# Solution to the [[Two Squares and Two Equilateral Triangles III]] Puzzle +-- {.image} [[TwoSquaresandTwoEquilateralTrianglesIII.png:pic]] > If the small square has area $2$, what’s the shaded area? =-- Note that as specified, the problem does not have a unique answer. The diagram implies that the right-hand vertices of the square and triangle are vertically in line and with that assumption then the problem is solvable. ## Solution by Properties of [[Equilateral Triangles]] and [[Pythagoras' Theorem]] +-- {.image} [[TwoSquaresandTwoEqTrianglesIIILabelled.png:pic]] =-- As the small square as area $2$, its side length is $\sqrt{2}$. This is half the side length of the [[equilateral triangle]], so its height is $\sqrt{3} \times \sqrt{2} = \sqrt{6}$. Therefore line segment $A B$ has length $2 \sqrt{2}$ and $B E$ has length $\sqrt{6} + \sqrt{2}$. Applying [[Pythagoras' theorem]] to triangle $A B E$ shows that the length of $A E$ is given by: $$ \sqrt{ (\sqrt{6} + \sqrt{2})^2 + (2\sqrt{2})^2 } = \sqrt{6 + 4\sqrt{3} + 2 + 8} = \sqrt{16 + 4 \sqrt{3}} $$ The large square therefore has area $16 + 4 \sqrt{3}$. The large equilateral triangle has area $\frac{\sqrt{3}}{4}$ times this, so the yellow region has area: $$ \left(1 - \frac{\sqrt{3}}{4}\right)\left(16 + 4\sqrt{3}\right) = 16 + 4 \sqrt{3} - 4 \sqrt{3} - 3 = 13 $$