>>11557526Define a coordinate system with the origin at the bottom left of the rectangle.
Let L be the length and H be the height of the outer rectangle.
Let (x_B, y_B) be the the center of the blue circle.
Let (x_G, y_G) be the the center of the green circle.
Let (x_R, y_R) be the the center of the red circle.
That's 8 variables so we need 8 equations:
>The intersections of the blue circle with the rectangle givex_B = 1.5
H - y_B = 1.5
>The intersection of the green circle with the rectangle givey_G = 2
>The intersections of the red circle with the rectangle giveL - x_R = 3
y_R = 3
H - y_R = 3
>The intersection of the blue and green circle gives(x_B - x_G)^2 + (y_B - y_G)^2 = 3.5^2
>The intersection of the red and green circle gives(x_R - x_G)^2 + (y_R - y_G)^2 = 5^2
Plugging this system into a CAS gives
L = 9/2 + 3sqrt(6)
H = 6
x_B = 3/2
y_B = 9/2
x_G = 3/2 + sqrt(6)
y_G = 2
x_R = 3/2 + 3sqrt(6)
y_R = 3
Therefore the line from A to B has the length
x_R - x_B = 3 sqrt(6)
Solving a geometry problem systematically like this is always more elegant than using some filthy geometry constructions.
