2008 AMC 12B Problem 13

Place $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1),$ so $\overline{AD}$ lies along the $y$-axis and distance from $\overline{AD}$ is the $x$-coordinate. The region lies in the strip $\tfrac13 \le x \le \tfrac23,$ which within the square has area $\tfrac13.$ Equilateral $\triangle ABE$ has $E = \left(\tfrac12, \tfrac{\sqrt3}{2}\right),$ with side $AE$ on $y = \sqrt3\,x$ and side $BE$ on $y = \sqrt3(1 - x).$ The area of the triangle inside the strip is $$\begin{aligned} &\int_{\frac{1}{3}}^{\frac{1}{2}} \sqrt3\,x\,dx \\ &\quad {}+ \int_{\frac{1}{2}}^{\frac{2}{3}} \sqrt3(1 - x)\,dx \\ &= 2\int_{\frac{1}{3}}^{\frac{1}{2}}\sqrt3\,x\,dx \\ &= \frac{5\sqrt3}{36}. \end{aligned}$$ Therefore $$[R] = \frac13 - \frac{5\sqrt3}{36} = \frac{12 - 5\sqrt3}{36}.$$ Thus, the correct answer is B.