# Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 5

## Solution:

From the formula of the area of a triangle, $\displaystyle A=\frac{1}{2}\cdot a\cdot b\cdot sin\left(\theta \right)$. Also, we know that an interior angle of an equilateral triangle is 60 degrees, and $\displaystyle sin\:60^{\circ} =\frac{\sqrt{3}}{2}$

$\displaystyle A=\frac{1}{2}\cdot x\cdot x\cdot sin\:60^{\circ}$

$\displaystyle A=\frac{1}{2}\cdot x^2\cdot \frac{\sqrt{3}}{2}$

$\displaystyle A=\frac{\sqrt{3}}{4}x^2$