Here is a little puzzle I got from my math teacher. I had a little trouble solving it; I tried to find the equations to lines by solving multiple systems, but to no avail. Could you guys please help me?
This is the given problem: "Let WXYZ be a square. Three parallel lines d, d' and d'' pass through X, Y and Z respectively. The distance between d and d' is 5 and the distance between d and d'' is 7. What is the area of the square?"
If the lines $d,d', d''$ make an angle of $\theta$ with $XY$ , then the side length $a$, satisfies
$ a \sin(\theta) = 5 $
and
$ a \cos(\theta) = 7 $
Hence, $\tan(\theta) = \dfrac{5}{7} $
and $\cos(\theta) = \dfrac{1 }{\sqrt{ 1 + \left(\dfrac{5}{7}\right)^2 }} = \dfrac{ 7 } {\sqrt{ 74 } } $
Hence $ a = \sqrt{74} $
And the area of the square $= a^2 = 74 $