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Segment through chained centers as radius multiples · 10 practice problems

3.OA.C.73.G.A.1

Generated variants — 10

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 answer: 44 cm

The figure on the right shows 1010 circles, each with a radius of 4 cm4\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

10 equal circles (radius 4 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 10 equal circles.
  • Each circle has radius 4 cm.
  • Each circle passes through the center of the neighboring circle, so neighboring centers are 4 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 (4 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 4 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 (4 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.
GC1=4,C1C2=4,,C10N=4G \to C_1 = 4,\quad C_1 \to C_2 = 4,\quad \dots,\quad C_{10} \to N = 4
The center-to-edge distance of any circle is its radius, so every piece of GN is the same 4 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 10 circles there are 10 + 1 = 11 equal pieces of 4 cm.
pieces=10+1=11\text{pieces} = 10 + 1 = 11
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 11 equal pieces, each 4 cm long, so multiply.
11×4=4411 \times 4 = 44
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 44 cm

Review

The answer is in centimeters, matching a length. 11 pieces of 4 cm give 44 cm, sensible for a row of 10 circles each 8 cm wide that heavily overlap.

Count the centers' span first: 10 centers have 9 gaps of 4 cm = 36 cm. Then add the two end radii (4 + 4 = 8 cm) to reach the outer edges: 36 + 8 = 44 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 4 cm.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Counting that 10 circles produce 11 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 11 x 4 = 44.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 2 answer: 110 cm

The figure on the right shows 2121 circles, each with a radius of 5 cm5\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
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.
GC1=5,C1C2=5,,C21N=5G \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.
pieces=21+1=22\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.
22×5=11022 \times 5 = 110
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 110 cm

Review

The answer is in centimeters, matching a length. 22 pieces of 5 cm give 110 cm, 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!
Variant 3 answer: 72 cm

The figure on the right shows 77 circles, each with a radius of 9 cm9\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

7 equal circles (radius 9 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 7 equal circles.
  • Each circle has radius 9 cm.
  • Each circle passes through the center of the neighboring circle, so neighboring centers are 9 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 (9 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 9 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 (9 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.
GC1=9,C1C2=9,,C7N=9G \to C_1 = 9,\quad C_1 \to C_2 = 9,\quad \dots,\quad C_{7} \to N = 9
The center-to-edge distance of any circle is its radius, so every piece of GN is the same 9 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 7 circles there are 7 + 1 = 8 equal pieces of 9 cm.
pieces=7+1=8\text{pieces} = 7 + 1 = 8
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 8 equal pieces, each 9 cm long, so multiply.
8×9=728 \times 9 = 72
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 72 cm

Review

The answer is in centimeters, matching a length. 8 pieces of 9 cm give 72 cm, sensible for a row of 7 circles each 18 cm wide that heavily overlap.

Count the centers' span first: 7 centers have 6 gaps of 9 cm = 54 cm. Then add the two end radii (9 + 9 = 18 cm) to reach the outer edges: 54 + 18 = 72 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 9 cm.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Counting that 7 circles produce 8 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 8 x 9 = 72.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 4 answer: 100 cm

The figure on the right shows 1919 circles, each with a radius of 5 cm5\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

19 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 19 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.
GC1=5,C1C2=5,,C19N=5G \to C_1 = 5,\quad C_1 \to C_2 = 5,\quad \dots,\quad C_{19} \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 19 circles there are 19 + 1 = 20 equal pieces of 5 cm.
pieces=19+1=20\text{pieces} = 19 + 1 = 20
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 20 equal pieces, each 5 cm long, so multiply.
20×5=10020 \times 5 = 100
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 100 cm

Review

The answer is in centimeters, matching a length. 20 pieces of 5 cm give 100 cm, sensible for a row of 19 circles each 10 cm wide that heavily overlap.

Count the centers' span first: 19 centers have 18 gaps of 5 cm = 90 cm. Then add the two end radii (5 + 5 = 10 cm) to reach the outer edges: 90 + 10 = 100 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 19 circles produce 20 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 20 x 5 = 100.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 5 answer: 100 cm

The figure on the right shows 2424 circles, each with a radius of 4 cm4\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

24 equal circles (radius 4 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 24 equal circles.
  • Each circle has radius 4 cm.
  • Each circle passes through the center of the neighboring circle, so neighboring centers are 4 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 (4 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 4 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 (4 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.
GC1=4,C1C2=4,,C24N=4G \to C_1 = 4,\quad C_1 \to C_2 = 4,\quad \dots,\quad C_{24} \to N = 4
The center-to-edge distance of any circle is its radius, so every piece of GN is the same 4 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 24 circles there are 24 + 1 = 25 equal pieces of 4 cm.
pieces=24+1=25\text{pieces} = 24 + 1 = 25
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 25 equal pieces, each 4 cm long, so multiply.
25×4=10025 \times 4 = 100
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 100 cm

Review

The answer is in centimeters, matching a length. 25 pieces of 4 cm give 100 cm, sensible for a row of 24 circles each 8 cm wide that heavily overlap.

Count the centers' span first: 24 centers have 23 gaps of 4 cm = 92 cm. Then add the two end radii (4 + 4 = 8 cm) to reach the outer edges: 92 + 8 = 100 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 4 cm.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Counting that 24 circles produce 25 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 25 x 4 = 100.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 6 answer: 45 cm

The figure on the right shows 88 circles, each with a radius of 5 cm5\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

8 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 8 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.
GC1=5,C1C2=5,,C8N=5G \to C_1 = 5,\quad C_1 \to C_2 = 5,\quad \dots,\quad C_{8} \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 8 circles there are 8 + 1 = 9 equal pieces of 5 cm.
pieces=8+1=9\text{pieces} = 8 + 1 = 9
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 9 equal pieces, each 5 cm long, so multiply.
9×5=459 \times 5 = 45
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 45 cm

Review

The answer is in centimeters, matching a length. 9 pieces of 5 cm give 45 cm, sensible for a row of 8 circles each 10 cm wide that heavily overlap.

