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Two concentric circles are of radii $6$ cm and $10$ cm. Find the length of the chord of the larger circle which touches the smaller circle.
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Solution:
$BC^2 = OB^2 - OC^2$
$\Rightarrow BC^2 = 10^2 - 6^2 = 64$
$\Rightarrow BC = 8$ cm
$AB = 8 \times 2 = 16$ cm
$BC^2 = OB^2 - OC^2$
$\Rightarrow BC^2 = 10^2 - 6^2 = 64$
$\Rightarrow BC = 8$ cm
$AB = 8 \times 2 = 16$ cm