# Solution to the [[Two Quarter Circles in a Square]] Puzzle +-- {.image} [[TwoQuarterCirclesinaSquare.png:pic]] > Two quarter circles inside a square. What fraction is shaded? =-- ## Solution by [[Area of a Circle]] and [[Area of a Triangle]] +-- {.image} [[TwoQuarterCirclesinaSquareAnnotated.png:pic]] =-- As this is a question about what fraction of the square is shaded, let us set the side length of the square to $1$. With the points labelled as above, the sector $A F B$ is an eighth of a circle with centre $B$ and radius $A B$, so its area is $\frac{\pi}{8}$. As $F$ is on the quarter circle with centre $B$, the line segment $B F$ is the same length as $A B$. Triangle $B F E$ is an [[isosceles]] [[right-angled triangle]] and so the length of $B E$ is $\frac{1}{\sqrt{2}}$. The area of the quarter circle $B E F$ is therefore: $$ \frac{1}{4} \times \pi \times \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{\pi}{8} $$ The area of triangle $B E F$ is then $\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{4}$ so the [[segment]] cut out from the smaller quarter circle by the [[chord]] $B F$ has area $\frac{\pi}{8} - \frac{1}{4}$. The shaded area is therefore: $$ \frac{\pi}{8} - \left( \frac{\pi}{9} - \frac{1}{4} \right) = \frac{1}{4} $$ And so the shaded area is one quarter of the square.