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

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PROBLEM:

Express the area AA of an equilateral triangle as a function of its side xx.

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SOLUTION:

From the formula of the area of a triangle, A=12absin⁡(θ)\displaystyle A=\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right). Also, we know that the interior angle of an equilateral triangle is 60 degrees, and sin⁡ 60∘=32\displaystyle \sin\:60^{\circ} =\frac{\sqrt{3}}{2}.

A=12absin⁡(θ)A=12⋅x⋅x⋅sin⁡ 60∘A=12⋅x2⋅32A=34x2  (Answer)\begin{align*} A & =\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right) \\ A & =\frac{1}{2} \cdot x\cdot x\cdot \sin\:60^{\circ} \\ A & =\frac{1}{2}\cdot x^2\cdot \frac{\sqrt{3}}{2} \\ A & =\frac{\sqrt{3}}{4}x^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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