# Solution to the [[Triangle in a Circle in a Quarter Circle in a Square]] Puzzle +-- {.image} [[TriangleinaCircleinaQuarterCircleinaSquare.png:pic]] > The area of the small square is $4$. What’s the area of the triangle? =-- ## Solution by [[Angle Between a Radius and Tangent]], Lengths in an [[Isosceles]] [[Right-Angled Triangle]] +-- {.image} [[TriangleinaCinaQCinaSquareAnnotated.png:pic]] =-- With the points labelled as above, let $r$ be the radius of the inner circle and $x$ the side length of the outer square. Then $B G$ has length $x$, and $B D$ is the [[diagonal]] of the square so has length $x \sqrt{2}$. Therefore, the length of $D G$ is $x (\sqrt{2} - 1)$. This is also the diagonal of the square of area $4$, so: $$ x(\sqrt{2} - 1) = 2 \sqrt{2} $$ and hence $x = 4 + 2 \sqrt(2)$. Since the [[angle between a radius and tangent]] is $90^\circ$, angles $O \hat{F} B$ and $O \hat{E} B$ are $90^\circ$, so $E B F O$ is a [[rectangle]], but since $O F$ and $O E$ are both [[radii]] of the circle they are the same length and so $E B F O$ is a [[square]]. Hence $O B$ has length $r \sqrt{2}$, so $B G$ has length $r(1 + \sqrt{2})$ and thus: $$ r(1 + \sqrt{2}) = x = 4 + 2 \sqrt{2} $$ and hence $r = 2 \sqrt{2}$, and then $O B$ has length $4$. Since $F E$ is also a diagonal of square $E B F O$ (and so also has length $4$), it [[bisects]] $O B$ and so $H O$ has length $2$. Therefore $H G$ has length $2 + 2 \sqrt{2}$, so triangle $E F G$ has area: $$ \frac{1}{2} (2 + 2 \sqrt{2}) \times 4 = 4 + 4 \sqrt{2} $$