Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle.

Let h be the height, then
Each of the equal sides of an isosceles triangle $=(h+2)cm$ and
Base $=12cm$

Area of a triangle $=\frac{1}{2}\times Base\times Height$
$=\frac{1}{2}\times 12\times h....(1)$

Area of isosceles triangle $=\frac{1}{4}\times b\times \sqrt{(4a^2-b^2)}$
$=\frac{1}{4}\times 12\times \sqrt{(4(h+2{)}^2-144)}...(2)$

From (1) and (2), we get
$\Rightarrow \frac{1}{2}\times 12\times h=\frac{1}{4}\times 12\sqrt{(4(h+2{)}^2-144)}$
$\Rightarrow 6h=3\sqrt{(4h^2+16h+16-144)}$
$\Rightarrow 2h=\sqrt{(4h^2+16h-128)}$
$\Rightarrow 4h^2=4h^2+16h-128$
$\Rightarrow 16h=128$
or $h=8$
Height is 8 cm.
Now, 
Area of a triangle $=\frac{1}{2}\times Base\times Height$
$=\frac{1}{2}\times 12cm\times 8cm$
$=\frac{1}{2}\times 96c{m}^2$
$=48c{m}^2.$