# Solution to the Multiple Semi-Circles III Puzzle +-- {.image} [[MultipleSemiCirclesIII.png:pic]] > The red diameter is $6$. What’s the total shaded area? =-- ## Solution by [[Circle Areas]] and [[Similar Triangles]] +-- {.image} [[MultipleSemiCirclesIIILabelled.png:pic]] =-- With the edges labelled as above, the largest semi-circle has area $\frac{1}{8} \pi (a + b)^2$, and the other two have areas $\frac{1}{8} \pi a^2$ and $\frac{1}{8} \pi b^2$. The shaded area therefore has area: $$ \begin{aligned} \frac{1}{8} \pi (a + b)^2 + \frac{1}{8} \pi a^2 - \frac{1}{8} \pi b^2 &= \frac{1}{8} \pi ( a^2 + 2 a b + b^2 + a^2 - b^2 ) \\ &= \frac{1}{8} \pi ( 2 a^2 + 2 a b) \\ &= \frac{1}{4} \pi a (a + b) \end{aligned} $$ The triangles $A B C$ and $D B A$ are similar, with $A B$ corresponding to $D B$ and $B C$ to $B A$, so $ A B : B C = D B : B A$. Hence $ \frac{6}{a} = \frac{a + b}{6}$. Rearranging this yields $a(a + b) = 36$ so the shaded area is $9 \pi$. The use of similar triangles can be replaced by [[Pythagoras' Theorem]]. Writing $c$ for the length of $A C$ and $d$ for the length of $A D$, the various [[right-angled triangles]] imply the following identities: $$ \begin{aligned} 6^2 &= a^2 + c^2 \\ (a + b)^2 &= 6^2 + d^2 \\ d^2 &= c^2 + b^2 \end{aligned} $$ Putting this together, $$ (a + b)^2 = 6^2 + d^2 = 6^2 + c^2 + b^2 = 2 \times 6^2 - a^2 + b^2 $$ so $$ (a + b)^2 + a^2 - b^2 = 2 \times 6^2 $$ Substituting this into the expression for the shaded area gives that that area is $9 \pi$ as before. ## Solution by [[Invariance Principle]] The point $C$ can slide on the red semi-circle. As it approaches $B$ then the diagram gets very large, but putting it at $A$ produces a diagram that makes the solution obvious. +-- {.image} [[MultipleSemiCirclesIIISpecial.png:pic]] =-- The radius of the circle is $3$ so its area is $9 \pi$.