Segment through chained centers as radius multiples

3.OA.C.73.G.A.1 · adapt · grade 3

Archetype: Radius and Diameter Relationships · step in a 11-type progression

The figure on the right shows $21$ circles, each with a radius of $5\ \text{cm}$ , drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

Show solution

Understand

Twenty-one equal circles (radius 5 cm) are lined up so each circle passes through the center of the next, with all centers on one straight line. Segment GN runs from the far-left edge of the first circle to the far-right edge of the last circle. We must find how long GN is.

Givens

There are 21 equal circles.
Each circle has radius 5 cm.
Each circle passes through the center of the neighboring circle, so neighboring centers are 5 cm apart.
All centers lie on one straight line; G is the left edge of the first circle and N is the right edge of the last circle.

Unknowns

The length of segment GN in centimeters.

Constraints

Because each circle passes through the next center, the distance between two neighboring centers equals one radius (5 cm).
From G to the first center is one radius, and from the last center to N is one radius.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem#5 Look for a Pattern

Sketch the line of centers and mark the equal hops. Trying a few circles first (2, then 3) reveals the pattern: GN is made of equal 5 cm pieces. Counting those pieces gives the total length.

Execute

#1 Draw a Diagram 3.G.A.1

Walk along segment GN from left to right. From G to the first center is one radius (5 cm). Each time we jump from one center to the next, that is also one radius, because each circle passes through the next center. Finally, from the last center to N is one more radius.

G \to C_1 = 5,\quad C_1 \to C_2 = 5,\quad \dots,\quad C_{21} \to N = 5

The center-to-edge distance of any circle is its radius, so every piece of GN is the same 5 cm length.

#9 Solve an Easier Related Problem 3.OA.A.3

Try a smaller version first. With 2 circles there is G->center1, center1->center2, center2->N: that is 3 pieces = (2 + 1). With 3 circles it is 4 pieces = (3 + 1). So with 21 circles there are 21 + 1 = 22 equal pieces of 5 cm.

\text{pieces} = 21 + 1 = 22

There is always one more gap than the number of jumps between centers, because we add the two end radii.

#5 Look for a Pattern 3.OA.C.7

There are 22 equal pieces, each 5 cm long, so multiply.

22 \times 5 = 110

Repeated equal lengths join into a total by multiplication, a Grade 3 fact within 100.

Answer: 110 cm

Review

The answer is in centimeters, matching a length. 22 pieces of 5 cm sit between 100 and 120, and 22 x 5 = 110 cm is sensible for a row of 21 circles each 10 cm wide that heavily overlap.

Count the centers' span first: 21 centers have 20 gaps of 5 cm = 100 cm. Then add the two end radii (5 + 5 = 10 cm) to reach the outer edges: 100 + 10 = 110 cm (Tool 7, Identify Subproblems).

Standards · min grade 3

3.G.A.1 Understand that shapes in different categories share attributes — Knowing that the distance from a circle's center to its edge is the radius, so each hop equals 5 cm.
3.OA.A.3 Solve multiplication and division word problems within 100 — Counting that 21 circles produce 22 equal pieces along GN.
3.OA.C.7 Fluently multiply and divide within 100 — Computing 22 x 5 = 110.

💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!