3. A hexagon $A B C D E F$ has $A B=18 \mathrm{~cm}, B C=8 \mathrm{~cm}, C D=10 \mathrm{~cm}, D E=15 \mathrm{~cm}$, $E F=20 \mathrm{~cm}, F A=1 \mathrm{~cm}, \angle F A B=90^{\circ}, \angle C D E=90^{\circ}$ and $B C$ is parallel to $E F$. Determine the area of this hexagon, in $\mathrm{cm}^{2}$.
Solution: First, we determine the length of $B F$ and $C E$. By the Pythagorean theorem, $B F=\sqrt{A B^{2}+A F^{2}}=\sqrt{18^{2}+1^{2}}=\sqrt{325} \mathrm{~cm}$ and $C E=\sqrt{C D^{2}+D E^{2}}=\sqrt{10^{2}+15^{2}}=$ $\sqrt{325} \mathrm{~cm}$. Furthermore, note that $B F=C E$.

Since $B C$ is parallel to $C F$ and $B F=C E, B C E F$ is an isosceles trapezoid with $C E=B F$.
The hexagon comprises of two right-angled triangles $F A B$ and $C D E$ and the isosceles trapezoid $B C E F$. The area of $F A B$ is $F A \times A B / 2=1 \times 18 / 2=9 \mathrm{~cm}^{2}$ and the area of $C D E$ is $C D \times D E / 2=10 \times 15 / 2=75 \mathrm{~cm}^{2}$. Now, we determine the area of the isosceles trapezoid $B C E F$.

Note that $B C$ is parallel to $E F$ and $B C<E F$ and $B F=C E$. Drop the perpendicular from $B, C$ to side $E F$, touching $E F$ at $X, Y$, respectively. Then $X Y=B C=8 \mathrm{~cm}, F X+$ $X Y+Y E=E F=20 \mathrm{~cm}$ and $F X=Y E$, since $B C E F$ is an isosceles trapezoid. Since $F X+Y E=20-X Y=20-8=12 \mathrm{~cm}, F X=Y E=6 \mathrm{~cm}$. Therefore, by the Pythagorean Theorem, $B X=\sqrt{B F^{2}-F X^{2}}=\sqrt{325-36}=\sqrt{289}=17 \mathrm{~cm}$. Therefore, the height of the trapezoid $B C E F$ is $17 \mathrm{~cm}$. Hence, the area of the trapezoid is $(B C+E F) \times B X / 2=$ $(8+20) \times 17 / 2=14 \times 17=238 \mathrm{~cm}^{2}$.

Summing all three pieces, the area of the hexagon is $9+75+238=322 \mathrm{~cm}^{2}$.