In the given figure, $O$ is the centre of circle and $A B = B C$ , $O D$ and $O E$ are perpendicular to $B C$ and $A B$ respectively. $O D$ is perpendicular to $O E$ also $O D = 5$ cm. Find $D E$ .

5

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5 \sqrt{2}

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10

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10 \sqrt{2}

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Solution

The correct option is B $5 \sqrt{2}$ cm

We know, equal chords are equidistance from the centre of the circle.
as, $A B = B C$
so, $O D = O E$

Now, in $△ D O E$ right angled at $O$ .
applying pythagoras theorem,
$D E^{2} = O D^{2} + O E^{2}$
$D E^{2} = 5^{2} + 5^{2}$
$D E^{2} = 25 + 25$
$D E = \sqrt{2 \times 25}$
$D E = 5 \sqrt{2}$ cm

Hence, correct answer is option $(b)$

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