Question

The sides of an equilateral triangle are 8 units long. What is the length of the altitude of the triangle?

Answer 1

$\bf \textit{height of an equilateral triangle}\\\\ h=\cfrac{s\sqrt{3}}{2}\qquad \begin{cases} s=\textit{length of a side}\\ ---------\\ s=8 \end{cases}\implies h=\cfrac{8\sqrt{3}}{2}$

Answer 2

Answer:  The answer is 4√3 units.

Step-by-step explanation: As shown in the attached figure, ABC is an equilateral triangle with AB = BC = CA = 8 units and AD is the altitude.

We know that every altitude of an equilateral triangle divides the opposite side in two equal parts.

So, in ΔABC, we have

BD = DC = half of BC.

So,

$BD=DC=\dfrac{8}{2}=4~\textup{units}.$

Since AD is perpendicular to BC, so ΔABD will be aright-angled triangle.

Using Pythagoras Theorem, we can write

$AB^2=AD^2+BD^2\\\\\Rightarrow 8^2=AD^2+4^2\\\\\Rightarrow 64=AD^2+16\\\\\Rightarrow AD^2=64-16\\\\\Rightarrow AD^2=48\\\\\Rightarrow AD=4\sqrt 3.$

Thus, the length of the altitude is 4√3 units.