Firstly notice that the two smaller triangles are identical, and that the height of each of them is the same as the height of the rectangle. If we call the radius of the blue circle ‘r’, then a small triangle is 4r tall. Let’s work out what the base is:

Using Pythagoras: (ar+3r)^2 = (ar+r)^2 + (4r)^2

Divide throughout by r^2 and multiply the brackets:

a^2+6a+9 = a^2+2a+1+16

4a = 8, a=2, therefore the base of the small triangle, and so also of the rectangle, is 3r.

Now let’s look at a small section within the rectangle:

Using Pythagoras again on the right-angle triangle in the middle of this figure:

r^2+18r+81 = r^2 + (9/4)r^2-27r+81

18r = (9/4)r^2-27r, since we know r is not equal to 0 we can divide throughout by r:

45 = 9r/4, r = 20.

So the radius of the blue circles is 20.