John wrote:

Areas of triangles using triangluar measure generate the square numbers
$1, 4, 9, 16, 25$.
So the two triangles $3$ and $4$ were a fairly special case as $3^2 + 4^2 = 5^2$

But there are others that work such as $5, 12,13$ - that is Pythagorean Triples.

In the original problem $a = 3$ and $b = 4$, so $3^2 + 4^2 = c^2$ giving $c = 5$.

This was essentially just another way of looking at Pythagoras's theorem.

In general:

The formula for the area of an equilateral triangle with side $x$ is

$\text{Area} = \frac{x^2\sqrt3}{4}$

So with the two triangles with sides a and b respectively, we are looking for a third triangle with area:
$$\frac{c^2\sqrt3}{4} = \frac{a^2\sqrt3}{4} + \frac{b^2\sqrt3}{4} $$