Question : AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle. If the distance between them is 2 cm, then the radius (in cm) of the circle is:
Option 1: $\frac{\sqrt{265}}{4}$
Option 2: $\frac{\sqrt{256}}{4}$
Option 3: $\frac{\sqrt{156}}{4}$
Option 4: $\frac{\sqrt{198}}{4}$
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Correct Answer: $\frac{\sqrt{265}}{4}$
Solution :
Given, AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle.
Let OM be perpendicular to AB and ON be perpendicular to CD.
Perpendicular from the centre to a chord bisects the chord.
So, AM = BM = 4 cm and CN = DN = 3 cm
Let the radius of the circle be r.
ON – OM = MN = 2 cm----(1)
In $\triangle$OAM,
OA
2
= OM
2
+ AM
2
⇒ r
2
= OM
2
+ 4
2
----(2)
In $\triangle$OCN,
OC
2
= ON
2
+ CN
2
⇒ r
2
= ON
2
+ 3
2
----(3)
From equation 2 and 3, we get,
OM
2
+ 4
2
= ON
2
+ 3
2
⇒ ON
2
– OM
2
= 16 – 9
⇒ (ON – OM)(ON + OM) = 7
⇒ (ON + OM) = $\frac{7}{2}$----(4)
Adding equation (1) and (4), we get:
⇒ ON = $\frac{11}{4}$
Substituting this value in equation (3), we get,
⇒ r
2
= $(\frac{11}{4})^2$ + 3
2
⇒ r
2
= $\frac{265}{16}$
$\therefore$ r = $\frac{\sqrt{265}}{4}$ cm
Hence, the correct answer is $\frac{\sqrt{265}}4$.
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