# Solution to the Two Quarter Circles and a Semi-Circle Puzzle +-- {.image} [[TwoQuarterCirclesandaSemiCircle.png:pic]] > Two quarter circles and a semicircle. What’s the shaded area? =-- ## Solution by [[Circle Area]] +-- {.image} [[TwoQuarterCirclesandaSemiCircleLabelled.png:pic]] =-- With the points labelled as above, let $a$ be the length of $A B$, $b$ the length of $B C$ and $c$ half the distance of $D E$. Then $a$, $b$, and $c$ are the radii of the three circles. The diagram gives $a + b = 12$, and $a + 2 c = b$. The shaded area is given by: $$ \frac{1}{4} \pi a^2 + \frac{1}{4} \pi b^2 - \frac{1}{2} c^2 = \frac{1}{4} \pi (a^2 + b^2 - 2 c^2) $$ From $a + 2 c = b$, $4 c^2 = (b - a)^2 = b^2 - 2 a b + a^2$. So: $$ a^2 + b^2 - 2 c^2 = a^2 + b^2 - \frac{1}{2} b^2 + a b - \frac{1}{2} a^2 =\frac{1}{2} a^2 + a b + \frac{1}{2} b^2 = \frac{1}{2} (a + b)^2 $$ So the shaded area is given by: $\frac{1}{8} \pi 12^2 = 18 \pi$. ## Solution by [[Invariance Principle]] There are two special cases,in the first the two quarter circles are the same size and the cut out semi-circle has no area. In the second, the smaller quarter circle has no size. +-- {.image} [[TwoQuarterCirclesandaSemiCircleSpecialA.png:pic]] =-- The area here is $\frac{1}{2} \pi 6^2 = 18\pi$. +-- {.image} [[TwoQuarterCirclesandaSemiCircleSpecialB.png:pic]] =-- The area here is $\frac{1}{4}\pi 12^2 - \frac{1}{2}\pi 6^2 = 18\pi$.