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.

Circle properties and circle theorems. Regular polygons and circles. Making and proving conjectures. Creating and manipulating expressions. Generalising. Sine, cosine, tangent. Sine rule & cosine rule. Mathematical reasoning & proof. Pythagoras' theorem. Area - triangles, quadrilaterals, compound shapes.