# Solution to the One Rectangle Inside Another With Semi-Circles Puzzle +-- {.image} [[OneRectangleInsideAnotherWithSemiCircles.png:pic]] > What fraction of the outer rectangle is shaded? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[OneRectangleInAnotherWithSemiCirclesLabelled.png:pic]] =-- With the points as labelled above, let $x$ be the length of $B C$ and $y$ of $C F$. Let $r$ be the radius of the smaller semi-circles, then $2 r$ is the radius of the larger ones. Triangle $A G F$ is [[right-angled triangle|right-angled]] with lengths $x$, $y - r$, and $r$. Applying [[Pythagoras' theorem]] gives the identity: $$ r^2 = x^2 + (y - r)^2 = x^2 + y^2 - 2 y r + r^2 $$ So $x^2 + y^2 - 2 y r = 0$. Triangle $F C D$ is also right-angled with lengths $y$, $2 r - x$, $2 r$. Applying [[Pythagoras' theorem]] gives the identity: $$ (2 r)^2 = y^2 + (2 r - x)^2 = y^2 + (2 r)^2 - 4 r x + x^2 $$ So $x^2 + y^2 - 4 r x = 0$. Putting these together shows that $y = 2 x$. Substituting back into the second then shows that $5 x^2 = 4 r x$ so $x = \frac{4}{5} r$ and then $y = \frac{8}{5} r$. The dimensions of the yellow rectangle can therefore be computed as $4 r - 2 x = \frac{12}{5} r$ and $2 y - 2 r = \frac{6}{5} r$. The area of the yellow rectangle is then $\frac{72}{25} r^2$ while the area of the outer rectangle is $2r \times 4 r = 8 r^2$. Therefore the fraction that is shaded is $\frac{9}{25}$.