Two parallel chords of length 8 units each are 6 units away from each other in a circle.
What is the radius of the circle in units?

The correct answer is 5 (Option 2).

Step-by-step Explanation:

1. Consider a circle with center $O$ and radius $r$.

2. We are given two parallel chords of equal length, each measuring 8 units. Let one of the chords be $AB$, so $AB=8$ units.

3. The perpendicular distance between the two parallel chords is 6 units. Since the two chords are equal in length, they must be at equal distances from the center of the circle. Therefore, the perpendicular distance from the center $O$ to each chord is:

$d=\frac{6}{2}=3$ units.

4. Draw a perpendicular line segment from the center $O$ to the chord $AB$, meeting $AB$ at point $M$. A perpendicular drawn from the center of a circle to a chord bisects the chord.

Hence, $AM=\frac{AB}{2}=\frac{8}{2}=4$ units.

5. Join the center $O$ to endpoint $A$ to form a right-angled triangle $OMA$, where $OA=r$ is the radius of the circle, $OM=3$ units, and $AM=4$ units.

6. Apply the Pythagorean theorem in right-angled triangle $OMA$:

$O{A}^2=O{M}^2+A{M}^2$

$r^2=3^2+4^2$

$r^2=9+16=25$

$r=\sqrt{25}=5$ units.

Thus, the radius of the circle is 5 units.