Challenge Level

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:

$$\frac{c^2\sqrt3}{4} = \frac{a^2\sqrt3}{4} + \frac{b^2\sqrt3}{4} $$

This simplifies to give $c^2 = a^2 + b^2$, which is Pythagoras's theorem. This also means that it is possible to find a triangle whose area is the sum of any two triangles, although the sides will not necessarily be integer lengths.