# Solution to the Two Circles and Two Semi-Circles Puzzle +-- {.image} [[TwoCirclesandTwoSemiCircles.png:pic]] > The yellow area is $16$. What’s the total black area? =-- ## Solution by [[Intersecting Chords Theorem]] or [[Pythagoras' Theorem]] +-- {.image} [[TwoCirclesandTwoSemiCirclesLabelled.png:pic]] =-- As in the above diagram, let $a$, $b$, and $c$ be the radii of the three sizes of circle. The relationship between them is given by the [[intersecting chords theorem]]. To apply this, the largest semi-circle needs to be completed to a full circle. The cross chord splits into two segments of length $b$, while the extension of the radius of the largest circle splits into $c - 2 a$ and $c + 2 a$. The intersecting chords theorem then shows that: $$ b^2 = (c - 2 a)(c + 2 a) = c^2 - 4 a^2 $$ This can also be established using [[Pythagoras' theorem]] applied to a right-angled triangle with sides $2a$, $b$, and $c$. The yellow and white regions combined form the middle sized semi-circle and two of the smallest circles, so this has area: $$ \frac{1}{2} \pi b^2 + 2 \pi a^2 = \frac{1}{2} \pi (b^2 + 4 a^2) $$ The black and white regions combined form the largest size semi-circle and so has area $\frac{1}{2} \pi c^2$. Since $c^2 = b^2 + 4 a^2$, these two regions have the same area. Therefore the yellow and black regions have the same area, which is $16$.