# Solution to the Six Circles Inside a Circle Inside a Square Puzzle +-- {.image} [[SixCirclesInsideaCircleInsideaSquare.png:pic]] > The six small circles each have radius $1$. What's the total shaded area? =-- ## Solution by [[Pythagoras' Theorem]] and [[circle|Lengths in a Circle]] +-- {.image} [[SixCirclesInsideaCircleInsideaSquareLabelled.png:pic]] =-- In the above diagram, points $A$ and $B$ are where the small circles touch the large circle. The line $A B$ is a diameter of the larger circle and passes through the centres of the lower left and upper right smaller circles, points $C$ and $D$ respectively. The lengths $A C$ and $D B$ are therefore radii of the small circles, and so have length $1$. The point $E$ is the centre of the lower right circle and so triangle $C E D$ is [[right-angled triangle|right-angled]]. The length of $D E$ is then $2$ and of $C E$ is $4$. Applying [[Pythagoras' theorem]] to this triangle shows that $C D$ has length $\sqrt{20} = 2 \sqrt{5}$ so the length of $A B$ is $2 + 2 \sqrt{5}$. The width of the square is $2 + 2 \sqrt{5}$ and the height of the shaded areas is $2 + 2 \sqrt{5} - 4 = 2 \sqrt{5} - 2$. The shaded area is therefore: $$ (2 + 2 \sqrt{5})(2\sqrt{5} - 2) = 20 - 4 = 16. $$