Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.4 Solutions

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.4 Solutions explain equal chords and their equal distances from the centre.
These solutions also prove the converse result: chords equidistant from the centre have equal lengths.

Chapter 5, I’m Up and Down and Round and Round Class 9, develops circle geometry through chords, radii, perpendicular distances and congruent right triangles. Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.4 Solutions focus on Theorem 6 and Theorem 7 from the section on distance of chords from the centre. Students prove that equal chords are equidistant from the centre, then prove the converse using Fig. 5.15 and the Baudhāyana Pythagoras theorem Class 9 method. These Class 9 Maths Chapter 5 Exercise 5.4 Solutions use copy-friendly equations, clear proof steps and the perpendicular from centre to chord property. Exercise Set 5.4 contains three questions based on equal chords and equal distances from the centre.

Key Takeaways

• Equal Chords: Chords with the same length are at equal distances from the centre.
• Equal Distances: Chords equidistant from the centre have equal lengths.
• Perpendicular Distance: Distance from the centre to a chord is measured along the perpendicular.
• Right Triangle Method: Radius, half-chord and perpendicular distance form a right triangle.

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.4 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.4 Solutions for Equal Chords

Exercise 5.4 has three proof-based questions on chord length and distance from centre. The exercise connects Theorem 6 and Theorem 7 using perpendiculars, half-chords and equal radii.

Q1. Use the Baudhāyana-Pythagoras theorem to show why Theorem 6 must be true.

Theorem 6 is true because equal chords form right triangles with equal hypotenuse and equal half-chord, so their perpendicular distances from the centre are equal.

Given:

Let AB and FG be equal chords of a circle with centre C.

CE is perpendicular to AB.

CH is perpendicular to FG.

E is the midpoint of AB.

H is the midpoint of FG.

To prove:

CE = CH

Proof:

1. AB and FG are equal chords.

Equation:

AB = FG

1. The perpendicular from the centre to a chord bisects the chord.

Equations:

AE = AB/2

FH = FG/2

1. Since AB = FG, their halves are also equal.

Equation:

AE = FH

1. CA and CF are radii of the same circle.

Equation:

CA = CF

1. In right triangle CEA, use the Baudhāyana-Pythagoras theorem.

Equation:

CA² = CE² + AE²

1. In right triangle CHF, use the Baudhāyana-Pythagoras theorem.

Equation:

CF² = CH² + FH²

1. Since CA = CF, their squares are equal.

Equation:

CA² = CF²

1. Substitute the right-triangle expressions.

Equation:

CE² + AE² = CH² + FH²

1. Since AE = FH, their squares are equal.

Equation:

AE² = FH²

1. Remove equal terms from both sides.

Equation:

CE² = CH²

1. Taking positive square roots:

Equation:

CE = CH

Answer:

Equal chords are equidistant from the centre because their perpendicular distances CE and CH are equal.

Ganita Manjari Class 9 Chapter 5 Exercise 5.4: Theorem 6 Class 9 Circles

Theorem 6 Class 9 circles states that chords of a circle having the same length are at the same distance from the centre. In Exercise 5.4, this result is proved through half-chords and the Baudhāyana-Pythagoras theorem.

Equal Chords Equidistant from Centre

If two chords of the same circle have equal lengths, their perpendicular distances from the centre are equal.

Copy-friendly proof result:

AB = FG

AE = FH

CA = CF

CA² = CE² + AE²

CF² = CH² + FH²

CE² = CH²

CE = CH

Conclusion:

Equal chords equidistant from centre means AB and FG are the same distance away from C.

Class 9 Maths Chapter 5 Exercise 5.4 Solutions for Chords Equidistant from Centre

The converse result is proved by starting with equal perpendicular distances from the centre. Once the right triangles are congruent, their half-chords become equal.

Q2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GF, and CE = CH, show that AB = GF.

AB = GF because equal perpendicular distances from the centre produce equal half-chords.

Given:

CE is perpendicular to AB.

CH is perpendicular to GF.

CE = CH

C is the centre of the circle.

