Ribbon around a box: four equal parts

2.MD.B.53.MD.D.8 · adapt · grade 3

Archetype: Perimeter by Tracing Every Side · step in a 11-type progression

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.)

Show solution

Understand

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.

Plan

#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.

Execute

#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.

#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.

#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; 335 cm regroups into 3 m and 35 cm.

Answer: 3 m 35 cm

Review

Each loop is a bit over 1 m, and two loops plus a 45 cm bow land near 3.4 m; 3 m 35 cm fits and the unit (length) matches the question.

Count equal parts (the box uses four 30 cm parts? no): group by edge type - 2 thirties (60) + 2 thirty-fives (70) + 4 heights (160) + bow 45 = 335 cm, the same total.

Standards · min 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.

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