# Solution to the [[Four Quarter Circles II]] Puzzle +-- {.image} [[FourQuarterCirclesII.jpg:pic]] > Four quarter circles. What fraction is shaded? =-- ## Solution by [[Pythagoras' Theorem]] and [[Area of a Circle]] +-- {.image} [[FourQuarterCirclesIILabelled.jpg:pic]] =-- With the points labelled as above, $O$ is the centre of the semi-circle and left-hand quarter circle, while $A$ is the centre of the right-hand quarter circle. Let $a$, $b$, and $c$ be the lengths of their respective radii, so $O B$ has length $a$, $O A$ has length $b$, and $A B$ has length $c$. Using the formula for the [[area of a circle]], the area of the shaded region is: \[ \label{shaded} \frac{1}{2} \pi a^2 - \frac{1}{4} \pi b^2 - \frac{1}{4} \pi c^2 = \frac{1}{4} \pi (2 a^2 - b^2 - c^2) \] Triangle $O A B$ is [[right-angled triangle|right-angled]] so [[Pythagoras' theorem]] applies, meaning that $a^2 = b^2 + c^2$. Substituting into Equation \eqref{shaded} shows that the shaded region has area: $$ \frac{1}{4} \pi (2 a^2 - b^2 - c^2) = \frac{1}{4} \pi a^2 $$ and this is half of the area of the full semi-circle. ## Solution by [[Invariance Principle]] +-- {.image} [[FourQuarterCirclesIIInvariance.jpg:pic]] =-- The inner quarter circles can be varied, and the configuration above shows that the shaded region comprises half of the overall diagram.