# Solution to the Multiple Semi-Circles Puzzle +-- {.image} [[MultipleSemiCircles.png:pic]] > The three coloured regions have the same area. Altogether, what fraction of the shape do they cover? =-- ## Solution by [[Angle in a Semi-circle]] +-- {.image} [[MultipleSemiCirclesLabelled.png:pic]] =-- In the above diagram, let $a$ be the length of $A B$, $b$ of $B C$, and $c$ of $C D$. From the formula for the [[area of a circle]], the area of a semi-circle can be found from its [[diameter]] as $\frac{1}{2} \pi \left( \frac{1}{2} d\right)^2 = \frac{1}{8} \pi d^2$. So the area of the purple semi-circle is $\frac{1}{8} \pi a^2$. The second largest semi-circle has area $\frac{1}{8} \pi (a + b)^2$, so the cyan region has area: $$ \frac{1}{8} \pi (a + b)^2 - \frac{1}{8} \pi a^2 - \frac{1}{8} \pi b^2 = \frac{1}{4} \pi a b $$ Since the purple and cyan regions have the same area, $\frac{1}{8} \pi a^2 = \frac{1}{4} \pi a b$ so $a = 2 b$. The diameter of the full semi-circle is $a + b + c = 3 b + c$. So the area of the yellow region is: $$ \frac{1}{8} \pi (3 b + c)^2 - \frac{1}{8} \pi (3 b)^2 - \frac{1}{8} c^2 = \frac{3}{4} \pi b c $$ This has the same area as the cyan region, so $\frac{3}{4} \pi b c = \frac{1}{4} \pi a b$ which means that $3 c = a$. The diameter of the full semi-circle is therefore $a + \frac{1}{2} a + \frac{1}{3} a = \frac{11}{3} a$ so its area is $\frac{1}{8} \pi \left( \frac{11}{3} a \right)^2 = \frac{121}{72} \pi a^2$. The unshaded regions have diameters $\frac{1}{2} a$ and $\frac{1}{3} a$ so have areas $\frac{1}{8} \pi \left( \frac{1}{2} a \right)^2 + \frac{1}{8} \pi \left( \frac{1}{3} a \right)^2 = \frac{13}{288} \pi a^2$. The fraction that is shaded is therefore: $$ \frac{ \frac{275}{288} \pi a^2 }{ \frac{121}{72} \pi a^2 } = \frac{275}{484} = \frac{25}{44} $$