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:

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

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

$\displaystyle A=36$