← 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

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Generated variants — 12

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

Variant 1 easy

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

Your answer

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.

\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) \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 \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 = 18.56

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

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

Your answer

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.

\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) \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 \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 = 33.12

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

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

Your answer

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.

\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) \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 \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 = 32.84

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

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

Your answer

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.

\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) \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 \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 = 49.4

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

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

Your answer

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.

\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) \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 \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 = 47.68

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

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

Your answer

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.

\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) \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 \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 = 66.24

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

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

Your answer

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.

\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) \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 \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 = 82.8

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

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

Your answer

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.

\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) \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 \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 = 99.36

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

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

Your answer

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.

\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) \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 \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 = 99.92

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

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

Your answer

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.

\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) \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 \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 = 130.48

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

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

Your answer

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.

\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) \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 \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 = 165.6

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

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

Your answer

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.

\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) \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 \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 = 200.72

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.

Where next?

← Back to the problem: The side that rolls into a rim is exactly as long as that rim