Angle Subtended by an Arc at the Center vs Circumference
What you will learn
Verify that the angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the remaining part of the circle (Angle at Center Theorem).
Materials needed
- Circular Geoboard with 36-peg graduated circumferenceWhat is it?Circular Geoboard with 36-peg graduated circumference₹50
Circular geometry pegboard with elastic cord runners and revolving degree protractor.
Specifications- Diameter
- 300 mm circular wooden base
- Pegs
- 36 boundary pegs at 10° intervals + center peg O
- Cords
- Color-coded elastic cords for forming inscribed rays
- Protractor
- Integrated 360° rotating pointer protractor
- Colored elastic rubber bands (red, blue, yellow) packWhat is it?Colored elastic rubber bands (red, blue, yellow) pack₹95
Set of clear transparent acrylic angle replicas (35°, 70°, 90°) for direct superimposition.
Specifications- Material
- 3 mm laser-cut transparent neon acrylic
- Angles
- 35.0° and 70.0° matched angle templates
- 360° Circular Full ProtractorWhat is it?360° Circular Full Protractor₹140
White drawing card and heavy compass with micron drafting pen.
Specifications- Card
- A3 size 250 GSM smooth Bristol board
- Pen
- 0.3 mm pigment technical drawing pen
- Tracing paper and Drawing Compass setWhat is it?Tracing paper and Drawing Compass set₹185
Standard laboratory apparatus used in Angle Subtended by an Arc at the Center vs Circumference.
Specifications- Category
- Mathematics Apparatus
- Grade
- CBSE / NEP 2020 Laboratory Standard
- Cardboard sheet with pinsWhat is it?Cardboard sheet with pins₹230
Standard laboratory apparatus used in Angle Subtended by an Arc at the Center vs Circumference.
Specifications- Category
- Mathematics Apparatus
- Grade
- CBSE / NEP 2020 Laboratory Standard
Tick items as you collect them. Tap + to see the product, specifications and buy options.
Stay safe
- Ensure center point O is accurately positioned before drawing radii.
- Point P must lie strictly on the remaining circumference, not on arc AB itself.
- Use sharp pencils and fine rubber bands on the geoboard to prevent angular reading errors.
- Verify the theorem across both minor arcs, major arcs, and semicircles (angle in a semicircle = 90°).
Steps
- 1Draw a circle of radius 10 cm on white cardboard and mark center O.Observation
Draw a circle of radius r = 8.0 cm with center O on white drawing sheet; mark arc AB subtending angle at center.
SafetyEnsure compass point is firmly anchored at center point O.
- 2Mark two points A and B on the circumference to define arc AB.Observation
Join OA and OB to form central angle ∠AOB; measure ∠AOB = 70.0° using circular protractor.
SafetyAlign 0° baseline of protractor exactly along radius OA.
- 3Join radius OA and radius OB to form central angle ∠AOB.Observation
Select point P1 on major arc; join AP1 and BP1 to form inscribed angle ∠AP1B; measure ∠AP1B = 35.0°.
SafetyUse sharp 2H pencil for razor-sharp geometric ray lines.
- 4Mark any point P on the remaining major arc circumference and join PA and PB to form inscribed angle ∠APB.Observation
Calculate ratio ∠AOB / ∠AP1B = 70.0° / 35.0° = 2.00, verifying the 2 : 1 theorem.
SafetyRecord readings in laboratory measurement table.
- 5Measure central angle ∠AOB accurately with the 360° protractor (e.g. ∠AOB = 80°).Observation
Choose alternative points P2 and P3 on arc; all measure exactly 35.0°, confirming angles in same segment are equal.
SafetyVerify that P does not cross into minor arc AB.
- 6Measure angle ∠APB with the protractor (measured ∠APB = 40°).Observation
Test semicircle case where AB is diameter (∠AOB = 180°); inscribed angle ∠APB measures exactly 90.0°.
SafetyMount tracing cutouts of angles to physically verify superimposition.
- 7Trace angle ∠APB onto transparent tracing paper twice. Place both traced angles side by side over ∠AOB to demonstrate that 2 × ∠APB fits perfectly over ∠AOB.
Tap + on a step to see what you will observe and any safety tips.
Record your observations
| Circle Trial | Position of Point P on Arc | Inscribed Angle ∠APB (°) | Central Angle ∠AOB (°) | Ratio ∠AOB / ∠APB |
|---|---|---|---|---|
| Trial 1 | P1 (Left upper circumference) | 35.0° | 70.0° | 2.00 (Exact 2 : 1) |
| Trial 2 | P2 (Center upper circumference) | 35.0° | 70.0° | 2.00 (Angles in same segment equal) |
| Trial 3 | P3 (Right upper circumference) | 35.0° | 70.0° | 2.00 (Angles in same segment equal) |
| Trial 4 (Larger Arc) | Arc AB subtending 110° at center | 55.0° | 110.0° | 2.00 (Exact 2 : 1) |
| Trial 5 (Semicircle) | Arc AB diameter (180° at center) | 90.0° (Right angle in semicircle) | 180.0° | 2.00 (Exact 2 : 1) |
Tap a highlighted cell to write your own observation.
Result and conclusion
For any chosen arc AB, angle ∠AOB at center measured 80° while inscribed angle ∠APB at circumference measured exactly 40°. Verified ∠AOB = 2 · ∠APB across all arc lengths.
| Observation | What it means |
|---|---|
| Fundamental Central Angle Theorem | The angle subtended by an arc at the center of a circle is strictly double the angle subtended by it at any point on the remaining part of the circle. |
| Invariance of Inscribed Angle | As point P slides along the major arc (P1 -> P2 -> P3), the inscribed angle ∠APB remains invariant at 35.0°, proving angles in the same segment of a circle are equal. |
| Thales Theorem as Special Case | When central angle is a straight line (180° diameter), inscribed angle ∠APB = 180°/2 = 90°, proving the angle in a semicircle is a right angle. |
| Cyclic Quadrilateral Connection | Opposite angles in cyclic quadrilaterals naturally sum to 180° as a direct consequence of this central angle theorem. |
Sample readings. Replace with your own results.