CD is the diameter which meets the chord AB at E. AE = BE = 4 cm. If DE = 3cm , then the radius of the circle is

One

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Two

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Infinite

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Not possible

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Solution

The correct option is D
Not possible

Given DE = 3 cm, AE = 4cm

Let radius be r cm $\Rightarrow O D = r c m$

$∴ O E = (r - 3) c m$

Consider $△ O A E$ , $∠ O E A = 90^{o}$

Applying Pythagoras theorem

$O A^{2} = O E^{2} + A E^{2}$

$r^{2} = (r - 3)^{2} + 4^{2}$

$r^{2} = r^{2} + 9 - 6 r + 16$

$\Rightarrow r = 4 \frac{1}{6}$

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CD is the diameter which meets the chord AB at E. AE = BE = 4 cm. If DE = 3cm , then the radius of the circle is

The correct option is D Not possible Given DE = 3 cm, AE = 4cm Let radius be r cm ⇒ OD=rcm ∴OE=(r−3) cm Consider △OAE, ∠OEA=90o Applying Pythagoras theorem OA2 = OE2 + AE2 r2 = (r−3)2 + 42 r2 = r2 + 9 − 6r + 16 ⇒r= 416