Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions explain the relation between a circle’s centre, a chord, and the chord’s midpoint.
These solutions cover the converse of Theorem 4, an inscribed isosceles triangle proof, and a parallel chords numerical problem.

Chapter 5, I’m Up and Down and Round and Round Class 9, moves from equal chords to the midpoint and perpendicular bisector of chord. Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions focus on three questions from the section “Midpoints and Perpendicular Bisectors of Chords”. Students prove that the perpendicular from centre to chord bisects the chord, apply the same result to an isosceles triangle inscribed in circle, and solve a numerical problem on parallel chords of a circle. These Class 9 Maths Chapter 5 Exercise 5.3 Solutions use RHS congruence, equal radii, midpoint of chord Class 9 facts, and the Baudhāyana-Pythagoras theorem. The textbook places Exercise Set 5.3 directly after Theorem 4 on the line joining the centre and midpoint of a chord.

Key Takeaways

• Perpendicular from centre to chord: It bisects the chord into two equal parts.
• Chord midpoint theorem Class 9: The line from the centre to a chord’s midpoint is perpendicular to the chord.
• Isosceles Triangle: The altitude from the unequal vertex also bisects the base.
• Parallel Chords: Chords on opposite sides of the centre have midpoint distance equal to the sum of their distances from the centre.

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions for Chord Midpoints

Exercise 5.3 has three questions based on the perpendicular relation between a circle’s centre and a chord. The first two questions are proof-based, while the third question uses right triangles inside a circle.

Q1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?

The perpendicular from the centre of a circle to a chord bisects the chord because the two right triangles formed are congruent by RHS.

Given:

Let O be the centre of the circle.

Let AB be a chord.

Let OM be perpendicular to AB.

To prove:

M is the midpoint of AB.

So:

AM = BM

Proof:

1. OA and OB are radii of the same circle.

Equation:

OA = OB

1. OM is common to both triangles.

Equation:

OM = OM

1. OM is perpendicular to AB.

Equations:

∠OMA = 90°

∠OMB = 90°

1. In right triangles OMA and OMB:

Equation:

OA = OB

OM = OM

∠OMA = ∠OMB = 90°

1. Therefore, by RHS congruence:

Equation:

ΔOMA ≅ ΔOMB

1. Corresponding parts of congruent triangles are equal.

Equation:

AM = BM

Answer:

The perpendicular from the centre to a chord bisects the chord because ΔOMA and ΔOMB are congruent by RHS.

Ganita Manjari Class 9 Chapter 5 Exercise 5.3: Perpendicular Bisector of Chord

A perpendicular drawn from the centre to a chord creates two equal halves of the chord. This is the converse of Theorem 4 used in Ganita Manjari Class 9 Chapter 5 Exercise 5.3.

Concept Used in Question 1

If O is the centre and OM ⟂ AB, then M divides AB into two equal parts.

Copy-friendly result:

AM = BM

Meaning:

M is the midpoint of chord AB.

The main proof rule is:

RHS congruence

This result is also called the perpendicular bisector of chord property.

Class 9 Maths Chapter 5 Exercise 5.3 Solutions for Inscribed Isosceles Triangle

An isosceles triangle inscribed in a circle gives a chord as its base. The altitude from the top vertex becomes the perpendicular bisector of the base chord.

Q2. 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.

The altitude from A to BC passes through the centre because it is the perpendicular bisector of chord BC.

Given:

ABC is an isosceles triangle inscribed in a circle.

AB = AC

Let AD be the altitude from A to BC.

So:

AD ⟂ BC

To prove:

AD passes through the centre of the circle.

Proof:

1. AB = AC is given.

Equation:

AB = AC

1. AD is the altitude from A to BC.

Equation:

AD ⟂ BC

1. In an isosceles triangle, the altitude from the unequal vertex bisects the base.

Equation:

BD = DC

1. Since BD = DC, D is the midpoint of BC.
2. Since AD is perpendicular to BC and passes through its midpoint, AD is the perpendicular bisector of BC.
3. BC is a chord of the circle.
4. The perpendicular bisector of a chord passes through the centre of the circle.

Conclusion:

AD passes through the centre of the circle.

Answer:

The altitude from A to BC passes through the centre because it is the perpendicular bisector of chord BC.

Class 9 Maths Ganita Manjari Chapter 5 Solutions: Parallel Chords of a Circle

The numerical question in Exercise 5.3 uses half-chord lengths and the radius to form right triangles. The Baudhāyana-Pythagoras theorem gives the distance of each chord from the centre.

Q3. 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.

The distance between the midpoints of the two chords is 7 cm.

Given:

Radius of circle = 5 cm

Length of first chord = 6 cm

Length of second chord = 8 cm

The chords are on opposite sides of the centre.

To find:

Distance between the midpoints of the chords.

Let the distance from centre to the 6 cm chord be x.

Let the distance from centre to the 8 cm chord be y.

For the 6 cm chord:

Half of 6 cm = 3 cm

Using Baudhāyana-Pythagoras theorem:

x² + 3² = 5²

x² + 9 = 25

x² = 25 - 9

x² = 16

x = 4 cm

For the 8 cm chord:

Half of 8 cm = 4 cm

Using Baudhāyana-Pythagoras theorem:

y² + 4² = 5²

y² + 16 = 25

y² = 25 - 16

y² = 9

y = 3 cm

Since the chords are on opposite sides of the centre:

Distance between midpoints = x + y

Distance between midpoints = 4 cm + 3 cm

Distance between midpoints = 7 cm

Answer:

The distance between the midpoints of the two parallel chords is 7 cm.

I’m Up and Down and Round and Round Class 9: Concepts Used in Exercise 5.3

Exercise 5.3 depends on the idea that the shortest distance from the centre to a chord is perpendicular to the chord. This perpendicular line also reaches the midpoint of the chord.

Perpendicular from Centre to Chord

If a perpendicular is drawn from the centre of a circle to a chord, it divides the chord equally.

Copy-friendly result:

If OM ⟂ AB, then AM = BM.

Here:

O = centre

AB = chord

M = midpoint of chord AB

Midpoint of Chord Class 9

The midpoint of chord Class 9 concept is used when a chord is divided into two equal parts.

Copy-friendly result:

AM = BM

So:

M is the midpoint of AB

This result helps solve both proof questions and numerical chord problems.

Distance Between Midpoints of Chords

For parallel chords on opposite sides of the centre, add their perpendicular distances from the centre.

Copy-friendly result:

Distance between midpoints = distance of first chord from centre + distance of second chord from centre

For Question 3:

Distance between midpoints = 4 cm + 3 cm

Distance between midpoints = 7 cm

Class 9 Maths Circles Solutions: Formula Pattern for Exercise 5.3

The numerical part of Exercise 5.3 uses a right triangle formed by the radius, half the chord and the perpendicular distance from the centre. This pattern is common in Class 9 Maths circles solutions.

Formula for Distance of a Chord from the Centre

Let:

r = radius of circle

l = length of chord

d = perpendicular distance from centre to chord

Then:

Half chord = l/2

Using Baudhāyana-Pythagoras theorem:

d² + (l/2)² = r²

So:

d² = r² - (l/2)²

d = √[r² - (l/2)²]

For the 6 cm chord:

d = √[5² - 3²]

d = √[25 - 9]

d = √16

d = 4 cm

For the 8 cm chord:

d = √[5² - 4²]

d = √[25 - 16]

d = √9

d = 3 cm

Quick Concept Table for Exercise 5.3

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5