Chapter 11.8, Problem 2PSA

To determine

To find: Area of an equilateral triangle having side 8 units.

Answer to Problem 2PSA

Area of the equilateral triangle with side 8 m is $16\sqrt{3}{\text{m}}^2$ .

Explanation of Solution

Given Information:

Side of an equilateral triangle $\left(a\right)=8$ units.

Formula used:

Using Heron’s formula,

Area of a triangle $\left(ar\Delta \right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Here, $s=\frac{a+b+c}{2}$ is the semi-perimeter of a triangle

  $a,b\text{and}c$ are the sides of a triangle

Calculation:

Let $a,b\text{and}c$ be the sides of the triangle.

The triangle is an equilateral triangle of side 8 m.

So, sides of the triangle will be equal to each other.

  $\therefore a=b=c=8\text{m}$

We know that, Area of a triangle $\left(ar\Delta \right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

  $"s"$ is the semi-perimeter of the triangle

Now, $s=\frac{a+b+c}{2}$

  $\begin{array}{l}s=\frac{8+8+8}{2}\text{m}\\ s=\frac{24}{2}\text{m}\\ s=12\text{m}\end{array}$

So, $ar\Delta =\sqrt{12\left(12-8\right)\left(12-8\right)\left(12-8\right)}{\text{m}}^2$

  $\begin{array}{l}ar\Delta =\sqrt{12\left(4\right)\left(4\right)\left(4\right)}{\text{m}}^2\\ ar\Delta =\sqrt{3\times 4\times 4\times 4\times 4}{\text{m}}^2\\ ar\Delta =4\times 4\sqrt{3}{\text{m}}^2\\ ar\Delta =16\sqrt{3}{\text{m}}^2\end{array}$

Hence, area of the equilateral triangle with side 8 m is $16\sqrt{3}{\text{m}}^2$