NCERT Solutions for Class 9th Maths Chapter 5 In-text Questions — Distance of Chords from the Centre (2026-27)

Q1.

Activity: Take a paper circle. Fold the circle from the boundary, inwards. Open the fold. The crease is now a chord (see Fig. 5.13 B). Now fold the paper again, so that the end points of the chord meet. Open the fold (see Fig. 5.13 C). Measure the lengths of the parts into which the chord is divided. Measure the angle between the creases. Measure the distance from the centre to the midpoint of the chord.

Answer

Three things come out of the folding, and each one is a theorem of this chapter in physical form.

The two parts of the chord are equal. The second fold makes the endpoints of the chord coincide, so the crease is the perpendicular bisector of the chord — it cuts the chord exactly in half at its midpoint.
The angle between the two creases is 90°. Fold the paper along crease 2 and the chord folds onto itself; the only way that happens is if crease 2 meets the chord at a right angle.
The second crease passes through the centre, and the distance from the centre to the midpoint of the chord is the distance from the centre to the chord. The second fold makes the boundary of the circle overlap itself, so crease 2 is a diameter and contains the centre. "Distance from a point to a line" always means the perpendicular distance, and here that perpendicular is exactly the segment from the centre to the midpoint.

Crease 1 = chord AB Crease 2 = perpendicular bisector of AB
Crease 2 passes through the centre O, meets AB at M
AM = MB, ∠OMA = ∠OMB = 90°, and OM = distance from O to the chord

Why the paper knows this: a fold is a reflection. The second fold is a reflection that swaps A and B and maps the circle to itself. A reflection that maps the circle to itself must fix the centre, so the crease passes through O; a reflection that swaps A and B must be the perpendicular bisector of AB. Those two statements together are Theorems 4 and 5.

Q2.

Now draw another chord of the same length. How will you do this? We will let you figure this out yourself. Join the centre to the midpoint of the new chord and measure its length. Is it the same as distance from the centre to the first chord?

Answer

How to draw a second chord of the same length: open a compass to the length of the first chord AB. Put the compass point anywhere on the circle, at a point P, and cut the circle at Q. Then PQ = AB. (Equivalently: trace the circle and the chord on tracing paper, then rotate the tracing paper about the centre — the traced chord lands on a new chord of the same length.)

Yes — the distance is the same. The two equal chords are equidistant from the centre.

Let M, N be the midpoints of AB and PQ.
In ΔOMA and ΔONP: OA = OP = r, AM = ½AB = ½PQ = PN, ∠OMA = ∠ONP = 90°
By RHS congruence, ΔOMA ≅ ΔONP ⇒ OM = ON

Why rotation makes this obvious: the circle has complete rotational symmetry about O. Rotating the chord AB about O gives a chord of the same length, and the perpendicular from O rotates with it — so its length cannot change. The congruence argument above is the same statement, written as a proof rather than an observation. That distinction matters: examples suggest, proofs settle. This result is Theorem 6.