# Solution to the Semi-Circle in a Circle Puzzle +-- {.image} [[SemiCircleinaCircle.png:pic]] > What fraction of the circle is shaded? =-- ## Solution by [[Intersecting Chords Theorem]] +-- {.image} [[SemiCircleinaCircleLabelled.png:pic]] =-- In the diagram above, let $r$ be the radius of the inner semi-circle. So $r$ is the length of each of $E A$, $E B$, and $E F$. The length of $D E$ is then $6 + r$ so applying the [[intersecting chords theorem]] to $A B$ and $C D$ results in: $$ r^2 = 1 \times (6 + r) = 6 + r $$ This has solution $r = 3$ (the other possible solution is $r = -2$ but as $r$ is a length this isn't possible). The diameter of the outer circle is $7 + r = 10$ so its area is $25 \pi$. The area of the inner semi-circle is $\frac{9}{2} \pi$. Therefore $\frac{41}{50}$ths of the circle is shaded.