Question

an equilateral triangle with side lengths equal to 12\sqrt{3} units is inscribed in a circle. half a side length of the equilateral triangle is 6\sqrt{3} units, so the apothem is blank units long and the radius of the circle is blank units long. each segment of the circle has an area equal to the difference between the areas of the sector and triangle, or (48 blank\pi - blank\sqrt{3}) units².

Explanation:

Step1: Find the apothem

In an equilateral triangle inscribed in a circle, if half - side length $a = 6\sqrt{3}$ and the central angle is $120^{\circ}$, using right - triangle trigonometry in the right - triangle formed by the apothem, half - side length and radius. The angle between the radius and half - side length is $30^{\circ}$. We know that $\tan30^{\circ}=\frac{\text{apothem}}{\text{half - side length}}$. Let the apothem be $h$. Then $h = 6$.

Step2: Find the radius

Using the Pythagorean theorem in the right - triangle with half - side length $a = 6\sqrt{3}$ and apothem $h = 6$. Let the radius be $r$. Then $r=\sqrt{(6\sqrt{3})^{2}+6^{2}}=\sqrt{108 + 36}=\sqrt{144}=12$.

Step3: Find the area of the sector

The central angle of the sector $\theta=120^{\circ}=\frac{2\pi}{3}$ radians and radius $r = 12$. The area of a sector $A_{s}=\frac{1}{2}r^{2}\theta=\frac{1}{2}\times12^{2}\times\frac{2\pi}{3}=48\pi$.

Step4: Find the area of the triangle

The area of an equilateral triangle with side length $s = 12\sqrt{3}$ is $A_{t}=\frac{\sqrt{3}}{4}s^{2}=\frac{\sqrt{3}}{4}\times(12\sqrt{3})^{2}=\frac{\sqrt{3}}{4}\times432 = 108\sqrt{3}$.

Step5: Find the area of the segment

The area of the segment $A = A_{s}-A_{t}=48\pi- 36\sqrt{3}$.

Answer:

Apothem: 6; Radius: 12; Area of segment: $48\pi-36\sqrt{3}$