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    Class 9 Maths Chapter 5 Exercise 5.5 Solutions – Chords and distance from the centre

    Class 9 Maths Chapter 5 Exercise 5.5 Solutions (Ganita Manjari 2026) – Chords and Distance from Centre

    R K Singh August 18, 2026 No Comments
    📘 NCERT Ganita Manjari (2026) 📚 CBSE 2026–27 🏆 Step-by-Step Solutions

    Class 9 Maths Chapter 5 Exercise 5.5 Solutions

    Chords and Distance from the Centre

    Prepare with complete Exercise 5.5 solutions from Chapter 5 – I’m Up and Down, and Round and Round based on the latest NCERT Ganita Manjari (2026). This exercise covers Theorem 8, the distance of a chord from the centre, Baudhāyana–Pythagoras theorem, and the relationship between chord length and its distance from the centre. Questions 1, 2 and 3 are explained in a clear, step-by-step CBSE answer-writing style.

    📖
    Exercise
    5.5
    ❓
    Questions
    3 Questions
    📐
    Main Concept
    Chords &
    Distance
    📏
    Key Result
    Theorem 8
    🧮
    Formula
    2√(r² − d²)
    ⭐
    Difficulty
    Moderate
    🎯
    Exam Importance
    ★★★★★
    🎯 By the End of Exercise 5.5, You Will Be Able To…
    ✅ Understand the Distance of a Chord from the Centre
    ✅ Apply Theorem 8 to Unequal Chords
    ✅ Use Baudhāyana–Pythagoras Theorem
    ✅ Find the Length of a Chord
    ✅ Understand the Chord-Length Formula
    ✅ Solve Questions 1, 2 and 3 Step by Step

    📑 Table of Contents

    📖 About Class 9 Maths Chapter 5 Exercise 5.5

    Class 9 Maths Chapter 5 Exercise 5.5 Solutions cover the fifth exercise of Chapter 5 – I’m Up and Down, and Round and Round from the latest NCERT Ganita Manjari (2026). This page is designed for students looking for complete and reliable NCERT Class 9 Maths solutions arranged according to the exercise questions.

    The page presents Exercise 5.5 in a clear, student-friendly CBSE answer-writing format and is useful for classroom practice, homework, self-study, revision, and examination preparation. Each question is organised separately so students can quickly find the solution they need.

    🎯 Exercise Snapshot

    📘 What You’ll Find
    • Complete coverage of Exercise Set 5.5.
    • NCERT-based solutions for all three questions.
    • Clear, question-wise presentation for Class 9 students.
    📚 Page Includes
    • Step-by-step NCERT solutions.
    • Relevant Chapter 5 exercise coverage.
    • Simple CBSE answer-writing approach.
    🏆 Best For
    • Students following Ganita Manjari (2026).
    • Homework and classroom practice.
    • Revision and CBSE examination preparation.
    📚 Learn Before You Solve
    Understand the basic concepts of circles before solving Exercise Set 5.5.

    Unequal Chords ⇄ Distance from the Centre

    Which of the two unequal chords is farther from the centre?

    🧪 NCERT Activity
    You have two chords on a circle. One is longer than the other. Which chord is closer to the centre? Can you guess?
    Activity
    1 Draw a circle.
    2 Draw chords of various lengths.
    3 Drop a perpendicular from the centre to each chord.
    4 Measure the length of each chord and its distance from the centre.
    5 Record your observations in Table 1.
    Class 9 Maths NCERT Activity — Unequal chords and their distances from the centre Circle with centre C and three chords of lengths 16 cm, 12 cm and 8 cm. Perpendicular distances from the centre to the chords are shown. The chord endpoints lie exactly on the circle. C 16 cm 12 cm 8 cm 6 cm 8 cm 9.17 cm
    Three chords of different lengths and their perpendicular distances from the centre
    Table 1 — Observation
    Length of Chord 16 cm 12 cm 8 cm
    Distance from Centre 6 cm 8 cm 9.17 cm
    📐 How are these distances obtained?
    Taking the radius as 10 cm and using Baudhāyana–Pythagoras theorem for the right triangle:
    Distance² = Radius² − (Half-chord)²

    For 16 cm chord: √(10² − 8²) = 6 cm
    For 12 cm chord: √(10² − 6²) = 8 cm
    For 8 cm chord: √(10² − 4²) = √84 ≈ 9.17 cm
    🔎 What do you observe?
    The longer the chord, the closer it is to the centre.
    Longer chord
    →
    Smaller distance from centre
    Let us try to understand why this is true.
    Theorem 8
    Let AB and DE be two chords of a circle with centre C. Suppose AB > DE. Then the distance from C to AB is less than the distance from C to DE.

