Octa - Gone!

Relevant wiki: Regular Polygons - Area

If the side length of the octagon is $s$ and the apothem(perpendicular distance from the centre to the side of the octagon) is $a$ , the area of the rectangle $ABEF$ will be $2as$ .

Now, we know that the area of a regular polygon of $n$ sides is $\dfrac{ n a s }{ 2 }.$

And the area of the octagon is given as​ 1.

$\therefore 1=\dfrac{8 as}{2} \\ \Rightarrow 2as=\dfrac 12.$

Hence the area of the rectangle $ABEF$ is $\dfrac 12. \square$

The octagon is formed by 8 isosceles triangles of the form OAB and OAH.

Consider triangle OAF. Since $AOE \parallel HF$ , by considering base $AO$ and point the point $F$ in a parallel manner, we see that $[OAF] = [ OAH]$ (Alternatively, these triangles have the same base $OA$ and the same height).
Similarly, we see that $[OBF]$ is equal to one of these isosceles triangles.

Thus, $[ABEF]$ comprises 4 isosceles triangles, hence the area is half that of the octagon.

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I found this result really surprising, and was looking for a more intuitive / non-calculation-intensive approach.

Relevant wiki: Regular Polygons - Problem Solving - Medium

Area of the four edged triangles is $4\cdot \dfrac12 \left(\dfrac a{\sqrt2} \right)^2 = a^2$ .

Area of a regular octagon is $s^2 - a^2$ .
$s = a \left( 1 + \dfrac2{\sqrt2} \right) = a (1+\sqrt2)$
$s^2 = a^2 (3+2\sqrt2)$
Area of regular octagon is $2 (1+\sqrt2) a^2 = 1$ .

$a^2 = \dfrac1{2(\sqrt2 +1)}$

Area of rectangle is $a\times s$
$= \dfrac1{2(\sqrt2 + 1)} \cdot (\sqrt2 + 1)$
$= \boxed{\dfrac12}$ .

Call the centre of the octagon O. The octagon is made up of 16 congruent triangles from O to the midpoint of each side and a vertex. ABEF is made up of 8 of these triangles so (8/16)*1=0.5