NCERT Solutions for Class 9th Maths Chapter 5 Exercise Set 5.5 — Which of the two unequal chords is farther from the centre?

Book page 105–106 Updated on2026-09-08

Q1.

Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

Answer

Chord = 2√13 cm ≈ 7.2 cm.

Let AB be the chord, O the centre, M the foot of the perpendicular from O.
By Theorem 5, M is the midpoint of AB, and ΔOMA is right-angled at M.

AM² = OA² − OM² = 7² − 6² = 49 − 36 = 13
AM = √13 cm
AB = 2 × AM = 2√13 cm ≈ 7.21 cm

Check it yourself: the answer must be less than the diameter 14 cm — and it is. Also, 6 cm is a fairly large distance for a radius of 7 cm, so a shortish chord is exactly what we should expect.

Q2.

Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r² − d²).

Answer

Given: circle with centre O, radius r; chord AB whose perpendicular distance from O is d. To show: AB = 2√(r² − d²).

Let M be the foot of the perpendicular from O to AB, so OM = d and ∠OMA = 90°.
By Theorem 5 (the perpendicular from the centre bisects the chord), AM = MB = ½ AB.

In the right triangle OMA, by Baudhāyana–Pythagoras:
OA² = OM² + AM²
r² = d² + AM²
AM² = r² − d²
AM = √(r² − d²)

AB = 2 AM = 2√(r² − d²)

One right triangle — radius, distance, half-chord — carries every chord calculation in the chapter.

Reading the formula: put d = 0 and you get AB = 2r, the diameter — the longest chord. Push d up towards r and the chord shrinks towards 0. And since r² − d² decreases as d increases, a bigger distance always means a shorter chord, which is Theorem 8 in one line.

Q3.

In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.

Answer

No, we cannot. The conclusion CD = 2 AB is false in general; it holds only in one very special case.

Let the distance of CD from the centre be d, so the distance of AB is 2d.
AB = 2√(r² − 4d²) and CD = 2√(r² − d²)

Take r = 10 cm and d = 3 cm:
CD = 2√(100 − 9) = 2√91 ≈ 19.08 cm
AB = 2√(100 − 36) = 2√64 = 16 cm
CD / AB ≈ 1.19, not 2.

When does CD = 2 AB actually happen?

2√(r² − d²) = 2 × 2√(r² − 4d²)
r² − d² = 4(r² − 4d²)
r² − d² = 4r² − 16d²
15d² = 3r² ⇒ r² = 5d², i.e. r = d√5

Check with d = 2, r = 2√5: CD = 2√(20 − 4) = 8, AB = 2√(20 − 16) = 4 ✓

Why the guess fails: distance and chord length are not proportional to each other. They are tied by d² + (½ chord)² = r², a squared relation, so doubling one quantity does not halve or double the other. All we may safely say from Theorem 8 is the direction: since AB is farther from the centre, AB is the shorter chord, so CD > AB. Whether CD is 1.1 times AB or exactly 2 times AB depends on r.

Was this helpful? Report an error