Area of an equilateral triangle 
Prove that if triangle $\triangle RST$ is equilateral, then the area of $\triangle RST$ is $\sqrt{\frac34}$ times the square of the length of a side.

My thoughts:
Let $s$ be the length of $RT$. Then $\frac s2$ is half the length of $\overline{RT}$. Construct the altitude from the $S$ to side $\overline{RT}$. Call the intersection point $P$. Now, you have a right triangle whose sides are $|\overline{RP}| = \frac s2$ and $|\overline{RS}| = s$. By the Pythagorean Theorem, $|\overline{SP}| = \sqrt{s^2 - \frac14 s^2} = \sqrt{ \frac34 s^2} = \frac{\sqrt{3}}{2} s$. The area of the triangle is $$\frac12 \left(|\overline{SP}|\right)\left(|\overline{RT}|\right) = \left(\frac12 s\right) \left(\frac{\sqrt{3}}{2} s\right) = \frac{\sqrt{3}}{4}s^2,$$ as suggested.

From AoPS wiki,

Method 1: Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is $\frac{s}{2}$. Using the Pythagorean theorem, we get $s^{2}=h^{2}+\frac{s^{2}}{4}$, where $h$ is the height of the triangle. Solving, $h=\frac{s \sqrt{3}}{2}$. (note we could use $30-60-90$ right triangles.)
We use the formula for the area of a triangle, $\frac{1}{2} b h$ (note that $s$ is the length of a base), so the area is
$$
\frac{1}{2}(s)\left(\frac{s \sqrt{3}}{2}\right)=\frac{s^{2} \sqrt{3}}{4}
$$
Method 2: (warning: uses trig.) The area of a triangle is $\frac{a b \sin C}{2}$. Plugging in $a=b=s$ and $C=\frac{\pi}{3}$ (the angle at each vertex, in radians), we get the area to be $\frac{s^{2} \sin c}{2}=\frac{s^{2} \frac{\sqrt{3}}{2}}{2}=\frac{s^{2} \sqrt{3}}{4}$

