Question:

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

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The perpendicular from the center of a circle to a chord bisects the chord. Then use Pythagoras theorem to find the radius.
  • \(5\text{ cm}\)
  • \(10\text{ cm}\)
  • \(15\text{ cm}\)
  • \(8\text{ cm}\)
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The Correct Option is B

Solution and Explanation

We are given a circle with center \(O\). The chord length is \[ 6\text{ cm}. \] The perpendicular distance from the center to the chord is \[ 4\text{ cm}. \] We know that the perpendicular from the center of a circle to a chord bisects the chord. Therefore, half of the chord is \[ \frac{6}{2}=3\text{ cm}. \] Now consider the right triangle formed by: \[ \text{radius of circle}, \] \[ \text{distance from center to chord}=4\text{ cm}, \] and \[ \text{half chord}=3\text{ cm}. \] Let the radius be \(r\). Using Pythagoras theorem: \[ r^2=3^2+4^2. \] \[ r^2=9+16. \] \[ r^2=25. \] \[ r=5\text{ cm}. \] The diameter is twice the radius: \[ \text{Diameter}=2r. \] \[ =2(5). \] \[ =10\text{ cm}. \] Therefore, the length of the diameter is \[ 10\text{ cm}. \]
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