A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN: 9780134753119
Author: Sheldon Ross
Publisher: PEARSON
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Concept explainers

Percentage
A percentage is a number indicated as a fraction of 100. It is a dimensionless number often expressed using the symbol %.
Algebraic Expressions
In mathematics, an algebraic expression consists of constant(s), variable(s), and mathematical operators. It is made up of terms.
Numbers
Numbers are some measures used for counting. They can be compared one with another to know its position in the number line and determine which one is greater or lesser than the other.
Subtraction
Before we begin to understand the subtraction of algebraic expressions, we need to list out a few things that form the basis of algebra.
Addition
Before we begin to understand the addition of algebraic expressions, we need to list out a few things that form the basis of algebra.
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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 \(6\sqrt{3}\).
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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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