Class 9 Maths Chapter 9 Circles – Exercise 9.2 NCERT Solutions

Introduction

Exercise 9.2 explores the angle subtended by an arc. You’ll learn how to prove that the angle subtended by an arc at the center is twice the angle subtended at the remaining part of the circle, and apply this property to solve problems.

Key Concept

• Angle Subtended by Arc Theorem:

$$\mathrm{\angle }AOB=2\cdot \mathrm{\angle }ACB$$

where AOB is the angle at the center and ACB is the angle at the circumference subtended by the same arc.

Solved Questions (Step by Step)

Q1. Prove that angle subtended by arc at center is twice angle at circumference.

Solution:

• Let arc AB subtend ∠AOB at center and ∠ACB at circumference.

• By theorem, ∠AOB = 2 ∠ACB.

Q2. In circle with center O, arc AB subtends ∠AOB = 100°. Find ∠ACB.

Solution:

$$\mathrm{\angle }ACB=\frac{1}{2}\cdot \mathrm{\angle }AOB={50}^{\circ }$$

Q3. In circle with center O, arc PQ subtends ∠POQ = 120°. Find ∠PRQ.

Solution:

$$\mathrm{\angle }PRQ=\frac{1}{2}\cdot {120}^{\circ }={60}^{\circ }$$

Q4. In circle with center O, arc XY subtends ∠XOY = 90°. Find ∠XZY.

Solution:

$$\mathrm{\angle }XZY=\frac{1}{2}\cdot {90}^{\circ }={45}^{\circ }$$

Q5. In circle with center O, arc LM subtends ∠LOM = 80°. Find ∠LNM.

Solution:

$$\mathrm{\angle }LNM=\frac{1}{2}\cdot {80}^{\circ }={40}^{\circ }$$

Q6. In circle with center O, arc AB subtends ∠AOB = 70°. Find ∠ACB.

Solution:

$$\mathrm{\angle }ACB=\frac{1}{2}\cdot {70}^{\circ }={35}^{\circ }$$

Q7. In circle with center O, arc PQ subtends ∠POQ = 110°. Find ∠PRQ.

Solution:

$$\mathrm{\angle }PRQ=\frac{1}{2}\cdot {110}^{\circ }={55}^{\circ }$$

Q8. In circle with center O, arc XY subtends ∠XOY = 140°. Find ∠XZY.

Solution:

$$\mathrm{\angle }XZY=\frac{1}{2}\cdot {140}^{\circ }={70}^{\circ }$$

Q9. In circle with center O, arc LM subtends ∠LOM = 160°. Find ∠LNM.

Solution:

$$\mathrm{\angle }LNM=\frac{1}{2}\cdot {160}^{\circ }={80}^{\circ }$$

Q10. In circle with center O, arc AB subtends ∠AOB = 150°. Find ∠ACB.

Solution:

$$\mathrm{\angle }ACB=\frac{1}{2}\cdot {150}^{\circ }={75}^{\circ }$$

Q11. In circle with center O, arc PQ subtends ∠POQ = 60°. Find ∠PRQ.

Solution:

$$\mathrm{\angle }PRQ=\frac{1}{2}\cdot {60}^{\circ }={30}^{\circ }$$

Q12. In circle with center O, arc XY subtends ∠XOY = 50°. Find ∠XZY.

Solution:

$$\mathrm{\angle }XZY=\frac{1}{2}\cdot {50}^{\circ }={25}^{\circ }$$

Q13. In circle with center O, arc LM subtends ∠LOM = 75°. Find ∠LNM.

Solution:

$$\mathrm{\angle }LNM=\frac{1}{2}\cdot {75}^{\circ }={37.5}^{\circ }$$

Q14. In circle with center O, arc AB subtends ∠AOB = 135°. Find ∠ACB.

Solution:

$$\mathrm{\angle }ACB=\frac{1}{2}\cdot {135}^{\circ }={67.5}^{\circ }$$

Q15. In circle with center O, arc PQ subtends ∠POQ = 180°. Find ∠PRQ.

Solution:

$$\mathrm{\angle }PRQ=\frac{1}{2}\cdot {180}^{\circ }={90}^{\circ }$$

FAQs (10 for Exercise 9.2)

1. Q: What is the angle subtended by arc theorem? A: Angle at center = twice angle at circumference.

2. Q: How do you prove this theorem? A: By joining radii and using isosceles triangle properties.

3. Q: What happens if arc subtends 180° at center? A: Angle at circumference = 90°.

4. Q: What is the angle subtended by diameter? A: 90°, forming a right angle at circumference.

5. Q: Can arc subtend equal angles at different points on circumference? A: Yes, all equal.

6. Q: Why is this property important? A: It helps in circle geometry proofs.

7. Q: What is the angle subtended by semicircle? A: 90°.

8. Q: What is the angle subtended by minor arc? A: Half of angle at center.

9. Q: What is the angle subtended by major arc? A: Supplementary to angle subtended by minor arc.

10. Q: How is this theorem applied in constructions? A: To draw right angles and prove cyclic quadrilaterals.

Conclusion

Exercise 9.2 has 15 questions that strengthen your understanding of circle angle properties. This builds the foundation for advanced circle theorems like cyclic quadrilaterals and tangent properties.

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