Question

In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the distance between the chords if they are
(i) on the same side of the centre
(ii) on the opposite sides of the centre.

Answer 1

We have:
(i)
Let AB and CD be two chords of a circle such that AB is parallel to CD on the same side of the circle.
Given: AB = 8 cm, CD = 6 cm and OB = OD = 5 cm
Join OL and OM.

The perpendicular from the centre of a circle to a chord bisects the chord.
∴ LB=$\frac{AB}{2}=\frac{8}{2}=4cm$
Now, in right angled ΔBLO, we have:
OB² = LB² + LO²
⇒ LO² = OB² − LB²
⇒ LO² = 5² − 4²
⇒ LO² = 25 − 16 = 9
∴ LO = 3 cm
Similarly, MD$=\frac{CD}{2}=\frac{6}{2}=3cm$
In right angled ΔDMO, we have:
OD² = MD² + MO²
⇒ MO² = OD² − MD²
⇒ MO² = 5² − 3²
⇒ MO² = 25 − 9 = 16
⇒MO = 4 cm
∴ Distance between the chords = (MO − LO) = (4 − 3) cm = 1 cm
(ii)
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the centre.
Given: AB = 8 cm and CD = 6 cm
Draw OL ⊥ AB and OM ⊥ CD.

Join OA and OC.
OA = OC = 5 cm (Radii of a circle)
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL=$\frac{AB}{2}=\frac{8}{2}=4cm$
Now, in right angled ΔOLA, we have:
OA² = AL² + LO²
⇒ LO² = OA² − AL²
⇒ LO² = 5² − 4²
⇒ LO² = 25 − 16 = 9
∴ LO = 3 cm
Similarly, CM=$\frac{CD}{2}=\frac{6}{2}=3cm$
In right angled ΔCMO, we have:
OC² = CM² + MO²
⇒ MO² = OC² − CM²
⇒ MO² = 5² − 3²
⇒ MO² = 25 − 9 = 16
∴ MO = 4 cm
Hence, distance between the chords = (MO + LO) = (4 + 3) cm = 7 cm