# Solution to the Three Circles and Two Semi-Circles in a Circle Puzzle +-- {.image} [[ThreeCirclesandTwoSemiCirclesinaCircle.png:pic]] > The three small circles are the same size. What fraction of the large circle is shaded? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[ThreeCirclesandTwoSemiCirclesinaCircleLabelled.png:pic]] =-- As in the above diagram, let $b$ be the radius of the smallest circle, $a$ of the middle, and $r$ of the outer. Then by considering the horizontal radius, $r = a + 2 b$. Applying [[Pythagoras' Theorem]] to the blue triangle yields $r^2 = a^2 + (a + b)^2$. Putting these together gives: $$ \begin{aligned} (a + 2 b)^2 &= a^2 + (a + b)^2 \\ a^2 + 4 a b + 4 b^2 &= 2 a^2 + 2 a b + b^2 \\ 0 &= a^2 - 2 a b - 3 b^2 \\ 0 &= (a + b)(a - 3 b) \end{aligned} $$ So either $a = - b$ or $a = 3 b$. As $a$ and $b$ are lengths, neither can be negative and so it must be the case that $a = 3b$. Then $r = 5 b$. The area of the large circle is $\pi r^2$ and the shaded area is $\pi a^2 + 3 \pi b^2$. So the area of the large circle is $25 \pi b^2$ and of the shaded area is $12 \pi b^2$. The fraction that is shaded is therefore $\frac{12}{25}$.