How do I find the original area of a square given that two 1 inch parallel cuts would divide it into three even piece? (Diagram included)

Question

If I take a perfect square, and make two parallel 1 inch cuts (as seen in diagram) this will divide the square into three perfectly equal sections. What is the original area of the square?

This is a question for my math seminar class, and I've really struggled with it. I've tried to make some triangles out of the square using the lines from the parallel cuts in it, and found that they were similar. Trying to relate the triangles and solve for the area of one third (given the value of 1 from the parallel cut), got me nowhere. All help is appreciated!

Answer 1

As shown in the above diagram, the side legnth is $s = a + b$

The area of one of the triangles is $ \frac{1}{2} a s = \frac{1}{3} s^2 $

Hence, $a = \dfrac{2}{3} s $ and $b = \frac{1}{3} s $

Now, we have $\tan \theta = \dfrac{a}{s} = \dfrac{2}{3} $

From which $\sec \theta = \sqrt{ \tan^2 \theta + 1 } = \dfrac{\sqrt{13}}{3} $

Finally, note that $b = \frac{1}{3} s = \sec \theta = \dfrac{\sqrt{13}}{3} $

Hence, $s = \sqrt{13} $ and the area is $s^2 = 13 $