← Ribbon runs match the box's parallel edges · Net and Solid Structure

Ribbon runs match the box's parallel edges · 12 practice problems

4.G.A.24.MD.A.24.MD.A.3

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Generated variants — 12

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

Variant 1 easy

The rectangular box shown at the right is tied with ribbon. If $15\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $10\ \text{cm}$ wide, $6\ \text{cm}$ deep, and $5\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 10 cm by 6 cm by 5 cm box is tied with ribbon going around it twice, in two directions, plus 15 cm for the bow. We need the total ribbon.

Givens

The box is 10 cm wide, 6 cm deep and 5 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 15 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 10 + 2 \times 5 = 20 + 10 = 30

30 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 6 + 2 \times 5 = 12 + 10 = 22

22 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

30 + 22 = 52

52 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 15 cm used for the bow (knot) on the top.

52 + 15 = 67

67 cm of ribbon in all.

Answer: 67 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 5 = 20), the width twice (2 x 10 = 20) and the depth twice (2 x 6 = 12). That is 52 cm, plus 15 for the bow = 67 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 2 easy

The rectangular box shown at the right is tied with ribbon. If $18\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $14\ \text{cm}$ wide, $10\ \text{cm}$ deep, and $7\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 14 cm by 10 cm by 7 cm box is tied with ribbon going around it twice, in two directions, plus 18 cm for the bow. We need the total ribbon.

Givens

The box is 14 cm wide, 10 cm deep and 7 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 18 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 14 + 2 \times 7 = 28 + 14 = 42

42 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 10 + 2 \times 7 = 20 + 14 = 34

34 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

42 + 34 = 76

76 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 18 cm used for the bow (knot) on the top.

76 + 18 = 94

94 cm of ribbon in all.

Answer: 94 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 7 = 28), the width twice (2 x 14 = 28) and the depth twice (2 x 10 = 20). That is 76 cm, plus 18 for the bow = 94 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 3 easy

The rectangular box shown at the right is tied with ribbon. If $20\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $16\ \text{cm}$ wide, $11\ \text{cm}$ deep, and $9\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 16 cm by 11 cm by 9 cm box is tied with ribbon going around it twice, in two directions, plus 20 cm for the bow. We need the total ribbon.

Givens

The box is 16 cm wide, 11 cm deep and 9 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 20 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 16 + 2 \times 9 = 32 + 18 = 50

50 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 11 + 2 \times 9 = 22 + 18 = 40

40 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

50 + 40 = 90

90 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 20 cm used for the bow (knot) on the top.

90 + 20 = 110

110 cm of ribbon in all.

Answer: 110 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 9 = 36), the width twice (2 x 16 = 32) and the depth twice (2 x 11 = 22). That is 90 cm, plus 20 for the bow = 110 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 4 easy

The rectangular box shown at the right is tied with ribbon. If $25\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $20\ \text{cm}$ wide, $14\ \text{cm}$ deep, and $10\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 20 cm by 14 cm by 10 cm box is tied with ribbon going around it twice, in two directions, plus 25 cm for the bow. We need the total ribbon.

Givens

The box is 20 cm wide, 14 cm deep and 10 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 25 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 20 + 2 \times 10 = 40 + 20 = 60

60 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 14 + 2 \times 10 = 28 + 20 = 48

48 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

60 + 48 = 108

108 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 25 cm used for the bow (knot) on the top.

108 + 25 = 133

133 cm of ribbon in all.

Answer: 133 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 10 = 40), the width twice (2 x 20 = 40) and the depth twice (2 x 14 = 28). That is 108 cm, plus 25 for the bow = 133 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 5 medium

The rectangular box shown at the right is tied with ribbon. If $28\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $22\ \text{cm}$ wide, $17\ \text{cm}$ deep, and $26\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 22 cm by 17 cm by 26 cm box is tied with ribbon going around it twice, in two directions, plus 28 cm for the bow. We need the total ribbon.

Givens

The box is 22 cm wide, 17 cm deep and 26 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 28 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 22 + 2 \times 26 = 44 + 52 = 96

96 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 17 + 2 \times 26 = 34 + 52 = 86

86 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

96 + 86 = 182

182 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 28 cm used for the bow (knot) on the top.

182 + 28 = 210

210 cm of ribbon in all.

