2004 Fermat Problem 23

Triangle ABC is isosceles with AB = AC and BC = 65 cm. P is a point on BC such that the perpendicular distances from P to AB and AC are 24 cm and 36 cm, respectively. The area of \triangle ABC, in \text{cm}^2, is

\textbf{(A)}\ 1254\quad \textbf{(B)}\ 1640\quad \textbf{(C)}\ 1950\quad \textbf{(D)}\ 2535\quad \textbf{(E)}\ 2942

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Fermat contests here.