In the adjoining figure, BD is the diameter of the circle which bisects the chord AC at point E. If `AC=8cm`, `BE=2cm`, then find the radius of the circle.

In the adjoining figure, BD is the diameter of the circle which bisect

1.	In the adjoining figure, BD is the diameter of the circle which bisects the chord AC at point E. If `AC=8cm`, `BE=2cm`, then find the radius of the circle.
Answer» Let radius `OA=OB =r` `OE=OB-EB` `=(r-2)cm` `AE=(AC)/(2)=(8)/(2)=4cm` Now, `angle OEA=90^@` (`because` The line segment joining the centre to mid-point of a chord is perpendicular to the chord) `therefore` In `DeltaOAE`, `OA^2=OE^2+AE^2` (by Pythagoras theorem) `rArrr^2=(r-2)^2+4^2` `rArrr^2=r^2-4r+4+16` `rArr4r=20 cm` `rArr=5cm` `therefore` Radius of circle =5 cm .

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In the adjoining figure, BD is the diameter of the circle which bisects the chord AC at point E. If `AC=8cm`, `BE=2cm`, then find the radius of the circle.

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