A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$, the length of the longer parallel side of each trapezoid. 


$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac  78$
Solution 1

Then $2[I]+2[II] = [I]+3[II] \Longrightarrow [I]=[II]$. Let the shorter side of $I$ be $m$ and the base of $II$ be $n$ such that $m+2n = x$; then $[I]=[II]$ implies that $2m=n$, and since $2m + 2n = 1$ it follows that $m = \frac 16$ and $x = \frac 56 \Longrightarrow \mathbf{(D)}$.

Solution 2
The area of the trapezoid is $\frac{1}{3}$, and the shorter base and height are both $\frac{1}{2}$.  Therefore, \[\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{\text{D}}\]