# Solution to the Circle in Triangle in Square Puzzle +-- {.image} [[CircleinTriangleinSquare.png:pic]] > A unit circle sits in an equilateral triangle inside a square. What's the total shaded area? =-- ## Solution by Properties of [[Equilateral Triangles]] +-- {.image} [[CircleinTriangleinSquareLabelled.png:pic]] =-- From [[lengths in an equilateral triangle]], the length of $O B$ is one third of $C B$, so $C B$ has length $3$. Then the length of $C B$ is $\sqrt{3}$ times the length of $B D$, so $B D$ has length $\sqrt{3}$. Triangle $A B D$ is [[isosceles]] and [[right-angled triangle|right-angled]], so $A B$ has the same length as $B D$. The length of the diagonal of the square is then $\sqrt{3} + 3$. The area of the square is $\frac{1}{2}(\sqrt{3} + 3)^2 = 6 + 3 \sqrt{3}$. Since the [[area of an equilateral triangle]] is $\frac{\sqrt{3}}{4}$ times the square of the length of one of its sides, the triangle has area $\frac{\sqrt{3}}{4} \times (2 \sqrt{3})^2 = 3\sqrt{3}$. The shaded area is therefore $6$.