$$ABCD$$ is a square of area 64 sq. units. $$AEFG$$ is another square of area 25 sq. units such that $$E$$, $$G$$ lie on the line segments $$AB$$, $$AD$$ respectively. If the line segments $$ED$$ and $$FG$$ meet in $$N$$, then find the ratio of the area of the region $$EBCDGNE$$ to the area of the region $$AGNE$$

Take $$A(0, 0)$$, $$B(8, 0)$$, $$D(0, 8)$$, so $$E(5, 0)$$, $$G(0, 5)$$ and $$FG$$ is the line $$y = 5$$. Line $$ED$$ is $$x = 5 - \frac{5y}{8}$$, which meets $$y = 5$$ at $$N\left(\frac{15}{8}, 5\right)$$. $$AGNE$$ is a trapezium with parallel sides $$GN = \frac{15}{8}$$ and $$AE = 5$$ and height 5, so its area is $$\frac{1}{2}\left(\frac{15}{8} + 5\right) \times 5 = \frac{275}{16}$$. The other region has area $$64 - \frac{275}{16} = \frac{749}{16}$$, so the ratio is $$749 \colon 275 \approx 2.72$$.