Question

Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between the chords.

Answer 1

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 = 30 cm and CD = 16 cm
Draw OL ⊥ AB and OM ⊥ CD.

Join OA and OC.
OA = OC = 17 cm (Radii of a circle)
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ $AL\mathit{=}\left(\frac{\mathit{A}\mathit{B}}{\mathit{2}}\right)=\left(\frac{30}{2}\right)=15\mathrm{cm}$AL=AB2=(82)=4cm
Now, in right angled ΔOLA, we have:
OA² = AL² + LO²
⇒ LO² = OA² − AL²
⇒ LO² = 17² − 15²
⇒ LO² = 289 − 225 = 64
∴ LO = 8 cm

Similarly, $CM\mathit{=}\left(\frac{\mathit{C}\mathit{D}}{\mathit{2}}\right)=\left(\frac{16}{2}\right)=8\mathrm{cm}$CM=CD2=(62)=3cm
In right angled ΔCMO, we have:
⇒OC² = CM² + MO²
⇒ MO² = OC² − CM²
⇒ MO² = 17² − 8²
⇒ MO² = 289 − 64 = 225
∴ MO = 15 cm

Hence, distance between the chords = (LO + MO) = (8 + 15) cm = 23 cm