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: Tangent Line to a Circle
#### Problem Statement:
Line BC is a tangent to circle A at point C. Given:
- \( B \) to \( C \) is 24 inches
- \( D \) to \( C \) is 16 inches
Find the length of segment \( A \) to \( C \).
#### Diagram Description:
- The diagram shows a circle with center \( A \).
- Line \( BC \) is tangent to the circle at point \( C \).
- Point \( D \) lies on the circle, perpendicular to line \( BC \), making \( \overline{AD} \) a radius of the circle.
- The length \( \overline{BC} = 24 \) inches.
- The length \( \overline{DC} = 16 \) inches.
#### Solution Approach:
To find the length of \( \overline{AC} \), we can use the Pythagorean theorem in the right triangle \( \Delta ADC \) where:
- \( \overline{AD} \) is the radius of the circle and is perpendicular to the tangent \( \overline{BC} \).
- \( \overline{DC} \) is the distance from point \( D \) (a point on the circle intersecting the radius) to the point of tangency \( C \).
Given:
- \( \overline{DC} = 16 \) inches (part of the radius perpendicular to the tangent)
- \( \overline{BC} = 24 \) inches (the entire tangent segment)
Since \( D \) is directly above \( B \) on the circle:
- The radius \( \overline{AC} \) can be calculated using Pythagoras' theorem in \( \Delta ADC \).
Using Pythagoras’ theorem:
\[ \overline{AD}^2 = \overline{AC}^2 - \overline{DC}^2 \]
\[ \overline{AD} = \sqrt{\overline{AC}^2 - \overline{DC}^2} \]
Plugging in the given values:
\[ \overline{AD} = \sqrt{24^2 - 16^2} \]
\[ \overline{AD} = \sqrt{576 - 256} \]
\[ \overline{AD} = \sqrt{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4052e509-76f5-4637-b3e2-bf83f94bbd00%2F17e4ff20-7479-4b64-8edf-58c2672a4973%2Fxabqn57_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Geometry Problem: Tangent Line to a Circle
#### Problem Statement:
Line BC is a tangent to circle A at point C. Given:
- \( B \) to \( C \) is 24 inches
- \( D \) to \( C \) is 16 inches
Find the length of segment \( A \) to \( C \).
#### Diagram Description:
- The diagram shows a circle with center \( A \).
- Line \( BC \) is tangent to the circle at point \( C \).
- Point \( D \) lies on the circle, perpendicular to line \( BC \), making \( \overline{AD} \) a radius of the circle.
- The length \( \overline{BC} = 24 \) inches.
- The length \( \overline{DC} = 16 \) inches.
#### Solution Approach:
To find the length of \( \overline{AC} \), we can use the Pythagorean theorem in the right triangle \( \Delta ADC \) where:
- \( \overline{AD} \) is the radius of the circle and is perpendicular to the tangent \( \overline{BC} \).
- \( \overline{DC} \) is the distance from point \( D \) (a point on the circle intersecting the radius) to the point of tangency \( C \).
Given:
- \( \overline{DC} = 16 \) inches (part of the radius perpendicular to the tangent)
- \( \overline{BC} = 24 \) inches (the entire tangent segment)
Since \( D \) is directly above \( B \) on the circle:
- The radius \( \overline{AC} \) can be calculated using Pythagoras' theorem in \( \Delta ADC \).
Using Pythagoras’ theorem:
\[ \overline{AD}^2 = \overline{AC}^2 - \overline{DC}^2 \]
\[ \overline{AD} = \sqrt{\overline{AC}^2 - \overline{DC}^2} \]
Plugging in the given values:
\[ \overline{AD} = \sqrt{24^2 - 16^2} \]
\[ \overline{AD} = \sqrt{576 - 256} \]
\[ \overline{AD} = \sqrt{
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