NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.4 (2026-27)

Question 1.
Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true.

Solution:
Given: AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendicular to AB and CD, respectively.

Therefore:

OP ⊥ AB and OQ ⊥ CD

To prove:

OP = OQ

Proof:

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

Therefore:

AP = AB/2

CQ = CD/2

Since AB = CD, we get:

AP = CQ

Also, OA and OC are radii of the same circle.

OA = OC = r

In right-angled ∆OPA, using the Baudhayana-Pythagoras theorem:

OA² = OP² + AP²

r² = OP² + AP² …(i)

In right-angled ∆OQC:

OC² = OQ² + CQ²

r² = OQ² + CQ² …(ii)

From equations (i) and (ii):

OP² + AP² = OQ² + CQ²

Since AP = CQ:

OP² = OQ²

Therefore:

OP = OQ

Hence, equal chords of a circle are equidistant from its centre.

Question 2.
Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.

Solution:
Given: AB and GF are chords of a circle with centre C.

CE ⊥ AB
CH ⊥ GF
CE = CH

To prove:

AB = GF

Proof:

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

Therefore:

AE = EB

and

GH = HF

Consider right-angled ∆CEA and ∆CHG.

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

CE = CH, as given.

∠CEA = ∠CHG = 90°

Therefore, by the RHS congruence criterion:

∆CEA ≅ ∆CHG

Hence, by CPCT:

AE = GH

Now:

AB = 2AE

and

GF = 2GH

Since AE = GH:

AB = GF

Therefore, the two chords are equal.

Question 3.
Solve the previous question using the Baudhayana-Pythagoras theorem.

Solution:
Given: AB and GF are chords of a circle with centre C.

CE ⊥ AB
CH ⊥ GF
CE = CH

To prove:

AB = GF

Proof:

Let:

CA = CG = r

because they are radii of the same circle.

Also, let:

CE = CH = d

In right-angled ∆CEA, using the Baudhayana-Pythagoras theorem:

CA² = CE² + AE²

AE² = CA² - CE²

AE² = r² - d² …(i)

In right-angled ∆CHG:

CG² = CH² + GH²

GH² = CG² - CH²

GH² = r² - d² …(ii)

From equations (i) and (ii):

AE² = GH²

Therefore:

AE = GH

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

So:

AB = 2AE

and

GF = 2GH

Since AE = GH:

AB = GF

Hence proved.

How to Use Class 9 Maths Chapter 5 Exercise 5.4 Solutions

• Try before checking: Solve each construction, proof, or numerical problem yourself before referring to the solution.

• Focus on the reason: Pay attention to why a theorem, congruence rule, or Baudhayana-Pythagoras theorem is used in each step.

• Revise difficult steps: If a proof feels confusing, go back to the corresponding figure and identify the given measurements and equal quantities.

• Use the PDF for revision: Students looking for Class 9 Maths Chapter 5 Exercise 5.4 solutions PDF can use the solution material for quick revision and practice.

• Practise beyond the exercise: The Class 9 Maths Chapter 5 Exercise 5.4 solutions worksheets can help you apply the same circle properties to similar questions.

• Keep the chapter connected: These solutions are part of Maths Class 9 Chapter 5 Exercise 5.4, so revise related circle concepts from the full chapter as needed.

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