In circle of radius $5 c m, A B$ and $A C$ are the two chords such that $A B = A C = 6 c m$ . Find the length of chord $B C$ .

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Solution

$A D$ is the bisector of $∠ B A C$ and perpendicular bisector of $B C$ passing through centre $O$ . $¯ ¯¯¯¯¯¯¯ ¯ A D$ and $¯ ¯¯¯¯¯¯ ¯ B C$ at midpoint $M$ .
$∴ ¯ ¯¯¯¯¯¯¯¯ ¯ B M = ¯ ¯¯¯¯¯¯¯¯ ¯ C M$
In right angled triangle, $△ A M C$
${M C}^{2} = {A C}^{2} - {A M}^{2} = 36 - {A M}^{2}$
In right angled triangle, $△ O M C$
${M C}^{2} = {O C}^{2} - {O M}^{2}$
$= 25 - {(A O - O M)}^{2} = 25 - {(5 - A M)}^{2}$
$∴ 36 - {A M}^{2} = 25 - {(5 - A M)}^{2}$
or $A M = 3.6 c m$
$¯ ¯¯¯¯¯¯¯¯ ¯ M C = \sqrt{36 - {(3.6)}^{2}}$
$= \sqrt{23.04} 4.8 c m$
$∵ B C = 2 M C$
$∴ B C = 2 \times 4.8 = 9.8 c m$

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