Let's break down the problem step-by-step:
First, we need to find the radius of the circumscribed circle of a regular hexagon with a side length of 3. In a regular hexagon, the radius of the circumscribed circle is given by:
[tex]\[ r = \frac{s \sqrt{3}}{2} \][/tex]
where [tex]\( s \)[/tex] is the side length of the hexagon. Substitute [tex]\( s = 3 \)[/tex]:
[tex]\[ r = \frac{3 \sqrt{3}}{2} = \frac{3 \sqrt{3}}{2} \][/tex]
Next, we need to calculate the area of the circle with this radius. The formula for the area of a circle is [tex]\( \pi r^2 \)[/tex]:
[tex]\[ \text{Area of the circle} = \pi \left( \frac{3 \sqrt{3}}{2} \right)^2 = \pi \left( \frac{9 \cdot 3}{4} \right) = \pi \left( \frac{27}{4} \right) = \frac{27 \pi}{4} \][/tex]
A regular hexagon can be divided into 6 equilateral triangles. The area of one equilateral triangle with a side length of 3 is given by:
[tex]\[ \text{Area of one equilateral triangle} = \frac{1}{4} s^2 \sqrt{3} = \frac{1}{4} (3)^2 \sqrt{3} = \frac{1}{4} \cdot 9 \sqrt{3} = \frac{9 \sqrt{3}}{4} \][/tex]
The area of one of the 6 sectors of the circle (each corresponding to one side of the hexagon) is:
[tex]\[ \text{Area of one sector} = \frac{1}{6} \cdot \text{Area of the circle} = \frac{1}{6} \cdot \frac{27 \pi}{4} = \frac{27 \pi}{24} = \frac{9 \pi}{8} \][/tex]
Now, the area of the segment formed by one side of the hexagon and the circle is the difference between the area of one sector and the area of one equilateral triangle:
[tex]\[ \text{Area of the segment} = \text{Area of the sector} - \text{Area of the equilateral triangle} \][/tex]
[tex]\[ \text{Area of the segment} = \frac{9 \pi}{8} - \frac{9 \sqrt{3}}{4} \][/tex]
To match the given template, we can factor out the [tex]\( \frac{9}{8} \)[/tex]:
[tex]\[ \text{Area of the segment} = \frac{9}{8} \left( \pi - \frac{2\sqrt{3}}{1} \right) \][/tex]
So, we place the numbers according to the expression [tex]\( -\pi-\frac{\square}{\square} \)[/tex]:
[tex]\[
\boxed{
\begin{array}{c}
A=\left(-\pi-\frac{2\sqrt{3}}{1}\right) \text{ inches}^2
\end{array}
}
\][/tex]
From the numbers provided:
[tex]\[ \text{Area of the segment} = -\pi - \frac{2 \sqrt{3}}{1} \][/tex]
Therefore, placing the numbers in the boxes gives us the correct segment area.
