Convex quadrilateral $ABCD$ has $AB=3$, $BC=4$, $CD=13$, $AD=12$, and $\angle ABC=90^{\circ}$, as shown. What is the area of the quadrilateral? $\textbf{(A)}\ 30\qquad$ $\textbf{(B)}\ 36\qquad$ $\textbf{(C)}\ 40\qquad$ $\textbf{(D)}\ 48\qquad$ $\textbf{(E)}\ 58.5$
Solution
Connect $AC$. Then $AC=5$.
$\triangle ACD$ has sides $5,12,13$, hence is a right triangle (theorem 19).
So the total area is $\frac12\cdot 3\cdot 4+\frac12\cdot 5\cdot 12=6+30=36$.
$\textbf{(A)}\ 30\qquad$ $\textbf{(B)}\ 36\qquad$ $\textbf{(C)}\ 40\qquad$ $\textbf{(D)}\ 48\qquad$ $\textbf{(E)}\ 58.5$ 
