Triangle KJL is inscribed in semicircle O shown below such that JK = 6 and JL =11. Which of the following is closest to the radius of the semicircle? (1) 5.68 (2) 6.26 (3) 10.61 (4) 12.53 K
Triangle KJL is inscribed in semicircle O shown below such that JK = 6 and JL =11. Which of the following is closest to the radius of the semicircle? (1) 5.68 (2) 6.26 (3) 10.61 (4) 12.53 K
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 Explanation
**Problem Statement:**
Triangle \( KJL \) is inscribed in a semicircle \( O \) shown below such that \( JK = 6 \) and \( JL = 11 \). Which of the following is closest to the radius of the semicircle?
1. 5.68
2. 6.26
3. 10.61
4. 12.53
**Diagram Description:**
The provided diagram illustrates a semicircle with center \( O \). A triangle \( KJL \) is inscribed within this semicircle. Here are the notable points and lengths:
- Point \( J \) is on the circumference of the semicircle.
- Points \( K \) and \( L \) are endpoints of the diameter of the semicircle, \( KL \).
- Segment \( JK \) is given as 6 units.
- Segment \( JL \) is given as 11 units.
- Line segment \( JO \) is also a radius of the semicircle.
**Step-by-Step Solution:**
Using the properties of the inscribed triangle in a semicircle, the hypotenuse of the right triangle \( KJL \) is the diameter, and the triangle \( KJL \) is right-angled at \( J \).
1. By the Pythagorean theorem, we know:
\[ KL = \sqrt{JK^2 + JL^2} = \sqrt{6^2 + 11^2} = \sqrt{36 + 121} = \sqrt{157} \]
2. The diameter of the semicircle is \( \sqrt{157} \).
3. The radius is half the diameter:
\[ \text{Radius} = \frac{\sqrt{157}}{2} \approx \frac{12.53}{2} = 6.26 \]
Therefore, the closest to the radius of the semicircle is:
\[ (2) 6.26 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8d84827-277a-47d3-aef5-d10d7e3c7bc7%2Fcbb2ff18-5248-46f9-a5c8-9e5156cad65e%2Fjmxlclj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Geometry Problem Explanation
**Problem Statement:**
Triangle \( KJL \) is inscribed in a semicircle \( O \) shown below such that \( JK = 6 \) and \( JL = 11 \). Which of the following is closest to the radius of the semicircle?
1. 5.68
2. 6.26
3. 10.61
4. 12.53
**Diagram Description:**
The provided diagram illustrates a semicircle with center \( O \). A triangle \( KJL \) is inscribed within this semicircle. Here are the notable points and lengths:
- Point \( J \) is on the circumference of the semicircle.
- Points \( K \) and \( L \) are endpoints of the diameter of the semicircle, \( KL \).
- Segment \( JK \) is given as 6 units.
- Segment \( JL \) is given as 11 units.
- Line segment \( JO \) is also a radius of the semicircle.
**Step-by-Step Solution:**
Using the properties of the inscribed triangle in a semicircle, the hypotenuse of the right triangle \( KJL \) is the diameter, and the triangle \( KJL \) is right-angled at \( J \).
1. By the Pythagorean theorem, we know:
\[ KL = \sqrt{JK^2 + JL^2} = \sqrt{6^2 + 11^2} = \sqrt{36 + 121} = \sqrt{157} \]
2. The diameter of the semicircle is \( \sqrt{157} \).
3. The radius is half the diameter:
\[ \text{Radius} = \frac{\sqrt{157}}{2} \approx \frac{12.53}{2} = 6.26 \]
Therefore, the closest to the radius of the semicircle is:
\[ (2) 6.26 \]
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