Exercise 12.4 — Arc and Angle
Angle subtended by an arc of a circle.
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Key Theorems for Exercise 12.4
All eight problems in this exercise use one or more of the following four fundamental results. Knowing which theorem applies and why is what separates full marks from partial marks in board examinations.
Theorem 1 — Central Angle is Double
The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining circle.
∠AOB = 2 × ∠APB
Theorem 2 — Same Segment Angles Equal
Angles subtended by the same arc in the same segment of a circle are all equal.
∠APB = ∠AQB = ∠ARB = ∠ASB
Theorem 3 — Cyclic Quadrilateral
Opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).
∠A + ∠C = 180°
∠B + ∠D = 180°
Theorem 4 — Equal Chords
Equal chords subtend equal angles at the centre. Chords at equal distances from the centre are equal.
AB = CD ⟺ ∠AOB = ∠COD
∠AOB (central) = 2 × ∠APB (inscribed)
Opposite angles of cyclic quad = 180°
Angles in Different Segments
< 90°
Major Segment
Angle inscribed in the major segment is an acute angle (less than 90°).
= 90°
Semicircle
Angle inscribed in a semicircle (on a diameter) is always exactly 90°.
> 90°
Minor Segment
Angle inscribed in the minor segment is an obtuse angle (greater than 90°).
🔑 Quick check: If you see an angle inscribed in a circle and you need to know if it's acute, right, or obtuse — just check whether the arc it subtends is minor (less than semicircle), a diameter, or major (more than semicircle). That tells you immediately.
Question 1 — Find ∠ADB Given ∠AOB = 100°
Question 1
O is the centre of the circle. ∠AOB = 100°. Find ∠ADB.
| Given | Circle with centre O; ∠AOB = 100° (central angle) |
| To Find | ∠ADB |
| Theorems | Central angle = 2 × inscribed angle; Opposite angles of cyclic quad = 180° |
∠AOB = 100° → ∠ACB = 50° → ∠ADB = 130°
Step 1 — Find ∠ACB (inscribed in same arc):
∠ACB = ½ × ∠AOB (central angle = 2 × inscribed angle)
∠ACB = ½ × 100° = 50°
Step 2 — Use cyclic quadrilateral property:
ABCD is a cyclic quadrilateral (all points on circle)
∠ACB + ∠ADB = 180° (opposite angles supplementary)
50° + ∠ADB = 180°
∠ADB = 180° − 50° = 130°
∴ ∠ADB = 130°
Question 2 — Find ∠BCD Given ∠BAD = 40°
Question 2
In the figure, ∠BAD = 40°. Find ∠BCD.
| Given | ∠BAD = 40°; A and C are on the same circle |
| To Find | ∠BCD |
| Theorem | Angles in the same segment of a circle are equal |
A and C subtend BD from same segment → ∠BAD = ∠BCD
BD is a chord of the circle. Join B and D.
A and C are both on the same arc (same side of chord BD).
∠BCD = ∠BAD (angles in the same segment of a circle are equal)
∠BCD = 40°
∴ ∠BCD = 40°
💡 This is the shortest proof in the exercise. Once you recognise that both A and C lie on the major arc of chord BD, same-segment theorem gives the answer immediately — no calculation needed.
Question 3 — Find ∠PQR and ∠PSR Given ∠POR = 120°
Question 3
O is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR.
| Given | Circle with centre O; ∠POR = 120° (central angle) |
| To Find | ∠PQR and ∠PSR |
| Theorems | Central = 2 × inscribed; Cyclic quad opposite angles = 180° |
∠POR=120° → ∠PQR=60° → ∠PSR=120°
Step 1 — Find ∠PQR:
∠PQR = ½ × ∠POR (central angle = 2 × inscribed angle)
∠PQR = ½ × 120° = 60°
Step 2 — Find ∠PSR (opposite angle in cyclic quad PQRS):
∠PQR + ∠PSR = 180° (opposite angles of cyclic quadrilateral)
60° + ∠PSR = 180°
∠PSR = 180° − 60° = 120°
∴ ∠PQR = 60° and ∠PSR = 120°
Question 4 — Find Radius Using Pythagoras (OM = 3 cm, AB = 8 cm)
Question 4
O is the centre; OM = 3 cm and AB = 8 cm. Find the radius of the circle.
