Equilateral Areas
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF .
![Equilateral Areas Equilateral Areas](/sites/default/files/styles/large/public/thumbnails/content-97-02-six5-feb6.gif?itok=InzLHb3b)
Does this work for any whole number side lengths?
If not, under what circumstances does it work?
What if the lengths of the sides of the triangles had been a and b instead of 3 and 4?
Can you construct an equilateral triangle whose area is the sum of the areas of ABC and DEF? What is the new area?
![Equilateral Areas Equilateral Areas](/sites/default/files/styles/large/public/thumbnails/content-97-02-six5-feb7.gif?itok=qCg5gNu5)
Drawing the triangles on isometric paper and using areas bsed on triangles rather than squares might help.
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:
$$\frac{c^2\sqrt3}{4} = \frac{a^2\sqrt3}{4} + \frac{b^2\sqrt3}{4} $$