9. If RS = 6, ST = 4, and RP = 15, what is the length 10. Find mZL, rounded to the nearest degree. of RQ? 25 0. 24 K. | 11. The center of circle Q has coordinates (6,4). If 12. If AF = 6, CD = 5, and BE = 4, what is the circle Q passes through Ã(10,7), what is the length perimeter of AABC ? of its diameter? %3D E 'C D F.
9. If RS = 6, ST = 4, and RP = 15, what is the length 10. Find mZL, rounded to the nearest degree. of RQ? 25 0. 24 K. | 11. The center of circle Q has coordinates (6,4). If 12. If AF = 6, CD = 5, and BE = 4, what is the circle Q passes through Ã(10,7), what is the length perimeter of AABC ? of its diameter? %3D E 'C D F.
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 Problems and Their Solutions
#### Problem 9
**Question:** If \( RS = 6 \), \( ST = 4 \), and \( RP = 15 \), what is the length of \( RQ \)?
**Explanation:**
The diagram shows a circle with center \( O \). A straight line forms a triangle \( RST \), with \( Q \) on the circle such that \( RQ \) is a radius. Given the distances:
- \( RP \) (from \( R \) to \( P \))
- \( RS \) (from \( R \) to \( S \))
- \( ST \) (from \( S \) to \( T \))
Make use of the properties of the circle along with these lengths to solve for \( RQ \).
#### Problem 10
**Question:** Find \( m ∠L \), rounded to the nearest degree.
**Explanation:**
The diagram represents a right-angled triangle \( JKL \) where:
- \( \overline{JK} = 24 \) units (base)
- \( \overline{KL} = 7 \) units (height)
- \( \overline{JL} = 25 \) units (hypotenuse)
Using trigonometric ratios (specifically the tangent function) to find the angle \( ∠L \):
\[ \tan(∠L) = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24} \]
Then, use the arctangent (inverse tangent) to calculate \( ∠L \).
#### Problem 11
**Question:** The center of circle \( Q \) has coordinates \( (6,4) \). If circle \( Q \) passes through \( A(10,7) \), what is the length of its diameter?
**Explanation:**
To determine the radius of the circle, calculate the distance between the center \( (6,4) \) and the point \( A(10,7) \) using the distance formula:
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]
\[ d = \sqrt{(10-6)^2 + (7-4)^2} \]
The diameter is twice the radius, so multiply the radius found by 2 to get the diameter.
#### Problem](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fccc6c07f-901f-4b9b-9990-bf05e0ed87d5%2F817ef756-a48a-41dc-b580-d13a5e4b17e6%2F0w91quq_processed.png&w=3840&q=75)
Transcribed Image Text:### Geometry Problems and Their Solutions
#### Problem 9
**Question:** If \( RS = 6 \), \( ST = 4 \), and \( RP = 15 \), what is the length of \( RQ \)?
**Explanation:**
The diagram shows a circle with center \( O \). A straight line forms a triangle \( RST \), with \( Q \) on the circle such that \( RQ \) is a radius. Given the distances:
- \( RP \) (from \( R \) to \( P \))
- \( RS \) (from \( R \) to \( S \))
- \( ST \) (from \( S \) to \( T \))
Make use of the properties of the circle along with these lengths to solve for \( RQ \).
#### Problem 10
**Question:** Find \( m ∠L \), rounded to the nearest degree.
**Explanation:**
The diagram represents a right-angled triangle \( JKL \) where:
- \( \overline{JK} = 24 \) units (base)
- \( \overline{KL} = 7 \) units (height)
- \( \overline{JL} = 25 \) units (hypotenuse)
Using trigonometric ratios (specifically the tangent function) to find the angle \( ∠L \):
\[ \tan(∠L) = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24} \]
Then, use the arctangent (inverse tangent) to calculate \( ∠L \).
#### Problem 11
**Question:** The center of circle \( Q \) has coordinates \( (6,4) \). If circle \( Q \) passes through \( A(10,7) \), what is the length of its diameter?
**Explanation:**
To determine the radius of the circle, calculate the distance between the center \( (6,4) \) and the point \( A(10,7) \) using the distance formula:
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]
\[ d = \sqrt{(10-6)^2 + (7-4)^2} \]
The diameter is twice the radius, so multiply the radius found by 2 to get the diameter.
#### Problem
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