Notes
subdivided decagon solution

Solution to the Subdivided Decagon Puzzle

Solution by Area of a Triangle

In the above diagram, point $O$ is the centre of the decagon. The diameter $C D$ is parallel to the edge $A B$ which means that the “height” of any point on $C D$ above the edge $A B$ is fixed (that is to say that the perpendicular distance from a point on $C D$ to the line through $A B$ is the same as from any other point on $C D$). In particular, the height of the triangle $A B C$ above $A B$ is the same as that of $A B O$ above $A B$, and so they have the same area. The area of the triangle $A B O$ is a tenth of that of the decagon, and so the shaded area is $\frac{8}{10} = \frac{4}{5}$ths of the area of the decagon.