### Instruction: 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 $\frac{a}{b}$?### Response: I start by drawing a diagram of the situation, labeling the sides of the triangle and the hexagon with $a$ and $b$, respectively.
<diagram>
I recall that the area of an equilateral triangle with side length $s$ is $\frac{s^2\sqrt{3}}{4}$, and the area of a regular hexagon with side length $s$ is $\frac{3s^2\sqrt{3}}{2}$.
I set the areas of the triangle and the hexagon equal, since they are given to be equal, and I plug in the expressions for the areas in terms of $a$ and $b$.
\[\frac{a^2\sqrt{3}}{4} = \frac{3b^2\sqrt{3}}{2}\]
I simplify the equation by dividing both sides by $\sqrt{3}$ and multiplying both sides by 4.
\[a^2 = 6b^2\]
I take the square root of both sides, noting that I only need the positive root since $a$ and $b$ are lengths.
\[a = \sqrt{6}b\]
I divide both sides by $b$ to get the ratio $\frac{a}{b}$.