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Sagot :
To find the area of a sector in a circle, you can use the formula:
[tex]\[ \text{Area of sector} = \frac{1}{2} \cdot r^2 \cdot \theta \][/tex]
where:
- [tex]\( r \)[/tex] is the radius of the circle
- [tex]\( \theta \)[/tex] is the central angle in radians
In this problem, we are given:
- Radius [tex]\( r = 8 \)[/tex]
- Central angle [tex]\( \theta = \frac{5 \pi}{3} \)[/tex]
Let's plug these values into the formula:
[tex]\[ \text{Area of sector} = \frac{1}{2} \cdot 8^2 \cdot \frac{5 \pi}{3} \][/tex]
First, calculate [tex]\( 8^2 \)[/tex]:
[tex]\[ 8^2 = 64 \][/tex]
Now, substitute this back into the formula:
[tex]\[ \text{Area of sector} = \frac{1}{2} \cdot 64 \cdot \frac{5 \pi}{3} \][/tex]
Next, simplify the term [tex]\( \frac{1}{2} \cdot 64 \)[/tex]:
[tex]\[ \frac{1}{2} \cdot 64 = 32 \][/tex]
Now, multiply this result by [tex]\( \frac{5 \pi}{3} \)[/tex]:
[tex]\[ 32 \cdot \frac{5 \pi}{3} = \frac{32 \cdot 5 \pi}{3} = \frac{160 \pi}{3} \][/tex]
Therefore, the area of the sector is:
[tex]\[ \frac{160 \pi}{3} \, \text{units}^2 \][/tex]
The correct answer is:
D. [tex]\(\frac{160 \pi}{3}\)[/tex] units[tex]\(^2\)[/tex]
[tex]\[ \text{Area of sector} = \frac{1}{2} \cdot r^2 \cdot \theta \][/tex]
where:
- [tex]\( r \)[/tex] is the radius of the circle
- [tex]\( \theta \)[/tex] is the central angle in radians
In this problem, we are given:
- Radius [tex]\( r = 8 \)[/tex]
- Central angle [tex]\( \theta = \frac{5 \pi}{3} \)[/tex]
Let's plug these values into the formula:
[tex]\[ \text{Area of sector} = \frac{1}{2} \cdot 8^2 \cdot \frac{5 \pi}{3} \][/tex]
First, calculate [tex]\( 8^2 \)[/tex]:
[tex]\[ 8^2 = 64 \][/tex]
Now, substitute this back into the formula:
[tex]\[ \text{Area of sector} = \frac{1}{2} \cdot 64 \cdot \frac{5 \pi}{3} \][/tex]
Next, simplify the term [tex]\( \frac{1}{2} \cdot 64 \)[/tex]:
[tex]\[ \frac{1}{2} \cdot 64 = 32 \][/tex]
Now, multiply this result by [tex]\( \frac{5 \pi}{3} \)[/tex]:
[tex]\[ 32 \cdot \frac{5 \pi}{3} = \frac{32 \cdot 5 \pi}{3} = \frac{160 \pi}{3} \][/tex]
Therefore, the area of the sector is:
[tex]\[ \frac{160 \pi}{3} \, \text{units}^2 \][/tex]
The correct answer is:
D. [tex]\(\frac{160 \pi}{3}\)[/tex] units[tex]\(^2\)[/tex]
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