Notes
arbelos solution

Arbelos

Solution by Angle in a Semi-Circle and Pythagoras' Theorem

In the above diagram, the angle $A \hat{D} C$ is a right-angle since it is the angle in a semi-circle. Therefore, triangle $A D C$ is a right-angled triangle.

Let the lengths of $A B$, $B C$, $A C$, $A D$, and $D C$ be $a$, $b$, $c$, $d$, and $e$ respectively. Applying Pythagoras' theorem to triangles $A D C$, $A B D$, and $B D C$ gives the following identities:

Therefore, $c^2 = a^2 + b^2 + 8$. The area of a semi-circle with diameter $p$ is $\frac{1}{8} \pi p^2$, so the area of the shaded region is:

Solution by Intersecting Chords Theorem

In the above diagram, the original diagram has been reflected in the diameter of the semi-circle. With the distances labelled as before, the intersecting chords theorem shows that $a b = 2 \times 2 = 4$. Since $c = a + b$, the area of the shaded region is: