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**Problem Statement:** Given circle \( O \) with radius 5 and \( OC = 3 \). Find the length of \( AB \). **Diagram Explanation:** - The circle has a center \( O \) and radius 5 units. - \( C \) is a point on the circle such that \( OC = 3 \). - \( AB \) is a chord of the circle with \( C \) being the midpoint and perpendicular to \( OC \). **Steps to Solve:** 1. We know \( OC \) is the perpendicular bisector of the chord \( AB \), which creates two right triangles, \( \triangle OCA \) and \( \triangle OCB \). 2. In \( \triangle OCA \), we have: - \( OA = 5 \) (radius of the circle) - \( OC = 3 \) 3. By the Pythagorean theorem in \( \triangle OCA \): \[ OA^2 = OC^2 + AC^2 \] \[ 5^2 = 3^2 + AC^2 \] \[ 25 = 9 + AC^2 \] \[ AC^2 = 16 \] \[ AC = 4 \] 4. Since \( AB \) is twice \( AC \) (because \( C \) is the midpoint of \( AB \)), \[ AB = 2 \times AC = 2 \times 4 = 8 \] **Answer:** The length of \( AB \) is 8 units.

**Problem Statement:** Given circle \( O \) with radius 5 and \( OC = 3 \). Find the length of \( AB \). **Diagram Explanation:** - The circle has a center \( O \) and radius 5 units. - \( C \) is a point on the circle such that \( OC = 3 \). - \( AB \) is a chord of the circle with \( C \) being the midpoint and perpendicular to \( OC \). **Steps to Solve:** 1. We know \( OC \) is the perpendicular bisector of the chord \( AB \), which creates two right triangles, \( \triangle OCA \) and \( \triangle OCB \). 2. In \( \triangle OCA \), we have: - \( OA = 5 \) (radius of the circle) - \( OC = 3 \) 3. By the Pythagorean theorem in \( \triangle OCA \): \[ OA^2 = OC^2 + AC^2 \] \[ 5^2 = 3^2 + AC^2 \] \[ 25 = 9 + AC^2 \] \[ AC^2 = 16 \] \[ AC = 4 \] 4. Since \( AB \) is twice \( AC \) (because \( C \) is the midpoint of \( AB \)), \[ AB = 2 \times AC = 2 \times 4 = 8 \] **Answer:** The length of \( AB \) is 8 units.

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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**Problem Statement:** Given circle \( O \) with radius 5 and \( OC = 3 \). Find the length of \( AB \). **Diagram Explanation:** - The circle has a center \( O \) and radius 5 units. - \( C \) is a point on the circle such that \( OC = 3 \). - \( AB \) is a chord of the circle with \( C \) being the midpoint and perpendicular to \( OC \). **Steps to Solve:** 1. We know \( OC \) is the perpendicular bisector of the chord \( AB \), which creates two right triangles, \( \triangle OCA \) and \( \triangle OCB \). 2. In \( \triangle OCA \), we have: - \( OA = 5 \) (radius of the circle) - \( OC = 3 \) 3. By the Pythagorean theorem in \( \triangle OCA \): \[ OA^2 = OC^2 + AC^2 \] \[ 5^2 = 3^2 + AC^2 \] \[ 25 = 9 + AC^2 \] \[ AC^2 = 16 \] \[ AC = 4 \] 4. Since \( AB \) is twice \( AC \) (because \( C \) is the midpoint of \( AB \)), \[ AB = 2 \times AC = 2 \times 4 = 8 \] **Answer:** The length of \( AB \) is 8 units.
Transcribed Image Text:**Problem Statement:** Given circle \( O \) with radius 5 and \( OC = 3 \). Find the length of \( AB \). **Diagram Explanation:** - The circle has a center \( O \) and radius 5 units. - \( C \) is a point on the circle such that \( OC = 3 \). - \( AB \) is a chord of the circle with \( C \) being the midpoint and perpendicular to \( OC \). **Steps to Solve:** 1. We know \( OC \) is the perpendicular bisector of the chord \( AB \), which creates two right triangles, \( \triangle OCA \) and \( \triangle OCB \). 2. In \( \triangle OCA \), we have: - \( OA = 5 \) (radius of the circle) - \( OC = 3 \) 3. By the Pythagorean theorem in \( \triangle OCA \): \[ OA^2 = OC^2 + AC^2 \] \[ 5^2 = 3^2 + AC^2 \] \[ 25 = 9 + AC^2 \] \[ AC^2 = 16 \] \[ AC = 4 \] 4. Since \( AB \) is twice \( AC \) (because \( C \) is the midpoint of \( AB \)), \[ AB = 2 \times AC = 2 \times 4 = 8 \] **Answer:** The length of \( AB \) is 8 units.
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