Class 9 Maths Chapter 9 Circles – Exercise 9.3 NCERT Solutions

Introduction

Exercise 9.3 explores the angle in a semicircle theorem and the properties of cyclic quadrilaterals. You’ll learn how to prove that the angle in a semicircle is a right angle and that opposite angles of a cyclic quadrilateral are supplementary.

Key Concepts

• Angle in a Semicircle:

$$\mathrm{\angle }ABC={90}^{\circ }$$

if AB is a diameter and C lies on the circle.

• Cyclic Quadrilateral Property: Opposite angles of a cyclic quadrilateral are supplementary.

$$\mathrm{\angle }A+\mathrm{\angle }C={180}^{\circ },\mathrm{\angle }B+\mathrm{\angle }D={180}^{\circ }$$

Solved Questions (Step by Step)

Q1. Prove that angle in a semicircle is a right angle.

Solution:

• Let AB be diameter, C a point on circle.

• ∠ACB = 90°.

Q2. In circle with diameter PQ, point R lies on circle. Prove ∠PRQ = 90°.

Solution:

• By semicircle theorem, ∠PRQ = 90°.

Q3. In circle with diameter XY, point Z lies on circle. Prove ∠XZY = 90°.

Solution:

• By semicircle theorem, ∠XZY = 90°.

Q4. In circle with diameter LM, point N lies on circle. Prove ∠LNM = 90°.

Solution:

• By semicircle theorem, ∠LNM = 90°.

Q5. In circle with diameter AB, point C lies on circle. Find ∠ACB.

Solution:

• ∠ACB = 90°.

Q6. In circle with diameter PQ, point R lies on circle. Find ∠PRQ.

Solution:

• ∠PRQ = 90°.

Q7. In circle with diameter XY, point Z lies on circle. Find ∠XZY.

Solution:

• ∠XZY = 90°.

Q8. In circle with diameter LM, point N lies on circle. Find ∠LNM.

Solution:

• ∠LNM = 90°.

Q9. Prove that opposite angles of cyclic quadrilateral are supplementary.

Solution:

• Let ABCD be cyclic quadrilateral.

• ∠A + ∠C = 180°, ∠B + ∠D = 180°.

Q10. In cyclic quadrilateral PQRS, prove ∠P + ∠R = 180°.

Solution:

• By cyclic quadrilateral property, ∠P + ∠R = 180°.

Q11. In cyclic quadrilateral LMNO, prove ∠L + ∠N = 180°.

Solution:

• By cyclic quadrilateral property, ∠L + ∠N = 180°.

Q12. In cyclic quadrilateral ABCD, prove ∠B + ∠D = 180°.

Solution:

• By cyclic quadrilateral property, ∠B + ∠D = 180°.

Q13. In cyclic quadrilateral PQRS, prove ∠Q + ∠S = 180°.

Solution:

• By cyclic quadrilateral property, ∠Q + ∠S = 180°.

Q14. In cyclic quadrilateral LMNO, prove ∠M + ∠O = 180°.

Solution:

• By cyclic quadrilateral property, ∠M + ∠O = 180°.

Q15. In cyclic quadrilateral ABCD, prove ∠A + ∠C = 180°.

Solution:

• By cyclic quadrilateral property, ∠A + ∠C = 180°.

FAQs (10 for Exercise 10.3)

1. Q: What is a semicircle theorem? A: Angle in a semicircle is a right angle.

2. Q: What is a cyclic quadrilateral? A: A quadrilateral whose vertices lie on a circle.

3. Q: What is the property of cyclic quadrilaterals? A: Opposite angles are supplementary.

4. Q: Can every quadrilateral be cyclic? A: No, only those whose vertices lie on a circle.

5. Q: What is the angle subtended by diameter? A: 90°.

6. Q: What is the sum of opposite angles in cyclic quadrilateral? A: 180°.

7. Q: Why is semicircle theorem important? A: It helps prove right angles in circle geometry.

8. Q: Why is cyclic quadrilateral property important? A: It simplifies angle calculations in circle geometry.

9. Q: What is the sum of angles in a quadrilateral? A: 360°.

10. Q: How do you prove cyclic quadrilateral property? A: By using angle subtended by arc theorem.

Conclusion

Exercise 9.3 has 15 questions that strengthen your understanding of semicircle and cyclic quadrilateral properties. This builds the foundation for advanced circle theorems involving tangents and cyclic figures.

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