Find the value of $x$ in the displayed figure Find $x$ in the following figure.

$AB,AC,AD,BC,BE,CD$ are straight lines.
$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$
$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$
NOTE: figure not to scale.

$∆BCE$ is right angle triangle.
Hence $BC^2 = BE^2 + EC^2$
 $EC = \sqrt{(BC^2 - BE^2})= \sqrt{(100 - (x-3)^2)}$
$∆ACD$ is right angle triangle.
Hence $AC^2 = CD^2 + AD^2$
$(AE + EC)^2 = CD^2 + AD^2$
Substitute the values,
       $(\sqrt{(100 - (x-3)^2)} + x)^2 = (x-3)^2 + (x+4)^2$
Then you can solve this equation easily for getting x value.
