# Solution to the Two Quarter Circles in a Rectangle in a Semi-Circle Puzzle +-- {.image} [[TwoQuarterCirclesinaRectangleinaSemiCircle.png:pic]] > Two quarter circles in a semicircle. What fraction is shaded? =-- ## Solution by [[Pythagoras' theorem]] and [[Area Scale Factor]] +-- {.image} [[TwoQuarterCirclesinaRectinaSemiCircleLabelled.png:pic]] =-- In the above diagram, point $O$ is the centre of the larger semi-circle. The line segment $D B$ passes through the point where the two quarter circles touch so its length is twice the radius of the quarter circles. Line segment $A D$ is a radius, so triangle $D A B$ is a [[right-angled triangle]] with hypotenuse twice the length of one of its sides. [[Pythagoras' theorem]] then shows that $A B$ has length $\sqrt{3}$ times the length of $A D$. Then $O A$ is half this, so applying [[Pythagoras' theorem]] to triangle $O A D$ shows that $O D$ is $\frac{\sqrt{7}}{2}$ times the length of $A D$. Since [[area]] scales as the square of the length scale factor, the outer semi-circle is then $\frac{7}{4}$ the area of the two quarter circles, so the shaded region is $\frac{4}{7}$ths of the total region.