---
title: "NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.3"
url: https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-3
dateModified: 2026-10-07T15:44:30+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.3

Chapter 5: I’m Up and Down, and Round and Round. Every question from Exercise 5.3, with full working and the final answer.

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## Exercise Set 5.3

### Question 1

*3 marks · Proof*

Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that $\angle CMA = \angle CMB = 90^\circ$. You need to show that $AM = BM$.)

**Solution**

1. Given: A circle with centre C, chord AB, and CM $\perp$ AB such that $\angle CMA = \angle CMB = 90^\circ$.
2. To show: AM = BM.
3. In $\Delta CMA$ and $\Delta CMB$, CA = CB (radii of the circle).
4. Side $\text{CM} = \text{CM}$ (common side).
5. Angle $\angle CMA = \angle CMB = 90^\circ$ (given).
6. By RHS congruence, $\Delta CMA \cong \Delta CMB$.
7. Hence, AM = BM (corresponding parts of congruent triangles).
8. Therefore, the perpendicular from the centre to a chord bisects the chord. Hence proved.

**Answer:** Hence proved.

> Common mistake: Using SAS instead of RHS congruence.

### Question 2

*4 marks · Proof*

An isosceles triangle ABC is inscribed in a circle, with $AB = AC$. Show that the altitude from A to BC passes through the centre of the circle.

**Solution**

1. Given: An isosceles triangle ABC inscribed in a circle with AB = AC, and AD as the altitude from A to BC.
2. To show: The altitude AD passes through the centre of the circle.
3. Since AB = AC, $\Delta ABC$ is isosceles with base BC.
4. The perpendicular bisector of the base of an isosceles triangle passes through the opposite vertex.
5. Therefore, the perpendicular bisector of BC passes through A and is the altitude AD.
6. The centre of the circumcircle of $\Delta ABC$ lies on the perpendicular bisector of BC.
7. Since AD is the perpendicular bisector of BC, the centre of the circle must lie on AD.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Assuming the centre lies on the altitude without stating that the altitude of an isosceles triangle is also the perpendicular bisector of the base.

### Question 3

*3 marks · Short answer*

Two parallel chords of lengths $6\text{ cm}$ and $8\text{ cm}$ are on opposite sides of the centre of a circle. If the radius of the circle is $5\text{ cm}$, find the distance between the midpoints of the chords.

**Solution**

1. Given: Radius $r = 5\text{ cm}$, lengths of parallel chords $AB = 6\text{ cm}$ and $CD = 8\text{ cm}$ on opposite sides of the centre.
2. Let O be the centre. Draw perpendiculars from O to both chords, meeting them at midpoints M and N.
3. Since the perpendicular from the centre bisects the chord, $\text{AM} = \frac{1}{2}AB = 3\text{ cm}$ and $\text{CN} = \frac{1}{2}CD = 4\text{ cm}$.
4. In right-angled triangle $\Delta OMA$, $\text{OM} = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{ cm}$.
5. In right-angled triangle $\Delta ONC$, $\text{ON} = \sqrt{OC^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3\text{ cm}$.
6. Since the chords are on opposite sides of the centre, the distance between their midpoints is $\text{OM} + \text{ON} = 4\text{ cm} + 3\text{ cm} = 7\text{ cm}$.

**Answer:** $7\text{ cm}$

> Common mistake: Subtracting the distances instead of adding them when chords are on opposite sides of the centre.

## Related pages

- [All Chapter 5 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions)
- [Exercise 5.1](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-1)
- [Exercise 5.2](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-2)
- [Exercise 5.4](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-4)
- [Exercise 5.5](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-5)
- [Exercise 5.6](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-6)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student