---
title: "NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.4"
url: https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-4
dateModified: 2026-10-07T15:44:30+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.4

Chapter 5: I’m Up and Down, and Round and Round. Every question from Exercise 5.4, with full working and the final answer.

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## Exercise Set 5.4

### Question 1

*4 marks · Proof*

Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.

**Solution**

1. Let a circle have centre C and chords AB and FG of equal length, so that $AB = FG$.
2. Let E and H be the midpoints of AB and FG respectively, so that $CE$ and $CH$ are perpendicular to AB and FG.
3. By Theorem 5, $AE = \frac{1}{2}AB$ and $FH = \frac{1}{2}FG$. Since $AB = FG$, we have $AE = FH$.
4. Consider right-angled triangles $\Delta CEA$ and $\Delta CHF$. By the Baudhāyana–Pythagoras theorem, $AC^2 = CE^2 + AE^2$ and $CF^2 = CH^2 + FH^2$.
5. Since $AC = CF$ (both are radii) and $AE = FH$, it follows that $CE^2 = CH^2$, giving $CE = CH$. Hence, chords of equal length are equidistant from the centre. Thus proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to state that radii of the same circle are equal.

### Question 2

*4 marks · Proof*

Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and $CE = CH$, show that $AB = GF$.

**Solution**

1. Given a circle with centre C, $CE \perp AB$, $CH \perp GH$ (where $GH$ is denoted as $FG$ in the theorem statement), and $CE = CH$.
2. From Theorem 5, the perpendicular from the centre bisects the chord, so $AE = \frac{1}{2}AB$ and $FH = \frac{1}{2}FG$.
3. Join CA and CF. Here $CA = CF$ since both are radii of the circle.
4. Consider right-angled triangles $\Delta CEA$ and $\Delta CHF$. We have $\angle CEA = \angle CHF = 90^\circ$, hypotenuse $CA = CF$, and side $CE = CH$.
5. By RHS congruence, $\Delta CEA \cong \Delta CHF$. Therefore, $AE = FH$. Since $AB = 2AE$ and $FG = 2FH$, it follows that $AB = FG$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Using SSS instead of RHS congruence when the hypotenuse and one side are known.

### Question 3

*4 marks · Proof*

Solve the previous question using the Baudhāyana–Pythagoras theorem.

**Solution**

1. Let the circle have centre C and radius $r$. We are given $CE = CH = d$, where $CE \perp AB$ and $CH \perp FG$.
2. By the Baudhāyana–Pythagoras theorem in right-angled triangle $\Delta CEA$, $AC^2 = CE^2 + AE^2$, which gives $r^2 = d^2 + AE^2$.
3. Similarly, in right-angled triangle $\Delta CHF$, $CF^2 = CH^2 + FH^2$, which gives $r^2 = d^2 + FH^2$.
4. Equating the expressions for $r^2$, we get $d^2 + AE^2 = d^2 + FH^2$, so $AE^2 = FH^2$, which means $AE = FH$.
5. Since $AB = 2AE$ and $FG = 2FH$ (perpendicular from centre bisects the chord), we obtain $AB = FG$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not explicitly stating that the perpendicular from the centre bisects the chord to relate $AE$ to $AB$.

## 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 5.5](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