AB A- 9.

A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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**8. Finding the Length of Chord AB:**

In this problem, we are given a circle with a center labeled as point \(O\). Within the circle, a chord \(AB\) is shown. The chord \(AB\) is divided into two segments, \(OA\) and \(OB\), forming two right triangles sharing the common vertex at point \(O\), the center of the circle.

Key Measurements:
- The distance from the center \(O\) to the midpoint of the chord (labeled as point \(M\)):
  \(OM = 3\)
- The radius of the circle:
  \(OB = 6\)

We are asked to find the length of the chord \(AB\).

*Diagram Explanation:*
- The circle has \(O\) as its center.
- The line segment \(AB\) is the chord of the circle.
- The line segment \(OM\) is perpendicular to \(AB\), thereby bisecting it at \(M\).
- The lengths indicated are \(OM = 3\) and \(OB = 6\).

To find the length of chord \(AB\), we note that \(OM\) is the perpendicular bisector of \(AB\).

We can use the Pythagorean theorem in the right triangle \(OBM\):

\[
OB^2 = OM^2 + MB^2
\]

\[
6^2 = 3^2 + MB^2
\]

\[
36 = 9 + MB^2
\]

Subtract 9 from both sides:

\[
27 = MB^2
\]

Taking the square root of both sides, we get:

\[
MB = \sqrt{27} = 3\sqrt{3}
\]

Since \(M\) is the midpoint of \(AB\), we have:

\[
AB = 2 \times MB = 2 \times 3\sqrt{3} = 6\sqrt{3}
\]

Thus, the length of the chord \(AB\) is \(6\sqrt{3}\).
Transcribed Image Text:**8. Finding the Length of Chord AB:** In this problem, we are given a circle with a center labeled as point \(O\). Within the circle, a chord \(AB\) is shown. The chord \(AB\) is divided into two segments, \(OA\) and \(OB\), forming two right triangles sharing the common vertex at point \(O\), the center of the circle. Key Measurements: - The distance from the center \(O\) to the midpoint of the chord (labeled as point \(M\)): \(OM = 3\) - The radius of the circle: \(OB = 6\) We are asked to find the length of the chord \(AB\). *Diagram Explanation:* - The circle has \(O\) as its center. - The line segment \(AB\) is the chord of the circle. - The line segment \(OM\) is perpendicular to \(AB\), thereby bisecting it at \(M\). - The lengths indicated are \(OM = 3\) and \(OB = 6\). To find the length of chord \(AB\), we note that \(OM\) is the perpendicular bisector of \(AB\). We can use the Pythagorean theorem in the right triangle \(OBM\): \[ OB^2 = OM^2 + MB^2 \] \[ 6^2 = 3^2 + MB^2 \] \[ 36 = 9 + MB^2 \] Subtract 9 from both sides: \[ 27 = MB^2 \] Taking the square root of both sides, we get: \[ MB = \sqrt{27} = 3\sqrt{3} \] Since \(M\) is the midpoint of \(AB\), we have: \[ AB = 2 \times MB = 2 \times 3\sqrt{3} = 6\sqrt{3} \] Thus, the length of the chord \(AB\) is \(6\sqrt{3}\).
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