# Solution to the Overlapping Rectangle and Semi-Circle Puzzle +-- {.image} [[OverlappingRectangleandSemiCircle.png:pic]] > What's the area of the rectangle? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[OverlappingRectangleandSemiCircleLabelled.png:pic]] =-- In the above diagram, the point labelled $O$ is the centre of the semi-circle. Let $r$ be the radius of the semi-circle, let $a$ be the length of $O D$ and $b$ of $C D$. The length of $A D$ is is $r + a$ so the area of the rectangle is $r(r + a) = r^2 + a r$. Applying [[Pythagoras' theorem]] to triangle $O D C$ gives the relationship: $$ r^2 = a^2 + b^2 $$ Applying it to triangle $C D A$ gives: $$ 6^2 = (a + r)^2 + b^2 = a^2 + 2 a r + r^2 + b^2 = 2 a r + 2 r^2 $$ Therefore the area of the rectangle is $18$.