In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is
- √12
- √14
- √13
- √11
The Correct Option is C
Solution and Explanation
To find the radius of a circle with two parallel chords of lengths 4 cm and 6 cm, separated by a distance of 1 cm, consider the following approach:
Let the circle's center be O. Assume the first chord of length 6 cm is AB and the second chord of length 4 cm is CD. Both chords are parallel and lie on the same side of the circle's diameter.
Let the perpendicular distance from the center O to the chord AB be d1, and to chord CD be d2. Then:
d2 = d1 + 1 cm.
The distance from the center to a chord is related to the radius and half of the chord length through the Pythagorean theorem:
For chord AB:
\( d1^2 + 3^2 = r^2 \)
(since half of 6 cm is 3 cm).
Thus,
\( d1^2 = r^2 - 9 \) (Equation 1)
For chord CD:
\( d2^2 + 2^2 = r^2 \)
(since half of 4 cm is 2 cm).
Thus,
\( (d1 + 1)^2 + 4 = r^2 \)
Expanding this equation:
\( d1^2 + 2d1 + 1 + 4 = r^2 \)
\( d1^2 + 2d1 + 5 = r^2 \) (Equation 2)
Substitute Equation 1 into Equation 2:
\( (r^2 - 9) + 2d1 + 5 = r^2 \)
\( r^2 - 9 + 2d1 + 5 = r^2 \)
\( 2d1 - 4 = 0 \)
\( 2d1 = 4 \)
\( d1 = 2 \)
Substitute \( d1 = 2 \) back into Equation 1:
\( 2^2 = r^2 - 9 \)
\( 4 = r^2 - 9 \)
\( r^2 = 13 \)
Thus, the radius of the circle is \( \sqrt{13} \) cm.
The correct answer is \( \sqrt{13} \).
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