Find the altitude to side AC of triangle with side AB = 20 cm, AC = 20 cm, BC = 30 cm.

This is an isosceles triangle.

$D C=15$
$A{D}^2+D{C}^2=A{C}^2$
$A{D}^2+{15}^2={20}^2$
$A{D}^2=400-225$
$A{D}^2=175$
$AD=5\sqrt{7}$

Area of the triangle $=\frac{1}{2}\times$ base $\times$ height
$=\frac{1}{2}\times BC\times AD\text{ or }\frac{1}{2}\times AC\times \left(h\right)$
$=\frac{1}{2}\times 30\times 5\sqrt{7}$
$=\frac{1}{2}\times (20\times h)$
$=\frac{3}{2}\times 5\sqrt{7}$
$h=7.5\sqrt{7}$