This MCQ module is based on: Chords, Arcs and Angles in a Circle
TOPIC 14 OF 25
Chords, Arcs and Angles in a Circle
🎓 Class 9
Mathematics
CBSE
Theory
Ch 5 — I’m Up and Down, and Round and Round
⏱ ~50 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Chords, Arcs and Angles in a Circle
Targeting Class 9 level in Geometry, with Intermediate difficulty.
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5.6 Angle Subtended by a Chord at the Centre
Take a chord PQ of a circle with centre O. The two radii OP and OQ together with the chord PQ form an isosceles triangle. The angle subtended? by the chord at the centre is the angle \(\angle POQ\).
Fig. 5.7 — Chord PQ subtends ∠POQ at the centre.
If we take a longer chord, the angle becomes larger. If the chord shrinks to nothing, the angle becomes zero. So the size of the angle and the length of the chord increase together. This suggests:
Theorem 2 — Equal chords ⇒ equal angles
Equal chords of a circle subtend equal angles at the centre.Proof. Let AB and CD be two equal chords of a circle with centre O. Consider \(\triangle\)AOB and \(\triangle\)COD. Then OA = OC (radii), OB = OD (radii) and AB = CD (given). By the SSS rule, \(\triangle\)AOB \(\cong\) \(\triangle\)COD. Hence \(\angle\)AOB = \(\angle\)COD. ∎
Theorem 3 — Converse
If the angles subtended by two chords at the centre are equal, then the chords themselves are equal.Proof sketch. Same triangles as above; this time SAS (two radii and the included angle) gives \(\triangle\)AOB \(\cong\) \(\triangle\)COD, so AB = CD.
5.7 Perpendicular from the Centre to a Chord
Fig. 5.8 — OM ⊥ AB and AM = MB.
Theorem 4
The perpendicular from the centre of a circle to a chord bisects the chord.Proof. Let OM be perpendicular to chord AB. In \(\triangle\)OMA and \(\triangle\)OMB:
- OA = OB (radii)
- OM = OM (common)
- \(\angle\)OMA = \(\angle\)OMB = 90°
By RHS, \(\triangle\)OMA \(\cong\) \(\triangle\)OMB. Hence AM = MB. ∎
Theorem 5 — Converse
The line through the centre that bisects a chord is perpendicular to the chord.This pair of theorems is extremely useful: any two of {centre, chord, midpoint, perpendicular} determine the fourth.
Activity: Discover the Bisector
L3 ApplyMaterials: circular disc, ruler, set-square.
Predict: When you fold a chord onto itself, where does the crease pass?
- Draw a chord AB on the disc.
- Fold the disc so that A lies exactly on B. Crease.
- Open and observe the crease.
The crease is perpendicular to AB and passes through its midpoint M. It also passes through the centre O of the disc — confirming Theorem 5.
5.8 Equal Chords Are Equidistant from the Centre
Theorem 6
Chords of equal length lie at equal distances from the centre of the circle. Conversely, chords equidistant from the centre are equal in length.Why? Let AB and CD be equal chords with feet of perpendiculars from O at E and F respectively. Drop OE ⊥ AB and OF ⊥ CD. Then by Theorem 4, AE = ½AB = ½CD = CF. Triangles OEA and OFC are right-angled with OA = OC (radii) and AE = CF. By RHS, they are congruent, so OE = OF.
Fig. 5.14 — Equal chords AB and CD lie at equal distances OE and OF from the centre.
5.9 Angles Subtended by an Arc
An arc of a circle is a connected portion of the circle. Two points A and B on a circle determine two arcs — a major arc and a minor arc. The angle subtended by an arc at the centre is the angle subtended by the chord AB on the side of the arc.
Fig. 5.17 — Two arcs determined by chord AB.
Theorem 7 — Equal arcs ⇔ equal chords
Two arcs of a circle are congruent (i.e. the same length) iff their chords are equal.5.10 Angle at the Centre vs Angle on the Arc
The most striking theorem about circles connects the angle a chord subtends at the centre with the angle it subtends at any point on the remaining arc.
Theorem 8 — Inscribed Angle Theorem
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.Fig. 5.21 — ∠AOB = 2∠ACB.
Proof. Join CO and produce it to D inside the circle. In \(\triangle\)OAC, OA = OC (radii), so the triangle is isosceles and \(\angle\)OAC = \(\angle\)OCA. The exterior angle at O equals the sum of the two interior opposite angles, so \(\angle\)AOD = 2\(\angle\)ACO. Similarly \(\angle\)BOD = 2\(\angle\)BCO. Adding,
\(\angle\)AOB = \(\angle\)AOD + \(\angle\)BOD = 2(\(\angle\)ACO + \(\angle\)BCO) = 2\(\angle\)ACB. ∎
Corollary — Angles in the same segment
Angles subtended by the same arc at any two points on the remaining part of the circle are equal.Reason: each is half of the same central angle.
