Exercise 12.2 — Chord and Angle
Angle subtended by a chord at a point on the circle.
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Key Concept — The Angle Subtended by a Chord at the Centre
When both endpoints of a chord are joined to the centre of the circle, they form an angle at the centre. This angle is called the angle subtended by the chord at the centre. Exercise 12.2 is built entirely on two important theorems about this angle.
∠AOB = angle subtended by chord AB at centre O
Longer chord → larger angle at centre
GH > EF > CD > AB means ∠GOH > ∠EOF > ∠COD > ∠AOB
The diagram above shows a key observation: as the length of a chord increases, the angle it subtends at the centre also increases. The longest possible chord (the diameter) subtends an angle of 180° at the centre.
The Two Fundamental Theorems
Both theorems in Exercise 12.2 work together as a pair — they are converses of each other:
🔷 Theorem 1 — Equal Chords → Equal Angles
Equal chords of a circle subtend equal angles at the centre.
If AB = CD → ∠AOB = ∠COD🔷 Theorem 2 (Converse) — Equal Angles → Equal Chords
If two chords subtend equal angles at the centre of a circle, then the chords are equal.
If ∠AOB = ∠COD → AB = CDAB = CD ⟹ ∠AOB = ∠COD
Equal chords subtend equal angles at centre
📌 How to remember: Think of a balance scale. If two chords are equal in weight (length), they "balance" at the centre and produce equal angles. If the angles are equal, the chords balance back — they must be equal too. These two theorems are if and only if — both directions hold.
Question 1 — Finding ∠COD when AB = CD and ∠AOB = 90°
Question 1
In the figure, AB = CD and ∠AOB = 90°. Find ∠COD.
AB = CD, ∠AOB = 90°
Given: AB = CD and ∠AOB = 90°. To find: ∠COD
Given: AB = CD and ∠AOB = 90°
Since AB = CD
⟹ ∠AOB = ∠COD (∵ Equal chords subtend equal angles at the centre — Theorem 1)
⟹ ∠COD = 90° (∵ ∠AOB = 90°)
∴ ∠COD = 90°
✅ Answer: ∠COD = 90°
Theorem used: Equal chords of a circle subtend equal angles at the centre. Since AB = CD, the angles they subtend at O must be equal, so ∠COD = ∠AOB = 90°. This is a direct, single-step application of Theorem 1.
Theorem used: Equal chords of a circle subtend equal angles at the centre. Since AB = CD, the angles they subtend at O must be equal, so ∠COD = ∠AOB = 90°. This is a direct, single-step application of Theorem 1.
💡 Why this works: The relationship between chord length and central angle is one-to-one in a given circle. If you know two chords are equal, you immediately know their central angles are equal — no calculation needed. This is why Theorem 1 makes Q1 a one-step answer.
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Question 2 — Finding ∠OPQ and ∠ROS when PQ = RS and ∠ORS = 48°
Question 2
In the figure, PQ = RS and ∠ORS = 48°. Find ∠OPQ and ∠ROS.
PQ = RS, ∠ORS = 48°
Given: PQ = RS and ∠ORS = 48°. To find: ∠OPQ and ∠ROS
This problem requires two separate sub-calculations. We find ∠ROS first (using triangle ORS), then use the equal-chord theorem to find ∠POQ, and finally find ∠OPQ (using triangle OPQ).
Step 1 — Find ∠ROS (in △ORS)
OR = OS (radii of same circle)
⟹ ∠OSR = ∠ORS (angles opp. equal sides)
⟹ ∠OSR = 48° (∵ ∠ORS = 48°)
In △ORS (angle sum = 180°):
∠ROS + ∠OSR + ∠ORS = 180°
∠ROS + 48° + 48° = 180°
∠ROS + 96° = 180°
∠ROS = 180° − 96°
∴ ∠ROS = 84°
Step 2 — Find ∠OPQ (in △OPQ)
PQ = RS (given)
⟹ ∠POQ = ∠ROS (equal chords → equal angles)
⟹ ∠POQ = 84°
In △OPQ: OP = OQ (radii)
⟹ ∠OPQ = ∠OQP (angles opp. equal sides)
∠OPQ + ∠OPQ + 84° = 180°
2∠OPQ = 180° − 84° = 96°
∠OPQ = 96° ÷ 2
∴ ∠OPQ = 48°
✅ Final Answers: ∠OPQ = 48° and ∠ROS = 84°
💡 Two properties used together:
- Isosceles triangle property: In triangles ORS and OPQ, two sides are radii (OR = OS and OP = OQ), so base angles are equal. This lets you find the missing angle.
- Equal chord theorem: Since PQ = RS, the central angles ∠POQ = ∠ROS. This connects the two triangles.
📐 Notice the symmetry: The base angle of △ORS is ∠ORS = 48°, and the final answer ∠OPQ is also 48°. This is not a coincidence — since PQ = RS, triangles OPQ and ORS are congruent (SSS), so all corresponding angles are equal.
