And yet more 15-75-90 fun

Continuing on the theme of 15-75-90 triangles (See: Last time and First Time) several interesting riffs on 15-75-90's in a box have come up recently.

Example dividing a square with four 15-75-90 triangles:

As is often the case, finding the relative area of the  triangles and square is straight forward using trigonometry:

Let s be the length of the  sides of the square:

The area of each triangle = \(\frac{1}{2} s^2 cos(15)sin(15) \) and using the double angle formulas
\(sin(30)=2sin(15)cos(15)\) so  after substitution and knowing sin(30) = \(\frac{1}{2}\) out pops the area = \(\frac{1}{8}s^2\)

But why is this happening? As usual a 30-60-90 triangle is usually lurking around that allows a Euclidean explanation.

What's particularly interesting about this is that it hints that dissections exist to transform a 1/4 or 1/8 of the larger square into the triangles and sure enough you slide the 1/4 triangle ABO until it becomes 2 15-75-90'!



But let's return to the original problem. There's another easy explanation of what's occurring that just uses the ratios of the triangle:

1. Note the  area of this triangle is \(\frac{1}{2}(2 - \sqrt{3})\)
2. Squaring the hypotenuse you get \(4(2 - \sqrt{3})\) which is 8 times the triangle area.
3. Or in other words each triangle is 1/8 of the square made on the hypotenuse.

And we've refound our original result.

Further questions: Are there other common triangles that divide the square into a unit or "simple" fraction.


I'll leave it to the reader to decide which problem based on this property is more fun (from @eylem and @sansu-seijin):

Given the square of length 6cm, how large is the shaded region?