jmc

Let $r$ be the original radius of the pizza. If the diameter increases by 2 inches, then the radius increases by 1 inch. We can calculate $A_{original}=\pi r^2$ and $A_{final}=\pi (r+1{)}^2$. Using the information given in the problem, $A_{final}=A_{original}\cdot 1.44$. Substituting, we get $$\begin{array}{rl}1.44\pi r^2& =\pi (r+1{)}^2\\ & =\pi (r^2+2r+1)\\ 1.44r^2& =r^2+2r+1\\ .44r^2-2r-1& =0\\ 25(.44r^2-2r-1)& =25(0)\\ 11r^2-50r-25& =0\\ (r-5)(11r+5)& =0\\ r& =5,-\frac{5}{11}.\end{array}$$$r$ cannot be negative, so $r=5$. This means the area of the original circle is $\pi \cdot 5^2=\boxed{25\pi }$.