Question

an equilateral triangle with side lengths equal to 12√3 units is inscribed in a circle. half a side length of the equilateral triangle is 6√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 (blankπ - blank√3) units². 4 48 144

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 of each of the three isosceles triangles formed at the center is $120^{\circ}$, and the right - angled triangle formed by the apothem, half - side length and the radius has a $30^{\circ}$ angle. Using the tangent function in the right - angled triangle, $\tan30^{\circ}=\frac{\text{apothem}}{6\sqrt{3}}$. Since $\tan30^{\circ}=\frac{1}{\sqrt{3}}$, the apothem $h = 6$.

Step2: Find the radius

Using the cosine function in the right - angled triangle, $\cos30^{\circ}=\frac{6\sqrt{3}}{r}$, where $r$ is the radius of the circle. Since $\cos30^{\circ}=\frac{\sqrt{3}}{2}$, we have $r = 12$.

Step3: Find the area of the sector

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

Step4: Find the area of the triangle

The area of an equilateral triangle with side length $s = 12\sqrt{3}$ is $A_{triangle}=\frac{\sqrt{3}}{4}s^{2}=\frac{\sqrt{3}}{4}\times(12\sqrt{3})^{2}=108\sqrt{3}$. The area of each of the three congruent isosceles triangles at the center is $\frac{1}{3}$ of the area of the equilateral triangle, so $A_{triangle1}=\frac{1}{3}\times108\sqrt{3}=36\sqrt{3}$.

Step5: Find the area of the segment

The area of each segment is $A = A_{sector}-A_{triangle1}=48\pi - 36\sqrt{3}$.

Answer:

Apothem: 6; Radius: 12; Area of segment: $48\pi-36\sqrt{3}$ (for the blanks in order: 6, 12, 48, 36)