figure it out

I know some of my readers run away screaming in horror when faced with math problems. To some extent, I understand: you haven't had to flex your brain in this way for years, and you've grown lazy and complacent—unwilling to exert yourself. But give these problems a try even if you fail. It's good to keep your brain working, and you're staving off dementia.

Here's the next math problem to solve, a week after the first one. Play the video for one second, then pause, so you can see the letters in the diagram (the thumbnail has no letters; you'll need them to follow along with my explanation):

My answer will be below, between the brackets. Highlight to see. Unlike the teacher in the video, I'm showing the only method I know to solve the problem.

[We're given a right triangle that's divided into smaller shapes. Let's call the big triangle △ABC. We've also got △ACE, △BCE, and △BDE. Note that △ABC, △BDE, and △ACE are all 30-60-90 triangles. You'll recall that a 30-60-90 triangle is basically half of an equilateral triangle, which means the triangle's short leg equals half the hypotenuse. The ratio of the sides of a 30-60-90 triangle, is the following:

hypotenuse : short leg : long leg = 2:1:√3 This is going to be important in a moment.

So, what can we reason out from the information we have? We deduce right away that ∠ACB is 60º. We're given that ∠AEC is 60º, so we can deduce that ∠ACE is 30º. Since we just said ∠ACB is 60º, then ∠BCE (or ∠DCE) must also be 30º to add up to 60º for the larger angle.

We're looking at a lot of 30-60-90 triangles, and we're given that segment BD has a length of 30, and our task is to find x, i.e., the length of segment AC. As we said above, for 30-60-90 triangles, the short leg is half the length of the hypotenuse. So for △BDE, the hypotenuse has a length of 30, which means the short leg (DE) has a length of 15, and the longer leg (BE) has a length of 15√3.

∠AEC is 60º; ∠AED is 90º, so ∠CED must be 30º. Since we already know that ∠DCE is 30º, we're look at an isosceles triangle with two 30º angles. We also now know that segment DC must be the same length as segment DE.

All of this means we know the length of the hypotenuse of the big triangle: 30 +15 = 45. Remembering that the short leg of a 30-60-90 triangle is half the length of the hypotenuse, we can now solve for x: half of 45 = 22.5. QED.]