<s>[INST] Proof for area of an equilateral triangle with respect to one side? I'm trying to find out the area of an equilateral triangle with respect to one side. Anything wrong with my proof?
An equilateral triangle with sides of length $a$ can be divided in half along the base to create two triangles with right angles. The sides would be $h$ (height), $y$ (hypotenuse) and $b$ (base). I know what some of these sides are:
$$y = a \\
b = 0.5a \\
h = ?$$
Using Pythagorean theorem, I can find the length of $h$:
$$\begin{align}c^2 &= a^2 + b^2 \\
h^2 &= b^2 + y^2 \\
 &= 0.5a^2 + a^2 \\
h &= \pm\frac{a\times\sqrt{3}}{2}
\end{align}$$
Since the length can only be positive, we're left with $h =\frac{a\times\sqrt{3}}{2}$.
The area $A$ of a triangle is $A = (b \times h) / 2$. We can plug in our numbers:
$$\begin{align}
A &= \frac{b \times h}{2} \\
&= \frac{\frac{a}{2} \times \frac{a\times\sqrt{3}}{2}}{2} \\
&= \frac{a^2\times\sqrt{3}}{8}
\end{align}$$
But since we halved the equilateral triangle together, the area of the equilateral is twice this number, since we recombine the two triangles:
$$\begin{align}
A &= 2\times\frac{a^2\times\sqrt{3}}{8} \\
&= \frac{a^2\times\sqrt{3}}{4}
\end{align}$$
This is indeed the area of an equilateral triangle with respect to side $a$.
 [/INST] Your proof is fine, but in the spirit of visual geometry, let me present a cute alternative: (You might want to break out paper/pencil to follow along with the diagram, but maybe not since the figure is so simple.)
Draw a regular hexagon made from 6 equilateral triangles, then draw the equilateral triangle formed by taking every other vertex of the hexagon.  Now, since all equilateral triangles are similar, the area of the large equilateral triangle is $ k^{2}A $, where $ A $ is the area of the smaller equilateral triangles, for some scale factor $k$.  On the other hand, the large equilateral triangle is half the area of the hexagon.  So, since the hexagon has area $ 6A $, we have that $ k^{2} = 3 $.  Now, we can compute the area of the hexagon as 3 times the area of the rhombuses formed by adjacent pairs of smaller equilateral triangles.  The area of each rhombus is $ \frac{ka \cdot a}{2}  $, if the side length of each small equilateral triangle is $ a $.  Thus, we have $ \frac{1}{2}ka^{2} = 6 A $ and $ k = \sqrt{3} $, so $ A = \frac{\sqrt{3}}{4}a^{2} $.
</s>