Chapter 2, Problem 7P

To determine

To find: The function that models area of equilateral triangle in terms of its length x of one of its sides.

Answer to Problem 7P

The function that models area of given equilateral triangle is $\frac{\sqrt{3}}{4}{x}^2$.

Explanation of Solution

Draw the figure of equilateral triangle as shown below.

Figure (1)

Area of triangle is,

$A=\frac{\text{1}}{\text{2}}\left(\text{base}\right)\left(\text{height}\right)$

The length of side of equilateral triangle is $x$.

In Figure (1) $AD$ is median, it bisects $CB$ so $CD$ is $\frac{x}{2}$.

Use Pythagorean theorem in triangle $ACD$ shown in Figure (1).

${\left(AC\right)}^2={\left(CD\right)}^2+{\left(AD\right)}^2$

Substitute x for $AC$, $\frac{x}{2}$ for $CD$ in above equation.

$\begin{array}{c}{\left(x\right)}^2={\left(\frac{x}{2}\right)}^2+{\left(AD\right)}^2\\ x^2=\frac{x^2}{4}+A{D}^2\end{array}$

Subtract $\frac{x^2}{4}$ from both sides of above equation.

$\begin{array}{c}{x}^2-\frac{x^2}{4}=A{D}^2\\ \frac{3}{4}{x}^2=A{D}^2\end{array}$

Take square root of above equation and switch sides.

$AD=\sqrt{\frac{3}{4}}x$

Summarize the all the information as shown in table below.

In Words	In Algebra
Base	$x$
Height	$\sqrt{\frac{3}{4}}x$

Use the information in the table and model the function.

$\begin{array}{c}A=\frac{\text{1}}{\text{2}}\left(\text{base}\right)\left(\text{height}\right)\\ =\frac{1}{2}\left(x\right)\left(\sqrt{\frac{3}{4}}x\right)\\ =\frac{1}{2}\left(x\right)\left(\frac{\sqrt{3}}{2}x\right)\\ =\frac{\sqrt{3}}{4}{x}^2\end{array}$

Thus, the function that models area of equilateral triangle is $A\left(x\right)=\frac{\sqrt{3}}{4}{x}^2$.