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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![### Geometry Problem: Finding FD
In the given diagram, we have a circle with center O. There are several points labeled on and around the circle. The task is to find the length of segment \( FD \).
#### Diagram Explanation:
- **Points**: The diagram includes points labeled \( D, E, F, G, T \).
- **Segments**:
- \( FD \): The segment from point \( F \) to point \( D \), which we need to find.
- \( FE \): A segment from point \( F \) to point \( E \) measuring 10 units.
- \( FG \): A segment from point \( F \) to point \( G \) measuring 9 units.
- \( GT \): A segment from point \( G \) to point \( T \), but its length is not given in the diagram.
- \( DT \): A chord of the circle passing through points \( D \) and \( T \).
#### Given Measurements:
- Segment \( FE \) is 10 units.
- Segment \( FG \) is 9 units.
- Segment \( ET \) is 11 units.
- Denote \( DE = x \)
Use the power of a point theorem that states:
\[ FG \cdot FE = ET \cdot DE \]
That is:
\[ 9 \cdot 10 = 11 \cdot x \]
Solving for \( x \):
\[ 90 = 11x \]
\[ x = \frac{90}{11} \approx 8.18 \]
Since \( FD = FE + DE \)
\[ FD = 10 + \frac{90}{11} \approx 10 + 8.18 = 18.18 \]
So, FD ≈ 18.18 units
#### Answer Box:
\[ \boxed{FD}\ =\ 18.18 \text{ units} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc32fdead-cf8a-44ab-a7f8-9f10e201d4c1%2Fec3ec8aa-d835-40e8-bad8-4e45dd454aa9%2Fegibrli_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Geometry Problem: Finding FD
In the given diagram, we have a circle with center O. There are several points labeled on and around the circle. The task is to find the length of segment \( FD \).
#### Diagram Explanation:
- **Points**: The diagram includes points labeled \( D, E, F, G, T \).
- **Segments**:
- \( FD \): The segment from point \( F \) to point \( D \), which we need to find.
- \( FE \): A segment from point \( F \) to point \( E \) measuring 10 units.
- \( FG \): A segment from point \( F \) to point \( G \) measuring 9 units.
- \( GT \): A segment from point \( G \) to point \( T \), but its length is not given in the diagram.
- \( DT \): A chord of the circle passing through points \( D \) and \( T \).
#### Given Measurements:
- Segment \( FE \) is 10 units.
- Segment \( FG \) is 9 units.
- Segment \( ET \) is 11 units.
- Denote \( DE = x \)
Use the power of a point theorem that states:
\[ FG \cdot FE = ET \cdot DE \]
That is:
\[ 9 \cdot 10 = 11 \cdot x \]
Solving for \( x \):
\[ 90 = 11x \]
\[ x = \frac{90}{11} \approx 8.18 \]
Since \( FD = FE + DE \)
\[ FD = 10 + \frac{90}{11} \approx 10 + 8.18 = 18.18 \]
So, FD ≈ 18.18 units
#### Answer Box:
\[ \boxed{FD}\ =\ 18.18 \text{ units} \]
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