Question

An equiangular octagon has four sides of length $1$ and four sides of length $\frac{\sqrt{2}}{2}$, arranged so that no two consecutive sides have the same length. What is the area of the octagon

Answer 1

If the corner cut has length 1, then each corner triangle has legs of length $\dfrac1{\sqrt2}$, and hence area $\dfrac12\cdot\dfrac1{\sqrt2}\cdot\dfrac1{\sqrt2}=\dfrac14$. The original square then has side lengths $\dfrac1{\sqrt2}+\dfrac1{\sqrt2}+\dfrac1{\sqrt2}=\dfrac3{\sqrt2}$, so its total area is $\left(\dfrac3{\sqrt2}\right)^2=\dfrac92$. Then subtract the triangles' areas from the square's area to get the octagon's area, $\dfrac92-4\cdot\dfrac14=\dfrac92-1=\boxed{\dfrac72}$.