Practice · Ribbon around a box: four equal parts · Sensim Math

Variant 1 easy

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $40\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $20\ \text{cm}$ (one side) and $25\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $30\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 20 cm and 25 cm and height 30 cm is tied with ribbon that wraps around once in each direction, plus 40 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 20 cm and 25 cm; the height (vertical edge) is 30 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 40 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 20 cm side and the height; the other circles the 25 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 20 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (20 cm), down one side (30 cm), under the bottom (20 cm), and up the other side (30 cm): two 20 cm edges and two heights.

20 + 30 + 20 + 30 = 100 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 25 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 25 cm edges and two heights of 30 cm.

25 + 30 + 25 + 30 = 110 \text{ cm}

Same loop idea, now around the 25 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 40 cm bow.

100 + 110 + 40 = 250 \text{ cm} = 2\,\text{m}\ 50\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 2 m 50 cm

4 · Reviewdoes it hold up?

Each loop is 100 cm and 110 cm, and adding a 40 cm bow gives 2 m 50 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 20 (40) + 2 of 25 (50) + 4 heights (120) + bow 40 = 250 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Variant 2 medium

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $45\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $35\ \text{cm}$ (one side) and $35\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $35\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 35 cm and 35 cm and height 35 cm is tied with ribbon that wraps around once in each direction, plus 45 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 35 cm and 35 cm; the height (vertical edge) is 35 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 45 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 35 cm side and the height; the other circles the 35 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 35 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (35 cm), down one side (35 cm), under the bottom (35 cm), and up the other side (35 cm): two 35 cm edges and two heights.

35 + 35 + 35 + 35 = 140 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 35 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 35 cm edges and two heights of 35 cm.

35 + 35 + 35 + 35 = 140 \text{ cm}

Same loop idea, now around the 35 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 45 cm bow.

140 + 140 + 45 = 325 \text{ cm} = 3\,\text{m}\ 25\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 3 m 25 cm

4 · Reviewdoes it hold up?

Each loop is 140 cm and 140 cm, and adding a 45 cm bow gives 3 m 25 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 35 (70) + 2 of 35 (70) + 4 heights (140) + bow 45 = 325 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Variant 3 medium

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $30\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $15\ \text{cm}$ (one side) and $45\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $35\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 15 cm and 45 cm and height 35 cm is tied with ribbon that wraps around once in each direction, plus 30 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 15 cm and 45 cm; the height (vertical edge) is 35 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 30 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 15 cm side and the height; the other circles the 45 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 15 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (15 cm), down one side (35 cm), under the bottom (15 cm), and up the other side (35 cm): two 15 cm edges and two heights.

15 + 35 + 15 + 35 = 100 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 45 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 45 cm edges and two heights of 35 cm.

45 + 35 + 45 + 35 = 160 \text{ cm}

Same loop idea, now around the 45 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 30 cm bow.

100 + 160 + 30 = 290 \text{ cm} = 2\,\text{m}\ 90\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 2 m 90 cm

4 · Reviewdoes it hold up?

Each loop is 100 cm and 160 cm, and adding a 30 cm bow gives 2 m 90 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 15 (30) + 2 of 45 (90) + 4 heights (140) + bow 30 = 290 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Variant 4 easy

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $45\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $30\ \text{cm}$ (one side) and $35\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $40\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 30 cm and 35 cm and height 40 cm is tied with ribbon that wraps around once in each direction, plus 45 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 30 cm and 35 cm; the height (vertical edge) is 40 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 45 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 30 cm side and the height; the other circles the 35 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 30 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (30 cm), down one side (40 cm), under the bottom (30 cm), and up the other side (40 cm): two 30 cm edges and two heights.

30 + 40 + 30 + 40 = 140 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 35 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 35 cm edges and two heights of 40 cm.

35 + 40 + 35 + 40 = 150 \text{ cm}

Same loop idea, now around the 35 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 45 cm bow.

140 + 150 + 45 = 335 \text{ cm} = 3\,\text{m}\ 35\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 3 m 35 cm

4 · Reviewdoes it hold up?

Each loop is 140 cm and 150 cm, and adding a 45 cm bow gives 3 m 35 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 30 (60) + 2 of 35 (70) + 4 heights (160) + bow 45 = 335 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Variant 5 medium

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $50\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $25\ \text{cm}$ (one side) and $35\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $20\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 25 cm and 35 cm and height 20 cm is tied with ribbon that wraps around once in each direction, plus 50 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 25 cm and 35 cm; the height (vertical edge) is 20 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 50 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 25 cm side and the height; the other circles the 35 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 25 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (25 cm), down one side (20 cm), under the bottom (25 cm), and up the other side (20 cm): two 25 cm edges and two heights.

25 + 20 + 25 + 20 = 90 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 35 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 35 cm edges and two heights of 20 cm.

35 + 20 + 35 + 20 = 110 \text{ cm}

Same loop idea, now around the 35 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 50 cm bow.

90 + 110 + 50 = 250 \text{ cm} = 2\,\text{m}\ 50\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 2 m 50 cm

4 · Reviewdoes it hold up?

