Problem
Two touching circles have centers separated by the sum of their radii. So each side is the sum of two neighboring radii.
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
3.G.A.1Draw A DiagramAdding the four sides counts every radius twice. So the perimeter equals twice the total of all four radii.
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
3.OA.D.9Solve An Easier Related ProblemAdding the four sides of the quadrilateral gives exactly twice the total of the four radii, because each radius is counted two times.
Each side equals the two radii at its ends added together, so the four sides together are the same as adding up eight radius lengths.
When two circles touch on the outside, the point where they meet lies on the straight line between their centers, splitting that side into one circle's radius and then the other's with nothing left over.
Among those eight radius lengths every circle's radius shows up exactly twice, because each circle touches the neighbor on each side of it.
The eight lengths can be added in any order, so the two copies of each radius can be slid next to each other.
Pairing them up leaves two matching copies of the whole four-radius total, and two equal copies of an amount is just that amount taken two times.
Since the perimeter is double the radius total, divide 60 by 2 to get the sum of the radii.
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
3.OA.A.2Solve An Easier Related ProblemThis only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!