# Arbelos +-- {.image} [[Arbelos.png:pic]] > The red line, of length $2$, is perpendicular to the bases of the three semicircles. What's the total shaded area? =-- ## Solution by [[Angle in a Semi-Circle]] and [[Pythagoras' Theorem]] +-- {.image} [[ArbelosLabelled.png:pic]] =-- 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: $$ \begin{aligned} c^2 &= d^2 + e^2 \\ d^2 &= a^2 + 4 \\ e^2 &= b^2 + 4 \end{aligned} $$ 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: $$ \frac{1}{8} \pi c^2 - \frac{1}{8} \pi a^2 - \frac{1}{8} \pi b^2 = \pi $$ ## Solution by [[Intersecting Chords Theorem]] +-- {.image} [[ArbelosChordsLabelled.png:pic]] =-- 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: $$ \frac{1}{8} \pi (a + b)^2 - \frac{1}{8} \pi a^2 - \frac{1}{8} \pi b^2 = \frac{2}{8} \pi a b = \pi $$