### 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 congruent sides and three congruent angles of 60 degrees each.
I also know that the altitude of an equilateral triangle bisects the base and the vertex angle, forming two right triangles with 30-60-90 angles.
I can use the properties of 30-60-90 triangles to find the length of the side of the equilateral triangle.
The altitude is the longer leg of the 30-60-90 triangle, and the shorter leg is half of the base.
The ratio of the longer leg to the shorter leg is $\sqrt3 : 1$, so I can set up a proportion and solve for the shorter leg.
$\frac{\sqrt6}{x} = \frac{\sqrt3}{1}$
Cross-multiply and simplify.
$\sqrt6 = x\sqrt3$
Divide both sides by $\sqrt3$.
$x = \frac{\sqrt6}{\sqrt3} = \frac{\sqrt6\sqrt3}{\sqrt3\sqrt3} = \frac{\sqrt{18}}{3} = \frac{\sqrt9\sqrt2}{3} = \frac{3\sqrt2}{3} = \sqrt2$
The shorter leg of the 30-60-90 triangle is $\sqrt2$ units, so the base of the equilateral triangle is twice that, or $2\sqrt2$ units.
The area of a triangle is half the product of the base and the height.
Plug in the values for the base and the height.
$A = \frac{1}{2}(2\sqrt2)(\sqrt6)$
Simplify by multiplying and rationalizing the denominator.