# Solution to the [[Two Quarter Circles and a Circle]] Puzzle +-- {.image} [[TwoQuarterCirclesandaCircle.jpeg:pic]] > The yellow circle has radius 4. What’s the total area of the two quarter circles? =-- ## Solution by [[Pythagoras' Theorem]] and Properties of an [[Isosceles]] [[Right-Angled Triangle]] +-- {.image} [[TwoQuarterCirclesandaCircleLabelled.jpeg:pic]] =-- In the diagram above, $O$ is the centre of the circle. Angle $A \hat{F} E$ is $45^\circ$, since triangle $A F C$ is an [[isosceles]] [[right-angled triangle]]. Therefore, as the [[angle at the centre is twice the angle at the circumference]], angle $A \hat{O} E$ is $90^\circ$. This means that triangle $A O E$ is also an [[isosceles]] [[right-angled triangle]], so the length of $A E$ is $\sqrt{2}$ times the length of $O A$, so is $4 \sqrt{2}$. Let the radii of the circles be $R$ and $r$, then applying [[Pythagoras' theorem]] to triangle $A C E$ shows that: $$ R^2 + r^2 = (4 \sqrt{2})^2 = 32 $$ The total area of the two quarter circles is therefore: $$ \frac{1}{4} \pi R^2 + \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (R^2 + r^2) = \frac{1}{4} \pi \times 32 = 8 \pi $$ ## Solution by [[Invariance Principle]] The sizes of the two quarter circles can vary. +-- {.image} [[TwoQuarterCirclesandaCircleInvarianceA.jpeg:pic]] =-- At one extreme, the smaller circle has shrunk down to zero size. In this case, $A F$ is a diameter of the circle so has length $8$. As triangle $A B F$ is [[isosceles]] and [[right-angled triangle|right-angled]], $F B$ has length $\frac{8}{\sqrt{2}} = 4 \sqrt{2}$ so the quarter circle has area: $$ \frac{1}{4} \pi (4 \sqrt{2} )^2 = 8 \pi $$ +-- {.image} [[TwoQuarterCirclesandaCircleInvarianceB.jpeg:pic]] =-- In this version, the two quarter circles have the same size. Then $A C F O$ is a square, so the quarter circles have the same radius as the circle, so have area: $$ \frac{1}{2} \pi 4^2 = 8 \pi $$