Question

One side of an equilateral triangle measures $\text{8 cm}$. Find its area using Heron's formula. What is its altitude?

Answer 1

Step 1: Finding the area of an equilateral triangle using heron's formula :

By heron's formula, $\text{A =}\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

where, a,b and c are sides of the triangle and $\text{s =}\frac{a+b+c}{2}$.

Length of the side of the equilateral triangle =$\text{8 cm}$

$s=\frac{8+8+8}{2}=\frac{24}{2}=12$

Area of the triangle = $\sqrt{12\left(12-8\right)\left(12-8\right)\left(12-8\right)}$

$$\begin{gathered} =\sqrt{12\times 4\times 4\times 4} \\ =\sqrt{3\times 4\times 4\times 4\times 4} \\ =4\times 4\sqrt{3} \\ =16\sqrt{3}c{m}^2 \end{gathered}$$

Step 2: Finding the altitude of the triangle :

We know that , Area = $\frac{1}{2}\times b\times h$

$\begin{array}{rcl}16\sqrt{3}& =& \frac{1}{2}\times 8\times h\\ h& =& \frac{16\sqrt{3}\times 2}{8}\\ h& =& 4\sqrt{3}\end{array}$

Hence, the area of triangle is$16\sqrt{3}c{m}^2$ and altitude is $4\sqrt{3}cm.$