Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5 Solutions

Chord length is the straight-line distance between two points on a circle, and it changes according to the chord’s perpendicular distance from the centre. Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5 connects this idea with the Baudhāyana-Pythagoras relation, where radius, perpendicular distance and half-chord form a right triangle.

In Exercise 5.5 of Chapter 5, students move from observing chords to calculating their exact lengths. The section is based on the fact that the chord nearer to the centre is longer, while the chord farther from the centre is shorter. Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5 Solutions cover three tasks: finding a chord when the radius and perpendicular distance are given, proving the formula chord length = 2√(r² - d²), and checking whether chord lengths change in direct proportion to their distances from the centre. Each answer uses the same diagram idea, where the radius is the hypotenuse and half the chord is found using the Baudhāyana-Pythagoras theorem.

Key Takeaways

• Chord Length: A chord of radius 7 cm and distance 6 cm from the centre has length 2√13 cm.
• Distance Rule: The longer chord is closer to the centre of the circle.
• Formula: If radius is r and distance from centre is d, chord length is 2√(r² - d²).
• Non-linear Relation: Doubling the distance from centre does not make the chord length half.

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5 Solutions for Chord Length

Exercise 5.5 has three questions based on the relation between a chord, its distance from the centre and the radius. The exercise uses the Baudhāyana-Pythagoras theorem to calculate or prove chord length.

Q1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

The length of the chord is 2√13 cm.

Given:

Radius = 7 cm

Perpendicular distance from centre to chord = 6 cm

Let AB be the chord.

Let O be the centre.

Let OM be perpendicular to AB.

So:

OM = 6 cm

OA = 7 cm

Since the perpendicular from the centre to a chord bisects the chord:

AM = MB

Using the Baudhāyana-Pythagoras theorem in right triangle OMA:

OA² = OM² + AM²

7² = 6² + AM²

49 = 36 + AM²

AM² = 49 - 36

AM² = 13

AM = √13 cm

Now:

AB = 2AM

AB = 2√13 cm

Answer:

The length of the chord is 2√13 cm.

Ganita Manjari Class 9 Chapter 5 Exercise 5.5: Distance of Chord from Centre

The distance of chord from centre is always measured along the perpendicular from the centre to the chord. This perpendicular also bisects the chord, so the full chord is twice the half-chord.

Concept Used in Question 1

Let:

r = radius

d = perpendicular distance from centre to chord

l = length of chord

Then:

Half chord = l/2

Right triangle relation:

r² = d² + (l/2)²

For Question 1:

7² = 6² + (AB/2)²

49 = 36 + (AB/2)²

(AB/2)² = 13

AB/2 = √13

AB = 2√13 cm

This is the main radius and chord relation used in Class 9 Maths circles solutions.

Class 9 Maths Chapter 5 Exercise 5.5 Solutions: Chord Length Formula Class 9

The chord length formula Class 9 comes directly from the right triangle formed by radius, perpendicular distance and half of the chord. The perpendicular from centre to chord gives the midpoint of the chord.

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).

The statement is true because half the chord, the perpendicular distance and the radius form a right triangle.

Given:

Radius = r

Perpendicular distance from centre to chord = d

Chord length = l

Let AB be the chord.

Let O be the centre.

Let OM be perpendicular to AB.

So:

OM = d

OA = r

AB = l

Since OM is perpendicular from centre to chord:

AM = AB/2

So:

AM = l/2

Using the Baudhāyana-Pythagoras theorem in right triangle OMA:

OA² = OM² + AM²

Substitute values:

r² = d² + (l/2)²

Now subtract d² from both sides:

r² - d² = (l/2)²

Take square root on both sides:

√(r² - d²) = l/2

Multiply both sides by 2:

l = 2√(r² - d²)

Answer:

The chord length is 2√(r² - d²) because half the chord equals √(r² - d²).

Class 9 Maths Ganita Manjari Chapter 5 Solutions: Longer Chord Closer to Centre

Theorem 8 states that if one chord is longer than another chord, then the longer chord is nearer to the centre. This does not mean chord length changes in direct proportion to distance from the centre.

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 = 2AB? Give reasons for your answer.

No, we cannot conclude that CD = 2AB.

Reason:

Chord length does not vary directly with the distance from the centre.

Let the radius of the circle be r.

Let the distance of chord AB from the centre be 2d.

Let the distance of chord CD from the centre be d.

Using the chord length formula:

AB = 2√(r² - (2d)²)

AB = 2√(r² - 4d²)

For chord CD:

CD = 2√(r² - d²)

Now compare this with 2AB:

2AB = 2 × 2√(r² - 4d²)

2AB = 4√(r² - 4d²)

For CD = 2AB, we would need:

2√(r² - d²) = 4√(r² - 4d²)

This is not true for all values of r and d.

Example:

Take r = 5 cm and d = 3 cm.

Then distance of AB from centre = 2d = 6 cm.

This is not possible because the distance from the centre to a chord cannot be greater than the radius.

So take r = 10 cm and d = 3 cm.

Then distance of AB from centre = 6 cm.

For chord AB:

AB = 2√(10² - 6²)

AB = 2√(100 - 36)

AB = 2√64

AB = 16 cm

For chord CD:

CD = 2√(10² - 3²)

CD = 2√(100 - 9)

CD = 2√91 cm

Now:

2AB = 2 × 16

2AB = 32 cm

But:

CD = 2√91 cm

Since √91 is about 9.54:

CD ≈ 19.08 cm

So:

CD ≠ 2AB

Answer:

No, CD is not necessarily equal to 2AB. Chord length depends on √(r² - d²), so it is not directly proportional to the distance from the centre.

I’m Up and Down and Round and Round Class 9: Concepts Used in Exercise 5.5

Exercise 5.5 uses the idea that a radius drawn to a chord endpoint and a perpendicular from the centre to the chord form a right triangle. This right triangle gives the chord length formula.

Perpendicular from Centre to Chord

The perpendicular from centre to chord bisects the chord.

Copy-friendly result:

If OM ⟂ AB, then AM = MB.

So:

AB = 2AM

This is why every chord length problem first finds half the chord.

Radius and Chord Relation

The radius and chord relation comes from a right triangle inside the circle.

Copy-friendly formula:

radius² = distance from centre² + half-chord²

Using symbols:

r² = d² + (l/2)²

So:

l = 2√(r² - d²)

This formula is used in Ganita Manjari Class 9 Chapter 5 Exercise 5.5.

Longer Chord Closer to Centre

A longer chord is closer to the centre.

Copy-friendly rule:

If AB > CD, then distance of AB from centre < distance of CD from centre.

Meaning:

The chord nearer to the centre is longer.

Class 9 Maths Circles Solutions: Formula Pattern for Exercise 5.5

Exercise 5.5 becomes simple when students identify the right triangle made inside the circle. The hypotenuse is the radius, and the two shorter sides are the perpendicular distance and half the chord.

Chord Length Formula Class 9

Let:

r = radius

d = perpendicular distance from centre to chord

l = chord length

Then:

Half chord = l/2

Using Baudhāyana-Pythagoras theorem:

r² = d² + (l/2)²

(l/2)² = r² - d²

l/2 = √(r² - d²)

l = 2√(r² - d²)

Distance of Chord from Centre Formula

From:

r² = d² + (l/2)²

We get:

d² = r² - (l/2)²

So:

d = √[r² - (l/2)²]

This form is useful when chord length and radius are given.

Quick Concept Table for Exercise 5.5

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5