11 \mathrm{cm} \\ {MB}&=2.5 \mathrm{cm} \\ {ND}&=5.5 \mathrm{cm}\end{align}\]

Let  \( OM = x \,\,\rm{cm} \)  

and  \( ON = 6 – x\,\,\rm{cm} \)

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

By Pythagoras theorem,

\[\begin{align} {OM}^{2}+{MB}^{2} &={OB}^{2} \\ x^{2}+2.5^{2} &={OB}^{2} \\ x^{2}+6.25 &={OB}^{2} \ldots .(1) \end{align}\]

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

By Pythagoras theorem,

\[\begin{align}{ON}^{2}+{ND}^{2}&={OD}^{2} \\ (6-x)^{2}+5.5^{2}&={OD}^{2} \\ \begin{Bmatrix} 36+x^{2} \\ -12 x+30.25 \end{Bmatrix} &={OD}^{2} \\ \begin{Bmatrix} x^{2}-12 x \\ +66.25 \end{Bmatrix} &={OD}^{2} \ldots .(2)\end{align}\]

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

Equating (\(1\)) and (\(2\)) we get,

\[\begin{align}x^{2}+6.25&=x^{2}-12 x+66.25 \\ 12 x&=60 \\ x&=5\end{align}\]

Substituting the value of \(\begin{align} x\end{align}\) in (\(1\)) or (\(2\)) we get the radius of circle \(= 5.59 \rm cm.\)

  
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