Chapter 1 - Section 1.1 - Four Ways to Represent a Function - 1.1 Exercises - Page 20: 67

$A=\frac{k^2*\sqrt 3}{4}$

Let $x$, $y$, $z$ be the sides of the equilateral triangle. All vertices are 60 degrees. Thus, we have the area of the triangle as $\frac{1}{2}base*height$: $A = 0.5 * x * m$ Using trigonometry, we have: $m = y * sin(60)$ Thus: $x=y$ and $sin(60) = \frac{\sqrt 3}{2}$ Finally, we get: $A=\frac{1}{2}*\frac{\sqrt 3}{2}*x^2 = \frac{x^2\sqrt 3}{4}$