Question

How do you express the area of an equilateral triangle as a function of the length $s$ of one of its sides?

Answer 1

Hint: Take an equilateral triangle with side ‘$s$’. Find the height of the triangle using Pythagoras’ theorem. Then use the area of triangle formula i.e. $\dfrac{1}{2}\times base\times height$ and put the values of base and height to obtain the required solution.

Complete step-by-step solution:

$\vartriangle ABC$ is an equilateral triangle of side ‘$s$’.
So in $\vartriangle ABC$, $AB=BC=AC=s$
‘AD’ is the perpendicular bisector of ‘BC’.
So, $BD=CD=\dfrac{s}{2}$
We know, in $\vartriangle ABD$, $AB=s$ and $BD=\dfrac{s}{2}$
To find the value of ‘AD’, we have to use the Pythagoras’ theorem.
in $\vartriangle ABD$ Pythagoras’ theorem can be applied as ${{\left( AB \right)}^{2}}={{\left( AD \right)}^{2}}+{{\left( BD \right)}^{2}}$
Putting the value of ‘AB’ and ‘BD’, we get
$\begin{align}
  & \Rightarrow {{\left( s \right)}^{2}}={{\left( AD \right)}^{2}}+{{\left( \dfrac{s}{2} \right)}^{2}} \\
 & \Rightarrow {{s}^{2}}={{\left( AD \right)}^{2}}+\dfrac{{{s}^{2}}}{4} \\
 & \Rightarrow {{\left( AD \right)}^{2}}={{s}^{2}}-\dfrac{{{s}^{2}}}{4} \\
 & \Rightarrow {{\left( AD \right)}^{2}}=\dfrac{4{{s}^{2}}-{{s}^{2}}}{4} \\
 & \Rightarrow AD=\sqrt{\dfrac{3{{s}^{2}}}{4}} \\
 & \Rightarrow AD=\dfrac{\sqrt{3}s}{2} \\
\end{align}$
Again we know, the area of triangle$=\dfrac{1}{2}\times base\times height$
So, the area of $\vartriangle ABC=\dfrac{1}{2}\times BC\times AD$
Putting the value of base ‘BC’ and height ‘AD’, we get
Area of $\vartriangle ABC=\dfrac{1}{2}\times s\times \dfrac{\sqrt{3}s}{2}=\dfrac{\sqrt{3}}{4}{{s}^{2}}$
This is the required solution of the given question.

Note: Finding the height of the triangle should be the first approach for solving this question as area is to be expressed which is equal to $\dfrac{1}{2}\times base\times height$. Hence, the height should be found using the $\vartriangle ABD$ and the Pythagoras’ theorem which states that in a right angle triangle the square of hypotenuse is the sum of the square of the base and the height.