In a circle of radius 13\text{ cm}, two parallel chords AB and CD are drawn on the same si… — JAMB UTME

In a circle of radius $13\text{ cm}$ , two parallel chords $AB$ and $CD$ are drawn on the same side of the center $O$ . If $AB = 24\text{ cm}$ and $CD = 10\text{ cm}$ , calculate the perpendicular distance between the two chords in centimeters.

Answer: 7 cm

Answer

The perpendicular distance between the two chords is

7\text{ cm}

The perpendicular line from the center

O

to a chord bisects the chord. Applying the Pythagorean theorem to the right triangles formed by the radius (

13\text{ cm}

) and half-chords (

12\text{ cm}

and

5\text{ cm}

) yields distances of

5\text{ cm}

and

12\text{ cm}

from the center, respectively. Since both parallel chords are on the same side of the center, the distance between them is

12\text{ cm} - 5\text{ cm} = 7\text{ cm}

Step-by-Step Solution

Find the perpendicular distance from center

O

to chord

AB

d_1 = 5\text{ cm}

A line drawn from the center of a circle perpendicular to a chord bisects the chord. Using the right triangle formed by the radius, half-chord (

12\text{ cm}

), and perpendicular distance:

d_1 = \sqrt{13^2 - 12^2} = 5\text{ cm}

Find the perpendicular distance from center

O

to chord

CD

d_2 = 12\text{ cm}

Using the perpendicular bisector property for chord

CD

(half-chord is

5\text{ cm}

d_2 = \sqrt{13^2 - 5^2} = 12\text{ cm}

Calculate the distance between the parallel chords

7\text{ cm}

Because both chords are on the same side of the center

O

, the distance between them is the difference of their individual distances from the center:

12\text{ cm} - 5\text{ cm} = 7\text{ cm}

Key Concept

Perpendicular from the center of a circle to a chord bisects the chord

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