B3. Regular decagon (10-sided polygon) $A B C D E F G H I J$ has area 2017 square units. Determine the area (in square units) of the rectangle $C D H I$.
Correct answer: 806.8 square units.
Solution 1: If $O$ is the centre of the decagon and we cut the decagon into equal isosceles triangles, as shown, we can see that the area of the decagon is 10 times the area of one triangle. Since diagonals $C H$ and $D I$ are equal in length and bisect each other, quadrilateral $C D H I$ is a rectangle, with the same base as each of the triangles, but twice the height. Thus the area of $C D H I$ is 4 times the area of each of the triangles and hence $40 \%$ of the area of the decagon or $0.4 \times 2017=806.8$ square units.

The area of the rectangle is 806.8 square units.
Solution 2: Let $s=|A B|$ be the side length of the regular decagon. Since it is a regular polygon with 10 sides, all its angles are equal, in particular $\angle A B C=\angle J A B=180 \times(10-2) / 10=144^{\circ}$. Let $M$ be the midpoint of $C I$. Draw $A M$, which will be perpendicular to $C I$. Place point $P$ on $A M$ and point $Q$ on $C M$ such that $B P$ is perpendicular to $A M$ and $B Q$ is perpendicular to $C M$.

The area of $A B C M$ could be found as a sum of areas of two right triangles $B P A$ and $B Q C$ and the rectangle $B P M Q$. Note that angle $B A P$ is $144 / 2=72^{\circ}$ and so angle $A B P$ is $90-72=18^{\circ}$. Now, angle $C B Q$ is $144-18-90=36^{\circ}$, and so angle $B C Q$ is $90-36=54^{\circ}$.
The area of the right triangle $B P A$ is $\frac{1}{2} s^{2} \sin 18^{\circ} \cos 18^{\circ}=\frac{1}{4} s^{2} \sin 36^{\circ}$.
The area of the right triangle $B Q C$ is $\frac{1}{2} s^{2} \sin 54^{\circ} \cos 54^{\circ}=\frac{1}{4} s^{2} \sin 108^{\circ}=\frac{1}{4} s^{2} \cos 18^{\circ}$.
The area of the rectangle $B P M Q$ is $s^{2} \sin 54^{\circ} \cos 18^{\circ}=\frac{1}{2} s^{2}\left(\sin 36^{\circ}+\sin 72^{\circ}\right)=\frac{1}{2} s^{2}\left(\sin 36^{\circ}+\cos 18^{\circ}\right)$. Therefore, the area of $A B C M$ is $\frac{3}{4} s^{2}\left(\sin 36^{\circ}+\cos 18^{\circ}\right) \equiv \Delta$.
Next, the area of $C D H I$ is $2 s^{2}\left(\sin 36^{\circ}+\cos 18^{\circ}\right)=\frac{8}{3} \Delta$.
The total area of the decagon is $4 \Delta+\frac{8}{3} \Delta=\frac{20}{3} \Delta=2017$.
Thus, the area of $C D H I$ is $\frac{8}{20} \times 2017=806.8$.