## Wednesday, May 20, 2015

### Cool Geometry Problem

I ran into another problem that I think demonstrates an unexpected and interesting fact

(from http://matharguments180.blogspot.com/2015/05/468-fields-of-green.html)

The following 3 squares are connected to form four triangles. What is the area of the whole hexagon shape?

The full solution requires the use of heron's formula as far as I can see which puts it out of reach for the kids. But the more fascinating part to me is the fact that the four triangles above all have the same area!

Not only that, it's not too hard to prove:

Let's look at the inner and lower triangle. They already share a base (let's pick the square with sides of length $S_2$).  So to show they have the same area we just need to see if they have the same heights. So I drew in the 2 height perpendicular lines $h_1$ on the interior and $h_2$ which is a continuation of the square for the exterior.  Let x be angle inside $\angle EDA$ then $\angle BAC$ must be 90  - x since they are on the same line and have a 90 degree corner of a square in the middle.  Since these are both right triangles we can find the 3rd angles which are just 90 and 90 - x again. So clearly the two triangles $\triangle ABC$ and $\triangle ADE$ are similar. But they also both have hypotenuses along the square with sides of  length $s_1$. So they are completely congruent and $h_1$ = $h_2$.  You can repeat this construction on every side to prove all the triangles have the same area.

Addendum: Another way of seeing the same thing is that you can rotate the inner triangle so its $s_2$  length edge lines up to the same edge on the outer one. These will form a larger triangle with the shared edge exactly a median so the 2 halves, the original triangles are the same size.

I'm saving this one for a future session. I think I'll start by having the kids cut out different size squares and measure the triangles to organically discover they appear to be of the same area regardless of the square sizes and then move onto to figuring out why this must be true. This will work especially well paired with Pythagorean Theorem work given its similarities.