Count the centers' span first: 8 centers have 7 gaps of 5 cm = 35 cm. Then add the two end radii (5 + 5 = 10 cm) to reach the outer edges: 35 + 10 = 45 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 8 circles produce 9 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 9 x 5 = 45.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 7 answer: 80 cm

The figure on the right shows 99 circles, each with a radius of 8 cm8\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

9 equal circles (radius 8 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 9 equal circles.
  • Each circle has radius 8 cm.
  • Each circle passes through the center of the neighboring circle, so neighboring centers are 8 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 (8 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 8 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 (8 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.
GC1=8,C1C2=8,,C9N=8G \to C_1 = 8,\quad C_1 \to C_2 = 8,\quad \dots,\quad C_{9} \to N = 8
The center-to-edge distance of any circle is its radius, so every piece of GN is the same 8 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 9 circles there are 9 + 1 = 10 equal pieces of 8 cm.
pieces=9+1=10\text{pieces} = 9 + 1 = 10
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 10 equal pieces, each 8 cm long, so multiply.
10×8=8010 \times 8 = 80
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 80 cm

Review

The answer is in centimeters, matching a length. 10 pieces of 8 cm give 80 cm, sensible for a row of 9 circles each 16 cm wide that heavily overlap.

Count the centers' span first: 9 centers have 8 gaps of 8 cm = 64 cm. Then add the two end radii (8 + 8 = 16 cm) to reach the outer edges: 64 + 16 = 80 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 8 cm.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Counting that 9 circles produce 10 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 10 x 8 = 80.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 8 answer: 72 cm

The figure on the right shows 1111 circles, each with a radius of 6 cm6\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

11 equal circles (radius 6 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 11 equal circles.
  • Each circle has radius 6 cm.
  • Each circle passes through the center of the neighboring circle, so neighboring centers are 6 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 (6 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 6 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 (6 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.
GC1=6,C1C2=6,,C11N=6G \to C_1 = 6,\quad C_1 \to C_2 = 6,\quad \dots,\quad C_{11} \to N = 6
The center-to-edge distance of any circle is its radius, so every piece of GN is the same 6 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 11 circles there are 11 + 1 = 12 equal pieces of 6 cm.
pieces=11+1=12\text{pieces} = 11 + 1 = 12
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 12 equal pieces, each 6 cm long, so multiply.
12×6=7212 \times 6 = 72
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 72 cm

Review

The answer is in centimeters, matching a length. 12 pieces of 6 cm give 72 cm, sensible for a row of 11 circles each 12 cm wide that heavily overlap.

Count the centers' span first: 11 centers have 10 gaps of 6 cm = 60 cm. Then add the two end radii (6 + 6 = 12 cm) to reach the outer edges: 60 + 12 = 72 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 6 cm.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Counting that 11 circles produce 12 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 12 x 6 = 72.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 9 answer: 91 cm

The figure on the right shows 1212 circles, each with a radius of 7 cm7\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

12 equal circles (radius 7 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 12 equal circles.
  • Each circle has radius 7 cm.
  • Each circle passes through the center of the neighboring circle, so neighboring centers are 7 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 (7 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 7 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 (7 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.
GC1=7,C1C2=7,,C12N=7G \to C_1 = 7,\quad C_1 \to C_2 = 7,\quad \dots,\quad C_{12} \to N = 7
The center-to-edge distance of any circle is its radius, so every piece of GN is the same 7 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 12 circles there are 12 + 1 = 13 equal pieces of 7 cm.
pieces=12+1=13\text{pieces} = 12 + 1 = 13
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 13 equal pieces, each 7 cm long, so multiply.
13×7=9113 \times 7 = 91
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 91 cm

Review

The answer is in centimeters, matching a length. 13 pieces of 7 cm give 91 cm, sensible for a row of 12 circles each 14 cm wide that heavily overlap.

Count the centers' span first: 12 centers have 11 gaps of 7 cm = 77 cm. Then add the two end radii (7 + 7 = 14 cm) to reach the outer edges: 77 + 14 = 91 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 7 cm.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Counting that 12 circles produce 13 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 13 x 7 = 91.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!
Variant 10 answer: 80 cm

The figure on the right shows 1515 circles, each with a radius of 5 cm5\ \text{cm}, drawn so that each circle passes through the center of the next one. How long is segment GN, in centimeters?

G N
Show solution

Understand

15 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 15 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.
GC1=5,C1C2=5,,C15N=5G \to C_1 = 5,\quad C_1 \to C_2 = 5,\quad \dots,\quad C_{15} \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 15 circles there are 15 + 1 = 16 equal pieces of 5 cm.
pieces=15+1=16\text{pieces} = 15 + 1 = 16
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 16 equal pieces, each 5 cm long, so multiply.
16×5=8016 \times 5 = 80
Repeated equal lengths join into a total by multiplication, a Grade 3 multiplication fact.
Answer: 80 cm

Review

The answer is in centimeters, matching a length. 16 pieces of 5 cm give 80 cm, sensible for a row of 15 circles each 10 cm wide that heavily overlap.

Count the centers' span first: 15 centers have 14 gaps of 5 cm = 70 cm. Then add the two end radii (5 + 5 = 10 cm) to reach the outer edges: 70 + 10 = 80 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 15 circles produce 16 equal pieces along GN.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 16 x 5 = 80.
💡 Every gap is one radius, so just count the gaps and multiply -- only Grade 3 multiplication you already know!