# Solution to the [[Three Quarter Circles II]] Puzzle +-- {.image} [[ThreeQuarterCirclesII.jpeg:pic]] > Three quarter circles. What’s the total shaded area? =-- ## Solution by [[Pythagoras' Theorem]] and Lengths in an [[Isosceles]] [[Right-angled Triangle]] +-- {.image} [[ThreeQuarterCirclesIIAnnotated.jpeg:pic]] =-- Let the points be labelled as above, where $D$ is the point on $C E$ such that $F D$ is [[perpendicular]] to $C E$ and $B$ on $A C$ likewise. Let $F D$ have length $a$ and $F B$ have length $b$. As $A C E$ is a [[quarter-circle]], angle $A \hat{C} E$ is a [[right angle]], and by construction so are $F \hat{D} C$ and $C \hat{B} F$, so $B C D F$ is a [[rectangle]]. Therefore $C D$ also has length $b$. Applying [[Pythagoras' theorem]] to triangle $F D C$ we have: $$ a^2 + b^2 = 8^2 = 64 $$ As $A C E$ is a quarter circle, angle $A \hat{E} C$ is $45^\circ$, as is angle $C \hat{A} E$. Therefore, triangles $F D E$ and $A B F$ are [[isosceles]] [[right-angled triangles]]. Hence $E F$ has length $\sqrt{2} a$ and $A F$ has length $\sqrt{2} b$. By the area of a [[circle]], the total shaded area is therefore: $$ \frac{1}{4} \left( \pi (\sqrt{2} a)^2 + \pi (\sqrt{2}b)^2 \right) = \frac{2\pi}{4} (a^2 + b^2) = \frac{\pi}{2} times 64 = 32 \pi $$ ## Solution by [[Invariance Principle]] The two shaded quarter circles can be drawn any size relative to each other. There are two special configurations. +-- {.image} [[ThreeQuarterCirclesIIInvarianceA.png:pic]] =-- In this configuration, the quarter circles are the same size making a [[semi-circle]]. The radius of this semi-circle is $8$, so the area is: $$ \frac{1}{2} \pi 8^2 = 32 \pi $$ +-- {.image} [[ThreeQuarterCirclesIIInvarianceB.png:pic]] =-- In this version, one quarter circle is shrunk to a point. The radius of the remaining quarter circle is $8 \sqrt{2}$, making its area: $$ \frac{1}{4} \pi (8 \sqrt{2})^2 = 32 \pi $$