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← 3-2 · Perimeter of shapes from tangent circles · Radius and Diameter Relationships

Perimeter of shapes from tangent circles · 10 practice problems

3.MD.D.83.OA.C.7

Generated variants — 10

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

Variant 1 answer: 36 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 12 in12\text{ in} and the small circles each have a diameter of 6 in6\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 12 in, small d = 6 in
Show solution

Understand

Two large circles (diameter 12 in each) sit on the left and right, and two small circles (diameter 6 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 12 in (so radius 6 in).
  • Two small circles, each with diameter 6 in (so radius 3 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 12 / 2 = 6 in; small radius is 6 / 2 = 3 in.
12÷2=6,6÷2=312 \div 2 = 6,\quad 6 \div 2 = 3
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
6+3=96 + 3 = 9
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 9 in. The perimeter is four of them.
4×9=364 \times 9 = 36
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 36 in

Review

Each side 9 in is between the large radius (6) and the full large diameter (12), which is sensible for a center-to-center distance. Four sides give 36 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 9 + 9 + 9 + 9 = 36 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (6 in) and small (3 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 9 = 36.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 2 answer: 64 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 20 in20\text{ in} and the small circles each have a diameter of 12 in12\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 20 in, small d = 12 in
Show solution

Understand

Two large circles (diameter 20 in each) sit on the left and right, and two small circles (diameter 12 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 20 in (so radius 10 in).
  • Two small circles, each with diameter 12 in (so radius 6 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 20 / 2 = 10 in; small radius is 12 / 2 = 6 in.
20÷2=10,12÷2=620 \div 2 = 10,\quad 12 \div 2 = 6
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
10+6=1610 + 6 = 16
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 16 in. The perimeter is four of them.
4×16=644 \times 16 = 64
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 64 in

Review

Each side 16 in is between the large radius (10) and the full large diameter (20), which is sensible for a center-to-center distance. Four sides give 64 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 16 + 16 + 16 + 16 = 64 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (10 in) and small (6 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 16 = 64.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 3 answer: 20 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 6 in6\text{ in} and the small circles each have a diameter of 4 in4\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 6 in, small d = 4 in
Show solution

Understand

Two large circles (diameter 6 in each) sit on the left and right, and two small circles (diameter 4 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 6 in (so radius 3 in).
  • Two small circles, each with diameter 4 in (so radius 2 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 6 / 2 = 3 in; small radius is 4 / 2 = 2 in.
6÷2=3,4÷2=26 \div 2 = 3,\quad 4 \div 2 = 2
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
3+2=53 + 2 = 5
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 5 in. The perimeter is four of them.
4×5=204 \times 5 = 20
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 20 in

Review

Each side 5 in is between the large radius (3) and the full large diameter (6), which is sensible for a center-to-center distance. Four sides give 20 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 5 + 5 + 5 + 5 = 20 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (3 in) and small (2 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 5 = 20.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 4 answer: 32 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 10 in10\text{ in} and the small circles each have a diameter of 6 in6\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 10 in, small d = 6 in
Show solution

Understand

Two large circles (diameter 10 in each) sit on the left and right, and two small circles (diameter 6 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 10 in (so radius 5 in).
  • Two small circles, each with diameter 6 in (so radius 3 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 10 / 2 = 5 in; small radius is 6 / 2 = 3 in.
10÷2=5,6÷2=310 \div 2 = 5,\quad 6 \div 2 = 3
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
5+3=85 + 3 = 8
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 8 in. The perimeter is four of them.
4×8=324 \times 8 = 32
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 32 in

Review

Each side 8 in is between the large radius (5) and the full large diameter (10), which is sensible for a center-to-center distance. Four sides give 32 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 8 + 8 + 8 + 8 = 32 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (5 in) and small (3 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 8 = 32.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 5 answer: 26 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 8 in8\text{ in} and the small circles each have a diameter of 5 in5\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 8 in, small d = 5 in
Show solution

Understand

Two large circles (diameter 8 in each) sit on the left and right, and two small circles (diameter 5 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 8 in (so radius 4 in).
  • Two small circles, each with diameter 5 in (so radius 2.5 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 8 / 2 = 4 in; small radius is 5 / 2 = 2.5 in.
8÷2=4,5÷2=2.58 \div 2 = 4,\quad 5 \div 2 = 2.5
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
4+2.5=6.54 + 2.5 = 6.5
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 6.5 in. The perimeter is four of them.
4×6.5=264 \times 6.5 = 26
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 26 in

Review

Each side 6.5 in is between the large radius (4) and the full large diameter (8), which is sensible for a center-to-center distance. Four sides give 26 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 6.5 + 6.5 + 6.5 + 6.5 = 26 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (4 in) and small (2.5 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 6.5 = 26.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 6 answer: 52 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 16 in16\text{ in} and the small circles each have a diameter of 10 in10\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 16 in, small d = 10 in
Show solution

Understand

Two large circles (diameter 16 in each) sit on the left and right, and two small circles (diameter 10 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 16 in (so radius 8 in).
  • Two small circles, each with diameter 10 in (so radius 5 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 16 / 2 = 8 in; small radius is 10 / 2 = 5 in.
16÷2=8,10÷2=516 \div 2 = 8,\quad 10 \div 2 = 5
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
8+5=138 + 5 = 13
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 13 in. The perimeter is four of them.
4×13=524 \times 13 = 52
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 52 in

Review

Each side 13 in is between the large radius (8) and the full large diameter (16), which is sensible for a center-to-center distance. Four sides give 52 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 13 + 13 + 13 + 13 = 52 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (8 in) and small (5 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 13 = 52.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 7 answer: 24 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 8 in8\text{ in} and the small circles each have a diameter of 4 in4\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 8 in, small d = 4 in
Show solution

