Problem
Touching circles have center distance equal to the sum of their radii. So each side is the sum of the two radii that meet there.
Each triangle side bridges two circles that touch, so it is just their two radii laid end to end.
3.G.A.1Draw A DiagramAdding all three sides counts every radius twice. So the perimeter equals twice the total of the three radii.
The repeated '+each radius twice' pattern turns the whole perimeter into a simple doubling.
3.OA.D.9Solve An Easier Related ProblemThe perimeter of triangle ABC equals twice the sum of the three radii.
The perimeter is the three sides added together, and each side is the two radii of the circles touching along it, so the perimeter is .
The perimeter is the whole triangle boundary broken into its three sides with no gaps, so the sides add back to it.
A side runs from one center to the next and the two circles touch at a point on it, so that side splits into one radius plus the other laid end to end.
In that total each radius appears in exactly two sides, so the six pieces can be reordered and regrouped as without changing the sum.
Moving the added pieces into a different order does not change their total.
Choosing which pieces to add together first does not change their total.
Each pair , , is that radius taken two times, so together they are .
Adding a radius to itself is two equal groups of that radius, which is 2 times it.
Taking two of each radius separately is the same as taking two of their combined total, so .
Two groups of the whole bundle can be split into two groups of each part inside it, and back again.
Since the perimeter is double the radius total, divide 44 by 2.
Undoing a doubling is just dividing by 2.
3.OA.A.2Solve An Easier Related ProblemThis only needs Grade 3 thinking: each radius shows up twice around the triangle, so halve the perimeter!