NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.3 2026-27

Question 1.
Can you explain why the converse of Theorem 4 is true, i.e., why the perpendicular from the centre of a circle to a chord bisects the chord?
(Hint: Use Figure. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM)

Solution:
Given: C is the centre of the circle, AB is a chord, and CM ⊥ AB.

Therefore:

∠CMA = ∠CMB = 90°

To prove:

AM = BM

Proof:

Consider ∆CMA and ∆CMB.

CA = CB, as they are radii of the same circle.

CM = CM, as it is the common side.

∠CMA = ∠CMB = 90°, as given.

Therefore, by the RHS congruence criterion:

∆CMA ≅ ∆CMB

Hence, by corresponding parts of congruent triangles:

AM = BM

Therefore, the perpendicular drawn from the centre of a circle to a chord bisects the chord.

Question 2.
An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.

Solution:

Given: ∆ABC is an isosceles triangle inscribed in a circle with centre O, where AB = AC. AM is the altitude drawn from A to BC.

To prove: AM passes through the centre O.

Proof:

Consider ∆AMB and ∆AMC.

AB = AC, as given.

AM = AM, as it is the common side.

∠AMB = ∠AMC = 90°, because AM is an altitude.

Therefore, by the RHS congruence criterion:

∆AMB ≅ ∆AMC

Hence:

BM = CM

Thus, M is the midpoint of chord BC.

The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. Therefore:

OM ⊥ BC

We also know that:

AM ⊥ BC

Since only one perpendicular can be drawn from point M to line BC, AM and OM must lie on the same line.

Therefore, A, O, and M are collinear. Hence, the altitude from A to BC passes through the circle's centre, O.

Question 3.
Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.

Solution:
Given:

Radius of the circle = 5 cm

Length of chord AB = 6 cm

Length of chord CD = 8 cm

Let N and M be the midpoints of chords AB and CD, respectively.

Since the line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord:

AN = AB/2 = 6/2 = 3 cm

In right-angled ∆ANO:

OA² = ON² + AN²

5² = ON² + 3²

25 = ON² + 9

ON² = 16

ON = 4 cm

Similarly:

CM = CD/2 = 8/2 = 4 cm

In right-angled ∆CMO:

OC² = OM² + CM²

5² = OM² + 4²

25 = OM² + 16

OM² = 9

OM = 3 cm

Since the chords lie on opposite sides of the centre:

MN = ON + OM

= 4 + 3

= 7 cm

Therefore, the distance between the midpoints of the two chords is 7 cm.

Key Points to Remember for Class 9 Maths Exercise 5.3 Chapter 5 I’m Up and Down, and Round and Round

When revising Chapter 5 maths class 9 exercise 5.3, keep these important ideas in mind:

• Identify the centre, chord, radius, midpoint, and perpendicular shown in the figure.

• Use equal radii and RHS congruence to establish relationships between triangles.

• Remember that a perpendicular from the centre bisects a chord.

• Apply the Pythagorean theorem when finding distances involving chords and the radius.

• For Maths Class 9, Chapter 5, Exercise 5.3, write proofs in a clear order: given information, reasoning, and conclusion.

Access Exercise Wise NCERT Solutions for Chapter 5 Maths Class 9

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