The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle?

We are given that AD = 8 cm and AB + AC + BC = 32 cm.
AB can be substituted by AC and BC can be written as 2*DC.
This gives, 2AC + 2DC = 32 
AC + DC = 16
In triangle ADC: 
$$AC^2-DC^2=AD^2$$
$$\left(AC-DC\right)\left(AC+DC\right)=8^2$$
$$\left(AC-DC\right)\left(16\right)=64$$
AC - DC = 4.
From these 2 equations, we get DC = 6 cm and BC = 2*6 = 12 cm.
Area of triangle ABC = $$\frac{1}{2}\times\ BC\times\ AD=\frac{1}{2}\times\ 12\times\ 8=48\ cm^2$$.