| Given | OM ⊥ AB; OM = 3 cm; chord AB = 8 cm |
| To Find | Radius OA |
| Theorems | Perpendicular from centre bisects the chord → Pythagoras theorem |
AM = BM = 4 cm (perp bisects chord)
Step 1 — OM bisects AB:
AM = ½ × AB = ½ × 8 = 4 cm (perp from centre bisects the chord)
Step 2 — Apply Pythagoras in △OAM (right angle at M):
OA² = OM² + AM²
OA² = 3² + 4²
OA² = 9 + 16 = 25
OA = √25 = 5 cm
∴ Radius of the circle = 5 cm
📌 This is the classic 3-4-5 right triangle (Pythagorean triple) hidden inside a circle problem. The perpendicular from the centre creates a right triangle where OM = 3, AM = 4, and radius OA = 5. Recognising Pythagorean triples (3,4,5 and 5,12,13) saves calculation time.
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Question 5 — Find RS Given OM = ON and PQ = 6 cm
Question 5
O is the centre. OM and ON are perpendiculars to chords PQ and RS respectively. If OM = ON and PQ = 6 cm, find RS.
| Given | OM ⊥ PQ, ON ⊥ RS; OM = ON; PQ = 6 cm |
| To Find | RS |
| Theorem | Chords equidistant from the centre are equal in length |
OM = ON → PQ = RS
OM = ON (given — equal perpendicular distances from centre)
By Theorem 4: Chords equidistant from the centre are equal in length
∴ PQ = RS
RS = PQ = 6 cm
∴ RS = 6 cm
Question 6 — Radius When A is Centre and ABCD is a Square, BD = 4 cm
Question 6
A is the centre of the circle and ABCD is a square. If BD = 4 cm, find the radius.
| Given | ABCD is a square; A is the centre of the circle; BD = 4 cm |
| To Find | Radius of the circle |
| Key Property | Diagonals of a square are equal — AC = BD |
AC = BD = 4 cm; radius = AC
In square ABCD: diagonals are equal → AC = BD
AC = 4 cm (∵ BD = 4 cm)
A is the centre of the circle; C lies on the circle
∴ Radius = AC = 4 cm
∴ Radius = 4 cm
📐 Why this works: Since ABCD is a square and A is the centre, the circle must pass through all four vertices (B, C, D are equidistant from A because they are all vertices of a square — and the diagonal AC equals BD). The radius is the distance from A to any vertex, which equals the diagonal length.
Question 7 — Draw Two Equidistant Chords
Question 7
Draw a circle with any radius and then draw two chords equidistant from the centre.
AB ∥ CD, both equidistant from O
Steps of Construction
- Step 1: Draw a circle with centre O and any convenient radius.
- Step 2: Draw a diameter of the circle (a straight line through O).
- Step 3: Set the compass to any radius r. Place the compass at O and draw two arcs — one above and one below the centre — cutting the diameter at two points equidistant from O.
- Step 4: At each intersection point, draw a chord perpendicular to the diameter. These are chords AB (above) and CD (below).
- Step 5: AB and CD are the required chords. They are equidistant from the centre (OM = ON) and therefore also equal in length (AB = CD).
✅ Why they're equal: By the converse of Theorem 4, since OM = ON (equal distances from centre), the chords AB = CD. Parallel chords on opposite sides of the centre that are symmetric about the centre are always equal and equidistant.
Question 8 — Angles of △OCD When AB = CD and ∠AOB = 70°
Question 8
O is the centre. AB and CD are equal chords. ∠AOB = 70°. Find all angles of △OCD.
| Given | AB = CD (equal chords); ∠AOB = 70° |
| To Find | ∠COD, ∠OCD, ∠ODC (all angles of △OCD) |
| Theorems | Equal chords subtend equal central angles; OC = OD (radii) → isosceles △OCD |
Equal chords → equal central angles → isosceles △OCD
Step 1 — Find ∠COD:
AB = CD → ∠AOB = ∠COD (equal chords subtend equal angles at centre)
∠COD = ∠AOB = 70°
Step 2 — △OCD is isosceles:
OC = OD = radius (both are radii of the same circle)
∴ ∠OCD = ∠ODC … (1) (base angles of isosceles △)
Step 3 — Angle sum in △OCD:
∠COD + ∠OCD + ∠ODC = 180°
70° + ∠OCD + ∠OCD = 180° (using eq.1)
70° + 2∠OCD = 180°
2∠OCD = 180° − 70° = 110°
∠OCD = 110° ÷ 2 = 55°
∠ODC = 55° (= ∠OCD, from eq.1)
∴ ∠COD = 70°, ∠OCD = 55°, ∠ODC = 55°
💡 Two radii always form an isosceles triangle. Whenever two radii of a circle are sides of a triangle, the triangle is isosceles, and the base angles are equal. This is used in Q8 (OC = OD) and is a pattern that recurs throughout Chapters 12 and 13.
