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.
 A: Let the lenght of the side be $s.$
In an equilateral triangle, the lengths of the sides are equal.
So, $RS=ST=TR=s.$ 
All angles are $\frac{π}{3}$ $radians.$
Area of a triangle as we know, is $$\frac{(RS)(ST)\sin(S)}{2}.$$
$$=\frac{s^2\sin(\frac{π}{3})}{2}$$
The area of the $ΔRST$ is thus,  $$\frac{s^2\sqrt3}{4}$$
$Quod  $   $erat $   $demonstrandum$
A: 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 $\frac{s^{2} \sin c}{2}=\frac{s^{2} \frac{\sqrt{3}}{2}}{2}=\frac{s^{2} \sqrt{3}}{4}$

