← The side that rolls into a rim is exactly as long as that rim · Net and Solid Structure

The side that rolls into a rim is exactly as long as that rim · 12 practice problems

7.G.B.46.G.A.47.G.B.6

Generated variants — 12

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

Variant 1 easy answer: 18.56 cm

The figure at the right is the net of a cylinder whose base has a circumference of 3.14 cm3.14\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

3 cm base circumference 3.14 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 3 cm tall between two circles, and the base is 3.14 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 3.14 cm.
  • The rectangle is 3 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=3.14\text{width} = 3.14
3.14 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(3.14+3)×2=12.28(3.14 + 3) \times 2 = 12.28
12.28 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
3.14×2=6.283.14 \times 2 = 6.28
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
12.28+6.28=18.5612.28 + 6.28 = 18.56
18.56 cm round the net.
Answer: 18.56 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 12.56 cm, and the two straight sides to 6 cm; together they are the same 18.56.

Another way: Treating the circles as if their contact hid part of the edge would give less than 18.56 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 2 easy answer: 33.12 cm

The figure at the right is the net of a cylinder whose base has a circumference of 6.28 cm6.28\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

4 cm base circumference 6.28 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 4 cm tall between two circles, and the base is 6.28 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 6.28 cm.
  • The rectangle is 4 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=6.28\text{width} = 6.28
6.28 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(6.28+4)×2=20.56(6.28 + 4) \times 2 = 20.56
20.56 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
6.28×2=12.566.28 \times 2 = 12.56
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
20.56+12.56=33.1220.56 + 12.56 = 33.12
33.12 cm round the net.
Answer: 33.12 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 25.12 cm, and the two straight sides to 8 cm; together they are the same 33.12.

Another way: Treating the circles as if their contact hid part of the edge would give less than 33.12 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 3 easy answer: 32.84 cm

The figure at the right is the net of a cylinder whose base has a circumference of 4.71 cm4.71\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

7 cm base circumference 4.71 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 7 cm tall between two circles, and the base is 4.71 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 4.71 cm.
  • The rectangle is 7 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=4.71\text{width} = 4.71
4.71 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(4.71+7)×2=23.42(4.71 + 7) \times 2 = 23.42
23.42 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
4.71×2=9.424.71 \times 2 = 9.42
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
23.42+9.42=32.8423.42 + 9.42 = 32.84
32.84 cm round the net.
Answer: 32.84 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 18.84 cm, and the two straight sides to 14 cm; together they are the same 32.84.

Another way: Treating the circles as if their contact hid part of the edge would give less than 32.84 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 4 easy answer: 49.4 cm

The figure at the right is the net of a cylinder whose base has a circumference of 7.85 cm7.85\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

9 cm base circumference 7.85 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 9 cm tall between two circles, and the base is 7.85 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 7.85 cm.
  • The rectangle is 9 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=7.85\text{width} = 7.85
7.85 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(7.85+9)×2=33.7(7.85 + 9) \times 2 = 33.7
33.7 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
7.85×2=15.77.85 \times 2 = 15.7
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
33.7+15.7=49.433.7 + 15.7 = 49.4
49.4 cm round the net.
Answer: 49.4 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 31.4 cm, and the two straight sides to 18 cm; together they are the same 49.4.

Another way: Treating the circles as if their contact hid part of the edge would give less than 49.4 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 5 medium answer: 47.68 cm

The figure at the right is the net of a cylinder whose base has a circumference of 9.42 cm9.42\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

5 cm base circumference 9.42 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 5 cm tall between two circles, and the base is 9.42 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 9.42 cm.
  • The rectangle is 5 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=9.42\text{width} = 9.42
9.42 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(9.42+5)×2=28.84(9.42 + 5) \times 2 = 28.84
28.84 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
9.42×2=18.849.42 \times 2 = 18.84
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
28.84+18.84=47.6828.84 + 18.84 = 47.68
47.68 cm round the net.
Answer: 47.68 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 37.68 cm, and the two straight sides to 10 cm; together they are the same 47.68.

Another way: Treating the circles as if their contact hid part of the edge would give less than 47.68 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 6 medium answer: 66.24 cm

The figure at the right is the net of a cylinder whose base has a circumference of 12.56 cm12.56\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

8 cm base circumference 12.56 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 8 cm tall between two circles, and the base is 12.56 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 12.56 cm.
  • The rectangle is 8 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=12.56\text{width} = 12.56
12.56 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(12.56+8)×2=41.12(12.56 + 8) \times 2 = 41.12
41.12 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
12.56×2=25.1212.56 \times 2 = 25.12
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
41.12+25.12=66.2441.12 + 25.12 = 66.24
66.24 cm round the net.
Answer: 66.24 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 50.24 cm, and the two straight sides to 16 cm; together they are the same 66.24.

Another way: Treating the circles as if their contact hid part of the edge would give less than 66.24 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 7 medium answer: 82.8 cm

The figure at the right is the net of a cylinder whose base has a circumference of 15.7 cm15.7\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

10 cm base circumference 15.7 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 10 cm tall between two circles, and the base is 15.7 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 15.7 cm.
  • The rectangle is 10 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=15.7\text{width} = 15.7
15.7 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(15.7+10)×2=51.4(15.7 + 10) \times 2 = 51.4
51.4 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
15.7×2=31.415.7 \times 2 = 31.4
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
51.4+31.4=82.851.4 + 31.4 = 82.8
82.8 cm round the net.
Answer: 82.8 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 62.8 cm, and the two straight sides to 20 cm; together they are the same 82.8.

