Question

The parallel chords, PQ and RS of circle of centre O are equal to 24 cm and 26 cm respectively. If the distance between parallel chords is equal to 10 cm, find the distance between the smaller chord and centre of circle.
1. 6.25 cm
2. 10 cm
3. 4.75 cm
4. 5.25 cm

Answer 1

Correct Answer - Option 1 : 6.25 cm

Given:

O is the centre of the circle.

PO and RO are the radii of the circle.

The parallel chords of the circle have values equal to 24 cm and 26 cm respectively.

Distance between parallel chords (AB) = 10 cm

Concepts used:

The line drawn from the centre on the chord of a circle is perpendicular to the chord and bisects the chord of the circle.

In right-angled triangle, Hypotenuse (H)² = Perpendicular (P)² + base (B)²

Calculation:

PQ and RS are the parallel chords of length 24 cm and 26 cm respectively.

Distance between parallel chords = 10 cm

In ΔPOA, line OA (length “a”) drawn from centre O meets chord PQ at A.

⇒ ∠PAO = 90°

⇒ PA = PQ/2 = 24/2 cm = 12 cm

In ΔPOA,

⇒ PO² = PA² + OA²

⇒ r² = 12² + a²

⇒ r² = a² + 144      ---- (1)

In ΔROB, line OB (length “10 -a”) drawn from centre O meets chord RS at B.

⇒ ∠RBO = 90°

⇒ RB = RS/2 = 26/2 cm = 13 cm

In ΔROB,

⇒ RO² = RB² + OB²

⇒ r² = 13² + (10 – a)²

⇒ 100 + a² - 20a = r² – 169

⇒ r² = a² – 20a + 269      ---- (2)

Equating (1) and (2),

 a² + 144 = a² – 20a + 269     

⇒ 20a = 269 – 144 cm

⇒ 20a = 125 cm

⇒ a = 6.25 cm

∴ Distance between small chord and centre of circle is equal to 6.25 cm.