A wire is cut into two pieces, one of length $a$ and the other of length $b$. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2}$
Using the area formulas for an equilateral triangle $\left(\frac{{s}^{2}\sqrt{3}}{4}\right)$ and regular hexagon $\left(\frac{3{s}^{2}\sqrt{3}}{2}\right)$ with side length $s$, plugging $\frac{a}{3}$ and $\frac{b}{6}$ into each equation, we find that $\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}$. Simplifying this, we get $\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}$