Ex 5.5 Class 9 Maths Ganita Manjari Solutions | Sid Classes

Class 9 Maths Ganita Manjari Exercise 5.6 solutions on the angle subtended by an arc, with theorem-based steps and diagrams. Sid Classes.

Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5. Designed to reflect how top-scoring students write their solutions and proofs in school examinations, this walkthrough makes chord length calculations, distance formulas, and geometric reasoning entirely clear.

🔑 Key Geometric Concepts Used in this Exercise:

  • Perpendicular Bisector Rule: The perpendicular drawn from the centre of a circle to a chord always bisects the chord.
  • Baudhāyana–Pythagoras Theorem: In any right-angled triangle formed by the radius, perpendicular distance, and half-chord, \(r^2 = d^2 + x^2\).
  • Chord Length Formula: The total length of a chord is given by \(2\sqrt{r^2 - d^2}\), where \(r\) is the radius and \(d\) is the perpendicular distance from the centre.

NCERT Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5 Solutions

Question 1

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

Solution

  1. Step 1: Let \(O\) be the centre of the circle, \(AB\) be the chord, and \(OM\) be the perpendicular from the centre to chord \(AB\).
  2. Step 2: Given data:
    • Radius (\(r = OA\) or \(OB\)) \(= 7\) cm
    • Perpendicular distance (\(d = OM\)) \(= 6\) cm
  3. Step 3: In right-angled \(\triangle OMA\), using the Baudhāyana–Pythagoras theorem:
    • \(OA^2 = OM^2 + AM^2\)
    • \(7^2 = 6^2 + AM^2\)
    • \(49 = 36 + AM^2\)
    • \(AM^2 = 49 - 36 = 13\)
    • \(AM = \sqrt{13}\) cm
  4. Step 4: Since the perpendicular from the centre bisects the chord, the total chord length \(AB = 2 \times AM\):
    • \(AB = 2\sqrt{13}\) cm

Answer: The length of the chord is \(\mathbf{2\sqrt{13}}\) cm (approximately \(7.21\) cm).

O A B M Fig: Chord Length Calculation (r=7, d=6)

Question 2

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 \(2\sqrt{r^2 - d^2}\).

Solution:

  1. Step 1: Let \(OM = d\) be the perpendicular distance from the centre \(O\) to chord \(AB\), and let \(OA = r\) be the radius of the circle.
  2. Step 2: In right-angled \(\triangle OMA\), the radius \(OA\) is the hypotenuse, \(OM\) is one leg equal to \(d\), and \(AM\) is the half-chord length equal to \(x\).
  3. Step 3: Applying the Baudhāyana–Pythagoras theorem:
    • \(OA^2 = OM^2 + AM^2\)
    • \(r^2 = d^2 + x^2\)
    • \(x^2 = r^2 - d^2 \implies x = \sqrt{r^2 - d^2}\)
  4. Step 4: Since the perpendicular from the centre bisects the chord, the total length of the chord \(AB = 2x\).
  5. Step 5: Substituting the value of \(x\), we get:
    • \(\text{Chord Length} = 2\sqrt{r^2 - d^2}\)

Conclusion: Hence, the formula \(2\sqrt{r^2 - d^2}\) correctly represents the length of any chord given its radius and perpendicular distance from the centre.

O A B M Fig: General Derivation of Chord Length

Question 3

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 = 2 AB? Give reasons for your answer.

Solution:

  1. Step 1: Let the perpendicular distance of chord \(AB\) from the centre be \(d_1\), and the distance of chord \(CD\) be \(d_2\). According to the question, \(d_1 = 2d_2\).
  2. Step 2: Let the radius of the circle be \(r\). Using the chord length formula from Question 2:
    • \(\text{Length of } AB = 2\sqrt{r^2 - d_1^2} = 2\sqrt{r^2 - (2d_2)^2} = 2\sqrt{r^2 - 4d_2^2}\)
    • \(\text{Length of } CD = 2\sqrt{r^2 - d_2^2}\)
  3. Step 3: Compare \(CD\) and \(2 \times AB\):
    • \(2 \times AB = 2 \times \left(2\sqrt{r^2 - 4d_2^2}\right) = 4\sqrt{r^2 - 4d_2^2}\)
    • \(CD = 2\sqrt{r^2 - d_2^2}\)
  4. Step 4: Clearly, \(CD \neq 2AB\). (Although \(CD\) is longer than \(AB\) because chords closer to the centre are longer, the relationship is non-linear due to the square root expression in the chord length formula).

Answer: No, we cannot conclude that \(CD = 2 AB\). The relationship between chord lengths and their distances from the centre is governed by square root terms in the Baudhāyana–Pythagoras theorem, not by simple direct proportion.

O d1 d2 Fig: Comparison of Chords at Different Distances

Frequently Asked Questions (FAQs)

Q1: How do you calculate chord length when given radius and perpendicular distance?
A: Use the formula \(2\sqrt{r^2 - d^2}\) derived directly from the Baudhāyana–Pythagoras theorem.

Q2: Are chords farther from the centre longer or shorter?
A: Chords farther from the centre are shorter, while chords closer to the centre are longer.

Q3: Why is chord length not directly proportional to distance from the centre?
A: Because of the square root operation in the Pythagorean relation, doubling the distance does not halve or double the chord length linearly.

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