Try it: An arc subtends \(80°\) at the centre. What angle does it subtend at a point on the major arc? 40°. And at a point on the minor arc? \(180° - 40° = 140°\) (because the points lie on the other side, giving the supplementary inscribed angle).
5.11 Worked Examples
Example 3. A chord of length 10 cm is at a distance of 12 cm from the centre. Find the radius.
Half-chord = 5 cm. Using right triangle (perpendicular from centre + half chord + radius): \(r^2 = 12^2 + 5^2 = 169\). So \(r = 13\) cm.
Example 4. AB and CD are two equal chords of a circle with centre O. If \(\angle\)AOB = 70°, find \(\angle\)COD.
Equal chords subtend equal angles at the centre (Theorem 2). So \(\angle\)COD = 70°.
Example 5. An arc PQ subtends an angle of 130° at the centre. P, Q, R lie on the circle, with R on the major arc. Find \(\angle\)PRQ.
By Theorem 8, \(\angle\)PRQ = ½(130°) = 65°.
Competency-Based Questions
Scenario: A clock-maker is engraving a circular medallion of radius 13 cm. She wants to inscribe a horizontal chord 24 cm long, then engrave a triangle with one vertex at a point P on the major arc and the chord as the base.
Q1. The perpendicular distance from the centre O to the 24 cm chord is:
L3 Apply(a) 5 cm. Half-chord = 12 cm, radius = 13 cm. \(\sqrt{13^2-12^2}=\sqrt{25}=5\).
Q2. The chord subtends an angle of \(2\theta\) at the centre. Show how \(\theta\) relates to the right-triangle obtained when the perpendicular OM is drawn, and compute \(2\theta\) numerically.
L4 AnalyseAnswer: \(\sin\theta = \tfrac{12}{13}\), so \(\theta \approx 67.38°\) and \(2\theta \approx 134.76°\).
Q3. Evaluate the claim: "The angle the chord subtends at any point P on the major arc is the same regardless of where P is."
L5 EvaluateTrue. By the corollary to the Inscribed Angle Theorem, all such angles equal half the central angle (\(\approx 67.38°\) in this case). The clock-maker can place P anywhere on the major arc and get an isosceles-or-otherwise triangle with the same apex angle.
Q4. The clock-maker wants to engrave a SECOND chord parallel to the first, but on the opposite side of O, of the same length. Argue (using a theorem) that the two chords are equidistant from the centre, and find that distance.
L6 CreateAnswer: By Theorem 6, equal chords are equidistant from the centre. Both perpendicular distances = 5 cm. So the two chords lie 5 cm above and 5 cm below O — separated by 10 cm.
Assertion–Reason Questions
A: Equal chords of a circle subtend equal angles at the centre.
R: Two triangles with the same three side lengths are congruent (SSS).
R: Two triangles with the same three side lengths are congruent (SSS).
(a) The proof of Theorem 2 uses SSS congruence of triangles formed by the two radii and the chord. R directly explains A.
A: The angle subtended by an arc at the centre equals the angle it subtends on the remaining arc.
R: All angles in the same circular segment are equal.
R: All angles in the same circular segment are equal.
(d) A is FALSE — the central angle is twice the inscribed angle, not equal. R is true.
A: The perpendicular from the centre of a circle to a chord bisects it.
R: Two right triangles with the same hypotenuse and one common leg are congruent (RHS).
R: Two right triangles with the same hypotenuse and one common leg are congruent (RHS).
(a) The proof uses RHS on triangles OMA and OMB.
Frequently Asked Questions
What is the perpendicular from centre to chord theorem in Class 9?
The theorem states that the perpendicular drawn from the centre of a circle to a chord bisects the chord. It is proved by drawing radii to the endpoints, noting two right triangles share a common side and equal hypotenuses (radii), and applying RHS congruence.
Why do equal chords of a circle subtend equal angles at the centre?
If two chords have the same length, the triangles formed with the centre share two sides equal to the radius and the third side equal to the chord. By SSS congruence, the triangles are congruent, so the angles at the centre are equal.
What is the converse: equal angles at centre imply equal chords?
If two chords subtend equal angles at the centre, the corresponding triangles share two equal radii and an equal included angle. By SAS congruence, the third sides — the chords — must be equal.
Are equal chords always equidistant from the centre of the circle?
Yes. Equal chords of a congruent or single circle are equidistant from the centre. Conversely, two chords equidistant from the centre must be equal in length. The result follows from RHS congruence of the right triangles formed by the perpendicular distances.
If a chord of length 24 cm is at a distance 5 cm from the centre, what is the radius?
The perpendicular from the centre bisects the chord, giving half-chord 12 cm. By Pythagoras, radius = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13 cm.
Do equal arcs of a circle correspond to equal chords?
Yes. In the same circle, equal arcs subtend equal angles at the centre, and equal angles at the centre cut off equal chords. So equal arcs always correspond to equal chords, and conversely.
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