Since $269 + 411 = 275 + 405,$ segment $EG$ passes through the center of the square; treat the trapezoid $ABFH$ the same way to locate $FH$

With side $s$ and rise $\delta$ of $EG,$ the length gives $s^2 + \delta^2 = 1156,$ and the ratio $269 : 1360$ for $AEPH$ pins down the product $s\delta$

Place $B = (0,0),$ $C = (s, 0),$ $D = (s, s),$ $A = (0, s),$ with $E = (0, e),$ $F = (f, 0),$ $G = (s, g),$ $H = (h, s).$ The regions $AEPH$ and $DHPG$ together form the trapezoid $AEGD,$ of area $\frac{s\,((s - e) + (s - g))}{2};$ since $\frac{269 + 411}{1360} = \frac{1}{2},$ this is half of $s^2,$ forcing $e + g = s,$ i.e. $EG$ passes through the center. Likewise $AEPH$ and $BFPE$ form the trapezoid $ABFH$ of area $\frac{s(f + h)}{2} = \frac{269 + 275}{1360}\,s^2 = \frac{2}{5}s^2,$ so $f + h = \frac{4s}{5}.$ Perpendicularity of the directions $(s, g - e)$ and $(h - f, s)$ gives $h - f = e - g.$ Writing $\delta = g - e,$ we get $E = \left(0, \frac{s - \delta}{2}\right),$ $G = \left(s, \frac{s + \delta}{2}\right),$ $F = \left(\frac{2s}{5} + \frac{\delta}{2}, 0\right),$ $H = \left(\frac{2s}{5} - \frac{\delta}{2}, s\right),$ and $EG = 34$ gives $s^2 + \delta^2 = 1156.$

Intersecting lines $EG$ and $FH$ (and simplifying with $s^2 + \delta^2 = 1156$) yields $$\begin{aligned} P &= \\ &\left(\frac{s}{2} - \frac{s^3}{11560},\; \frac{s}{2} - \frac{s^2\delta}{11560}\right), \end{aligned}$$ and the shoelace formula on $A,$ $E,$ $P,$ $H$ then gives $[AEPH] = \frac{s^2}{5} - \frac{s^3\delta}{231200}.$ Setting this equal to $\frac{269}{1360}s^2$ leaves $\frac{3s^2}{1360} = \frac{s^3\delta}{231200},$ so $s\delta = 510.$

Now $s^2 + \delta^2 = 1156$ and $s^2\delta^2 = 260100,$ so $s^2$ and $\delta^2$ are the roots of $t^2 - 1156t + 260100 = 0,$ which are $\frac{1156 \pm 544}{2} = 850$ and $306.$ Since $|\delta| = |g - e| \lt s,$ the area is $s^2 = 850.$