Let \(\displaystyle s\) be the length of a side of an equilateral triangle. Then the formula for the area of an equilateral triangle with side \(\displaystyle s\) is

 \(\displaystyle \frac{s^2\sqrt{3}}{4}\)

So solving \(\displaystyle 1.732 = \frac{s^2\sqrt{3}}{4}\) 

we get \(\displaystyle s=2\).

Alternative Solution:

Without knowing this formula you can still use the Pythagorean Theorem to solve this. By drawing the height of the triangle, you split the triangle into 2 right triangles of equal size. The sides are the height, \(\displaystyle \frac{s}{2}\) and \(\displaystyle s\). Letting \(\displaystyle h\) stand for the unknown height, we solve 

\(\displaystyle (\frac{s}{2})^2 + h^2= s^2\) solving for \(\displaystyle h\) we get

 \(\displaystyle h=\sqrt(s^2-(\frac{s}{2})^2) = \sqrt(s^2-\frac{s^2}{4}) = \sqrt(\frac{3s^2}{4}) = \frac{s\sqrt(3)}{2}\)

The area for any triangle is the base times the height divided by 2. So

 \(\displaystyle 1.732=\frac{sh}{2} = \frac{s(s\sqrt(3))}{2*2} = \frac{s^2\sqrt(3)}{4}\) or \(\displaystyle s=2\).

To find the area of a traingle, we need the height and base lengths. Plug the given values into the following formula:

\(\displaystyle A= \frac{bh}{2}\)

\(\displaystyle = \frac{(3*2)}2 {}\)

\(\displaystyle =3\)

The area of an equilateral triangle is given by the following formula:

 \(\displaystyle \frac{s^{2}\sqrt{3}}{4}\), where \(\displaystyle s\) is the length of a side.

Since we know the length of the side, we can simply plug it in the formula and we have \(\displaystyle \frac{4\sqrt{3}}{4}\) or \(\displaystyle \sqrt{3}\), which is the final answer.

Since we are given a radius for the circle, we should be able to find the length of the height of the equilateral triangle, indeed, the center of the circle is \(\displaystyle \frac{2}{3}\) of the length of the height from any vertex.

Therefore, the height is \(\displaystyle 3=\frac{2}{3}\cdot h\) where \(\displaystyle h\) is the length of the height of the triangle. Therefore \(\displaystyle h=\frac{9}{2}\).

We can now plug in this value in the formula of the height of an equilateral triangle\(\displaystyle h=s\frac{\sqrt{3}}{2}\), where \(\displaystyle s\) is the length of the side of the triangle.

Therefore, \(\displaystyle s=\frac{9}{\sqrt{3}}\) or \(\displaystyle 3\sqrt{3}\).

Now we should plug in this value into the formula for the area of an equilateral triangle \(\displaystyle a=s^{2}\frac{\sqrt{3}}{4}\) where \(\displaystyle a\) is the value of the area of the equilateral triangle. Therefore \(\displaystyle a= 27\frac{\sqrt{3}}{4}\), which is our final answer. 

For a given equilateral triangle with a side length \(\displaystyle b\) and a height \(\displaystyle h\), the area \(\displaystyle A\) is 

\(\displaystyle A=\frac{1}{2}bh\). Plugging in the values provided:

\(\displaystyle A=\frac{1}{2}(4)(7)\)

\(\displaystyle A=\frac{1}{2}(28)\)

\(\displaystyle A=14\)

For a given right triangle with a side length \(\displaystyle b\) and a height \(\displaystyle h\), the area \(\displaystyle A\) is 

\(\displaystyle A=\frac{1}{2}bh\). Plugging in the values provided:

\(\displaystyle A=\frac{1}{2}(5)(4)\)

\(\displaystyle A=\frac{1}{2}(20)\)

\(\displaystyle A=10\)

For a given right triangle with a side length \(\displaystyle b\) and a height \(\displaystyle h\), the area \(\displaystyle A\) is 

\(\displaystyle A=\frac{1}{2}bh\). Plugging in the values provided:

\(\displaystyle A=\frac{1}{2}(9)(8)\)

\(\displaystyle A=\frac{1}{2}(72)\)

\(\displaystyle A=36\)