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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Question
![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa242a625-2826-4f00-ad8f-039422d3d327%2Ff048e36f-2b7e-489a-99a8-b8552ec6cc1b%2Fkvxxl1p_processed.jpeg&w=3840&q=75)
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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