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Sagot :
The radius is given from the area of the sector which is half the area of
the triangle.
Response:
- The radius of the circle is approximately 10.16 units
Which is the method by which the radius can be calculated?
Given parameter;
Angle at the vertex of the triangle that is the center of the circle = 60°
Length of the adjacent sides of the 60° angle = 12, and 12·√3
The arc of the circle inside the triangle divides it into two equal parts
Required:
To find the radius of the circle that has a center at the vertex of the
triangle.
Solution:
[tex]The \ area \ of \ a \ triangle\ is, \ A = \mathbf{ \dfrac{1}{2} \cdot a \cdot b \cdot sin(C)}[/tex]
Where;
a, and b, are the adjacent sides to the angle C, therefore;
a = 12
b = 12·√3
C = 60°
Which gives;
- [tex]A = \mathbf{ \dfrac{1}{2} \times 12 \times 12 \cdot \sqrt{3} \times sin(60^{\circ})} = \dfrac{1}{2} \times 12 \times 12 \cdot \sqrt{3} \times \dfrac{\sqrt{3} }{2} = 108[/tex]
The area of the triangle, A = 108 square units
[tex]Area \ of \ the \ sector, \ A_{sector} = \mathbf{ \dfrac{\pi \cdot r^2}{360^{\circ}} \times 60^{\circ}} = \dfrac{A}{2}[/tex]
Which gives;
[tex]A_{sector} = \dfrac{\pi \cdot r^2}{360^{\circ}} \times 60^{\circ} = \dfrac{108}{2} = 54[/tex]
Therefore;
[tex]\dfrac{\pi \cdot r^2}{360^{\circ}} \times 60^{\circ} = \dfrac{\pi \cdot r^2}{6} = 54[/tex]
[tex]r^2= \dfrac{54 \times 6}{\pi} = \dfrac{324}{\pi}[/tex]
[tex]r= \sqrt{ \dfrac{324}{\pi}} = \dfrac{18}{\sqrt{\pi} } = \dfrac{18 \cdot \sqrt{\pi} }{\pi} \approx \mathbf{10.16}[/tex]
- The radius of the circle, r ≈ 10.16 units
Learn more about the area of a triangle here:
https://brainly.com/question/10572314
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