In the following figure, if chords AB and CD are equal to 6 cm, then the distances of the chords from the centre of the circle of radius 5 cm are _ and _ respectively.

4 cm, 4 cm

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4 cm, 8 cm

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8 cm, 4 cm

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8 cm, 8 cm

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Solution

The correct option is A
4 cm, 4 cm

Given that chords AB and CD are equal to 6 cm.
Draw perpendiculars from the centre of the circle to these chords.
We know the property that the perpendicular drawn from the centre of a circle to the chord bisects the chord.
$∴ A E = B E = D F = C F = \frac{6}{2} = 3 c m .$

Considering $△ O E B$ and $△ O F D,$ we have
OB=OD, (Radii of the same circle)

$∠ O E B = ∠ O F D = 90^{\circ}$

BE=DF.

$Thus, by RHS criterion, △ O E B ≅ △ O F D .$

$⟹ O E = O F$ (By C.P.C.T.)
But, using Pythagoras' theorem in $△ O E B,$ we have
$O E^{2} + B E^{2} = O B^{2} .$
$⟹ O E^{2} + 3^{2} = 5^{2} ⟹ O E = 4 c m$
Thus, $O E = O F = 4 c m .$

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