Question

Find the altitude of an equilateral triangle of side $8cm$.

Answer 1

Calculate the altitude of the equilateral triangle using Pythagoras theorem

In the figure, $ABC$ is an equilateral triangle, where, side $AB=BC=AC=8cm$ and $AD$ is the altitude.

Let the altitude $AD$ be $x$

As the altitude $AD$ divides the base $BC$ in two equal halves (since the triangle is equilateral) we have $BD=DC=4cm$

Now, according to the Pythagoras theorem, ${\left(H\right)}^2={\left(B\right)}^2+{\left(P\right)}^2$

where, hypotenuse $\left(H\right)=AC$, base $\left(B\right)=DC$ and perpendicular $\left(P\right)=AD$

$$\begin{gathered} \Rightarrow {\left(AC\right)}^2={\left(DC\right)}^2+{\left(AD\right)}^2 \\ \Rightarrow 8^2=4^2+x^2 \\ \Rightarrow 64=16+x^2 \\ \Rightarrow x^2=64-16 \\ \Rightarrow x=\sqrt{48} \\ \Rightarrow x=4\sqrt{3}cm \end{gathered}$$

Hence, the altitude of the equilateral triangle is $4\sqrt{3}cm$.