### 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 an equilateral triangle has three equal sides and three equal angles of 60 degrees.
I also know that the altitude of an equilateral triangle divides it into two congruent 30-60-90 right triangles, which have a special relationship among their side lengths.
In a 30-60-90 triangle, the hypotenuse is twice the length of the shorter leg, and the longer leg is $\sqrt3$ times the length of the shorter leg.
Since the altitude of the equilateral triangle is $\sqrt6$, that means it is the longer leg of one of the 30-60-90 triangles, and the shorter leg is $\sqrt6 / \sqrt3 = \sqrt2$.
The hypotenuse of the 30-60-90 triangle is then twice the shorter leg, which is $2\sqrt2$.
This means that the side length of the equilateral triangle is also $2\sqrt2$.
To find the area of the equilateral triangle, I need to use the formula $A = \frac{1}{2} bh$, where b is the base and h is the height.
I can choose any side of the equilateral triangle as the base, so I'll use $2\sqrt2$ as b, and $\sqrt6$ as h.
Plugging these values into the formula, I get $A = \frac{1}{2} (2\sqrt2)(\sqrt6)$.
Simplifying, I get $A = \sqrt2 \sqrt6 = \sqrt{12}$.
However, this is not the simplest radical form, because 12 has a perfect square factor of 4.
I can rewrite $\sqrt{12}$ as $\sqrt{4 \times 3} = \sqrt4 \sqrt3 = 2\sqrt3$.
This is the simplest radical form of the area.