# Solution to the Multiple Semi-Circles IV Puzzle +-- {.image} [[MultipleSemiCirclesIV.png:pic]] > The three smallest semicircles are the same size. The area of the red semicircle is $10$. What’s the total yellow area? =-- ## Solution by [[Pythagoras' Theorem]] and the [[Area of a Circle]] +-- {.image} [[MultipleSemiCirclesIVLabelled.png:pic]] =-- Let $a$, $b$, and $c$ be the radii of the circles, in ascending order. So with the points labelled as above then $O R$ has length $a$, $Q P$ and $Q R$ both have length $b$, and $O P$ has length $c$. As the three smallest semi-circles are the same size, and span across the largest semi-circle, then $c = 3 a$. Let $h$ be the length of $O Q$. Applying [[Pythagoras' theorem]] to triangle $P Q O$ shows that $$ c^2 = b^2 + h^2 $$ Applying it to triangle $Q O R$ shows that $$ b^2 = a^2 + h^2 $$ Eliminating $h^2$ from these gives $c^2 - b^2 = b^2 - a^2$, so $c^2 + a^2 = 2 b^2$. The yellow region has area: $$ \frac{1}{2} \pi c^2 - \frac{1}{2} \pi a^2 + 2 \times \frac{1}{2} \pi a^2 = \frac{1}{2} \pi c^2 + \frac{1}{2} \pi a^2 = \pi b^2 $$ The information given in the puzzle shows that $\frac{1}{2} \pi b^2 = 10$, so the yellow region has area $2 \times 10 = 20$.