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Question

Two concentric circles are of radii $$ 5 cm $$ and $$ 3 cm $$. Find the length of the chord of the larger circle which touches the smaller circle.
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Solution

Let $$ O $$ be the common center of two concentric circles and let $$ AB $$ be a chord of larger circle touching the smaller circle at $$ P $$ join $$ OP $$

Since $$ OP $$ is the radius of the smaller circle to any chrod of the circle bisects the chord.

$$\therefore$$ $$ AP =BP $$

In right $$ \Delta APO  $$ we have
$$ OA^2 = AP^2 + OP^2 \Rightarrow 25 - 9 = AP^2 $$
$$ \Rightarrow AP^2 = 16 \Rightarrow AP = 4  $$

Now $$ AB = 2 , AP  = 2 \times 4 = 8 [ \because AP =PB ] $$
hence the length of the chord of the larger circle which touches the smaller circle is $$  8 cm $$.

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