Circle Answer key

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Answer Key & Marking Scheme

Class 9 Mathematics - Chapter: Circles

Total Marks: 80
Note: These are the model solutions. Marks are awarded for stepwise working.
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SECTION A: Very Short Answer (1 Mark)

  • 1. ∠PRQ (The angle subtended at point R).
  • 2. Zero (0). The centre lies on the diameter.
  • 3. 180°.
  • 4. 90° (Right Angle).
  • 5. Bisects.
  • 6. They are equal in length.
  • 7. The corresponding chords are equal.
  • 8. A quadrilateral where all 4 vertices lie on a circle.
  • 9. Distance.
  • 10. If angles subtended by chords at the centre are equal, the chords are equal.

SECTION B: Short Answer (2 Marks)

11. Length of AM:
Answer: 3 cm.
Reason: Perpendicular from centre bisects the chord. AM = ½ × 6 = 3 cm.

12. Relation between PQ and RS:
Answer: They are Equal (PQ = RS).
Theorem: Chords equidistant from the centre are equal in length.

13. Angle on remaining part:
Answer: 75°.
Calculation: Angle at circumference is half of angle at centre. 150° / 2 = 75°.

14. Angle between line and chord:
Answer: 90°.
Reason: Line joining centre to the mid-point of a chord is perpendicular to the chord.

15. Measure of ∠D:
Answer: 65°.
Calculation: Opposite angles of cyclic quad sum to 180°. ∠D = 180° - 115° = 65°.

16. Measure of ∠DAB:
Answer: 100°.
Step 1: ∠CAD = ∠DBC = 55° (Angles in same segment).
Step 2: ∠DAB = ∠CAD + ∠BAC = 55° + 45° = 100°.

17. Conclusion about points:
Answer: They are Concyclic.
(Meaning: Points P, Q, R, S lie on the same circle).

18. Relation between ∠AOB and ∠COD:
Answer: They are Equal.
Justification: ΔAOB ≅ ΔCOD by SSS Rule (Radii are equal, Chords given equal).

19. Relationship:
Answer: AB = CD.

20. Find ∠BDC:
Answer: 80°.
Step 1: In ΔABC, ∠A = 180° - (69°+31°) = 80°.
Step 2: ∠BDC = ∠BAC (Angles in same segment).

SECTION C: Detailed Solutions (3 Marks)

21. Proof (Theorem 9.2):
Given ∠AOB = ∠COD.
In ΔAOB and ΔCOD:
1. OA = OC (Radii)
2. OB = OD (Radii)
3. ∠AOB = ∠COD (Given)
∴ ΔAOB ≅ ΔCOD (SAS Rule).
By CPCT, Chord AB = Chord CD.
23. Cyclic Parallelogram is a Rectangle:
1. Since it's a parallelogram, ∠A = ∠C.
2. Since it's cyclic, ∠A + ∠C = 180°.
3. substituting (1) in (2): 2∠A = 180° ⇒ ∠A = 90°.
A parallelogram with one 90° angle is a Rectangle.
24. Find ∠ADC:
1. Central Angle ∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°.
2. Angle at remaining part is half of central angle.
3. ∠ADC = ½ × 90° = 45°.

SECTION D: Long Answer (5 Marks)

32. Proof: Intersecting Chords
Given: ∠AEQ = ∠DEQ.
Construction: Draw OL ⊥ AB and OM ⊥ CD.
Step 1: In ΔOLE and ΔOME:
- ∠OLE = ∠OME = 90°
- OE = OE (Common)
- ∠LEO = ∠MEO (Given)
Step 2: ΔOLE ≅ ΔOME (AAS Rule).
Step 3: By CPCT, OL = OM.
Conclusion: Since chords are equidistant from the centre (OL=OM), the chords are equal. AB = CD.
34. Find ∠BAC:
Step 1: Find ∠CED (Linear Pair).
∠CED = 180° - 130° = 50°.

Step 2: In ΔCDE, find ∠D.
∠D = 180° - (50° + 20°) = 110°.

Step 3: Angles in same segment are equal.
Since ∠BAC and ∠BDC (∠D) are subtended by the same arc BC...
∠BAC = 110°.