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
Related questions
Question
Need work shown
![**Problem 3: Geometry - Problem Solving**
In the diagram below of triangle \( \triangle PAO \), \( \overline{AP} \) is tangent to circle \( O \) at point \( A \). Given that \( OB = 7 \) and \( BP = 18 \), determine the length of \( AP \).
**Diagram Description:**
The diagram depicts a circle with center \( O \) and a tangent line \( \overline{AP} \) touching the circle at point \( A \). A triangle \( \triangle PAO \) is formed with the points \( P, A, \) and \( O \). Point \( B \) is on \( \overline{OP} \) such that \( OB = 7 \) and \( BP = 18 \).
Given:
- \( OB = 7 \)
- \( BP = 18 \)
Find:
- The length of \( \overline{AP} \)
**Explanation:**
1. Identify that \( OB \) and \( BP \) form a straight line segment \( OP \) with \( OB + BP = OP = 7 + 18 \).
2. Since \( \overline{AP} \) is tangent to the circle at point \( A \), \( \angle OAP \) is a right angle.
Use the Pythagorean theorem in \( \triangle OAP \):
\[ OP^2 = OB^2 + AP^2 \]
3. Calculate the total length of segment \( OP \):
\[ OP = OB + BP = 7 + 18 = 25 \]
4. Thus, \( OP = 25 \).
Use the Pythagorean theorem:
\[ OP^2 = OA^2 + AP^2 \]
\[ 25^2 = 7^2 + AP^2 \]
\[ 625 = 49 + AP^2 \]
Isolate \( AP^2 \):
\[ AP^2 = 625 - 49 \]
\[ AP^2 = 576 \]
Find \( AP \):
\[ AP = \sqrt{576} \]
\[ AP = 24 \]
Therefore, the length of \( \overline{AP} \) is \( 24 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae3ab45a-d390-4040-bd9a-ae2bbd819782%2Fc4ee730e-3a0f-44bf-bed2-b710f20aab2a%2F8ivsbmk_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 3: Geometry - Problem Solving**
In the diagram below of triangle \( \triangle PAO \), \( \overline{AP} \) is tangent to circle \( O \) at point \( A \). Given that \( OB = 7 \) and \( BP = 18 \), determine the length of \( AP \).
**Diagram Description:**
The diagram depicts a circle with center \( O \) and a tangent line \( \overline{AP} \) touching the circle at point \( A \). A triangle \( \triangle PAO \) is formed with the points \( P, A, \) and \( O \). Point \( B \) is on \( \overline{OP} \) such that \( OB = 7 \) and \( BP = 18 \).
Given:
- \( OB = 7 \)
- \( BP = 18 \)
Find:
- The length of \( \overline{AP} \)
**Explanation:**
1. Identify that \( OB \) and \( BP \) form a straight line segment \( OP \) with \( OB + BP = OP = 7 + 18 \).
2. Since \( \overline{AP} \) is tangent to the circle at point \( A \), \( \angle OAP \) is a right angle.
Use the Pythagorean theorem in \( \triangle OAP \):
\[ OP^2 = OB^2 + AP^2 \]
3. Calculate the total length of segment \( OP \):
\[ OP = OB + BP = 7 + 18 = 25 \]
4. Thus, \( OP = 25 \).
Use the Pythagorean theorem:
\[ OP^2 = OA^2 + AP^2 \]
\[ 25^2 = 7^2 + AP^2 \]
\[ 625 = 49 + AP^2 \]
Isolate \( AP^2 \):
\[ AP^2 = 625 - 49 \]
\[ AP^2 = 576 \]
Find \( AP \):
\[ AP = \sqrt{576} \]
\[ AP = 24 \]
Therefore, the length of \( \overline{AP} \) is \( 24 \).
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, geometry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Elementary Geometry For College Students, 7e
Geometry
ISBN:
9781337614085
Author:
Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:
Cengage,
Elementary Geometry for College Students
Geometry
ISBN:
9781285195698
Author:
Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:
Cengage Learning
Elementary Geometry For College Students, 7e
Geometry
ISBN:
9781337614085
Author:
Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:
Cengage,
Elementary Geometry for College Students
Geometry
ISBN:
9781285195698
Author:
Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:
Cengage Learning