Reasoning · Grade 3-2 Circles and Length

Problem

Arcs centered at polygon vertices

Three mutually touching circles are centred at a triangle's vertices. Two are equal large circles and one is small, with the large diameter twice the small. Each side equals the sum of the two radii at its ends. With a perimeter of 30 in, find the small diameter.
Your answer
How to solve
Strategy Draw a Diagram — Mark each radius on the diagram so I can read off every triangle side as a sum of two radii (Draw a Diagram). Because the perimeter is built from three separate sides, I split the work into finding each side in terms of the small radius (Identify Subproblems), add them to one total, then find the size that makes the total 30 in (Guess and Check confirms the single value).
1STEP 1

Name the small diameter and find all four radii

With small diameter d, the radii are d/2 and d.

r_small=d/2, r_large=2d/2=d
2STEP 2

Write each side as a sum of two radii

The side joining the two large circles is 2d.

AC=d+d=2d, AB=BC=d+d/2
3STEP 3

Add the three sides to get the perimeter

The other two are d + d/2 each, so the perimeter is 5d.

P=2d+(d+d/2)+(d+d/2)=4d+d=5d
4STEP 4

Set the perimeter equal to 30 and solve

5d = 30 makes the small diameter 6 in.

5d=30 in → d=30/5=6 in
Answer
6 in
5d = 30
With d = 6 in: small radius 3 in, large radius 6 in. Sides are AC = 12 in, AB = BC = 9 in, total 12 + 9 + 9 = 30 in. That matches the given perimeter exactly, and the small diameter (6 in) is sensibly smaller than a large diameter (12 in).
Takeaway

Every side is just two touching radii added together, so the whole trip around the triangle is 5 small diameters — split 30 into 5 equal parts to get 6 in!

  • Name the small diameter and find all four radii
  • Write each side as a sum of two radii
  • Add the three sides to get the perimeter
  • Set the perimeter equal to 30 and solve