An equilateral triangle having a side of 9 cm. Find it's altitude.

In this case, AD is the triangle's altitude and ABC is an equilateral triangle.

Now, in $∆ADB$

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

$9^2=A{D}^2+4.5^2$

$A{D}^2=60.75$

$AD=\sqrt{6}0.75=\frac{9\sqrt{3}}{2} \mathrm{cm}=7.794 \mathrm{cm}$