# Solution to the [[Two Squares and Two Equilateral Triangles II]] Puzzle +-- {.image} [[TwoSquaresandTwoEquilateralTrianglesII.png:pic]] > Two squares and two equilateral triangles. If the area of the smaller square is $8$, what’s the shaded area? =-- ## Solution by Properties of [[Equilateral Triangles]] and [[Pythagoras' Theorem]] +-- {.image} [[TwoSquaresandTwoEquilateralTrianglesIILabelled.png:pic]] =-- In the diagram above, the point $D$ is such that $C D$ is [[perpendicular]] to the base of the square. As the square has area $8$, its side length is $\sqrt{8} = 2\sqrt{2}$, so the length of line segment $A D$ is $\sqrt{2}$. From the lengths in an [[equilateral triangle]], line segment $C D$ has length $\frac{\sqrt{3}}{2} \times 2 \sqrt{2} + 2 \sqrt{2} = \sqrt{6} + 2 \sqrt{2}$. Applying [[Pythagoras' theorem]] to triangle $A C D$ shows that the length of $A C$ is: $$ \sqrt{ (\sqrt{6} + 2 \sqrt{2})^2 + (\sqrt{2})^2 } = \sqrt{ 6 + 8 \sqrt{3} + 8 + 2 } = \sqrt{16 + 8 \sqrt{3}} $$ The area of the large square is therefore $16 + 8 \sqrt{3}$, and the area of the large [[equilateral triangle]] is $\frac{\sqrt{3}}{4}$ times this, so the area of the pink region is: $$ \left(1 - \frac{\sqrt{3}}{4}\right)(16 + 8 \sqrt{3}) = 16 + 8 \sqrt{3} - 4\sqrt{3} - 6 = 10 + 4 \sqrt{3} $$