Using the radii of the circles, we can create an equation for the box using length (l) and width (w).
l = r + r + (r - 5) + (r - 5) = 4r - 10
w = r + r = 2r
As the area of a rectangle is length times width (l*w), we can find the total area (A_"rectangle") of the rectangle in terms of r:
A_"rectangle" = l*w
A_"rectangle" = (4r - 10) * 2r = 8r^2 - 20r
The area of the shaded region is 250cm^2, so the area of the circles (A_"circles") would be the area of the rectangle minus the area of the shaded region:
A_"circles" = A_"rectangle" - 250
A_"circles" = 8r^2 - 20r - 250
Another way of expressing the area of the circles would be to just use their given radii and the formula A_"circle" = πr^2:
A_"circle1" = πr^2
A_"circle2" = π(r-5)^2 = πr^2 - 10πr + 25π
As A_"circles" is just the area of the two circles added together,
A_"circles" = A_"circle1" + A_"circle2"
A_"circles" = πr^2 + (πr^2 - 10πr + 25π)
A_"circles" = 2πr^2 - 10πr + 25π
Now we have two equations for A_"circles" in terms of r, which means we can solve for r (aka what the problem is asking for):
A_"circles" = 8r^2 - 20r - 250
A_"circles" = 2πr^2 - 10πr + 25π
8r^2 - 20r - 250 = 2πr^2 - 10πr + 25π
8r^2 - 20r - 250 - 2πr^2 + 10πr - 25π = 0
(8-2π)r^2 +(-20+10π)r + (-250-25π) = 0
Using the quadratic formula we can solve for r (the values in the parentheses are a, b, and c respectively):
(-(-20+10π) +- sqrt((-20+10π)^2 - 4(8-2π)(-250-25π)))/(2(8-2π))
We can "simplify" this to get:
((20-10π +- sqrt(400 - 400π + 100π^2 + 8000 - 1200π - 200π^2))/(16-4π))
((20-10π +- sqrt(8400 - 1600π -100π^2))/(16-4π))
As you see this is still very complicated... So I would suggest using the magic of the calculator at this point and solving for r. When you do so, you get:
r = 10.9, - 17.6
As r can't be negative because the problem is talking about geometry, r must equal 10.9cm.
The radius of the smaller circle is then r - 5 (as shown in the picture):
r_"small" = r - 5 = 10.9 - 5 = 5.9
