---
title: "NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.5"
url: https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-5
dateModified: 2026-10-07T15:44:30+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.5

Chapter 5: I’m Up and Down, and Round and Round. Every question from Exercise 5.5, with full working and the final answer.

Free PDF (26 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-5-i-m-up-and-down-and-round-and-round-3f5dbc3f8b.pdf

## Exercise Set 5.5

### Question 1

*3 marks · Short answer*

Find the length of the chord of a circle where the radius is $7\text{ cm}$ and perpendicular distance is $6\text{ cm}$.

**Solution**

1. Given: Radius $r = 7\text{ cm}$, perpendicular distance from centre $d = 6\text{ cm}$.
2. By the Baudhāyana--Pythagoras theorem applied to half the chord, radius, and perpendicular distance: half chord length $= \sqrt{r^2 - d^2} = \sqrt{7^2 - 6^2} = \sqrt{49 - 36} = \sqrt{13}\text{ cm}$.
3. The total length of the chord is twice the length of half the chord.
4. Result: $2\sqrt{13}\text{ cm}$.

**Answer:** $2\sqrt{13}\text{ cm}$

> Common mistake: Forgetting to multiply the half-chord length by 2 to get the full chord length.

### Question 2

*3 marks · Proof*

Explain why the following statement is true: If the perpendicular distance of a chord from the centre is $d$ and the radius is $r$, then the chord length is $2\sqrt{r^2 - d^2}$.

**Solution**

1. Let the circle have centre $C$, radius $r$, and let AB be a chord.
2. Let $M$ be the midpoint of AB, so that CM is perpendicular to AB with length $d$.
3. Join CA to form right-angled triangle CMA, where $\angle CMA = 90^\circ$ and hypotenuse $CA = r$.
4. By the Baudhāyana--Pythagoras Theorem in $\Delta CMA$, $AM^2 + CM^2 = CA^2$.
5. Substitute $CM = d$ and $CA = r$: $AM^2 + d^2 = r^2$, which gives $AM = \sqrt{r^2 - d^2}$.
6. Since the perpendicular from the centre bisects the chord, $AB = 2 \times AM = 2\sqrt{r^2 - d^2}$.
7. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not stating that the perpendicular from the centre bisects the chord.

### Question 3

*3 marks · Short answer*

In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that $CD = 2 AB$? Give reasons for your answer.

**Solution**

1. Let the distance of chord AB from the centre be $d_1$ and the distance of chord CD be $d_2$. We are given that $d_1 = 2d_2$.
2. From the formula for chord length, $AB = 2\sqrt{r^2 - d_1^2}$ and $CD = 2\sqrt{r^2 - d_2^2}$.
3. Substituting $d_1 = 2d_2$, we get $AB = 2\sqrt{r^2 - 4d_2^2}$, whereas $2CD = 4\sqrt{r^2 - d_2^2}$.
4. Since $\sqrt{r^2 - 4d_2^2} \neq 2\sqrt{r^2 - d_2^2}$, we cannot conclude that $CD = 2AB$.

**Answer:** No, we cannot conclude that $CD = 2AB$.

> Common mistake: Assuming linear proportionality between chord length and its distance from 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 5.3](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-3)
- [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 5.6](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