[[!redirects a semi-circle and a quarter circle in a rectangle solutions]] # Solution to the [[A Semi-Circle and a Quarter Circle in a Rectangle]] Puzzle +-- {.image} [[ASemiCircleandaQuarterCircleinaRectangle.jpeg:pic]] > What’s the shaded area? =-- ## Solution by [[Pythagoras' Theorem]] and the [[Area of a Circle]] +-- {.image} [[ASCircleandaQCircleinaRectangleLabelled.jpeg:pic]] =-- Label the points as above, where $B$ is the centre of the semi-circle and $F$ is the point where the circles meet. Then $D F B$ is a straight line. Let $r$ be the radius of the semi-circle, so the vertical side of the rectangle has length $2r$, and this is then the radius of the quarter circle. In terms of $r$, the shaded area is: $$ \frac{1}{2} \pi r^2 + \frac{1}{4} \pi (2 r)^2 = \frac{3}{2} \pi r^2 $$ Triangle $B C D$ is [[right-angled triangle|right-angled]], so it satisfies [[Pythagoras' theorem]]. Its side lengths are $4$, $r$, and $3r$, so: $$ 4^2 + r^2 = (3 r)^2 = 9 r^2 $$ so $8 r^2 = 16$ and hence $r^2 = 2$. The shaded area is therefore $3 \pi$.