Quick Summary — All 8 Questions at a Glance
| Q | Given | Find / Task | Theorem Used | Answer |
|---|---|---|---|---|
| 1 | ∠AOB = 100° | ∠ADB | Central = 2× inscribed; Cyclic quad | 130° |
| 2 | ∠BAD = 40° | ∠BCD | Same segment angles equal | 40° |
| 3 | ∠POR = 120° | ∠PQR and ∠PSR | Central = 2× inscribed; Cyclic quad | 60° and 120° |
| 4 | OM = 3, AB = 8 cm | Radius | Perp bisects chord + Pythagoras | 5 cm |
| 5 | OM = ON, PQ = 6 cm | RS | Equal distances → equal chords | 6 cm |
| 6 | Square ABCD, BD = 4 cm, A is centre | Radius | Diagonals of square are equal | 4 cm |
| 7 | — | Draw 2 equidistant chords | Symmetric arcs from O; perp chords | Construction |
| 8 | AB = CD, ∠AOB = 70° | All angles of △OCD | Equal chords → equal central angles; isosceles △ | 70°, 55°, 55° |
Common Mistakes to Avoid in Board Exams
- Applying central angle theorem in the wrong direction: The central angle is twice the inscribed angle — not the other way round. ∠inscribed = ½ × ∠central. Many students write ∠AOB = ½ × ∠APB and get the wrong sign direction.
- Same-segment theorem requires same arc: The theorem ∠BAD = ∠BCD only works when A and C are on the same arc (same side of chord BD). If they are on opposite arcs, the angles are supplementary, not equal.
- Forgetting that opposite angles of cyclic quad sum to 180°, not equal each other: ∠A + ∠C = 180° (supplementary), not ∠A = ∠C. A very common error under exam pressure.
- Not stating "OC = OD (radii)" before claiming the triangle is isosceles in Q8: The reason for the isosceles property must be explicitly stated. Examiners require justification.
- Skipping the Pythagoras step in Q4: After finding AM = 4 using the bisector theorem, you still need the Pythagoras theorem to find OA. Writing OA = AM + OM = 4 + 3 = 7 is a major error — you must use OA² = OM² + AM².
⚠️ Most common error (Q1 and Q3): After correctly finding ∠ACB = 50° from the central angle, some students write ∠ADB = 180° − 100° = 80° (subtracting the central angle from 180°). That is wrong. You must first find the inscribed angle (50°) and then subtract from 180° to get the opposite angle in the cyclic quadrilateral (130°). Always use the inscribed angle — not the central angle — in the cyclic quadrilateral step.
What This Exercise Prepares You For
The central angle theorem (∠AOB = 2∠APB) and the cyclic quadrilateral property (opposite angles = 180°) are two of the most widely applied circle theorems throughout secondary mathematics. They reappear in Exercise 12.5 on cyclic quadrilaterals, where you prove more advanced results about cyclic polygons.
The Pythagoras-in-circle technique from Q4 is a template used throughout Chapter 12 and again in Class 10 Chapter 10 (Tangent to a circle), where right triangles formed by the radius and tangent are solved the same way. Question 8's isosceles triangle technique (two radii → base angles equal) is used in nearly every circle proof.
For Telangana and Andhra Pradesh board exams, Q1/Q3 type (find angle using central angle + cyclic quad) appear in almost every year's paper as 2–3 mark questions. Q4 type (Pythagoras to find radius) appears as a 3–4 mark question. Q8 type (angles of isosceles triangle formed by radii) is a reliable 4-mark problem.
✅ Board Exam Strategy (CBSE, Telangana & AP): For every angle problem in Circles, write down the theorem you are using as a reason in brackets — "(central angle = 2 × inscribed angle)" or "(opposite angles of cyclic quadrilateral are supplementary)". These bracketed reasons earn dedicated marks, and skipping them turns a 4-mark problem into a 2-mark one.