In two concentric circles, a chord length 80 cm of larger circle becomes a tangent to the smaller circle whose radius is 9 cm. The radius of the larger circle will be

13 cm

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41 cm

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52 cm

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75 cm

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Solution

The correct option is B
41 cm
$R = \sqrt{40^{2} + 9^{2}}$

$R = 41 c m$

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In two concentric circles, a chord length 80 cm of larger circle becomes a tangent to the smaller circle whose radius is 9 cm. The radius of the larger circle will be

The correct option is B 41 cm R=√402+92 R=41 cm

The correct option is B
41 cm
$R = \sqrt{40^{2} + 9^{2}}$

$R = 41 c m$