Chapter 11.2, Problem 20PSB

To determine

To find the area of equilateral triangle ABC with a perimeter of $45$ m.

Answer to Problem 20PSB

The area of equilateral triangle ABC is $97.42{\text{ m}}^2$

Explanation of Solution

Given:

  $AB=BC=CA=15$

Formula Used:

Altitude of Equilateral Triangle:

  $h=\frac{1}{2}\ast \sqrt{3}\ast a$

Where

a = 15, is the side of the triangle.

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

By drawing an altitude $AD$ , $\Delta ADC$ is a ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}a=15\\ \text{AD}=\frac{1}{2}\ast \sqrt{3}\ast 15\\ =\frac{15}{2}\sqrt{3}\end{array}$

So, Altitude of Equilateral Triangle is $\frac{15}{2}\sqrt{3}$ .

  $\begin{array}{l}A(\Delta ABC)=\frac{1}{2}*BC*AD\\ =\frac{1}{2}*(15)*(\frac{15}{2}\sqrt{3})\\ =\frac{225}{4}\sqrt{3}\end{array}$

  $=97.42m$