# Five Overlapping Circles +-- {.image} [[FiveOverlappingCircles.png:pic]] > What's the area of the smallest circle? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[FiveOverlappingCirclesLabelled.png:pic]] =-- Let $a$ be the radius of the cyan circle and $b$ of the dark blue. Then the length of $A B$ is $2 a$, of $B C$ is $2 a - b$, and of $A C$ is $2 a + b$. Applying [[Pythagoras' theorem]] to triangle $A B C$ shows that: $$ (2 a + b)^2 = (2 a - b)^2 + (2 a)^2 = 8 a^2 - 4 a b + b^2 $$ This simplifies to $8 a b = 4 a^2$, so $a = 2 b$. This means that the area of the small circle is a quarter of the area of the cyan circle, so has area $9$.