Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.5 Solutions – I’m Up and Down, and Round and Round

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.5

Exercise Set 5.5 focuses on the relation between radius, perpendicular distance from centre and chord length.

Question 1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

Answer:
Radius = 7 cm
Distance from centre to chord = 6 cm

Half chord = √(r² - d²)

= √(7² - 6²)
= √(49 - 36)
= √13

Chord length = 2√13 cm

Final answer:
2√13 cm

Question 2. Explain why, if the perpendicular distance of a chord from the centre is d and the radius is r, then chord length is 2√(r² - d²).

Answer:
Let AB be the chord and O be the centre.

Drop perpendicular OM to AB.

Then:

OM = d
OA = r

Since perpendicular from centre to chord bisects the chord:

AM = AB/2

In right triangle OMA:

OA² = OM² + AM²

r² = d² + AM²

AM² = r² - d²

AM = √(r² - d²)

Therefore:

AB = 2AM
= 2√(r² - d²)

Final answer:
Chord length = 2√(r² - d²).

Question 3. In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, can we conclude that CD = 2AB?

Answer:
No.

Chord length does not vary directly with distance from the centre. The formula is:

Chord length = 2√(r² - d²)

If one chord is at distance d and another is at distance 2d, their lengths are:

2√(r² - d²) and 2√(r² - 4d²)

These are not in the ratio 1:2.

Final answer:
No, CD = 2AB cannot be concluded. Chord length depends on √(r² - d²), not directly on d.