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 $\boldsymbol{\square}$ units long and the radius of the circle is $\boldsymbol{\square}$ units long. each segment of the circle has an area equal to the difference between the areas of the sector and triangle, or $(\boldsymbol{\square}\pi - \boldsymbol{\square}\sqrt{3})$ units$^2$.

Explanation:

Step1: Find the apothem

In an equilateral triangle inscribed in a circle, the apothem (distance from center to a side, forming a 30 - 60 - 90 triangle with half the side and the radius). We know half the side is \(6\sqrt{3}\), and the angle at the center is \(30^{\circ}\) for the right - triangle. Let the apothem be \(a\). In a 30 - 60 - 90 triangle, \(\tan(30^{\circ})=\frac{a}{6\sqrt{3}}\), and \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(a = 6\sqrt{3}\times\frac{1}{\sqrt{3}}=6\).

Step2: Find the radius of the circle

In the 30 - 60 - 90 triangle, the radius \(r\) (hypotenuse) and the apothem \(a\) (adjacent side to \(30^{\circ}\) angle) have the relationship \(r = 2a\) (since in 30 - 60 - 90 triangle, hypotenuse is twice the shorter leg, and the apothem is the shorter leg here as the angle is \(30^{\circ}\)). Since \(a = 6\), \(r=12\).

Step3: Find the area of the sector

The central angle of the sector is \(120^{\circ}\) (or \(\frac{2\pi}{3}\) radians). The area of a sector is \(A_{sector}=\frac{\theta}{360^{\circ}}\times\pi r^{2}\), where \(\theta = 120^{\circ}\) and \(r = 12\). So \(A_{sector}=\frac{120^{\circ}}{360^{\circ}}\times\pi\times12^{2}=\frac{1}{3}\times\pi\times144 = 48\pi\).

Step4: Find the area of the triangular segment of the sector

The area of the triangle part of the sector: The triangle is an isosceles triangle with two sides equal to the radius (\(r = 12\)) and included angle \(120^{\circ}\). The area of a triangle with two sides \(a,b\) and included angle \(\theta\) is \(A=\frac{1}{2}ab\sin\theta\). Here \(a = b=12\) and \(\theta = 120^{\circ}\), \(\sin(120^{\circ})=\frac{\sqrt{3}}{2}\). So \(A=\frac{1}{2}\times12\times12\times\frac{\sqrt{3}}{2}=36\sqrt{3}\).

Step5: Find the area of the circular segment

The area of the circular segment is \(A_{segment}=A_{sector}-A_{triangle}=48\pi - 36\sqrt{3}\).

Answer:

The apothem is \(\boldsymbol{6}\) units long and the radius of the circle is \(\boldsymbol{12}\) units long. Each segment of the circle has an area equal to \((\boldsymbol{48}\pi-\boldsymbol{36}\sqrt{3})\) units\(^{2}\).