    Method 1 — NCERT Book Method

    Complete proof using Baudhāyana–Pythagoras theorem

    Given: AB > DE
    To prove: CF < CG
    Theorem 8 Class 9 Maths — Unequal Chords and Their Distances from the Centre of a Circle Class 9 Mathematics geometry diagram showing a circle with centre C and two unequal chords AB and DE. Chord AB is longer than chord DE. CF is perpendicular to AB at its midpoint F, and CG is perpendicular to DE at its midpoint G. Radii AC and DC are drawn from the centre to the endpoints A and D. The diagram illustrates that the longer chord is closer to the centre of the circle than the shorter chord. C A B D E F G CF CG
    Fig. 5.16 — Unequal chords and their perpendicular distances from the centre
    ✍️ Proof
    Since CF ⟂ AB and CG ⟂ DE, the perpendiculars from the centre to the chords bisect the chords.
    AF = FB
    DG = GE
    Therefore, F and G are the midpoints of AB and DE.
    Since
    AB > DE
    we get
    AF > DG
    AF² > DG²
    In right-angled triangle ACF, by Baudhāyana–Pythagoras theorem,
    AC² = AF² + CF²
    Similarly, in right-angled triangle DCG,
    DC² = DG² + CG²
    But AC = DC, since both are radii of the same circle.
    AF² + CF² = DG² + CG²
    Since AF² > DG², it follows that
    CF² < CG²
    ∴ CF < CG
    ✓ Hence Proved
    The longer chord AB is nearer to the centre than the shorter chord DE.

    Method 2 — Easy Method

    Important: Direct inequality approach

    Given: AB > DE
    To prove: CF < CG
    Theorem 8 Class 9 Maths — Unequal Chords and Their Distances from the Centre of a Circle Class 9 Mathematics geometry diagram showing a circle with centre C and two unequal chords AB and DE. Chord AB is longer than chord DE. CF is perpendicular to AB at its midpoint F, and CG is perpendicular to DE at its midpoint G. Radii AC and DC are drawn from the centre to the endpoints A and D. The diagram illustrates that the longer chord is closer to the centre of the circle than the shorter chord. C A B D E F G CF CG
    Fig. 5.16 — Unequal chords with perpendicular distances CF and CG
    ✍️ Proof
    Since F and G are the midpoints of AB and DE,
    AF = ½ AB
    DG = ½ DE
    Given
    AB > DE
    ½ AB > ½ DE
    AF > DG
    AF² > DG²
    From the right-angled triangles ACF and DCG, by Baudhāyana–Pythagoras theorem,
    AF² = AC² − CF²
    DG² = DC² − CG²
    Since AC = DC,
    AC² − CF² > AC² − CG²
    Subtracting AC² from both sides,
    −CF² > −CG²
    Multiplying both sides by −1, the inequality reverses:
    CF² < CG²
    ∴ CF < CG
    ✓ Hence Proved
    The longer chord AB is nearer to the centre than the shorter chord DE.

    Consequences of Theorem 8

    Two important ideas to remember

    Theorem 8 tells us that a longer chord is closer to the centre of a circle.
    Now let us see what happens at the two extreme positions.

    1️⃣ Diameter — The Greatest Chord

    Diameter is the greatest chord of a circle — Class 9 Maths Exercise 5.5 A circle with centre C and diameter AB passing through the centre. Since the diameter passes through the centre, its distance from the centre is zero and it is the greatest chord. A B C Distance from centre = 0
    A diameter passes through the centre of the circle.
    A chord that passes through the centre of a circle is called a diameter.

    Since the diameter passes through the centre, its distance from the centre is:
    0

    This is the smallest possible distance. Therefore, the corresponding chord has the greatest possible length.
    ⭐ Remember
    Diameter → Distance from centre = 0
    Diameter = Greatest Chord
    💡 Think
    Can the distance of a chord from the centre be less than zero?
    No. Therefore, the chord passing through the centre gives the maximum chord length.

