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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![### Problem Statement:
In circle \( N \) with \( m \angle M N P = 60^\circ \) and \( M N = 8 \) units, find the area of sector MNP. Round to the nearest hundredth.
### Diagram Explanation:
The image is a diagram of a circle centered at point \( N \). The points \( M \) and \( P \) are on the circumference of the circle. The line segment \( M N \) forms a radius of the circle. The angle \( \angle M N P \) is given as \( 60^\circ \).
### Solution:
To find the area of sector \( MNP \), we can use the formula for the area of a sector in a circle:
\[ \text{Area of sector} = \left(\frac{\theta}{360^\circ}\right) \pi r^2 \]
where \( \theta \) is the central angle in degrees, and \( r \) is the radius of the circle.
Given that \( \theta = 60^\circ \) and \( r = 8 \) units:
\[ \text{Area of sector MNP} = \left(\frac{60^\circ}{360^\circ}\right) \pi (8)^2 \]
First, simplify the fraction:
\[ \frac{60^\circ}{360^\circ} = \frac{1}{6} \]
Then, calculate the area:
\[ \text{Area of sector MNP} = \frac{1}{6} \pi (8)^2 = \frac{1}{6} \pi (64) = \frac{64}{6} \pi \]
\[ \text{Area of sector MNP} = \frac{32}{3} \pi \approx 10.67 \pi \]
Now, using the approximate value of \( \pi \approx 3.14 \):
\[ \text{Area of sector MNP} \approx 10.67 \times 3.14 \approx 33.51 \]
So, the area of sector MNP is approximately \( 33.51 \) square units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03ee4b44-fd9b-4d43-bc23-25b15a4cdf81%2Fe8f8b2cb-22a0-45b3-a86c-54bc2f3b8425%2Fxroi0fp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
In circle \( N \) with \( m \angle M N P = 60^\circ \) and \( M N = 8 \) units, find the area of sector MNP. Round to the nearest hundredth.
### Diagram Explanation:
The image is a diagram of a circle centered at point \( N \). The points \( M \) and \( P \) are on the circumference of the circle. The line segment \( M N \) forms a radius of the circle. The angle \( \angle M N P \) is given as \( 60^\circ \).
### Solution:
To find the area of sector \( MNP \), we can use the formula for the area of a sector in a circle:
\[ \text{Area of sector} = \left(\frac{\theta}{360^\circ}\right) \pi r^2 \]
where \( \theta \) is the central angle in degrees, and \( r \) is the radius of the circle.
Given that \( \theta = 60^\circ \) and \( r = 8 \) units:
\[ \text{Area of sector MNP} = \left(\frac{60^\circ}{360^\circ}\right) \pi (8)^2 \]
First, simplify the fraction:
\[ \frac{60^\circ}{360^\circ} = \frac{1}{6} \]
Then, calculate the area:
\[ \text{Area of sector MNP} = \frac{1}{6} \pi (8)^2 = \frac{1}{6} \pi (64) = \frac{64}{6} \pi \]
\[ \text{Area of sector MNP} = \frac{32}{3} \pi \approx 10.67 \pi \]
Now, using the approximate value of \( \pi \approx 3.14 \):
\[ \text{Area of sector MNP} \approx 10.67 \times 3.14 \approx 33.51 \]
So, the area of sector MNP is approximately \( 33.51 \) square units.
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