Question 1199330
<pre>

{{{drawing(400,400,-4,4,-4,4, 

green(
line(sqrt(2),sqrt(2),1+sqrt(2),1+sqrt(2)),
line(sqrt(2),sqrt(2),1+sqrt(2),sqrt(2)),
line(1+sqrt(2),1+sqrt(2),1+sqrt(2),sqrt(2))),

red(
line(sqrt(2),sqrt(2),sqrt(2),0),
line(sqrt(2),sqrt(2),0,0),
line(sqrt(2),0,0,0)),
locate(1.79,1.46,1),locate(2.47,2.0,1),locate(1.6,2.2,sqrt(2)),

locate(1.5,.9,sqrt(2)),



circle(0,0,2+sqrt(2)), circle(sqrt(2),sqrt(2),sqrt(2)), graph(400,400,-4,4,-4,4)
)}}} 

To find the area of the larger circle, we need to find its radius.

The red and green triangles are similar 45-45-90 isosceles right triangles.
Since the green hypotenuse is given to be {{{sqrt(2)}}}, it is a standard
1-1-{{{sqrt(2)}}} isosceles right triangle, and its legs are 1 each.

The radius of the larger circle is the sum of the red and green hypotenuses.
The green hypotenuse is {{{sqrt(2)}}} and is a radius of the small circle, So, the 
vertical red side is also {{{sqrt(2)}}}, because it is also a radius of the smaller circle. 

We set up a proportion between the sides of the two similar triangles:

{{{matrix(1,2,green,hypotenuse)/matrix(1,2,red,hypotenuse)}}}{{{""=""}}}{{{matrix(1,3,green,vertical,side)/matrix(1,3,red,vertical,side)}}}

{{{sqrt(2)/matrix(1,2,red,hypotenuse)}}}{{{""=""}}}{{{1/sqrt(2)}}}

Cross-multiply and get

{{{matrix(1,2,red,hypotenuse)}}}{{{""=""}}}{{{2}}}

{{{matrix(1,4,radius,of,larger,circle)}}}{{{""=""}}}{{{matrix(1,2,red,hypotenuse)}}}{{{""+""}}}{{{matrix(1,2,green,hypotenuse)}}}

{{{matrix(1,4,radius,of,larger,circle)}}}{{{""=""}}}{{{2}}}{{{""+""}}}{{{sqrt(2)}}}

Using {{{Area}}}{{{""=""}}}{{{pi*radius^2}}},

{{{Area}}}{{{""=""}}}{{{pi(2+sqrt(2))^2}}} cm<sup>2</sup>

Edwin</pre>