On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$.
Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\overline{AD}$. 

The fraction of the area of the square $ABCD$ which is occupied by trapezoid $BCGE$ is \[\frac{275+405}{269+275+405+411}=\frac 12,\]so $Q$ is the center of $ABCD$. Thus $R$, $Q$, $S$ are collinear, and $RT=QS=\tfrac 12 s$. Similarly, the fraction of the area occupied by trapezoid $CDHF$ is $\tfrac 35$, so $RS=\tfrac 35s$ and $RQ=\tfrac{1}{10}s$. 
Because $\triangle QSG \cong \triangle RTH$, the area of $DHPG$ is the sum \[[DHPG]=[DTRS]+[RPQ].\] Rectangle $DTRS$ has area $RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2$. If $\angle QRP = \theta$ , then $\triangle RPQ$ has area \[[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.\]Therefore the area of $[DHPG]$ is $s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)$. Because the area of trapezoid $CDHF$ is $\tfrac 35 s^2$, the area of $CGPF$ is $s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)$.
Because these areas are in the ratio $411:405=(408+3):(408-3)$, it follows that \[\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},\]from which we get $\sin 2\theta = \tfrac {15}{17}$. Note that $\theta =\angle RHT > \angle QAT = 45^\circ$, so $\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}$ and $\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}$. Then \[[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.\]