Question

The length of an altitude of an equilateral triangle of side a is:
(a) $\dfrac{2a}{\sqrt{3}}$
(b) $\dfrac{\sqrt{3}}{2a}$
(c) $\dfrac{a\sqrt{3}}{2}$
(d) $\dfrac{a}{2\sqrt{3}}$

Answer 1

Hint: In this question, we will use equilateral triangle properties and Pythagoras theorem to find the length of altitude.
Complete step-by-step answer:
Let us consider an equilateral triangle of side a as given below:

Here, ABC is an equilateral triangle with all sides of length ‘a’ and altitude AD.
We know that the altitude of a triangle is perpendicular to the base. So, here, AD is perpendicular to BC. Also, for an equilateral triangle, the altitude of a triangle is equivalent to the median of a triangle.
So, AD will also be the median of triangle ABC.
Now, we know that the median of a triangle divides the base of the triangle in two equal parts. therefore, AD will divide BC in the two equal parts BD and CD.
That is, $BD=CD$
Also, $BC=a$
   $\Rightarrow BD+CD=a$
Using above equality, we can write,
$\begin{align}
  & BD+BD=a \\
 & \Rightarrow 2BD=a \\
\end{align}$
Dividing 2 on both sides of the equation, we get,
$BD=\dfrac{a}{2}$
Therefore, $CD=\dfrac{a}{2}$
Now, we have, ADB is a right-angle triangle, right angles at D.
Therefore, we can apply Pythagoras theorem here,
So, applying Pythagoras theorem, we get,
Putting value of CD, we get,
$\begin{align}
  & {{\left( \dfrac{a}{2} \right)}^{2}}+A{{D}^{2}}={{a}^{2}} \\
 & \Rightarrow \dfrac{{{a}^{2}}}{4}+A{{D}^{2}}={{a}^{2}} \\
\end{align}$
Subtracting $\dfrac{{{a}^{2}}}{4}$ from both sides of the equation, we get,
$A{{D}^{2}}={{a}^{2}}-\dfrac{{{a}^{2}}}{4}$
Taking LCM, we get,
$A{{D}^{2}}=\dfrac{4{{a}^{2}}-{{a}^{2}}}{4}=\dfrac{3{{a}^{2}}}{4}$
Taking root on both sides of the equation, we get,
$\begin{align}
  & A{{D}^{2}}=\sqrt{\dfrac{4{{a}^{2}}-{{a}^{2}}}{4}}=\dfrac{\sqrt{3{{a}^{2}}}}{\sqrt{4}} \\
 & \Rightarrow AD=\dfrac{a\sqrt{3}}{2} \\
\end{align}$
Hence, the length of altitude of the equilateral triangle is $\dfrac{a\sqrt{3}}{2}$.
Therefore, (c) is the correct option.

Note: Take a note that this rule cannot be applied to any other triangle as for other triangles median and altitude are not the same.