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← 3-2 · Polygon side equals sum of two radii · Radius and Diameter Relationships

Polygon side equals sum of two radii · 10 practice problems

3.MD.D.83.G.A.1

Generated variants — 10

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

Variant 1 answer: 18 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 36 cm36\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 36 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 36 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 36 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
36=2×(rA+rB+rC+rD)36 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 36 by 2 to get the sum of the radii.
36÷2=1836 \div 2 = 18
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 18 cm

Review

The radius total (18 cm) is exactly half the 36-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 2 answer: 25 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 50 cm50\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 50 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 50 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 50 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
50=2×(rA+rB+rC+rD)50 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 50 by 2 to get the sum of the radii.
50÷2=2550 \div 2 = 25
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 25 cm

Review

The radius total (25 cm) is exactly half the 50-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 3 answer: 40 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 80 cm80\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 80 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 80 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 80 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
80=2×(rA+rB+rC+rD)80 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 80 by 2 to get the sum of the radii.
80÷2=4080 \div 2 = 40
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 40 cm

Review

The radius total (40 cm) is exactly half the 80-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 4 answer: 50 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 100 cm100\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 100 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 100 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 100 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
100=2×(rA+rB+rC+rD)100 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 100 by 2 to get the sum of the radii.
100÷2=50100 \div 2 = 50
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 50 cm

Review

The radius total (50 cm) is exactly half the 100-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 5 answer: 30 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 60 cm60\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 60 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 60 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 60 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
60=2×(rA+rB+rC+rD)60 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 60 by 2 to get the sum of the radii.
60÷2=3060 \div 2 = 30
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 30 cm

Review

The radius total (30 cm) is exactly half the 60-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 6 answer: 12 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 24 cm24\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 24 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 24 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 24 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
24=2×(rA+rB+rC+rD)24 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 24 by 2 to get the sum of the radii.
24÷2=1224 \div 2 = 12
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 12 cm

Review

The radius total (12 cm) is exactly half the 24-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 7 answer: 20 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 40 cm40\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 40 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 40 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 40 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
40=2×(rA+rB+rC+rD)40 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 40 by 2 to get the sum of the radii.
40÷2=2040 \div 2 = 20
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 20 cm

Review

The radius total (20 cm) is exactly half the 40-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 8 answer: 60 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 120 cm120\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 120 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 120 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 120 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
120=2×(rA+rB+rC+rD)120 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 120 by 2 to get the sum of the radii.
120÷2=60120 \div 2 = 60
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 60 cm

Review

The radius total (60 cm) is exactly half the 120-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 9 answer: 8 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 16 cm16\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 16 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 16 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 16 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
16=2×(rA+rB+rC+rD)16 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 16 by 2 to get the sum of the radii.
16÷2=816 \div 2 = 8
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 8 cm

Review

The radius total (8 cm) is exactly half the 16-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!
Variant 10 answer: 100 cm

The figure on the right shows four circles of different sizes drawn so that they touch one another in a ring, with the centers of the circles joined to form quadrilateral ABCD. If the perimeter of quadrilateral ABCD is 200 cm200\ \text{cm}, what is the sum of the radii of the four circles, in centimeters?

A B C D perimeter of ABCD = 200 cm
Show solution

Understand

Four circles of different sizes touch one another in a ring. Joining their centers makes a quadrilateral ABCD. Each side connects two touching circles, so each side length equals the sum of those two circles' radii. The quadrilateral's perimeter is 200 cm, and I need the total of all four radii.

Givens
  • Four circles touch (are tangent) in a ring; neighboring circles touch.
  • Centers joined form quadrilateral ABCD.
  • Each side equals the sum of the radii of the two circles meeting along it.
  • Perimeter of ABCD is 200 cm.
Unknowns
  • The sum of the radii of the four circles, in cm.
Constraints
  • When two circles touch on the outside, the distance between their centers equals the sum of their radii.
  • Every radius is counted in exactly two sides (each circle touches two neighbors).

Plan

#9 Solve an Easier Related Problem · also uses: #1 Draw a Diagram

Write each side as a radius sum, then add all four sides. The sum reveals each radius appears exactly twice, turning the perimeter into double the radius total.

Execute

#1 Draw a Diagram 3.G.A.1
Two touching circles have their centers separated by the sum of their radii. So AB = rA + rB, BC = rB + rC, CD = rC + rD, and DA = rD + rA.
AB+BC+CD+DA=(rA+rB)+(rB+rC)+(rC+rD)+(rD+rA)AB+BC+CD+DA = (r_A+r_B)+(r_B+r_C)+(r_C+r_D)+(r_D+r_A)
The figure shows each side bridging two circles that touch, so the side is just both radii laid end to end.
#9 Solve an Easier Related Problem 3.OA.D.9
When the four sides are added, every circle's radius appears exactly two times, because each circle touches two neighbors. So the perimeter equals twice the total of all four radii.
200=2×(rA+rB+rC+rD)200 = 2 \times (r_A+r_B+r_C+r_D)
Spotting the repeated pattern of '+each radius twice' makes the big sum collapse into a simple doubling.
#9 Solve an Easier Related Problem 3.OA.A.2
Since the perimeter is double the radius total, divide 200 by 2 to get the sum of the radii.
200÷2=100200 \div 2 = 100
Undoing a doubling is just dividing by 2, a basic grade-3 division fact.
Answer: 100 cm

Review

The radius total (100 cm) is exactly half the 200-cm perimeter, which makes sense because the perimeter counts every radius twice. Units stay in centimeters.

Guess and check (tool 6): pick any four radii that sum to the answer; their four pairwise sides add to the given perimeter, confirming the rule.

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Recognizing each side of the center-quadrilateral as a sum of two touching circles' radii.
  • 3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Seeing that each radius appears twice when all sides are added, so the perimeter is double the radius total.
  • 3.OA.A.2 Interpret whole-number quotients of whole numbers — Dividing the perimeter by 2 to recover the sum of the radii.
💡 This only needs Grade 3 thinking: every radius gets counted twice, so just split the perimeter in half!