### Instruction: In trapezoid $ABCD$ the lengths of the bases $AB$ and $CD$ are 8 and 17 respectively. The legs of the trapezoid are extended beyond $A$ and $B$ to meet at point $E$. What is the ratio of the area of triangle $EAB$ to the area of trapezoid $ABCD$? Express your answer as a common fraction.### Response: I'll start by drawing a picture of the trapezoid and labeling the given information.
I notice that triangle $EAB$ and trapezoid $ABCD$ share the same height, which is the perpendicular distance from $AB$ to $CD$.
Let's call this height $h$.
I also notice that the area of triangle $EAB$ is half the product of its base $AB$ and its height $h$, while the area of trapezoid $ABCD$ is the average of its bases $AB$ and $CD$ times its height $h$.
So, the ratio of the area of triangle $EAB$ to the area of trapezoid $ABCD$ is

\[\frac{\frac{1}{2}AB \cdot h}{\frac{1}{2}(AB + CD) \cdot h} = \frac{AB}{AB + CD}.\]
Plugging in the given values of $AB$ and $CD$, we get

\[\frac{8}{8 + 17} = \frac{8}{25}.\]