Answer
$A = \frac{s^2}{4}~\sqrt{3}$
Work Step by Step
Theorem 11.4.1: The area of an acute triangle equals one-half the product of the lengths of two sides and the sine of the included angle.
In an equilateral triangle, each side has the same length of $s$ and each angle has the same measure of $60^{\circ}$.
We can find the area:
$A = \frac{1}{2}~s \times s \times sin(\theta)$
$A = \frac{1}{2}~s^2 sin(\theta)$
$A = \frac{1}{2}~s^2 sin(60^{\circ})$
$A = \frac{1}{2}~s^2 ~(\frac{\sqrt{3}}{2})$
$A = \frac{s^2}{4}~\sqrt{3}$
