DS2320

I have a few things to say about this problem.

1. Yes, it is a real problem.

2. THE TRAP OF C SHOULD BE SCREAMING AT YOU. In this problem, they want the area of a rectangular piece. If you have the length, and you have the width, you can answer the question. Putting both of the answers together, THAT'S EXACTLY WHAT YOU GET! No way am I ever going to choose C in that condition. Too easy. F.U. ETS, not fallin' for that one :fu . I'm going deeper. If I had to guess right off the starting block, I'd say D and move on, because there's no difference what each one is telling us, meaning, whatever the solution, we should be able to apply it in the same with with the width and the length.

3. Any perfect hexagon is, well, perfect. All the sides are equal and all the angles are 120 degrees. If we divide one up, it'll be composed of a bunch of perfect equilateral triangels, where all the sides are equal. Take a look at this:



So the line down the middle is exactly twice the length of the side, and from the blue outside triangle I drew, we can know the sides of the spaces outside the hexagons, all in terms of x.

Since it's all in terms of x, we can add them all up the pieces of the length, or of the width, and either way, if we knew either side, we'd have everything we need to solve.