Solution:

Here $\overline{EM}$ is a side of the square. Let $h$ be the altitude of triangle $GEM.$ If $h \le 10,$ the triangle would lie entirely inside the square, and its area $\frac{1}{2} \cdot 10 \cdot h = 80$ would force $h = 16,$ a contradiction. So $h \gt 10$ and the apex $G$ lies outside the square; the opposite side $\overline{AI}$ cuts off a smaller triangle similar to $GEM$ with height $h - 10$ and base $\frac{10(h - 10)}{h}.$

The common region is triangle $GEM$ minus that small triangle: $$80 = 5h - \frac{1}{2} \cdot \frac{10(h - 10)}{h} \cdot (h - 10) = 5h - \frac{5(h - 10)^2}{h}.$$ Multiplying by $h$ gives $80h = 5h^2 - 5(h - 10)^2 = 5(20h - 100) = 100h - 500,$ so $20h = 500$ and $h = 25.$