Question 574556
{{{drawing(360,150,0,36,0,15,


line(0.5,0.5,35.5,0.5),
line(0.5,0.5,0.5,14.5),
line(35.5,0.5,35.5,14.5),
line(35.5,14.5,0.5,14.5),
line(0.5,0.5,35.5,14.5),

circle(6.2,9,5.8),
circle(36-6.2, 15-9, 5.8),
line(6.2,9.5,6.2,14.5),
line(6.2,9.5,0.5,9.5),
locate(3.1,9.5,r),
locate(6.4,12,r),
line(6.2,9.5, 36-6.2, 6),

locate(2,2,A),
locate(34,2,B),
locate(34,13,C),
locate(1.5,13.5,D)
)
}}}


An easy way to find this distance is to assume that A is the origin of some xy coordinate system. Therefore we may assume that A = (0,0), B = (12,0), C = (12,5), D = (0,5).


First we should find r. There are several ways to do this, but the easiest way is to use the area formula for triangle ACD:


*[tex \LARGE Area = rs]


where r is the inradius (shown above) and s is the semi-perimeter of ACD ((5+12+13)/2 = 15) Plugging in, we have


*[tex \LARGE 30 = 15r \Rightarrow r = 2]


Since the two radii meet at right angles with AD and DC, the x-coordinate of the center of the upper-left circle is simply 2. The y-coordinate is 5-2, or 3, so the coordinates of this circle are (2,3).


Similarly, the coordinates of the center of the lower-right circle are (12-2, 2) or (10,2) (by symmetry). We use the distance formula to find the distance between the two centers:


*[tex \LARGE d = \sqrt{(10-2)^2 + (2-3)^2} = \sqrt{65}].