Question 1133427
.
<pre>

Let A, B, C, D, E, F, G, and H be eight consecutive vertices of the octagon.


Draw the segments OA, OB, OC, OD, OE, OF, OG and OH, connecting the vertices with the common center O of the squares.


The segments divide the octagon in 8 triangles.


In these triangles, AB, BC, CD, DE, EF, FG, GH and HA are the bases; the height is  {{{1/2}}}  of a unit in all the triangles.


Therefore the area of the octagon is


    Area = {{{(1/2)*(1/2)*(AB + BC + CD + DE + EF + FG + GH + HA)}}} =

         = {{{(1/4)*Perimeter_of_the_octagon}}} = {{{(1/4)*3.5}}} = {{{3.5/4}}} = 0.875 of a square unit.     <U>ANSWER</U>
</pre>

Solved.


----------------


The solution above is complete and goes with minimal wording.


In order for to understand better the situation, you can also derive the following facts related to this configuration.


<pre>
    (a)  Due to symmetry, all the sides of the octagon  AB, BC, CD, DE, EF, FG, GH and HA  have <U>the same length</U>.


    (b)  This octagon is <U>circumscribed about the circle</U> of the radius 0.5 of a unit with the center at the shared center of the squares.


    (c)  It is not necessary a regular octagon; but its sides all have equal lengths.


    (d)  As for any polygon circumscribed about a circle, its area is half the product of the perimeter and the radius of the circle.
</pre>


It is a nice Geometry problem of a Math circle level.