Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN: 9781337614085
Author: Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher: Cengage,
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### Problem Statement:

**Given:**
- Circle \( O \) with radius \( OT = 17 \) and \( OX = 8 \).

**Determine:**
- The length of \( NT \).

**Instructions:**
- Use mathematics to explain how you determined the answer.

### Diagram Explanation:

The given image contains a diagram of a circle with the following labeled points and lines:
- \( O \): Center of the circle.
- \( T \): A point on the circumference such that the radius \( OT = 17 \).
- \( X \): A point inside the circle such that \( OX = 8 \).
- Line segments \( TN \) and \( PX \) intersect at point \( X \) within the circle.
- \( P \) and \( N \) are other points of interest on the circle.

### Steps to Determine \( NT \):

1. **Understanding the Triangle:**
   - From the given information, \( OT \) is a radius of the circle, so \( OT = 17 \).
   - \( OX = 8 \), which is the distance from the circle’s center \( O \) to point \( X \) inside the circle.

2. **Using the Pythagorean Theorem:**
   - Triangle \( OTX \) is a right-angled triangle since \( OT \) is the hypotenuse, \( OX \) a leg, and \( XT \) the other leg.
   - Applying the Pythagorean theorem:
     \[
     OT^2 = OX^2 + XT^2
     \]
     \[
     17^2 = 8^2 + XT^2
     \]
     \[
     289 = 64 + XT^2
     \]
     \[
     XT^2 = 225
     \]
     \[
     XT = \sqrt{225} = 15
     \]

3. **Finding \( NT \):**
   - Notice that \( X \) lies between \( N \) and \( T \) on the circle’s diameter if extended.
   - Since the given problem usually applies the properties of secants and the intersecting chords in a circle, and \( XT \) being a leg of a right triangle, it is straightforward.
   - \( XT \) is partly the length from \( N \) to \( T \), so if
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Transcribed Image Text:### Problem Statement: **Given:** - Circle \( O \) with radius \( OT = 17 \) and \( OX = 8 \). **Determine:** - The length of \( NT \). **Instructions:** - Use mathematics to explain how you determined the answer. ### Diagram Explanation: The given image contains a diagram of a circle with the following labeled points and lines: - \( O \): Center of the circle. - \( T \): A point on the circumference such that the radius \( OT = 17 \). - \( X \): A point inside the circle such that \( OX = 8 \). - Line segments \( TN \) and \( PX \) intersect at point \( X \) within the circle. - \( P \) and \( N \) are other points of interest on the circle. ### Steps to Determine \( NT \): 1. **Understanding the Triangle:** - From the given information, \( OT \) is a radius of the circle, so \( OT = 17 \). - \( OX = 8 \), which is the distance from the circle’s center \( O \) to point \( X \) inside the circle. 2. **Using the Pythagorean Theorem:** - Triangle \( OTX \) is a right-angled triangle since \( OT \) is the hypotenuse, \( OX \) a leg, and \( XT \) the other leg. - Applying the Pythagorean theorem: \[ OT^2 = OX^2 + XT^2 \] \[ 17^2 = 8^2 + XT^2 \] \[ 289 = 64 + XT^2 \] \[ XT^2 = 225 \] \[ XT = \sqrt{225} = 15 \] 3. **Finding \( NT \):** - Notice that \( X \) lies between \( N \) and \( T \) on the circle’s diameter if extended. - Since the given problem usually applies the properties of secants and the intersecting chords in a circle, and \( XT \) being a leg of a right triangle, it is straightforward. - \( XT \) is partly the length from \( N \) to \( T \), so if
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