Question 462860
{{{drawing(300,300, -4,4,-4,4,


line(-3,-3,3,-3),
line(3,-3,3,3),
line(3,3,-3,3),
line(-3,3,-3,-3),
line(-1,-3,-1,3),
line(1,-3,1,3),
line(3,-1,-3,-1),
line(3,1,-3,1),
circle(0,0,2sqrt(2)),
line(0,0,2,2),
locate(2,2,A),
locate(0,0,O)
)
}}}


All we need to do is find the radius of the circle. We know that the radius passes through the center and the center of the corner square (denoted A). If we draw the 45-45-90 triangle with hypotenuse OA, we will see that the "leg" has length 2 and the hypotenuse (radius) is therefore 2sqrt(2). This is because "leg" has the same side length as the length of one of the smaller squares, or 2. Hence the area of the circle is


*[tex \LARGE A = \pi (2\sqrt{2})^2 = 8\pi]