    2️⃣ Chord Becomes a Point

    Chord reduced to a point on the circumference — Class 9 Maths Exercise 5.5 A circle with centre C and a point P on the circumference. CP is a radius. At the extreme position the chord length becomes zero and the distance from the centre becomes equal to the radius. C P r Chord length = 0
    At the extreme position, the chord reduces to a point.
    As a chord moves farther away from the centre, its length becomes smaller.

    At the extreme position, the chord reduces to a single point on the circle.

    Therefore:
    Chord length = 0

    The distance of this point from the centre is equal to the radius of the circle.
    ⭐ Remember
    Chord length = 0
    Distance from centre = Radius
    💡 Think
    What happens to the chord when it moves farther and farther from the centre?
    Its length keeps decreasing.
    🧠 Quick Revision
    Longer Chord
    →
    Closer to Centre
    →
    Diameter = Greatest Chord
    As the distance from the centre increases, the chord becomes shorter.
    📝 Exam Tip: If a question asks for the greatest chord of a circle, immediately think of the diameter. The diameter passes through the centre, so its distance from the centre is zero.

    Finding the Length of a Chord

    From radius and distance from the centre to chord length

    How can we find the length of a chord when the radius of the circle and the chord’s distance from the centre are known? We first derive the relationship using a right-angled triangle and then apply it to a fresh example.

    📐 Understanding the Relationship

    Chord length derivation using radius and perpendicular distance Class 9 Maths diagram showing a circle with centre C, chord AB, perpendicular CM to chord AB, midpoint M, radius AC, distance CM and half-chord AM. The diagram is used to derive the chord length relationship using the Baudhāyana–Pythagoras theorem. A B C M Radius Distance Half-chord
    The perpendicular from the centre bisects the chord.
    Step 1 — Bisect the chord
    Since CM ⟂ AB, the perpendicular from the centre to a chord bisects the chord. Therefore,
    AM = MB
    So, AM represents half of the chord AB.
    Step 2 — Use the right triangle
    In right-angled triangle ACM:

    AC = Radius
    CM = Distance of chord from centre
    AM = Half-chord
    Step 3 — Apply Pythagoras
    By the Baudhāyana–Pythagoras theorem,
    AC² = CM² + AM²
    Therefore,
    Radius² = Distance² + (Half-chord)²
    Step 4 — Find the chord
    (Half-chord)² = Radius² − Distance²
    Half-chord = √(Radius² − Distance²)
    Chord = 2√(Radius² − Distance²)

