\begin{equation}\begin{array}{l}{57-61 \text { Find a formula for the described function and state its }} \\ {\text { domain. }}\end{array}\end{equation}Express the area of an equilateral triangle as a function of the length of a side.

Key Concepts: Area Of Equilateral Triangle, Function, Domain Explanation: To express the area of an equilateral triangle as a function of its side length, we first recall that the formula for the area of any triangle is given by:\n$A=\frac{1}{2}bh$\nwhere $b$ is the base and $h$ is the corresponding height. In an equilateral triangle, all sides are congruent and all angles are equal. Let's denote the side length by $s$. We can divide the equilateral triangle into two right triangles, where the base is $s/2$ and the height is:$h=\sqrt{s^2-(s/2{)}^2}=\sqrt{3}/2s$\nTherefore, the area of the equilateral triangle can be expressed as:\n$A(s)=\frac{1}{2}\times s\times \frac{\sqrt{3}}{2}s=\frac{1}{4}\sqrt{3}{s}^2$ Step by Step Solution: Step 1. Divide the equilateral triangle into two right triangles by drawing the altitude from any vertex. Step 2. Using Pythagoras' Theorem, the height $h$ can be expressed as $\sqrt{s^2-(s/2{)}^2}=\sqrt{3}/2s$ Step 3. The area of the equilateral triangle can be expressed as $A(s)=\frac{1}{2}bh=\frac{1}{2}\times s\times \frac{\sqrt{3}}{2}s=\frac{1}{4}\sqrt{3}{s}^2$ Final Answer: The area of the equilateral triangle can be expressed as $A(s)=\frac{1}{4}\sqrt{3}{s}^2$, for $s>0$.