<p>Its a grid in, can someone explain the fastest way to go about doing this.</p>
<p><a href="http://i44.tinypic.com/27y7ift.jpg%5B/url%5D">http://i44.tinypic.com/27y7ift.jpg</a></p>
<p>What I would do is come up with a rectangle that has an area between 8 and 18, so I’ll have one with a length of 4 and a width of 3 (area = 12). That means FE = 3, meaning AB and AF are also 3 since they are all equal.</p>
<p>Before plugging in random values of FE and FC such that 8 < FE*FC < 18, let’s try to understand the actual problem. Let AF = FE = AB = x. Triangle AFB is a right isosceles triangle, and by some simple angle chasing we see that triangle FBC is also right isosceles. This means that FB = BC = x sqrt(2) and FC = 2x.</p>
<p>Since FE = x and FC = 2x, we see that 8 < 2x^2 < 18, so 4 < x^2 < 9. Therefore any value of x strictly between 2 and 3 works (note that 2 and 3 are incorrect answers!).</p>
<p>Yup, exactly what MITer said.</p>
<p>@MITer94 – So that explains why they provided another triangle. Thanks for the correction.</p>
<p>On a recent thread, I mentioned that they often draw the figure not-to-scale to keep you from noticing something that would make the problem really easy if you did in fact notice it. This problem illustrates that idea perfectly:</p>
<p>Take a moment to redraw it, making that triangle at the bottom left isosceles as it should be. Keep your drawing neat and it will be immediately apparent that you are looking at a square with the top left corner cut off. And if the right half of a square has area that ranges form 8 to 18, then the full square ranges from 16 to 36. So the full length of a side ranges from 4 to 9 and half ranges from 2 to 3…</p>