    ✏️ Try a Fresh Example

    Example Question
    A circle has radius 13 cm. A chord is 5 cm from the centre. Find the length of the chord.
    Class 9 Maths solved example — chord length when radius is 13 cm and perpendicular distance is 5 cm Class 9 Mathematics geometry diagram showing a circle with centre C, chord AB, midpoint M, perpendicular CM equal to 5 cm, radius AC equal to 13 cm, and half-chord AM equal to 12 cm. Points A and B lie exactly on the circumference. A B C M 13 cm 5 cm 12 cm
    Here AM is the half-chord and AB is the complete chord.
    Given
    Radius = 13 cm
    Distance from centre = 5 cm
    Solution
    Using:
    Radius² = Distance² + (Half-chord)²
    13² = 5² + (Half-chord)²
    169 = 25 + (Half-chord)²
    (Half-chord)² = 169 − 25
    (Half-chord)² = 144
    Half-chord = 12 cm
    The complete chord is twice its half.
    Chord = 2 × 12 = 24 cm
    ✅ Answer
    Length of the chord = 24 cm
    🧠 Remember the Method
    Radius → Distance → Half-chord → Full chord
    Radius² = Distance² + (Half-chord)²
    Question 1
    Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
    ✍️ Solution
    Given,
    Radius of the circle = 7 cm
    Perpendicular distance of the chord from the centre = 6 cm
    To find,
    Length of the chord
    📐 Figure
    Class 9 Maths Exercise 5.5 Question 1 — Chord Length Using Radius and Perpendicular Distance Class 9 Mathematics geometry diagram showing a circle with centre C and chord AB. CM is perpendicular to chord AB and bisects the chord at M. AC is the radius of 7 cm and CM is the perpendicular distance of 6 cm from the centre to the chord. The diagram is used to find the length of the chord using the Baudhāyana–Pythagoras theorem. A B M C 7 cm 6 cm
    CM ⟂ AB and M is the midpoint of chord AB.
    ✍️ Solution
    Since CM is perpendicular to chord AB, the perpendicular from the centre of a circle to a chord bisects the chord.
    AM = MB
    Therefore, AM is half of the chord AB.
    In right-angled triangle ACM,
    AC = 7 cm
    CM = 6 cm
    By the Baudhāyana–Pythagoras theorem,
    AC² = AM² + CM²
    Substituting the values,
    7² = AM² + 6²
    49 = AM² + 36
    49 − 36 = AM²
    13 = AM²
    AM = √13 cm
    Remember: AM is only half of chord AB because M bisects AB.
    Therefore,
    AB = AM + MB
    AB = √13 + √13
    AB = 2√13 cm
    Hence, the length of the chord is approximately 7.21 cm.
    ✅ Final Answer
    Length of the chord = 2√13 cm
    ≈ 7.21 cm
    ② Question
    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).
    ✍️ Solution
    Given,
    Radius of the circle = r
    Perpendicular distance of chord AB from centre C = d
    To show,
    Length of chord AB = 2√(r² − d²)
    📐 Figure
    Class 9 Maths chord length derivation diagram Class 9 Mathematics diagram showing a circle with centre C, chord AB, midpoint M, perpendicular CM to chord AB, radius AC, distance d from the centre to the chord, and half-chord AM. The diagram illustrates the derivation of the chord length relationship using the Baudhāyana–Pythagoras theorem. A B C M r d AM
    CM ⟂ AB and M is the midpoint of chord AB.
    ✍️ Explanation
    Draw CM perpendicular to chord AB, where C is the centre of the circle.
    Since the perpendicular from the centre of a circle to a chord bisects the chord,
    AM = MB
    Therefore, AM is half of chord AB.
    In right-angled triangle ACM:
    AC = r   (radius)
    CM = d   (distance from centre)
    AM = half of the chord
    By the Baudhāyana–Pythagoras theorem,
    AC² = AM² + CM²
    Substituting AC = r and CM = d,
    r² = AM² + d²
    Therefore,
    AM² = r² − d²
    AM = √(r² − d²)
    Since M bisects AB,
    AB = AM + MB
    AB = 2AM
    Therefore,
    AB = 2√(r² − d²)
    Hence, the statement is true.
    ✅ Hence Proved
    Chord length = 2√(r² − d²)
    ③ Question
    In a circle, if the distance of a chord from the centre is twice the distance of another chord from the centre, can we conclude that the first chord is half the length of the second chord? Give reasons for your answer.
    ✍️ Solution
    Given,
    Two chords AB and CD lie in the same circle.
    Distance of AB from the centre = 2d
    Distance of CD from the centre = d
    To find,
    Whether AB = ½ CD
    📐 Figure
    Two chords at different distances from the centre of a circle with two right-angled triangles Class 9 Mathematics diagram showing a circle with centre O, two chords AB and CD, perpendiculars OM and ON from the centre to the chords, and two right-angled triangles OMA and ONC used to compare the chord lengths. A B C D O M N 2d d r
    OM ⟂ AB and ON ⟂ CD, forming two right-angled triangles.
    ✍️ Solution
    Let the radius of the circle be r. Let the distance of AB from the centre be 2d and the distance of CD from the centre be d.
    For chord AB:
    Since OM ⟂ AB, OM bisects AB. Therefore, AM is half of AB.
    r² = AM² + (2d)²
    AM² = r² − 4d²
    AB = 2√(r² − 4d²)
    For chord CD:
    Since ON ⟂ CD, ON bisects CD. Therefore, CN is half of CD.
    r² = CN² + d²
    CN² = r² − d²
    CD = 2√(r² − d²)
    Compare the Chord Lengths
    r² − 4d² < r² − d²
    Therefore,
    AB < CD

    Thus, the first chord is shorter, but this does not mean it is half the second chord.
    Now take a numerical example:
    Let r = 5 cm and d = 2 cm.
    Therefore,
    Distance of AB = 2d = 4 cm
    Distance of CD = d = 2 cm.
    For AB,
    AB = 2√(5² − 4²)
    = 2√(25 − 16)
    = 2√9
    AB = 6 cm
    For CD,
    CD = 2√(5² − 2²)
    = 2√(25 − 4)
    = 2√21
    CD ≈ 9.17 cm
    Compare the Actual Values
    AB = 6 cm
    CD ≈ 9.17 cm
    Therefore,
    ½ CD ≈ 4.59 cm
    but
    AB = 6 cm

    Hence,
    AB ≠ ½ CD
    Thus, the numerical example confirms the general result.
    ✅ Final Answer
    No, we cannot conclude that
    AB = ½ CD.

