A chord $A B$ of a circle whose center is $O,$ is bisected at $E$ by a diameter $C D .$ $O C = O D = 15 c m$ and $O E = 9 c m .$ Find the length of $B C .$

$10 \sqrt{3} c m$

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$6 \sqrt{10} c m$

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$12 c m$

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$12 \sqrt{5} c m$

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Solution

The correct option is A: $12 \sqrt{5} c m$

Given: $O E = 9 c m, O A = O C = O D = 15 c m$

Consider $△ O A E$ , $∠ O E A = 90^{o}$
[Since the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord]

Applying Pythagoras' theorem in $△ A O E,$ we have

$A E^{2} = O A^{2} - O E^{2}$

$A E^{2} = 15^{2} - 9^{2}$

$A E^{2} = 225 - 81 = 144$

$\Rightarrow A E = 12 c m$ and $B E = 12 c m$ [ $∵ E$ is the midpoint of $A B$ ]

$∴ C E = O C + O E = 15 + 9 = 24 c m$

Consider $△ C E B$ , $∠ C E B = 90^{\circ}$

Applying Pythagoras theorem

$B C^{2} = C E^{2} + E B^{2}$

$B C^{2} = 24^{2} + 12^{2}$

$B C^{2} = 576 + 144 = 720$

$\Rightarrow B C = \sqrt{720} = \sqrt{2 \times 2 - ---- - \times 2 \times 2 - ---- - \times 3 \times 3 - ---- - \times 5}$
$∴ B C = 12 \sqrt{5} c m$

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