The side of a regular octagon inscribed in a circle is 8 cm. Find the perimeter of a square inscribed in the same circle.

Given: a regular octagon inscribed in a circle with side a = 8 cm;
Find the perimeter of a square inscribed in the same circle -?
Decision:
1) The vertices of the square will coincide with the vertices of the octagon every other time, so we get 4 equal isosceles triangles with sides of 8 cm;
2) Degree measure of each corner of the hexagon = 180 (8-2)) / 8 = 180 * 6/8 = 135 degrees
By the cosine theorem, the side of the square = √ (8 ^ 2 + 8 ^ 2 – 2 * 8 * 8 * cos135) = √ (64 (1 + 1-2 * (- cos45))) = √64 (2 + 2 * √ 2/2) = √64 (2 + √2) = 8 √ ((2 + √2)
P of the square = 4 * 8 * √ ((2 + √2) = 32√ ((2 + √2) see.
Answer: 32√ ((2 + √2) see.