Question:medium

In a circle with center \(O\), a \(6\) cm long chord is at a distance \(4\) cm from the center. Then the length of diameter is

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The perpendicular from the center to a chord bisects the chord.
  • \(5\) cm
  • \(10\) cm
  • \(15\) cm
  • \(8\) cm
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This is a classic geometry problem involving the relationship between the radius, a chord, and the perpendicular distance from the center to the chord. The key is to form a right-angled triangle.
Step 2: Key Formula or Approach:
1. A line drawn from the center of a circle perpendicular to a chord bisects the chord. 2. The Pythagorean theorem in a right-angled triangle: $a^2 + b^2 = c^2$, where c is the hypotenuse.
Step 3: Detailed Explanation:
Let the circle have center O and radius r. Let AB be the chord with length 6 cm. Let M be the midpoint of AB. The distance from the center O to the chord AB is the length of the perpendicular segment OM, which is given as 4 cm. Since the perpendicular from the center bisects the chord, the length of AM is half the length of AB. \[ AM = \frac{AB}{2} = \frac{6}{2} = 3 \text{ cm} \] Now, consider the triangle $\triangle OMA$. It is a right-angled triangle with the right angle at M. The sides are:
OM = 4 cm (distance from center to chord)
AM = 3 cm (half the chord length)
OA = r (the radius of the circle, which is the hypotenuse)
Using the Pythagorean theorem: \[ OA^2 = OM^2 + AM^2 \] \[ r^2 = 4^2 + 3^2 \] \[ r^2 = 16 + 9 = 25 \] \[ r = \sqrt{25} = 5 \text{ cm} \] The question asks for the length of the diameter. The diameter (d) is twice the radius. \[ d = 2r = 2 \times 5 = 10 \text{ cm} \] Step 4: Final Answer:
The length of the diameter of the circle is 10 cm. Therefore, option (B) is correct.
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