Understand

Two large circles (diameter 8 in each) sit on the left and right, and two small circles (diameter 4 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 8 in (so radius 4 in).
  • Two small circles, each with diameter 4 in (so radius 2 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 8 / 2 = 4 in; small radius is 4 / 2 = 2 in.
8÷2=4,4÷2=28 \div 2 = 4,\quad 4 \div 2 = 2
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
4+2=64 + 2 = 6
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 6 in. The perimeter is four of them.
4×6=244 \times 6 = 24
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 24 in

Review

Each side 6 in is between the large radius (4) and the full large diameter (8), which is sensible for a center-to-center distance. Four sides give 24 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 6 + 6 + 6 + 6 = 24 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (4 in) and small (2 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 6 = 24.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 8 answer: 28 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 10 in10\text{ in} and the small circles each have a diameter of 4 in4\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 10 in, small d = 4 in
Show solution

Understand

Two large circles (diameter 10 in each) sit on the left and right, and two small circles (diameter 4 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 10 in (so radius 5 in).
  • Two small circles, each with diameter 4 in (so radius 2 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 10 / 2 = 5 in; small radius is 4 / 2 = 2 in.
10÷2=5,4÷2=210 \div 2 = 5,\quad 4 \div 2 = 2
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
5+2=75 + 2 = 7
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 7 in. The perimeter is four of them.
4×7=284 \times 7 = 28
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 28 in

Review

Each side 7 in is between the large radius (5) and the full large diameter (10), which is sensible for a center-to-center distance. Four sides give 28 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 7 + 7 + 7 + 7 = 28 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (5 in) and small (2 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 7 = 28.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 9 answer: 40 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 12 in12\text{ in} and the small circles each have a diameter of 8 in8\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 12 in, small d = 8 in
Show solution

Understand

Two large circles (diameter 12 in each) sit on the left and right, and two small circles (diameter 8 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 12 in (so radius 6 in).
  • Two small circles, each with diameter 8 in (so radius 4 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 12 / 2 = 6 in; small radius is 8 / 2 = 4 in.
12÷2=6,8÷2=412 \div 2 = 6,\quad 8 \div 2 = 4
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
6+4=106 + 4 = 10
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 10 in. The perimeter is four of them.
4×10=404 \times 10 = 40
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 40 in

Review

Each side 10 in is between the large radius (6) and the full large diameter (12), which is sensible for a center-to-center distance. Four sides give 40 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 10 + 10 + 10 + 10 = 40 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (6 in) and small (4 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 10 = 40.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!
Variant 10 answer: 48 in

The figure on the right is made of circles of two different sizes drawn so that neighboring circles touch. The large circles each have a diameter of 14 in14\text{ in} and the small circles each have a diameter of 10 in10\text{ in}. The center of each circle is joined to the centers of its neighbors to form a quadrilateral. What is the perimeter of the quadrilateral, in inches? (The two large circles are the same size as each other, and the two small circles are the same size as each other.)

large d = 14 in, small d = 10 in
Show solution

Understand

Two large circles (diameter 14 in each) sit on the left and right, and two small circles (diameter 10 in each) sit on top and bottom, all touching around a central gap. Joining the centers of neighboring circles makes a four-sided figure (a rhombus). Each side connects a large circle's center to a small circle's center. We must find the perimeter of the quadrilateral.

Givens
  • Two large circles, each with diameter 14 in (so radius 7 in).
  • Two small circles, each with diameter 10 in (so radius 5 in).
  • Neighboring circles touch (are tangent).
  • Each of the four sides joins the center of a large circle to the center of a small circle.
Unknowns
  • The perimeter of the quadrilateral in inches.
Constraints
  • When a large and a small circle touch on the outside, the distance between their centers is the large radius plus the small radius.
  • All four sides join a large center to a small center, so all four sides are equal.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Find one side as large radius + small radius using the touching rule, then multiply by the four equal sides to get the perimeter.

Execute

#1 Draw a Diagram 3.G.A.1
Radius is half the diameter. Large radius is 14 / 2 = 7 in; small radius is 10 / 2 = 5 in.
14÷2=7,10÷2=514 \div 2 = 7,\quad 10 \div 2 = 5
Diameter is twice the radius, so halving gives each radius.
#7 Identify Subproblems 3.MD.D.8
Where a large and small circle touch, the touch point is one large radius from the large center and one small radius from the small center. So the center-to-center distance is large radius + small radius.
7+5=127 + 5 = 12
Two radii laid end to end across the touch point give the distance between the two centers.
#7 Identify Subproblems 3.OA.C.7
All four sides each join a large center to a small center, so each side is 12 in. The perimeter is four of them.
4×12=484 \times 12 = 48
Four equal sides are totaled by multiplication, a Grade 3 idea.
Answer: 48 in

Review

Each side 12 in is between the large radius (7) and the full large diameter (14), which is sensible for a center-to-center distance. Four sides give 48 in. Units are inches, correct for a perimeter.

Add the sides one at a time: 12 + 12 + 12 + 12 = 48 in (Tool 2, Make a Systematic List).

Standards · min grade 3

  • 3.G.A.1 Understand that shapes in different categories share attributes — Using diameter = twice radius to get the large (7 in) and small (5 in) radii.
  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding one side as large radius + small radius and totaling the perimeter.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Computing 4 x 12 = 48.
💡 Each side is a big radius plus a small radius -- add them, then multiply by 4 sides with Grade 3 math!