A square is cut into three equal area regions by two parallel lines find area of square. A square is cut into three equal area regions by two parallel lines that are 1 cm apart, each one passing through exactly one of two diagonally opposed vertices. What is the area of the square ? 


Let $AE=x$. By symmetry (equal area etc.) $FC=x$. Let the side of square be $a$. Then $DE=BF=\sqrt{a^2+x^2}$.
Area of the parallelogram $FBED$ = $\sqrt{a^2+x^2} \cdot 1=\sqrt{a^2+x^2}$.
Area of $\triangle DAE$ = Area of $\triangle FCB$ = $\frac{ax}{2}$.
$$\frac{ax}{2}=\sqrt{a^2+x^2} \implies \color{red}{x^2=\frac{4a^2}{a^2-4}}.$$
And area of square is thrice the area of the triangle gives us
$$a^2=\frac{3ax}{2} \implies 2a=3x.$$
Solve these to get $a^2=13$ (area of the square).
