pls help

Shown below is rectangle EFGH . Its diagonals meet at Y .
Let X be the foot if an altitude is dropped from E to line FH . If
EX=24 and GY = 25, find the perimeter of rectangle .

$\text{Let $EH$ = x }\\ \text{Let $EF$ = y }\\ \text{Let $EY=HY$ = 25 }\\ \text{Let $HF$ = 50 }$

1.

$\begin{array}{|rcll|} \hline EY^2 &=& EX^2+XY^2 \quad & | \quad XY=HY-H = 25 - HX \\ 25^2 &=& 24^2+(25-HX)^2 \\ 25^2-24^2 &=& (25-HX)^2 \\ \sqrt{25^2-24^2} &=& 25-HX \\ HX &=& 25 - \sqrt{25^2-24^2} \\ HX &=& 25 - 7 \\ \mathbf{ HX} & \mathbf{=} & \mathbf{18} \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline x^2 &=& 24^2 + HX^2 \quad & | \quad HX=18\\ x^2 &=& 24^2 + 18^2 \\ x &=& \sqrt{24^2 + 18^2} \\ \mathbf{ x} & \mathbf{=} & \mathbf{30} \\ \hline \end{array}$

3.

$\begin{array}{|rcll|} \hline x^2+y^2 &=& HF^2 \quad & | \quad x=30, ~ HF=50 \\ 30^2+y^2 &=& 50^2 \\ y^2 &=& 50^2-30^2 \\ y &=& \sqrt{50^2-30^2} \\ \mathbf{ y} & \mathbf{=} & \mathbf{40} \\ \hline \end{array}$

The perimeter of rectangle:

$\begin{array}{|rcll|} \hline && 2x+2y \\ &=& 2\cdot 30 + 2\cdot 40 \\ &=& 60 + 80 \\ &\mathbf{=}& \mathbf{140} \\ \hline \end{array}$

The perimeter of rectangle is 140