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In a circle of radius $13$ cm, $PQ$ and $RS$ are two parallel chords of length $24$ cm and 10 cm respectively. The chords are on the same side of the centre then the distance between the chords is?

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7

12

5

10

answer is A.

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Detailed Solution

Consider the given question,
In a circle of radius

13

cm,

PQ

and

RS

are two parallel chords of length

24

cm and 10 cm respectively.
Draw the diagram according to the given data,
Question Image

Here,

OP = OR = 13 cm

PQ = 24 cm

RS = 10 cm

PM = 12 cm

and

RN = 5 cm

[Perpendicular from the centre bisects the chord]
In right angled triangle OMP and ONR ,
By pythagoras theorem,

{OP}^{2} = {OM}^{2} + {MP}^{2}

\Rightarrow {OM}^{2} = {OP}^{2} - {MP}^{2}

\Rightarrow OM = \sqrt{{OP}^{2} - {M P}^{2}}

\Rightarrow OM = \sqrt{13^{2} - 12^{2}}

\Rightarrow OM = \sqrt{169 - 144}

\Rightarrow OM = \sqrt{25} cm

\Rightarrow OM = 5 cm

Similarly,

{OR}^{2} = {ON}^{2} + {RN}^{2}

\Rightarrow {ON}^{2} = {OR}^{2} - {RN}^{2}

\Rightarrow ON = \sqrt{{OR}^{2} - {RN}^{2}}

\Rightarrow ON = \sqrt{13^{2} - 5^{2}}

\Rightarrow ON = \sqrt{169 - 25}

\Rightarrow ON = \sqrt{144}

\Rightarrow ON = 12 cm

Then, the distance between the chords is

MN = ON - OM

\Rightarrow MN = 12 - 5

\Rightarrow MN = 7 cm

Hence, distance between the chords is

7 cm

.
Therefore, option 1 is correct.

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