To prove:

AB = GF

Proof:

1. CE is perpendicular to AB.

Equation:

CE ⟂ AB

1. CH is perpendicular to GF.

Equation:

CH ⟂ GF

1. A perpendicular from the centre to a chord bisects the chord.

Equations:

AE = EB

GH = HF

1. CA and CF are radii of the same circle.

Equation:

CA = CF

1. CE = CH is given.

Equation:

CE = CH

1. ∠CEA and ∠CHF are right angles.

Equations:

∠CEA = 90°

∠CHF = 90°

1. In right triangles CEA and CHF:

Equations:

CA = CF

CE = CH

∠CEA = ∠CHF = 90°

1. Therefore, by RHS congruence:

Equation:

ΔCEA ≅ ΔCHF

1. Corresponding parts of congruent triangles are equal.

Equation:

AE = FH

1. Since the perpendicular from centre bisects each chord:

Equations:

AB = 2AE

GF = 2FH

1. Since AE = FH:

Equation:

AB = GF

Answer:

AB = GF because the right triangles formed by the radii and perpendicular distances are congruent by RHS.

Class 9 Maths Ganita Manjari Chapter 5 Solutions: Theorem 7 Class 9 Circles

Theorem 7 Class 9 circles says that chords of a circle equidistant from the centre have equal length. Exercise 5.4 Question 2 proves this result using congruent right triangles.

Chords Equidistant from Centre Are Equal

If two chords have the same perpendicular distance from the centre, their half-lengths are equal.

Copy-friendly result:

CE = CH

CA = CF

ΔCEA ≅ ΔCHF

AE = FH

AB = 2AE

GF = 2FH

Therefore:

AB = GF

Conclusion:

Chords equidistant from centre are equal.

Class 9 Maths Chapter 5 Exercise 5.4 Solutions Using Baudhāyana-Pythagoras Theorem

The Baudhāyana-Pythagoras theorem gives a second proof for Question 2. This method compares the two right triangles made by radius, perpendicular distance and half-chord.

Q3. Solve the previous question using the Baudhāyana-Pythagoras theorem.

AB = GF because equal radii and equal perpendicular distances give equal half-chords by the Baudhāyana-Pythagoras theorem.

Given:

CE is perpendicular to AB.

CH is perpendicular to GF.

CE = CH

C is the centre of the circle.

To prove:

AB = GF

Proof:

1. CA and CF are radii of the same circle.

Equation:

CA = CF

1. Square both sides.

Equation:

CA² = CF²

1. In right triangle CEA:

Equation:

CA² = CE² + AE²

1. In right triangle CHF:

Equation:

CF² = CH² + FH²

1. Since CA² = CF²:

Equation:

CE² + AE² = CH² + FH²

1. CE = CH is given.

Equation:

CE² = CH²

1. Remove equal terms from both sides.

Equation:

AE² = FH²

1. Taking positive square roots:

Equation:

AE = FH

1. Since CE and CH bisect the chords:

Equations:

AB = 2AE

GF = 2FH

1. Since AE = FH:

Equation:

AB = GF

Answer:

AB = GF because the half-chords AE and FH are equal by the Baudhāyana-Pythagoras theorem.

I’m Up and Down and Round and Round Class 9: Concepts Used in Exercise 5.4

Exercise 5.4 uses one diagram idea repeatedly: a radius drawn to a chord endpoint and a perpendicular from the centre to the chord form a right triangle. This triangle connects chord length with distance from the centre.

Perpendicular from Centre to Chord

The perpendicular from centre to chord divides the chord into two equal parts.

Copy-friendly result:

If CE ⟂ AB, then AE = EB.

If CH ⟂ GF, then GH = HF.

This result helps convert chord length into half-chord length.

Perpendicular Bisector of Chord

The perpendicular bisector of chord passes through the centre of the circle.

In Exercise 5.4, CE and CH work as perpendicular distances from the centre to the chords.

Copy-friendly result:

Distance from C to AB = CE

Distance from C to GF = CH

Baudhāyana Pythagoras Theorem Class 9 Pattern

In each right triangle, the radius is the hypotenuse.

Copy-friendly formula:

Radius² = perpendicular distance² + half-chord²

For triangle CEA:

CA² = CE² + AE²

For triangle CHF:

CF² = CH² + FH²

This is the main equation pattern in Class 9 Maths circles solutions.

Quick Concept Table for Exercise 5.4

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5