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
`\frac1{\sqrt2}`, and hence area
`\frac12\cdot\frac1{\sqrt2}\cdot\frac1{\sqrt2}=\frac14`. The original square then has side lengths
`\frac1{\sqrt2}+\frac1{\sqrt2}+\frac1{\sqrt2}=\frac3{\sqrt2}`, so its total area is
`\left(\frac3{\sqrt2}\right)^2=\frac92`. Then subtract the triangles' areas from the square's area to get the octagon's area,
`\frac92-4\cdot\frac14=\frac92-1=\boxed{\frac72}`.