In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$ (diagram not to scale). The area of $ABCD$ is $\text{(A)}\ 182 \qquad$ $ \text{(B)}\ 195 \qquad$ $ \text{(C)}\ 210 \qquad$ $ \text{(D)}\ 234 \qquad$ $ \text{(E)}\ 260$
Solution
Place $A=(0,0), B=(52,0)$. Let $D=(x,h), C=(x+39,h)$ (since $CD=39$, parallel to $AB$).
$\text{(A)}\ 182 \qquad$ $ \text{(B)}\ 195 \qquad$ $ \text{(C)}\ 210 \qquad$ $ \text{(D)}\ 234 \qquad$ $ \text{(E)}\ 260$ 
