# Solution to the Triangle Overlapping a Hexagon Puzzle +-- {.image} [[TriangleOverlappingaHexagon.png:pic]] > If the area of the regular hexagon is $24$, what's the area of the equilateral triangle? =-- ## Solution by [[Pythagoras' Theorem]] and Properties of [[Regular Hexagons]] and [[Equilateral Triangles]] +-- {.image} [[TriangleOverlappingaHexagonLabelled.png:pic]] =-- Let $x$ be the length of one of the sides of the hexagon and let $y$ be the length of one of the sides of the triangle. Then using [[lengths in a regular hexagon]], the length of $E C$ in the diagram is $\sqrt{3} x$. Applying [[Pythagoras' theorem]] to triangle $E B C$ shows that: $$ y^2 = \left(\frac{1}{2} x\right)^2 + \left( \sqrt{3} x\right)^2 = \frac{13}{4} x^2 $$ From [[regular hexagon]], the area of the hexagon is $\frac{3\sqrt{3}}{2} x^2$ and from [[equilateral triangle]], the area of the triangle is $\frac{\sqrt{3}}{4} y^2$. So $x^2 = \frac{16}{\sqrt{3}}$ and $$ \frac{\sqrt{3}}{4} y^2 = \frac{13\sqrt{3}}{16} x^2 = 13 $$ Hence the area of the equilateral triangle is $13$.