We derive the equation of base of a triangle from the area of a triangle formula:

$\displaystyle A=\frac{bh}{2}$

$\displaystyle b=\frac{2A}{h}$

$\displaystyle b=\frac{2*12}{5}$

$\displaystyle b=4.8$

Similarily, we can use the same formula for the height to find the length of the side of an equilateral triangle, which is given by 

$\displaystyle s=\frac{2h}{\sqrt{3}}$, where $\displaystyle h$ is the length of the height.

Therefore, the final answer is

 $\displaystyle \frac{2\cdot 5}{\sqrt3}=\frac{10}{\sqrt{3}}$. 

Since we are given the radius, we should be able to find the height of the equilateral triangle. Indeed, the center of the circle is at the intersections of the heights of the triangle, and is located $\displaystyle \frac{2}{3}$ away from the edge of a given height.

Therefore 5, the radius of the circle is $\displaystyle \frac{2}{3}$ of the height.

Therefore, the height must be $\displaystyle \frac{15}{2}$.

From here, we can use the formula for the height of the equilateral triangle $\displaystyle h=s\frac{\sqrt{3}}{2}$, where $\displaystyle h$ is the length of the height and $\displaystyle s$ is the length of a side of the equilateral triangle.

Therefore, $\displaystyle s=2\frac{h}{\sqrt{3}}$, then $\displaystyle \frac{15}{\sqrt{3}}$ is the final answer.

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:

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

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

$\displaystyle A=36$