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

Given: An equilateral triangle of side $8\ cm$.

To do: To find the length of altitude of an equilateral triangle. 

$\vartriangle ABC$ is an equilateral triangle.         [Given]

$AD\perp BC$                  [Given]

Now In $\vartriangle ADB$ and $\vartriangle ADC$,

$AB=BC$           [SIdes of an equilateral triangle]

$\angle 1=\angle 2=90^o$

 

$AD=AD$                       [Common]

$\therefore \vartriangle ADB\cong\vartriangle ADC$            [By RHS congruence criterion]

$\Rightarrow BD=DC$                     [CPCT]

$\therefore BD=DC=\frac{BC}{2}=\frac{8}{2}=4\ cm$

$\therefore $ By pythagoras theorem, In right angled triangle $ADB$, we have

$AD^2+BD^2=AB^2$

$\Rightarrow AD^2+(4)=(8)^2$

$\Rightarrow AD^2=64−16$

$\Rightarrow AD^2=48$

$\Rightarrow AD=4\sqrt{3}\ cm$

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