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Exercise 5.5

Chords and Distance from the Centre • Class 9 Mathematics • Ganita Manjari • Chapter 5

Practice Mode: पहले question को खुद solve करने की कोशिश करें। समझ न आए तो हिंदी में समझें पर click करें। उसके बाद Solution देखें से पूरा diagram और step-by-step solution check करें।
Question 1
Find the length of the chord of a circle where the radius is 7 cm and the perpendicular distance of the chord from the centre is 6 cm.
हिंदी में समझें:Centre से chord पर डाला गया perpendicular chord को दो बराबर भागों में बाँटता है। इसलिए पहले chord का आधा भाग निकालेंगे। Radius 7 cm और perpendicular distance 6 cm है, इसलिए एक right triangle बनेगा और Baudhāyana–Pythagoras theorem से half-chord मिलेगा। फिर उसे 2 से multiply करके पूरा chord मिलेगा।
Step-by-Step Solution:
Radius, Perpendicular Distance and Chord
OABM OM = 6 cmOB = 7 cmAB = chord
Given: Radius OA = OB = 7 cm, OM = 6 cm and OM ⟂ AB.

Since perpendicular from the centre to a chord bisects the chord, AM = MB.

In right-angled ΔOMA:
OA² = OM² + AM²
7² = 6² + AM²
49 = 36 + AM²
AM² = 13
AM = √13 cm
Now, AB = 2 × AM.
AB = 2 × √13 = 2√13 cm
Final Answer: The length of the chord is 2√13 cm (approximately 7.21 cm).
Question 2
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√(r² − d²).
हिंदी में समझें:यह Question पहले वाले result को general form में साबित करने के लिए है। Radius को r, centre से chord की perpendicular distance को d और पूरे chord को L मान लें। Perpendicular centre से chord को दो बराबर हिस्सों में बाँटता है, इसलिए half-chord L/2 होगा। Right triangle में Pythagoras लगाकर formula मिलेगा।
Step-by-Step Proof:
General Chord-Length Diagram
OABM drL/2
Let the chord be AB = L. Let OM ⟂ AB, where OM = d and radius OA = r.

Perpendicular from centre to chord bisects the chord, so AM = L/2.

In right-angled ΔOMA, by Baudhāyana–Pythagoras theorem:
OA² = OM² + AM²
r² = d² + (L/2)²
Therefore,
(L/2)² = r² − d²
L/2 = √(r² − d²)
L = 2√(r² − d²)
Hence proved: If the radius is r and perpendicular distance of the chord from the centre is d, then the chord length is 2√(r² − d²).
Question 3
A circle has a radius of 10 cm. Two chords of lengths 12 cm and 16 cm are drawn in the circle. Find the distance of each chord from the centre and determine which chord is nearer to the centre.
हिंदी में समझें:दोनों chords के लिए radius समान है। प्रत्येक chord का आधा भाग लेकर centre से chord की perpendicular distance निकालेंगे। जिस chord की distance कम होगी, वही centre के ज्यादा पास होगी। यह chapter के result “longer chord is closer to the centre” को भी verify करता है।
Step-by-Step Solution:
Comparing Two Chords
O12 cm chord16 cm chordd₁d₂
Case 1: Chord = 12 cm
Half-chord = 6 cm.
Radius r = 10 cm.
r² = d₁² + 6²
10² = d₁² + 36
100 − 36 = d₁²
d₁² = 64
d₁ = 8 cm
Case 2: Chord = 16 cm
Half-chord = 8 cm.
Radius r = 10 cm.
r² = d₂² + 8²
10² = d₂² + 64
100 − 64 = d₂²
d₂² = 36
d₂ = 6 cm
Since 6 cm < 8 cm, the 16 cm chord is nearer to the centre.

Longer chord → smaller distance from centre
Final Answer:
Distance of 12 cm chord = 8 cm
Distance of 16 cm chord = 6 cm
Therefore, the 16 cm chord is nearer to the centre.

Exercise 5.5 Complete

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आपने Exercise 5.5 के सभी questions complete कर लिए हैं। अब Exercise 5.6 पर जा सकते हैं।

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