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Sagot :
To determine the area of a segment of a circle with a [tex]\(120^\circ\)[/tex] arc and a chord of [tex]\(8\sqrt{3}\)[/tex] inches, follow these steps:
1. Convert the arc angle from degrees to radians:
[tex]\[ \theta = 120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3} \text{ radians} \][/tex]
2. Calculate the radius ([tex]\(r\)[/tex]) of the circle:
The formula for the chord length [tex]\(c\)[/tex] in terms of the radius [tex]\(r\)[/tex] and the angle [tex]\(\theta\)[/tex] is:
[tex]\[ c = 2r \sin\left(\frac{\theta}{2}\right) \][/tex]
Given the chord length [tex]\(c = 8\sqrt{3}\)[/tex]:
[tex]\[ 8\sqrt{3} = 2r \sin\left(\frac{2\pi}{6}\right) \][/tex]
[tex]\[ 8\sqrt{3} = 2r \sin\left(\frac{\pi}{3}\right) \][/tex]
Since [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ 8\sqrt{3} = 2r \cdot \frac{\sqrt{3}}{2} \][/tex]
Simplifying:
[tex]\[ 8\sqrt{3} = r\sqrt{3} \][/tex]
[tex]\[ r = 8 \][/tex]
3. Calculate the area of the sector ([tex]\(A_{\text{sector}}\)[/tex]):
The area of a sector of a circle is given by:
[tex]\[ A_{\text{sector}} = \frac{1}{2} r^2 \theta \][/tex]
Plugging in [tex]\(r = 8\)[/tex] and [tex]\(\theta = \frac{2\pi}{3}\)[/tex]:
[tex]\[ A_{\text{sector}} = \frac{1}{2} \cdot 8^2 \cdot \frac{2\pi}{3} \][/tex]
[tex]\[ A_{\text{sector}} = \frac{1}{2} \cdot 64 \cdot \frac{2\pi}{3} \][/tex]
[tex]\[ A_{\text{sector}} = \frac{128\pi}{6} \][/tex]
[tex]\[ A_{\text{sector}} = \frac{64\pi}{3} \][/tex]
4. Calculate the area of the triangle formed by the chord and the radii ([tex]\(A_{\text{triangle}}\)[/tex]):
Using trigonometry, for an isosceles triangle, we can find the height ([tex]\(h\)[/tex]) of the triangular portion:
[tex]\[ h = r - r \cos\left(\frac{\theta}{2}\right) \][/tex]
Since [tex]\(r = 8\)[/tex] and [tex]\(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)[/tex]:
[tex]\[ h = 8 - 8 \cdot \frac{1}{2} \][/tex]
[tex]\[ h = 8 - 4 \][/tex]
[tex]\[ h = 4 \][/tex]
Therefore, the area of the triangle is:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Here the base is the chord length [tex]\(8\sqrt{3}\)[/tex]:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times 8\sqrt{3} \times 4 \][/tex]
[tex]\[ A_{\text{triangle}} = 16 \sqrt{3} \][/tex]
5. Calculate the area of the segment ([tex]\(A_{\text{segment}}\)[/tex]):
The area of the segment is the area of the sector minus the area of the triangle:
[tex]\[ A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} \][/tex]
[tex]\[ A_{\text{segment}} = \frac{64\pi}{3} - 16\sqrt{3} \][/tex]
So, the area of the segment is:
[tex]\[ A = \left( \frac{64\pi}{3} - 16\sqrt{3} \right) \text{ square inches} \][/tex]
Expressed in the required format:
[tex]\[ A = \left( 64\pi/3 - \sqrt{4096} \right) \text{ inches}^2 \][/tex]
1. Convert the arc angle from degrees to radians:
[tex]\[ \theta = 120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3} \text{ radians} \][/tex]
2. Calculate the radius ([tex]\(r\)[/tex]) of the circle:
The formula for the chord length [tex]\(c\)[/tex] in terms of the radius [tex]\(r\)[/tex] and the angle [tex]\(\theta\)[/tex] is:
[tex]\[ c = 2r \sin\left(\frac{\theta}{2}\right) \][/tex]
Given the chord length [tex]\(c = 8\sqrt{3}\)[/tex]:
[tex]\[ 8\sqrt{3} = 2r \sin\left(\frac{2\pi}{6}\right) \][/tex]
[tex]\[ 8\sqrt{3} = 2r \sin\left(\frac{\pi}{3}\right) \][/tex]
Since [tex]\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ 8\sqrt{3} = 2r \cdot \frac{\sqrt{3}}{2} \][/tex]
Simplifying:
[tex]\[ 8\sqrt{3} = r\sqrt{3} \][/tex]
[tex]\[ r = 8 \][/tex]
3. Calculate the area of the sector ([tex]\(A_{\text{sector}}\)[/tex]):
The area of a sector of a circle is given by:
[tex]\[ A_{\text{sector}} = \frac{1}{2} r^2 \theta \][/tex]
Plugging in [tex]\(r = 8\)[/tex] and [tex]\(\theta = \frac{2\pi}{3}\)[/tex]:
[tex]\[ A_{\text{sector}} = \frac{1}{2} \cdot 8^2 \cdot \frac{2\pi}{3} \][/tex]
[tex]\[ A_{\text{sector}} = \frac{1}{2} \cdot 64 \cdot \frac{2\pi}{3} \][/tex]
[tex]\[ A_{\text{sector}} = \frac{128\pi}{6} \][/tex]
[tex]\[ A_{\text{sector}} = \frac{64\pi}{3} \][/tex]
4. Calculate the area of the triangle formed by the chord and the radii ([tex]\(A_{\text{triangle}}\)[/tex]):
Using trigonometry, for an isosceles triangle, we can find the height ([tex]\(h\)[/tex]) of the triangular portion:
[tex]\[ h = r - r \cos\left(\frac{\theta}{2}\right) \][/tex]
Since [tex]\(r = 8\)[/tex] and [tex]\(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)[/tex]:
[tex]\[ h = 8 - 8 \cdot \frac{1}{2} \][/tex]
[tex]\[ h = 8 - 4 \][/tex]
[tex]\[ h = 4 \][/tex]
Therefore, the area of the triangle is:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Here the base is the chord length [tex]\(8\sqrt{3}\)[/tex]:
[tex]\[ A_{\text{triangle}} = \frac{1}{2} \times 8\sqrt{3} \times 4 \][/tex]
[tex]\[ A_{\text{triangle}} = 16 \sqrt{3} \][/tex]
5. Calculate the area of the segment ([tex]\(A_{\text{segment}}\)[/tex]):
The area of the segment is the area of the sector minus the area of the triangle:
[tex]\[ A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} \][/tex]
[tex]\[ A_{\text{segment}} = \frac{64\pi}{3} - 16\sqrt{3} \][/tex]
So, the area of the segment is:
[tex]\[ A = \left( \frac{64\pi}{3} - 16\sqrt{3} \right) \text{ square inches} \][/tex]
Expressed in the required format:
[tex]\[ A = \left( 64\pi/3 - \sqrt{4096} \right) \text{ inches}^2 \][/tex]
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