Answer: 210 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 26 = 104), the width twice (2 x 22 = 44) and the depth twice (2 x 17 = 34). That is 182 cm, plus 28 for the bow = 210 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 6 medium

The rectangular box shown at the right is tied with ribbon. If $30\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $12\ \text{cm}$ wide, $8\ \text{cm}$ deep, and $15\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 12 cm by 8 cm by 15 cm box is tied with ribbon going around it twice, in two directions, plus 30 cm for the bow. We need the total ribbon.

Givens

The box is 12 cm wide, 8 cm deep and 15 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 30 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 12 + 2 \times 15 = 24 + 30 = 54

54 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 8 + 2 \times 15 = 16 + 30 = 46

46 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

54 + 46 = 100

100 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 30 cm used for the bow (knot) on the top.

100 + 30 = 130

130 cm of ribbon in all.

Answer: 130 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 15 = 60), the width twice (2 x 12 = 24) and the depth twice (2 x 8 = 16). That is 100 cm, plus 30 for the bow = 130 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 7 medium

The rectangular box shown at the right is tied with ribbon. If $32\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $28\ \text{cm}$ wide, $19\ \text{cm}$ deep, and $13\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 28 cm by 19 cm by 13 cm box is tied with ribbon going around it twice, in two directions, plus 32 cm for the bow. We need the total ribbon.

Givens

The box is 28 cm wide, 19 cm deep and 13 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 32 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 28 + 2 \times 13 = 56 + 26 = 82

82 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 19 + 2 \times 13 = 38 + 26 = 64

64 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

82 + 64 = 146

146 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 32 cm used for the bow (knot) on the top.

146 + 32 = 178

178 cm of ribbon in all.

Answer: 178 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 13 = 52), the width twice (2 x 28 = 56) and the depth twice (2 x 19 = 38). That is 146 cm, plus 32 for the bow = 178 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 8 medium

The rectangular box shown at the right is tied with ribbon. If $35\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $25\ \text{cm}$ wide, $20\ \text{cm}$ deep, and $16\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 25 cm by 20 cm by 16 cm box is tied with ribbon going around it twice, in two directions, plus 35 cm for the bow. We need the total ribbon.

Givens

The box is 25 cm wide, 20 cm deep and 16 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 35 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 25 + 2 \times 16 = 50 + 32 = 82

82 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 20 + 2 \times 16 = 40 + 32 = 72

72 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

82 + 72 = 154

154 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 35 cm used for the bow (knot) on the top.

154 + 35 = 189

189 cm of ribbon in all.

Answer: 189 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 16 = 64), the width twice (2 x 25 = 50) and the depth twice (2 x 20 = 40). That is 154 cm, plus 35 for the bow = 189 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 9 hard

The rectangular box shown at the right is tied with ribbon. If $40\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $18\ \text{cm}$ wide, $9\ \text{cm}$ deep, and $12\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 18 cm by 9 cm by 12 cm box is tied with ribbon going around it twice, in two directions, plus 40 cm for the bow. We need the total ribbon.

Givens

The box is 18 cm wide, 9 cm deep and 12 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 40 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 18 + 2 \times 12 = 36 + 24 = 60

60 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 9 + 2 \times 12 = 18 + 24 = 42

42 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

60 + 42 = 102

102 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 40 cm used for the bow (knot) on the top.

102 + 40 = 142

142 cm of ribbon in all.

Answer: 142 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 12 = 48), the width twice (2 x 18 = 36) and the depth twice (2 x 9 = 18). That is 102 cm, plus 40 for the bow = 142 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 10 hard

The rectangular box shown at the right is tied with ribbon. If $45\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $24\ \text{cm}$ wide, $15\ \text{cm}$ deep, and $20\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 24 cm by 15 cm by 20 cm box is tied with ribbon going around it twice, in two directions, plus 45 cm for the bow. We need the total ribbon.

Givens

The box is 24 cm wide, 15 cm deep and 20 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 45 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 24 + 2 \times 20 = 48 + 40 = 88

88 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 15 + 2 \times 20 = 30 + 40 = 70

70 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

88 + 70 = 158

158 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 45 cm used for the bow (knot) on the top.

158 + 45 = 203

203 cm of ribbon in all.

