A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is
A. 12 cm
B. 14 cm
C. 16 cm
D. 18 cm
Answers
Given : A chord of length 14 cm is at a distance of 6 cm from the center of a circle and another chord is at a distance of 2 cm from the centre of the circle.
To find : The length of another chord at a distance of 2 cm from the center of the circle.
Solution :
Let AB and CD be two chords of the circle.
Draw OM ⊥ AB and ON ⊥ CD
We have , AB = 14 cm , OM = 6 cm and ON = 2 cm
Let CD = x
In ∆AOM,
By using Pythagoras theorem,
AO² = AM² + OM²
AO² = 7² + 6²
AO² = 49 + 36
AO² = 85 ………(1)
In ∆CON,
CO² = ON² + CN²
CO² = 2² + (x/2)²
[CN = ½ CD , CN = ½ x]
CO² = 4 + x²/4 ………..(2)
Since radius of the circle are equal :
∴ AO = CO
AO² = CO²
85 = 4 + x²/4
[From eq 1 & 2]
85 - 4 = x²/4
81 = x²/4
x² = 81 × 4
x = √81 × 4
x = 9 × 2
x = 18 cm
CD = 18 cm
Hence, the length of another chord at a distance of 2 cm from the center of the circle is 18 cm.
Among the given options option (D) 18 cm is correct.
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Answer:
Step-by-step explanation:
Let AB and CD be two chords of the circle.
Draw OM ⊥ AB and ON ⊥ CD
We have , AB = 14 cm , OM = 6 cm and ON = 2 cm
Let CD = x
In ∆AOM,
By using Pythagoras theorem,
AO² = AM² + OM²
AO² = 7² + 6²
AO² = 49 + 36
AO² = 85 ………(1)
In ∆CON,
CO² = ON² + CN²
CO² = 2² + (x/2)²
[CN = ½ CD , CN = ½ x]
CO² = 4 + x²/4 ………..(2)
Since radius of the circle are equal :
∴ AO = CO
AO² = CO²
85 = 4 + x²/4
[From eq 1 & 2]
85 - 4 = x²/4
81 = x²/4
x² = 81 × 4
x = √81 × 4
x = 9 × 2
x = 18 cm
CD = 18 cm