Question

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.

• A. 4 cm, 4 cm
• B. 4 cm, 8 cm
• C. 8 cm, 4 cm
• D. 8 cm, 8 cm

Answer 1

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.

$\therefore AE=BE=DF=CF=\frac{6}{2}=3cm.$

Now, using Pythagoras' theorem in $△OEB,$ we have

$⟹O{E}^2+B{E}^2=O{B}^2$

$⟹O{E}^2+3^2=5^2$

$⟹O{E}^2=25-9$

$⟹O{E}^2=16$

$⟹O{E}^2=4^2$

$⟹OE=4cm$

Since equal chords are equidistant from the center, therefore,

$OE=OF=4cm.$