1957 AHSME Problem 25

The vertices of \triangle PQR have coordinates as follows: P(0,a),\,Q(b,0),\,R(c,d), where a,\,b,\,c and d are positive. The origin and point R lie on opposite sides of PQ. The area of \triangle PQR may be found from the expression:

\textbf{(A)}\ \frac{ab + ac + bc + cd}{2} \qquad \textbf{(B)}\ \frac{ac + bd - ab}{2}\qquad \textbf{(C)}\ \frac{ab-ac-bd}{2}\qquad \textbf{(D)}\ \frac{ac+bd+ab}{2}\qquad \textbf{(E)}\ \frac{ac+bd-ab-cd}{2}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.