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NCERT Class 9 Mathematics · Chapter 5 · CBSE Board Pattern

Class 9 Mathematics Chapter 5 Question Bank CBSE Board Pattern

Section A — MCQs (10 questions, 1 mark each)

  1. A circle is defined as the set of all points in a plane that are
    (a) at a fixed distance from a given line
    (b) equidistant from a given point called the centre
    (c) at a fixed distance from two given points
    (d) lying on a straight line

  2. The longest chord of a circle is called
    (a) radius (b) diameter (c) arc (d) tangent

  3. The line segment joining the centre of a circle to the midpoint of a chord is
    (a) parallel to the chord (b) perpendicular to the chord
    (c) equal to the chord (d) a diameter

  4. Equal chords of a circle subtend
    (a) equal angles at the centre (b) supplementary angles at the centre
    (c) right angles at the centre (d) no definite relation

  5. The angle subtended by a diameter at any point on the circle is
    (a) 45° (b) 60° (c) 90° (d) 180°

  6. If three points are collinear, the number of circles passing through them is
    (a) one (b) two (c) infinitely many (d) zero

  7. In a cyclic quadrilateral, the sum of each pair of opposite angles is
    (a) 90° (b) 180° (c) 270° (d) 360°

  8. The locus of the centres of all circles passing through two fixed points A and B is
    (a) the line segment AB (b) the perpendicular bisector of AB
    (c) a circle with diameter AB (d) any straight line

Assertion-Reason Questions

  1. Assertion (A): The perpendicular from the centre of a circle to a chord bisects the chord.
    Reason (R): The line joining the centre to the midpoint of a chord is perpendicular to the chord.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  2. Assertion (A): The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining part of the circle.
    Reason (R): Equal chords subtend equal angles at the centre.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

Section B — Very Short Answer (6 questions, 2 marks each)

  1. Define a circle and state the meaning of its radius and centre.
  2. What is the difference between a chord and a diameter?
  3. How many circles can pass through two distinct points? Where do their centres lie?
  4. State the condition under which three distinct points determine a unique circle.
  5. If two chords of a circle are equal, what can be said about their distances from the centre?
  6. What is a cyclic quadrilateral? State one property of its opposite angles.

Section C — Short Answer (5 questions, 3 marks each)

  1. In a circle of radius 7 cm, the perpendicular distance of a chord from the centre is 6 cm. Find the length of the chord.
  2. Two parallel chords of lengths 6 cm and 8 cm lie on opposite sides of the centre of a circle of radius 5 cm. Find the distance between their midpoints.
  3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn. Find the distance of the chord from the centre.
  4. A circle has radius 15 cm. A chord is drawn at a distance of 9 cm from the centre. Find the length of the chord.
  5. In a circle of radius 13 cm, a chord is 5 cm away from the centre. Calculate the length of the chord.

Section D — Long Answer (3 questions, 5 marks each)

  1. Prove that equal chords of a circle subtend equal angles at the centre. (Use SSS congruence and justify each step.)
  2. Construct the circumcircle of △ABC where AB = 6 cm, BC = 7 cm and CA = 7 cm. Locate the circumcentre O and state whether it lies inside, outside or on the triangle. Justify your answer with reference to the type of triangle.
  3. A chord of length 16 cm is at a distance of 6 cm from the centre of a circle. Find the radius of the circle. Then prove that the perpendicular bisector of any chord passes through the centre.

Section E — Case/Source-Based (2 questions, 4 marks each)

Case 1: Designing a Circular Flower Bed

A gardener is marking a circular flower bed of radius 5 m on level ground. He fixes two points A and B, 8 m apart, as reference points on the boundary and draws several chords.

(i) How many circles can pass through A and B? Where do their centres lie?
(ii) If a third non-collinear point C is marked on the boundary, how many circles pass through A, B and C? Name the centre.
(iii) Two equal chords are drawn inside the bed. What can be said about their distances from the centre?
(iv) One chord is longer than another. Which chord lies closer to the centre? Give reason.

Case 2: Bicycle Wheel

A bicycle wheel is a perfect circle of radius 35 cm. A spoke is tied with a thread forming a chord of length 48 cm. The wheel rotates about its centre.

(i) Does the wheel have rotational symmetry? Through what angles?
(ii) The thread (chord) subtends an angle at the centre. If another equal chord is formed after rotation, what can be said about the angles they subtend at the centre?
(iii) The perpendicular from the centre to the chord bisects it. Verify this property using the given chord length and radius.
(iv) If the angle subtended by the chord at the centre is 60°, find the length of the chord (use the data given).

▶ Answer Key Attempt all questions first,
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Section A

  1. (b)
  2. (b)
  3. (b)
  4. (a)
  5. (c)
  6. (d)
  7. (b)
  8. (b)
  9. (a) — Both true; R explains why the perpendicular bisects the chord (Theorem 5).
  10. (b) — Both true but R is a separate result (Theorems 2 & 9).

Section B

  1. Circle = locus of points equidistant from a fixed point (centre); radius = that fixed distance.
  2. Chord joins any two points on circle; diameter is a chord passing through the centre.
  3. Infinitely many; centres lie on the perpendicular bisector of AB.
  4. The three points must be non-collinear (Theorem 1).
  5. They are equidistant from the centre (Theorem 6).
  6. Quadrilateral inscribed in a circle; opposite angles sum to 180° (Theorem 11).

Section C

  1. Chord length = \(2\sqrt{7^2-6^2}=2\sqrt{13}\) cm (Pythagoras on right triangle formed by radius, distance and half-chord).
  2. Let distances from centre be \(x\) and \(y\); then \(x^2+3^2=5^2\), \(y^2+4^2=5^2\) → distances 4 cm and 3 cm; separation = 7 cm.
  3. Half-chord = 12 cm → distance = \(\sqrt{13^2-12^2}=5\) cm.
  4. Half-chord = \(\sqrt{15^2-9^2}=12. 5\) cm → chord = 24 cm.
  5. Half-chord = \(\sqrt{13^2-5^2}=12\) cm → chord = 24 cm.

Section D

  1. Let AB = DE. Triangles CAB and CDE are congruent by SSS (CA = CD = radius, CB = CE = radius, AB = DE). Hence ∠ACB = ∠DCE.
  2. Draw perpendicular bisectors of AB and BC; they intersect at unique point O (circumcentre). For isosceles triangle with AB = AC? Wait, sides 6,7,7 → acute, O inside.
  3. Radius = \(\sqrt{8^2+6^2}=10\) cm. Proof: join centre to endpoints; two right triangles congruent by Hypotenuse-leg (radii equal) or by SAS after showing mid-point property.

Section E

Case 1

(i) Infinitely many; centres on perpendicular bisector of AB.
(ii) Exactly one (Theorem 1); circumcentre.
(iii) Equal chords are equidistant from centre.
(iv) Longer chord is closer to centre (Theorem 8).

Case 2

(i) Yes, rotational symmetry of any angle about centre.
(ii) Equal chords subtend equal angles at centre (Theorem 2).
(iii) Let M be midpoint; CM ⊥ chord (Theorem 4); half-chord = 24 cm, radius 35 cm → distance = \(\sqrt{35^2-24^2}=7\sqrt{17}\) cm.
(iv) Chord length = 2 × 35 × sin(30°) = 35 cm.

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All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.