Answer: 203 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 20 = 80), the width twice (2 x 24 = 48) and the depth twice (2 x 15 = 30). That is 158 cm, plus 45 for the bow = 203 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 11 hard

The rectangular box shown at the right is tied with ribbon. If $50\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $30\ \text{cm}$ wide, $22\ \text{cm}$ deep, and $18\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 30 cm by 22 cm by 18 cm box is tied with ribbon going around it twice, in two directions, plus 50 cm for the bow. We need the total ribbon.

Givens

The box is 30 cm wide, 22 cm deep and 18 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 50 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 30 + 2 \times 18 = 60 + 36 = 96

96 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 22 + 2 \times 18 = 44 + 36 = 80

80 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

96 + 80 = 176

176 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 50 cm used for the bow (knot) on the top.

176 + 50 = 226

226 cm of ribbon in all.

Answer: 226 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 18 = 72), the width twice (2 x 30 = 60) and the depth twice (2 x 22 = 44). That is 176 cm, plus 50 for the bow = 226 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Variant 12 hard

The rectangular box shown at the right is tied with ribbon. If $60\ \text{cm}$ of ribbon was used to tie the bow (knot), find the total length, in $\text{cm}$ , of ribbon used to wrap the box.

The box is a rectangular prism that is $35\ \text{cm}$ wide, $28\ \text{cm}$ deep, and $22\ \text{cm}$ tall. The ribbon goes once around the box in each of two directions, over the top and bottom and along the sides, and is finished with a bow on the top face.

Your answer

Show solution

1 · Understandwhat's really being asked

A 35 cm by 28 cm by 22 cm box is tied with ribbon going around it twice, in two directions, plus 60 cm for the bow. We need the total ribbon.

Givens

The box is 35 cm wide, 28 cm deep and 22 cm tall.
The ribbon goes once around in each of two perpendicular directions.
The bow uses 60 cm.

Unknowns

The total length of ribbon.

Constraints

Each straight run of ribbon lies flat against the box.

2 · Planchoose the strategy

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

Follow the ribbon on a drawing. Every straight piece runs beside one of the box's edges, so its length is that edge's length -- which turns the whole thing into two rectangle perimeters plus the bow.

3 · Execute5 carry out the plan

1See that ribbon pieces match box edges

#1 Draw a Diagram 4.G.A.2

Draw the box and follow the ribbon. Each straight portion lies flat against the box and runs parallel to one of the box's edges, so each piece is exactly as long as that edge.

No ribbon length is a mystery -- each is an edge.

2Length of the first loop

#7 Identify Subproblems 4.MD.A.3

The first loop wraps around the face made of the width and the height. Going over the top, down one side, across the bottom, and up the other side, it covers two widths and two heights.

2 \times 35 + 2 \times 22 = 70 + 44 = 114

114 cm for the first time around.

3Length of the second loop

#7 Identify Subproblems 4.MD.A.3

The second loop is perpendicular to the first and wraps around the face made of the depth and the height. It covers two depths and two heights.

2 \times 28 + 2 \times 22 = 56 + 44 = 100

100 cm for the second time around.

4Add the two loops

#7 Identify Subproblems 4.MD.A.2

Add the ribbon used by the two loops to get the length that actually goes around the box.

114 + 100 = 214

214 cm wrapped around.

5Add the bow

#7 Identify Subproblems 4.MD.A.2

Finally, add the 60 cm used for the bow (knot) on the top.

214 + 60 = 274

274 cm of ribbon in all.

Answer: 274 cm

4 · Reviewdoes it hold up?

Group by direction instead: the ribbon crosses the height four times (4 x 22 = 88), the width twice (2 x 35 = 70) and the depth twice (2 x 28 = 56). That is 214 cm, plus 60 for the bow = 274 cm.

Another way: A common slip is to count the height only twice. The height belongs to both loops, which is why it appears four times and not two.

Standardsmin grade 4

4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Matching each ribbon run to the box edge it runs parallel to.
4.MD.A.2 Solve word problems involving distances and measurement quantities — Adding the loops and the bow into one total length.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reading each loop as the perimeter of a rectangle.

💡Takeaway. Ribbon lying flat on a box is just the box's edges in disguise, and the height gets crossed by both loops.

Where next?

← Back to the problem: Ribbon runs match the box's parallel edges