Question

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

Answer 1

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

$AD\perp BC$ [Given]

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

Refer image.

$AB=BC$ [SIdes of an equilateral $\Delta$]

$\angle 1=\angle 2={90}^0$

$AD=AD$ [Common]

$\therefore \Delta ADB=\Delta ADC$ [By RHS congruence criterion]

$\Rightarrow BD=DC$ [CPCT]

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

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

$A{D}^2+B{D}^2=A{B}^2$

$\Rightarrow A{D}^2+\left(4\right)=(8{)}^2$

$\Rightarrow A{D}^2=64-16$

$\Rightarrow A{D}^2=48$

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