Question 3 — Proving PQ = RS when PR and QS are Two Diameters
Question 3
In the figure, PR and QS are two diameters. Is PQ = RS? Justify your answer.
PR and QS are diameters; O is centre
PQRS forms a quadrilateral — is it a parallelogram?
Given: PR and QS are two diameters of a circle with centre O. To prove: PQ = RS
Key observation: Since PR and QS are both diameters, they both pass through the centre O.
So O is the intersection point of PR and QS.
Now consider quadrilateral PQRS (formed by joining P, Q, R, S in order).
The diagonals of PQRS are PR and QS.
OP = OR (∵ both are radii of the same circle)
OQ = OS (∵ both are radii of the same circle)
⟹ Diagonals PR and QS bisect each other at O.
(Each diagonal is cut into two equal halves at O)
⟹ PQRS is a parallelogram. (∵ If diagonals of a quadrilateral bisect each other, it is a parallelogram)
⟹ PQ = RS (∵ Opposite sides of a parallelogram are equal)
∴ Yes, PQ = RS ✓
✅ Conclusion: Yes, PQ = RS
The logic chain is: Both diameters pass through O → OP = OR and OQ = OS (radii) → diagonals of PQRS bisect each other → PQRS is a parallelogram → opposite sides PQ = RS.
The logic chain is: Both diameters pass through O → OP = OR and OQ = OS (radii) → diagonals of PQRS bisect each other → PQRS is a parallelogram → opposite sides PQ = RS.
🔵 The Parallelogram Connection: This problem beautifully links circle geometry with quadrilateral properties. Whenever two diameters of a circle are drawn, their four endpoints always form a parallelogram. If the two diameters are also equal in length (which they always are — both equal the diameter of the circle) AND perpendicular to each other, the parallelogram becomes a rectangle. If additionally all sides are equal, it becomes a square.
Quick Reference — All Answers at a Glance
| Question | Given | To Find / Prove | Theorem / Property Used | Answer |
|---|---|---|---|---|
| Q1 | AB = CD, ∠AOB = 90° | ∠COD | Equal chords → equal angles at centre | ∠COD = 90° |
| Q2 | PQ = RS, ∠ORS = 48° | ∠OPQ and ∠ROS | Isosceles △ (radii equal) + equal chords theorem | ∠OPQ = 48°, ∠ROS = 84° |
| Q3 | PR and QS are diameters | Prove PQ = RS | Radii equal → diagonals bisect each other → PQRS is a parallelogram | PQ = RS ✓ |
Common Mistakes to Avoid in Exercise 12.2
- Forgetting to justify Q1 with the theorem name: Simply writing "∠COD = 90°" without mentioning "equal chords subtend equal angles at the centre" loses the reasoning mark in board exams. Always state the theorem.
- In Q2, skipping the isosceles triangle step: Many students jump straight to the angle sum without first establishing that ∠OSR = ∠ORS = 48°. The isosceles step (OR = OS = radii) is essential and must be written explicitly.
- In Q2, not connecting the two triangles: After finding ∠ROS = 84°, students sometimes try to find ∠OPQ directly from scratch. The shortcut is to use PQ = RS → ∠POQ = ∠ROS = 84° first, which makes the second triangle calculation straightforward.
- In Q3, not identifying the quadrilateral: The key insight is recognising that P, Q, R, S form a quadrilateral whose diagonals are the two diameters. Without this identification, the proof becomes very difficult.
- Confusing "bisect" with "bisect at right angles": In Q3, the diagonals bisect each other (making PQRS a parallelogram), but they are not necessarily perpendicular. Do not assume it is a rectangle unless the diameters are perpendicular.
⛔ Board Exam Trap: In Q2, the two final answers ∠OPQ = 48° and ∠ORS = 48° are the same. This coincidence (a result of the congruence of the two isosceles triangles) can tempt students to think they made an error. They have not — it is correct.
What Exercise 12.2 Prepares You For
The central angle theorem from Exercise 12.2 is a stepping stone to the most important result in circle geometry: the Inscribed Angle Theorem, which states that the angle subtended by a chord at any point on the major arc is exactly half the angle it subtends at the centre. Once you have mastered Exercise 12.2, that theorem becomes much easier to grasp.
The parallelogram reasoning in Q3 directly connects to the Quadrilaterals chapter — specifically the theorem that a quadrilateral whose diagonals bisect each other is a parallelogram. Revisiting the Introduction to Circles and Exercise 12.1 will also help reinforce the vocabulary used throughout this exercise.
📐 Telangana & AP Board Exam Tip: Exercise 12.2 problems appear as 2-mark or 4-mark questions. For 2-mark questions (like Q1), full marks require both the answer and the theorem name. For 4-mark questions (like Q2 and Q3), every logical step must be written with a reason in brackets — examiners mark each step individually.
Chapter 12 — Circles
Exercise 12.2 Complete
3 Questions Solved
Central Angle Theorem
CBSE · TS · AP Board