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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 of a unit in all the triangles.
Therefore the area of the octagon is
Area = =
= = = = 0.875 of a square unit. ANSWER
Solved.
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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.
(a) Due to symmetry, all the sides of the octagon AB, BC, CD, DE, EF, FG, GH and HA have the same length.
(b) This octagon is circumscribed about the circle 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.
It is a nice Geometry problem of a Math circle level.