Let's solve this step-by-step:
1. Given Values:
- The side length of the equilateral triangle is [tex]\(12 \sqrt{3}\)[/tex] units.
2. Half the Side Length:
- Half a side length of the equilateral triangle is [tex]\(6 \sqrt{3}\)[/tex] units.
3. Calculating the Apothem:
- In an equilateral triangle, the apothem [tex]\(a\)[/tex] can be calculated using the formula [tex]\(a = \frac{s}{2 \sqrt{3}}\)[/tex], where [tex]\(s\)[/tex] is the side length.
- Given the side length [tex]\(s = 12 \sqrt{3}\)[/tex]:
[tex]\[
a = \frac{12 \sqrt{3}}{2 \sqrt{3}} = 6
\][/tex]
- Thus, the apothem is [tex]\(6\)[/tex] units long.
4. Radius of the Circumscribed Circle:
- The radius [tex]\(R\)[/tex] of the circumscribed circle of an equilateral triangle can be calculated using the formula [tex]\(R = \frac{s \sqrt{3}}{3}\)[/tex].
- Given the side length [tex]\(s = 12 \sqrt{3}\)[/tex]:
[tex]\[
R = \frac{12 \sqrt{3} \cdot \sqrt{3}}{3} = \frac{12 \cdot 3}{3} = 12
\][/tex]
- Thus, the radius of the circle is [tex]\(12\)[/tex] units long.
5. Area of the Equilateral Triangle:
- The area [tex]\(A_T\)[/tex] of an equilateral triangle can be calculated using the formula [tex]\(A_T = \frac{\sqrt{3}}{4} s^2\)[/tex].
- Given the side length [tex]\(s = 12 \sqrt{3}\)[/tex]:
[tex]\[
A_T = \frac{\sqrt{3}}{4} (12 \sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 432 = 108 \sqrt{3}
\][/tex]
- Numerically, this is approximately [tex]\(187.06148721743872\)[/tex] units².
6. Area of the Sector:
- Recall that the central angle for an equilateral triangle inscribed in a circle is [tex]\(120^\circ\)[/tex], which is [tex]\(\frac{1}{3}\)[/tex] of the full circle.
- The area of the sector [tex]\(A_S\)[/tex] can be found using [tex]\(\frac{1}{3} \pi R^2\)[/tex].
- Given the radius [tex]\(R = 12\)[/tex] units:
[tex]\[
A_S = \frac{1}{3} \pi (12^2) = \frac{1}{3} \pi \cdot 144 = 48 \pi
\][/tex]
- Numerically, this is approximately [tex]\(150.79644737231007\)[/tex] units².
7. Area of Each Segment of the Circle:
- The segment area is the area of the sector minus the area of the triangle.
- Given the area of the sector ([tex]\(48 \pi\)[/tex] units² or approximately [tex]\(150.79644737231007\)[/tex] units²) and the area of the triangle ([tex]\(108 \sqrt{3}\)[/tex] units² or approximately [tex]\(187.06148721743872\)[/tex] units²):
[tex]\[
\text{Segment Area} = A_S - A_T = 48 \pi - 108 \sqrt{3}
\][/tex]
- Numerically, this is approximately:
[tex]\[
150.79644737231007 - 187.06148721743872 = -36.265039845128655 \text{ units}^2
\][/tex]
Therefore, the final values are:
- The apothem is [tex]\(6\)[/tex] units long.
- The radius of the circle is [tex]\(12\)[/tex] units long.
- The area of each arc segment of the circle is [tex]\(48 \pi - 108 \sqrt{3}\)[/tex] units². Numerically, this is approximately [tex]\(-36.265039845128655\)[/tex] units².
