# Five Semi-Circles Around a Rectangle +-- {.image} [[FiveSemiCirclesAroundaRectangle.png:pic]] > Three of these semicircles have area $36$. What's the total area of the other two? =-- ## Solution by [[Pythagoras' Theorem]] and [[Area of a Circle]] +-- {.image} [[FiveSemiCirclesAroundaRectangleLabelled.png:pic]] =-- In the above diagram, point $E$ is the midpoint of $A B$ and so is the centre of the largest semi-circle. Point $F$ is the midpoint of $C D$, and point $G$ is the centre of the smallest semi-circle. Let $a$ be the radius of the largest semi-circle, $b$ of the smallest, and $c$ of the three semi-circles with area $36$. Then $\frac{1}{2} \pi c^2 = 36$. As the lengths of $A B$ and $C D$ are the same, $2 a = 2 b + 2 c$. The length of $F D$ is half this, so is $b + c$, and then the length of $F G$ is $c$. The length of $C B$ is $2 c$, and of $G E$ is $a + b$. Applying [[Pythagoras' theorem]] to triangle $E F G$ shows that: $$ (a + b)^2 = (2 c)^2 + c^2 = 5 c^2 $$ Rearranging the identity $2a = 2 b + 2 c$ yields $c = a - b$ and so $c^2 = (a - b)^2$. So combining this with the above gives: $$ 6 c^2 = (a + b)^2 + (a - b)^2 = 2 a^2 + 2 b^2 $$ The total area of the purple semi-circles is therefore: $$ \frac{1}{2} \pi a^2 + \frac{1}{2} \pi b^2 = \frac{1}{2} \pi (a^2 + b^2) = 3 \times \frac{1}{2} \pi c^2 = 3 \times 36 = 108 $$