Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter44: Solution Of Equations By The Subtraction, Addition, And Division Principles Of Equality
Section: Chapter Questions
Problem 38A
Related questions
Question
![**Question 13: Find the length of arc \( \overset{\frown}{JK} \). Round your answer to the nearest tenth.**
**Diagram Explanation:**
The diagram provides a circle with center \(M\), and the radius is labeled as 10 cm. There is an angle formed at the center \(M\) by the radii \(MJ\) and \(MK\), which measures 140 degrees. The points \(J\) and \(K\) lie on the circumference of the circle.
**Solution Steps:**
1. **Identify the given values:**
- Radius \(r = 10\) cm
- Central angle \(\theta = 140^\circ\)
2. **Arc length formula:**
The length \(L\) of an arc of a circle with radius \(r\) and central angle \(\theta\) (measured in degrees) is given by:
\[
L = 2\pi r \left(\frac{\theta}{360}\right)
\]
3. **Substitute the values into the formula:**
\[
L = 2\pi (10) \left(\frac{140}{360}\right)
\]
4. **Calculate the arc length:**
\[
L = 20\pi \left(\frac{140}{360}\right)
\]
Simplify the fraction:
\[
\frac{140}{360} = \frac{7}{18}
\]
\[
L = 20\pi \left(\frac{7}{18}\right)
\]
\[
L = \frac{140\pi}{18}
\]
5. **Perform the division and multiply by \(\pi\):**
\[
L \approx \frac{140 \times 3.14159}{18}
\]
\[
L \approx \frac{439.8226}{18}
\]
\[
L \approx 24.4346 \text{ cm}
\]
6. **Round the answer to the nearest tenth:**
\[
L \approx 24.4 \text{ cm}
\]
Therefore, the length of \( \overset{\frown}{JK} \) is approximately 24.4 cm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c8a549a-ce13-4c73-8e46-ff62e8d935c2%2Febe8e101-14f6-4c0f-a989-0240db27fc2d%2Fo7k69t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 13: Find the length of arc \( \overset{\frown}{JK} \). Round your answer to the nearest tenth.**
**Diagram Explanation:**
The diagram provides a circle with center \(M\), and the radius is labeled as 10 cm. There is an angle formed at the center \(M\) by the radii \(MJ\) and \(MK\), which measures 140 degrees. The points \(J\) and \(K\) lie on the circumference of the circle.
**Solution Steps:**
1. **Identify the given values:**
- Radius \(r = 10\) cm
- Central angle \(\theta = 140^\circ\)
2. **Arc length formula:**
The length \(L\) of an arc of a circle with radius \(r\) and central angle \(\theta\) (measured in degrees) is given by:
\[
L = 2\pi r \left(\frac{\theta}{360}\right)
\]
3. **Substitute the values into the formula:**
\[
L = 2\pi (10) \left(\frac{140}{360}\right)
\]
4. **Calculate the arc length:**
\[
L = 20\pi \left(\frac{140}{360}\right)
\]
Simplify the fraction:
\[
\frac{140}{360} = \frac{7}{18}
\]
\[
L = 20\pi \left(\frac{7}{18}\right)
\]
\[
L = \frac{140\pi}{18}
\]
5. **Perform the division and multiply by \(\pi\):**
\[
L \approx \frac{140 \times 3.14159}{18}
\]
\[
L \approx \frac{439.8226}{18}
\]
\[
L \approx 24.4346 \text{ cm}
\]
6. **Round the answer to the nearest tenth:**
\[
L \approx 24.4 \text{ cm}
\]
Therefore, the length of \( \overset{\frown}{JK} \) is approximately 24.4 cm.
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