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A chord of a circle is $12 cm$ in length and its distance from the centre is $8 cm$ . The length of the chord of the same circle which is at a distance of $6 cm$ from the centre

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30 cm

24 cm

16 cm

18 cm

answer is C.

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Detailed Solution

From the question we have,
Question Image

OA

is the radius of a circle.
Perpendicular drawn from the centre bisects the chord.
Therefore,

AM = \frac{1}{2} AB = \frac{12}{2} = 6 cm

AM = 6 cm

∠ OMA = 90^{\circ}

From Pythagoras theorem, we know that

∵ H^{2} = P^{2} + B^{2} {OA}^{2} = {OM}^{2} + {MA}^{2}

OA = \sqrt{{OM}^{2} + {MA}^{2}}

OA = \sqrt{8^{2} + 6^{2}}

OA = \sqrt{100} cm

OA = 10 cm

Radius of a circle is

10 cm

.
Therefore,

OD = 10 cm

Now, we will have to find the length of the chord CD.
Similarly OP is perpendicular to

CD

so it bisects

CD

Triangle

OPD

is right angled at

P

∠ OPD = 90^{\circ}

By Pythagoras theorem,

∵ H^{2} = P^{2} + B^{2}

{OD}^{2} = {OP}^{2} + {PD}^{2}

PD = \sqrt{{OD}^{2} - {OP}^{2}}

PD = \sqrt{10^{2} - 6^{2}}

PD = \sqrt{64} cmPD = 8 cm

CD = 2 . PD = 16 cm .

Hence, the length of chord (CD)

= 16 cm

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