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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: $\frac{\sqrt{198}}{4}$


Team Careers360 22nd Jan, 2024
Answer (1)
Team Careers360 23rd Jan, 2024

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,
OA2 = OM2 + AM2
⇒ r2 = OM2 + 4----(2)
In $\triangle$OCN,
OC2 = ON2 + CN2
⇒ r2 = ON2 + 32----(3)
From equation 2 and 3, we get,
OM2 + 42 = ON2 + 32
⇒ ON– OM= 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,
⇒ r2 = $(\frac{11}{4})^2$ + 32
⇒ r2 = $\frac{265}{16}$
$\therefore$ r = $\frac{\sqrt{265}}{4}$ cm
Hence, the correct answer is $\frac{\sqrt{265}}4$.

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