Question

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

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

Answer 1

The correct option is A

4cm, 4cm

Given chords are AB and CD. Draw perpendicular from centre to the chords.

Perpendicular drawn from the centre bisects the chord. Therefore,

AE = BE = 3cm and DF = CF= 3 cm

Apply Pythagoras theorem in $\Delta$OEB and $\Delta$OFD

In $\Delta$OEB

${OE}^2$ + ${BE}^2$ = ${OB}^2$

${OE}^2$ = 25 - 9 = 16

$\Rightarrow$OE = 4 cm

Similarly, In $\Delta$OFD

${OF}^2$ + ${FD}^2$ = ${OD}^2$

${OF}^2$ = 25 - 9 = 16

$\Rightarrow$OF = 4 cm

So we see that equal chords are equidistant from the centre of the circle.