Ooooooooh!! Special triangles!
There are two speials triangles. The first is a "45-45-90" triangle (referring to the 3 angles of the triangle) and the second is a "30-60-90" triangles (also referring to the angle measures).
Problem 1 uses "45-45-90" triangles. We know this because it has a right angle (this is the angle with 90 degrees) and the other two angles are congruent and must sum to 180 - 90 = 90 (because there are 180 degrees in a triangle) which means that they are both 45 degrees.
"45-45-90" triangles are isosceles triangles (if two angles are congruent, the sides opposite of them are congruent). All "45-45-90" triangles are in the ratio: \(x : x : x\sqrt{2}\) (where the side lengths are each 'x' and the hypotenuse (the long side) is x root 2). We are told that the hypotenuse is 4, so:
\(x\sqrt{2} = 4\) so \(x = \frac{4}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}\).
'X' is the side length of the big triangle, but the hypotenuse of the next triangle. Since this next triangle is also a "45-45-90" triangle, you can do the same thing. Then you can do this again and again to get the 'x' you are looking for. You should end up getting x = 1. Make sure you get it, understand it, and are able to show your work. "45-45-90" triangles are very common and important.
Problem 2 uses the "30-60-90" triangle. On the side with the 90 degree angle and the 30 degree angle, flip the triangle over. You should now have two triangles forming one big triangle. Well, that "one big triangle" is a equilateral triangle! (The 30 degree angle, when flipped, should have created a 60 degree angle). Using this knowledge, try to find the ratio of the legs of this triangle (where x is the line connecting the 60 degree angle and 90 degree angle). You should get \(x : x\sqrt{3}:2x\) (where x root 3 is the line connecting the 90 degree angle and the 30 degree angle, and the "2x" is the hypotenuse of the triangle).
Once you have this, plug in x = 1. The hypotenuse of this triangle will be a leg of the next. Do this several times until you find the 'x' in the problem. Again, I will tell you the answer, but try to find it yourself so you can show your work and understand it better. You should get 32/9.
One more hint (because I know this is a hard problem for those unfamiliar with these special triangles and this will help you figure it out and know if you are correct), is that the hypotenuses of the 5 triangles (starting with the smallest) are:
\(2, \frac{4}{\sqrt{3}}, \frac{8}{3}, \frac{16}{3\sqrt{3}}, \frac{32}{9}\).
I hope that helped!!
