Find the area of square $ABCD$

Notice that $269+411=275+405$ . This means $\overline{EG}$ passes through the centre of the square.

Draw $\overline{IJ} \parallel \overline{HF}$ with $I$ on $\overline{AD}$ , $J$ on $\overline{BC}$ such that $\overline{IJ}$ and $\overline{EG}$ intersects at the centre of the square $O$ .

Let the area of the square be $1360a$ . Then the area of $HPOI=71a$ and the area of $FPOJ=65a$ . This is because $\overline{HF}$ is perpendicular to $\overline{EG}$ (given in the problem), so $\overline{IJ}$ is also perpendicular to $\overline{EG}$ . These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.

Let the side length of the square be $d=\sqrt{1360a}$ .

Draw $\overline{OK}\parallel \overline{HI}$ and intersects $\overline{HF}$ at $K$ . $OK=d\cdot\frac{HFJI}{ABCD}=\frac{d}{10}$ .

The area of $HKOI=\frac12\cdot HFJI=68a$ , so the area of $POK=3a$ .

Let $\overline{PO}=h\$. Then \$KP=\frac{6a}{h}$

Consider the area of $PFJO$ . $\frac12(PF+OJ)(PO)=65a$ $(17-\frac{3a}{h})h=65a$ $h=4a$ Thus, $KP=1.5$ .

Solving $(4a)^2+1.5^2=(\frac{d}{10})^2=13.6a$ , we get $a=\frac58$ .

Therefore, the area of $ABCD=1360a=\boxed{850}$