[Geometry]
Problem: Hexagon $A B C D E F$ has vertices $A(0,0), B(4,0), C(7,2), D(7,5), E(3,5)$, $F(0,3)$. What is the area of hexagon $A B C D E F$ ?

Solution: Let $P$ be the point with coordinates $(7,0)$ and let $Q$ be the point with coordinates $(0,5)$.

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Then $A P D Q$ is a rectangle with width 7 and height 5 , and so it has area $7 \cdot 5=35$.

Hexagon $A B C D E F$ is formed by removing two triangles from rectangle $A P D Q$, namely $\triangle B P C$ and $\triangle E Q F$.

Each of $\triangle B P C$ and $\triangle E Q F$ is right-angled, because each shares an angle with rectangle $A P D Q$.

Each of $\triangle B P C$ and $\triangle E Q F$ has a base of length 3 and a height of 2.

Thus, their combined area is $2 \cdot \frac{1}{2} \cdot 3 \cdot 2=6$.

This means that the area of hexagon $A B C D E F$ is $35-6=29$.

Final answer: 29