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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![**Mathematics Problem:**
**Question:**
4. In the diagram below of circle \( O \) with radius \(\overline{OA}\), tangent \(\overline{CA}\) and secant \(\overline{COB}\) are drawn. If \( AC = 20 \) cm and \( OA = 7 \) cm, what is the length of \(\overline{OC}\), to the **nearest centimeter**?
**Diagram Explanation:**
- There is a circle centered at \( O \).
- Radius \( \overline{OA} \) extends from the center \( O \) to point \( A \) on the circumference.
- A tangent line \( \overline{CA} \) touches the circle at point \( A \).
- A secant line \( \overline{COB} \) intersects the circle at points \( B \) and \( O \), extending beyond the circle to point \( C \).
- Given measurements are \( AC = 20 \) cm and \( OA = 7 \) cm.
**Solution:**
To solve for the length of \( \overline{OC} \), we use the Tangent-Secant Theorem, which states that for a tangent \( \overline{CA} \) and secant \( \overline{COB} \) intersecting at point \( C \):
\[ CA^2 = CO \cdot (CO - OB) \]
Given:
- \( AC = 20 \) cm
- \( OA = 7 \) cm (since \( O \) is the center and \( A \) is on the circle, \( OA \) is the radius)
Therefore, \( OB = 7 \) cm.
Let \( x = CO \).
According to the theorem:
\[ (20)^2 = x \cdot (x - 7) \]
\[ 400 = x^2 - 7x \]
\[ x^2 - 7x - 400 = 0 \]
Solve the quadratic equation:
\[ x = \frac{7 \pm \sqrt{49 + 1600}}{2} \]
\[ x = \frac{7 \pm \sqrt{1649}}{2} \]
\[ x \approx \frac{7 \pm 40.6}{2} \]
Choose the positive solution:
\[ x \approx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae3ab45a-d390-4040-bd9a-ae2bbd819782%2F021b7876-2970-4612-abc1-e16d47139b10%2Fib54mgd_processed.png&w=3840&q=75)
Transcribed Image Text:**Mathematics Problem:**
**Question:**
4. In the diagram below of circle \( O \) with radius \(\overline{OA}\), tangent \(\overline{CA}\) and secant \(\overline{COB}\) are drawn. If \( AC = 20 \) cm and \( OA = 7 \) cm, what is the length of \(\overline{OC}\), to the **nearest centimeter**?
**Diagram Explanation:**
- There is a circle centered at \( O \).
- Radius \( \overline{OA} \) extends from the center \( O \) to point \( A \) on the circumference.
- A tangent line \( \overline{CA} \) touches the circle at point \( A \).
- A secant line \( \overline{COB} \) intersects the circle at points \( B \) and \( O \), extending beyond the circle to point \( C \).
- Given measurements are \( AC = 20 \) cm and \( OA = 7 \) cm.
**Solution:**
To solve for the length of \( \overline{OC} \), we use the Tangent-Secant Theorem, which states that for a tangent \( \overline{CA} \) and secant \( \overline{COB} \) intersecting at point \( C \):
\[ CA^2 = CO \cdot (CO - OB) \]
Given:
- \( AC = 20 \) cm
- \( OA = 7 \) cm (since \( O \) is the center and \( A \) is on the circle, \( OA \) is the radius)
Therefore, \( OB = 7 \) cm.
Let \( x = CO \).
According to the theorem:
\[ (20)^2 = x \cdot (x - 7) \]
\[ 400 = x^2 - 7x \]
\[ x^2 - 7x - 400 = 0 \]
Solve the quadratic equation:
\[ x = \frac{7 \pm \sqrt{49 + 1600}}{2} \]
\[ x = \frac{7 \pm \sqrt{1649}}{2} \]
\[ x \approx \frac{7 \pm 40.6}{2} \]
Choose the positive solution:
\[ x \approx
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