    In general, AB < CD, but AB is not necessarily half of CD.

    📚 Continue Learning

    Congratulations! You have completed the Class 9 Maths chapter 5 Exercise 5.5 Solutions. You have now worked through Exercise 5.5 of Chapter 5 – I’m Up and Down, and Round and Round. Continue your learning journey by moving to the previous or next exercise, or revisiting Exercise 5.5 whenever you need a quick revision.

    📖 Explore More Class 9 Maths Chapters

    ❓ Frequently Asked Questions

    Find quick answers to common questions about Class 9 Maths Chapter 5 Exercise 5.5 Solutions. These FAQs help students understand what is covered in the exercise, how the questions are approached, and how to use the solutions effectively for learning, revision, and examination preparation.

    What is the main focus of Class 9 Maths Exercise 5.5?

    Exercise 5.5 of Class 9 Maths Chapter 5 focuses on chords and their distance from the centre of a circle. The exercise includes Theorem 8 and three NCERT questions that help students apply the ideas developed in this part of the chapter.

    How many questions are there in Exercise 5.5?

    There are three questions in Exercise 5.5. The solutions on this page are arranged question-wise so that students can move directly to Question 1, Question 2, or Question 3 and follow the required steps clearly.

    What is Theorem 8 in Class 9 Maths Chapter 5?

    Theorem 8 states the relationship between the lengths of chords of a circle and their distances from the centre. In particular, a longer chord lies closer to the centre than a shorter chord. This theorem is important for understanding and solving the questions in Exercise 5.5.

    What should I know before attempting Exercise 5.5?

    Students should be comfortable with basic properties of circles and chords, perpendicular distance, right-angled triangles, and the Baudhāyana–Pythagoras theorem. The earlier exercises of Chapter 5 also provide useful preparation for Exercise 5.5.

    How is the length of a chord found in Exercise 5.5?

    The chord-length questions are approached by considering the right-angled triangle formed using the radius, the perpendicular distance from the centre to the chord, and half of the chord. The required steps are shown clearly in the step-by-step solutions for Exercise 5.5.

    Why is the distance from the centre important when comparing chords?

    The distance from the centre helps compare the positions of chords within the same circle. Exercise 5.5 uses this relationship to help students understand why chords of different lengths occur at different distances from the centre.

    What are common mistakes students make in Exercise 5.5?

    Common mistakes include confusing the chord with its distance from the centre, forgetting that the perpendicular distance is measured from the centre to the chord, and missing intermediate steps while applying Pythagoras. Writing each step clearly helps avoid these errors.

    Are these Class 9 Maths Exercise 5.5 Solutions based on Ganita Manjari 2026?

    Yes. These Class 9 Maths Chapter 5 Exercise 5.5 Solutions are prepared for the latest NCERT Ganita Manjari (2026) textbook and presented in a student-friendly, step-by-step format suitable for classroom practice, self-study, revision, and CBSE examination preparation.

    📚 Useful Learning Resources

    Continue your preparation with more Class 9 Maths resources from Maths Gurukulam, or visit the official NCERT and CBSE websites for the latest textbooks, syllabus, and academic updates.

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    👨‍🏫 Reviewed & Prepared By

    Rakesh Kumar Singh

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    Teaching CBSE Mathematics Since 2006

    These Class 9 Maths Chapter 5 Exercise 5.5 Solutions are carefully prepared according to the latest NCERT Ganita Manjari (2026) and the CBSE curriculum. The solutions follow a clear, step-by-step approach to help students understand the questions in Exercise 5.5 and build confidence in solving circle-based problems.

    📘 NCERT Ganita Manjari (2026) 🎯 CBSE Aligned 📝 Step-by-Step Solutions 💡 Concept-Based Learning
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