<s>[INST] 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.
 [/INST] 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 $   $demonstrandum$
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