<p><a href="http://i.imgur.com/euUo8NI.png%5B/url%5D">http://i.imgur.com/euUo8NI.png</a></p>
<p>Answer is 5</p>
<p>Think about how you would find a single side length of the square. Each side of the sqaure is the hypotenuse of what triangle?</p>
<p>Premise: It’s a square, so all side lengths are the same.
Premise: The area of a square is x^2, where x is side length. </p>
<p>Now look at line segment AD. It is the hypotenuse of a right triangle with legs 1 and 2. By the Pythagorean theorem, 1+4=x^2. Since x^2 is the area, area = 5.</p>
<p>Ah, I got it, thank you. But how are you supposed to know that the lines of the square form right triangles?</p>
<p>Sorry, purpleacorn, didn’t see that post before I posted. Nice use of the Socratic method.</p>
<p>You eyeball it. If it quacks like a duck… If it doesn’t say “<em>Not drawn to scale</em>”, then it’s probably drawn to scale and you can assume what you see.</p>
<p>^^Exactly. I went into this problem thinking, “I need area, and in order to get area, I need to know the side length. How do I find a side length?”</p>
<p>Well, the problem mentioned smaller squares, so I may use that. Well, AB is made up of a little square length-wise and two little squares height-wise. Hey look, CD is the same way… etc.</p>
<p>Just for fun, another way to look at it, no pythag thm required:</p>
<p>There is a big square that is 3x3 =9. The little tilted square is what remains when you remove 4 triangles from the big one. </p>
<p>Each of the four triangles has area of (1/2)(1)(2)=1. And there are four of them…</p>
<p>9 - 4 =5.</p>
<p>BTW, that’s one way to prove the pythag thm…</p>