# Multiple Semi-Circles II +-- {.image} [[MultipleSemiCirclesII.png:pic]] > Is more of this design red or yellow? =-- ## Solution by [[Pythagoras' Theorem]] and the [[Area of a Circle]] Consider first this diagram where all the arcs are [[semi-circles]] with parallel diameters and centres at $O$, $B$, and $C$. +-- {:.image} [[SemiCircularLune.png:pic]] =-- Since the diameters are parallel, $B$ is the [[midpoint]] of the [[chord]] and so triangle $O B A$ is [[right-angled triangle|right-angled]]. Therefore, [[Pythagoras' theorem]] applies to triangle $A B O$. Writing the radii of the circles as $a$, $b$, $c$ with $a$ the length of $A B$, $b$ of $B O$, and $c$ of $O A$ then this means that: $$ c^2 = a^2 + b^2 $$ The area of a [[semi-circle]] is $\frac{1}{2}\pi$ times the square of its radius, so from the above then: $$ \frac{1}{2} \pi c^2 = \frac{1}{2} \pi a^2 + \frac{1}{2} \pi b^2 $$ which uses the fact that $C D$ and $B O$ have the same length. This means that the area of the largest semi-circle is the sum of the areas of the two smaller ones. In terms of the coloured regions, this means that: $$ \text{green} + \text{orange} + \text{blue} = \text{blue} + \text{purple} + \text{green} $$ And hence the purple region (which is known as a semi-circular [[lune]]) has the same area as the orange. This applies to the problem as follows. +-- {.image} [[MultipleSemiCirclesIILabelled.png:pic]] =-- Using the above calculation of the area of a semi-circular lune, the area of region $A$ is equal to the area of the union of regions $B$ and $C$. Similarly, the area of region $D$ is equal to the area of the union of regions $E$, $B$, and $C$, $F$. Therefore, the yellow and red regions have the same area.