Another way: Treating the circles as if their contact hid part of the edge would give less than 82.8 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 8 medium answer: 99.36 cm

The figure at the right is the net of a cylinder whose base has a circumference of 18.84 cm18.84\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

12 cm base circumference 18.84 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 12 cm tall between two circles, and the base is 18.84 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 18.84 cm.
  • The rectangle is 12 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=18.84\text{width} = 18.84
18.84 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(18.84+12)×2=61.68(18.84 + 12) \times 2 = 61.68
61.68 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
18.84×2=37.6818.84 \times 2 = 37.68
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
61.68+37.68=99.3661.68 + 37.68 = 99.36
99.36 cm round the net.
Answer: 99.36 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 75.36 cm, and the two straight sides to 24 cm; together they are the same 99.36.

Another way: Treating the circles as if their contact hid part of the edge would give less than 99.36 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 9 hard answer: 99.92 cm

The figure at the right is the net of a cylinder whose base has a circumference of 21.98 cm21.98\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

6 cm base circumference 21.98 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 6 cm tall between two circles, and the base is 21.98 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 21.98 cm.
  • The rectangle is 6 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=21.98\text{width} = 21.98
21.98 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(21.98+6)×2=55.96(21.98 + 6) \times 2 = 55.96
55.96 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
21.98×2=43.9621.98 \times 2 = 43.96
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
55.96+43.96=99.9255.96 + 43.96 = 99.92
99.92 cm round the net.
Answer: 99.92 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 87.92 cm, and the two straight sides to 12 cm; together they are the same 99.92.

Another way: Treating the circles as if their contact hid part of the edge would give less than 99.92 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 10 hard answer: 130.48 cm

The figure at the right is the net of a cylinder whose base has a circumference of 25.12 cm25.12\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

15 cm base circumference 25.12 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 15 cm tall between two circles, and the base is 25.12 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 25.12 cm.
  • The rectangle is 15 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=25.12\text{width} = 25.12
25.12 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(25.12+15)×2=80.24(25.12 + 15) \times 2 = 80.24
80.24 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
25.12×2=50.2425.12 \times 2 = 50.24
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
80.24+50.24=130.4880.24 + 50.24 = 130.48
130.48 cm round the net.
Answer: 130.48 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 100.48 cm, and the two straight sides to 30 cm; together they are the same 130.48.

Another way: Treating the circles as if their contact hid part of the edge would give less than 130.48 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 11 hard answer: 165.6 cm

The figure at the right is the net of a cylinder whose base has a circumference of 31.4 cm31.4\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

20 cm base circumference 31.4 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 20 cm tall between two circles, and the base is 31.4 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 31.4 cm.
  • The rectangle is 20 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=31.4\text{width} = 31.4
31.4 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(31.4+20)×2=102.8(31.4 + 20) \times 2 = 102.8
102.8 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
31.4×2=62.831.4 \times 2 = 62.8
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
102.8+62.8=165.6102.8 + 62.8 = 165.6
165.6 cm round the net.
Answer: 165.6 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 125.6 cm, and the two straight sides to 40 cm; together they are the same 165.6.

Another way: Treating the circles as if their contact hid part of the edge would give less than 165.6 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.
Variant 12 hard answer: 200.72 cm

The figure at the right is the net of a cylinder whose base has a circumference of 37.68 cm37.68\ \text{cm}. What is the perimeter of this net, in cm\text{cm}?

25 cm base circumference 37.68 cm
Show solution
1 · Understandwhat's really being asked

A cylinder's net has a rectangle 25 cm tall between two circles, and the base is 37.68 cm round. We want the distance round the whole net.

Givens
  • The base's circumference is 37.68 cm.
  • The rectangle is 25 cm tall.
  • A circle is attached above the rectangle and another below.
Unknowns
  • The perimeter of the net.
Constraints
  • The net folds into a cylinder, so its parts fit each other exactly.
2 · Planchoose the strategy

#17 Visualize Spatial Relationships · also uses: #1 Draw a Diagram#7 Identify Subproblems

Two lengths have to be read off the folding rather than measured: the rectangle's width, which becomes the rim, and how much of each circle is really on the outside.

3 · Execute4 carry out the plan

1Find the rectangle's width

#17 Visualize Spatial Relationships 6.G.A.4
Rolled up, that edge wraps once round the base, so it is the circumference.
width=37.68\text{width} = 37.68
37.68 cm wide.

2Go round the rectangle

#7 Identify Subproblems 7.G.B.6
Two widths and two heights.
(37.68+25)×2=125.36(37.68 + 25) \times 2 = 125.36
125.36 cm so far.

3Add the circles in full

#1 Draw a Diagram 7.G.B.4
Each circle meets the rectangle at a single point, so nothing of its edge is hidden.
37.68×2=75.3637.68 \times 2 = 75.36
Both circles count whole.

4Add the two parts

#7 Identify Subproblems 7.G.B.6
The rectangle's outline plus the two circles.
125.36+75.36=200.72125.36 + 75.36 = 200.72
200.72 cm round the net.
Answer: 200.72 cm
4 · Reviewdoes it hold up?

The four curved stretches -- two circles and the two widths that become rims -- come to 150.72 cm, and the two straight sides to 50 cm; together they are the same 200.72.

Another way: Treating the circles as if their contact hid part of the edge would give less than 200.72 cm, but a circle touching a line meets it at one point only, which has no length.

Standardsmin grade 7
  • 7.G.B.4 Know the formulas for area and circumference of a circle — Using the base's circumference for the circles and the width.
  • 6.G.A.4 Surface area using nets of 3D figures — Reading the net's widths from how it folds.
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume — Totalling the net's boundary.
💡Takeaway. On a net, a side that rolls into a rim is exactly as long as that rim -- and a circle touching a line loses nothing.