Correct Answer:
B. 8 cm
The correct answer is 8 cm. This problem is a classic application of the perpendicular from the center to a chord theorem in circle geometry. According to this theorem, a perpendicular drawn from the center of a circle to any chord bisects that chord, meaning it divides the chord into two equal halves.
Step‑by‑Step Solution
- Step 1: Visualize the right triangle. Draw a circle with center O. Let AB be a chord of length 12 cm. Draw a perpendicular from O to AB, meeting AB at point M. Since OM ⟂ AB, M is the midpoint of AB.
- Step 2: Find the half‑chord length. AM = MB = 12 ÷ 2 = 6 cm.
- Step 3: Identify the hypotenuse. In right triangle OMA, OA is the radius of the circle, so OA = 10 cm (the hypotenuse).
- Step 4: Apply the Pythagorean theorem. OM² + AM² = OA² → OM² + 6² = 10² → OM² + 36 = 100 → OM² = 64 → OM = √64 = 8 cm.
Thus, the perpendicular distance from the center to the chord is 8 cm.
Why the Other Options Are Incorrect
- 6 cm (Option 1): This is the length of the half‑chord (AM), not the distance from the center. Many confuse the two values.
- 10 cm (Option 3): This is the radius of the circle. The radius is the hypotenuse, not the distance from the center to the chord (which must be shorter than the radius).
- 12 cm (Option 4): This is the total length of the chord itself. The distance from the center to the chord is always less than the radius for any non‑diameter chord, so 12 cm (greater than 10) is geometrically impossible.
Hence, applying the perpendicular‑bisector theorem and the Pythagorean theorem confirms that the distance is exactly 8 cm.