Question

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

• A. √12
• B. √14
• C. √13
• D. √11

Answer 1

The Correct Option is C

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 \( \sqrt{13} \).