Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions ā€“ Iā€™m Up and Down, and Round and Round

The Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions provide step-by-step explanations to help students understand and solve the questions easily. These solutions are designed to strengthen mathematical concepts and support effective exam preparation. A PDF of the solutions is also available for convenient access and revision.

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.3

Exercise Set 5.3 focuses on the perpendicular from the centre to a chord and the line joining the centre to the midpoint of a chord.

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.3 Solutions ā€“ Iā€™m Up and Down, and Round and Round

Question 1. Explain why the perpendicular from the centre of a circle to a chord bisects the chord.

Answer:
Let AB be a chord of a circle with centre C. Let CM be perpendicular to AB.

Join CA and CB.

Since A and B lie on the circle:

CA = CB

Also:

CM = CM

And:

āˆ CMA = āˆ CMB = 90Ā°

Therefore, by RHS congruence:

Ī”CMA ā‰… Ī”CMB

So:

AM = BM

Final answer:
The perpendicular from the centre to a chord bisects the chord.

Question 2. 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.

Answer:
Since AB = AC, triangle ABC is isosceles.

The altitude from A to BC also bisects BC.

Let the altitude meet BC at M. Then:

BM = CM
AM āŸ‚ BC

Since M is the midpoint of chord BC, the line joining the centre of the circle to M is perpendicular to BC.

But only one perpendicular can be drawn to BC at M. Therefore, the centre of the circle lies on AM.

Final answer:
The altitude from A to BC passes through the centre of the circle.

Question 3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius is 5 cm, find the distance between the midpoints of the chords.

Answer:
For a chord of length 6 cm:

Half chord = 3 cm
Radius = 5 cm

Distance from centre:

dā‚ = āˆš(5Ā² - 3Ā²)
= āˆš(25 - 9)
= āˆš16
= 4 cm

For a chord of length 8 cm:

Half chord = 4 cm

Distance from centre:

dā‚‚ = āˆš(5Ā² - 4Ā²)
= āˆš(25 - 16)
= āˆš9
= 3 cm

Since the chords are on opposite sides of the centre, the distance between their midpoints is:

4 + 3 = 7 cm

Final answer:
The distance between the midpoints is 7 cm.

NCERT Solutions for Class 9 Maths Chapter 5 - Related Links