Since the vertices of the square containing the circle are on 0,-2\sqrt{2} root 2 and 2\sqrt{2},0, it means that if a line is drawn to the two points we will have one edge of the square.

The length between each of the two vertices to the origin are both equal and are 2\sqrt{2}.

That means that they are two equal sides of an isosceles 1,1,\sqrt{2} triangle where the length from the vertices to the origin are the 1,1, equal sides.

Hence the \sqrt{2} side of the 1,1, \sqrt{2} triangle is 2\sqrt{2} * \sqrt{2} which is 4.

Hence all the sides of the square have length, L = 4.

Since the circle is inscribed in the square, then the diameter, D = 4 and the radius, r = 2.

Area of the square = L raised to power 2 =
L * L = 4*4

Area of the circle is pi * r raised to power 2 =
pi * r * r = 3.142 * 2 * 2 = 3.142 * 4

Area of the region outside the circle but in the square =

4*4 - 3.142 * 4 = (4 - 3.142) * 4 = 0.858*4

= 3.432 approximately 3.4 to the nearest tenth.


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