Equations And A Square!

As we can see in the picture above, the Figure could be like this.

Now a very general method of solving this question, involves usage of the Pythagoras Theorem at the very first level. But I thought that such questions could be solved using Coordinate Geometry and Trignometry within seconds.

So, here's the trick-

Let Side of the Square be X

Let $Angle\quad EGH\quad be\quad \alpha$ . Since opposite sides of a square are parallel to each other. $Therefore,\quad Angle\quad GCA\quad also\quad is\quad equal\quad to\quad \alpha \quad (Corresponding\quad Angles)$ .

Now, Slope of the line X=Y, is $\tan { \alpha }$ . When $\alpha =Angle\quad GCA$ , we get-

$\tan { \quad \alpha } = \frac { x }{ 8-x }$

On the Other hand When, $\alpha =Angle\quad EGH$ , we get-

$tan {\alpha } = \frac { 8-x }{ x }$

Equating Both, we get-

$\frac { x }{ 8-x } =\frac { 8-x }{ x }$

$\Rightarrow { \quad x }^{ 2 }=64-16x+{ x }^{ 2 }$

$\Rightarrow 64=16x$

$\Rightarrow x=4$

${ x }^{ 2 }=16$ Which is the answer.

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