# Circle in a Quarter Circle +-- {.image} [[CircleinaQuarterCircle.png:pic]] > What's the total shaded area? =-- ## Solution by [[Pythagoras' Theorem]] and [[Area of a Circle]] +-- {.image} [[CircleinaQuarterCircleLabelled.png:pic]] =-- Let $R$ be the radius of the quarter circle, which is the length of $O B$, and let $r$ be the radius of the smaller circle, so the length of $C D$ is $2 r$. The lengths of $C D$ and $A O$ are the same, so applying [[Pythagoras' theorem]] to [[right-angled triangle]] $O A B$ shows that: $$ R^2 = (2 r)^2 + 12^2 = 4 r^2 + 12^2 $$ The area of the quarter circle is $\frac{1}{4} \pi R^2$ and the area of the inner circle is $\pi r^2$, so the area of the shaded region is: $$ \frac{1}{4} \pi R^2 - \pi r^2 = \frac{1}{4} \pi (4 r^2 + 12^2) - \pi r^2 = 36 \pi $$ So the area of the shaded region is $36 \pi$.