### Instruction: The altitude of an equilateral triangle is $\sqrt6$ units.  What is the area of the triangle, in square units? Express your answer in simplest radical form.### Response: I know that the altitude of an equilateral triangle divides it into two congruent 30-60-90 right triangles.
In a 30-60-90 triangle, the ratio of the sides is 1:$\sqrt3$:2, where the shortest side is opposite the 30 degree angle, the longest side is the hypotenuse, and the middle side is opposite the 60 degree angle.
Since the altitude is $\sqrt6$, it must be the middle side of the 30-60-90 triangles.
That means the shortest side is $\sqrt6/\sqrt3$, which simplifies to $\sqrt2$.
The shortest side is also the half of the base of the equilateral triangle, so the full base is 2$\sqrt2$ units.
To find the area of the equilateral triangle, I can use the formula $A = \frac{1}{2}bh$, where b is the base and h is the altitude.
Plugging in the values, I get $A = \frac{1}{2}(2\sqrt2)(\sqrt6) = \sqrt2\sqrt6 = \sqrt{12}$.
However, this is not in simplest radical form, because 12 has a perfect square factor of 4.
So I can rewrite $\sqrt{12}$ as $\sqrt{4}\sqrt{3} = 2\sqrt{3}$.
This is the final answer.