Each loop is 90 cm and 110 cm, and adding a 50 cm bow gives 2 m 50 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 25 (50) + 2 of 35 (70) + 4 heights (80) + bow 50 = 250 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Variant 6 medium

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $55\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $50\ \text{cm}$ (one side) and $30\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $45\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 50 cm and 30 cm and height 45 cm is tied with ribbon that wraps around once in each direction, plus 55 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 50 cm and 30 cm; the height (vertical edge) is 45 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 55 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 50 cm side and the height; the other circles the 30 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 50 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (50 cm), down one side (45 cm), under the bottom (50 cm), and up the other side (45 cm): two 50 cm edges and two heights.

50 + 45 + 50 + 45 = 190 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 30 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 30 cm edges and two heights of 45 cm.

30 + 45 + 30 + 45 = 150 \text{ cm}

Same loop idea, now around the 30 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 55 cm bow.

190 + 150 + 55 = 395 \text{ cm} = 3\,\text{m}\ 95\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 3 m 95 cm

4 · Reviewdoes it hold up?

Each loop is 190 cm and 150 cm, and adding a 55 cm bow gives 3 m 95 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 50 (100) + 2 of 30 (60) + 4 heights (180) + bow 55 = 395 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Variant 7 hard

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $60\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $40\ \text{cm}$ (one side) and $40\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $50\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 40 cm and 40 cm and height 50 cm is tied with ribbon that wraps around once in each direction, plus 60 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 40 cm and 40 cm; the height (vertical edge) is 50 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 60 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 40 cm side and the height; the other circles the 40 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 40 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (40 cm), down one side (50 cm), under the bottom (40 cm), and up the other side (50 cm): two 40 cm edges and two heights.

40 + 50 + 40 + 50 = 180 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 40 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 40 cm edges and two heights of 50 cm.

40 + 50 + 40 + 50 = 180 \text{ cm}

Same loop idea, now around the 40 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 60 cm bow.

180 + 180 + 60 = 420 \text{ cm} = 4\,\text{m}\ 20\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 4 m 20 cm

4 · Reviewdoes it hold up?

Each loop is 180 cm and 180 cm, and adding a 60 cm bow gives 4 m 20 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 40 (80) + 2 of 40 (80) + 4 heights (200) + bow 60 = 420 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Variant 8 hard

Mia wants to tie up a gift box with ribbon, as shown on the right. If the ribbon used for the bow (knot) is $70\ \text{cm}$ long, what is the total length of ribbon needed, in m and cm? (The ribbon wraps around the box exactly once in each direction.)

The top face edges are labeled $60\ \text{cm}$ (one side) and $55\ \text{cm}$ (the other side), and the vertical edge (height) is labeled $40\ \text{cm}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A box with top edges 60 cm and 55 cm and height 40 cm is tied with ribbon that wraps around once in each direction, plus 70 cm of ribbon for the bow. Find the total ribbon length in meters and centimeters.

Givens

The box top edges are 60 cm and 55 cm; the height (vertical edge) is 40 cm.
The ribbon wraps around the box exactly once in each of the two directions.
The bow (knot) uses 70 cm of ribbon.

Unknowns

The total length of ribbon needed, in meters and centimeters.

Constraints

1 m = 100 cm.
Each wrap loop is a rectangle around the box, so it equals 2 of one edge plus 2 of the height.

2 · Planchoose the strategy

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

Picture each wrap as a loop around the box. One loop circles the 60 cm side and the height; the other circles the 55 cm side and the height. Each loop uses two of its base edge and two heights. Add both loops, then add the bow.

3 · Execute3 carry out the plan

1Length of the loop in the 60 cm direction

#7 Identify Subproblems 3.MD.D.8

This loop goes over the top (60 cm), down one side (40 cm), under the bottom (60 cm), and up the other side (40 cm): two 60 cm edges and two heights.

60 + 40 + 60 + 40 = 200 \text{ cm}

Wrapping a loop around the box traces a rectangle, so it uses each edge of that rectangle twice.

2Length of the loop in the 55 cm direction

#7 Identify Subproblems 3.MD.D.8

The other loop uses two 55 cm edges and two heights of 40 cm.

55 + 40 + 55 + 40 = 190 \text{ cm}

Same loop idea, now around the 55 cm cross-section of the box.

3Add both loops and the bow

#1 Draw a Diagram 2.MD.B.5

Sum the two wrap loops, then add the 70 cm bow.

200 + 190 + 70 = 460 \text{ cm} = 4\,\text{m}\ 60\,\text{cm}

Total ribbon is all wrapping plus the extra for the bow; the centimeter total regroups into meters and centimeters.

Answer: 4 m 60 cm

4 · Reviewdoes it hold up?

Each loop is 200 cm and 190 cm, and adding a 70 cm bow gives 4 m 60 cm; the unit (length) matches the question and the total is larger than either loop.

Another way: Group by edge type: 2 of 60 (120) + 2 of 55 (110) + 4 heights (160) + bow 70 = 460 cm, the same total.

Standardsmin grade 3

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Computing each wrap loop as the perimeter of a rectangle around the box.
2.MD.B.5 Solve word problems involving lengths using same units — Adding the two loops and the bow length, then writing it in m and cm.

💡Takeaway. Each ribbon wrap is just a rectangle's perimeter — add both loops and the bow, plain Grade 3 length work!

Ribbon around a box: four equal parts · 8 practice problems

Generated variants — 8

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