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← 3-2 · Find a rectangle side via one unknown · Find Two Unknowns from Sum and Difference

Find a rectangle side via one unknown · 12 practice problems

3.MD.D.83.OA.C.73.OA.A.3

Generated variants — 12

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

Variant 1 answer: 33 inches

Liam used a piece of wire 8888 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 33 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 88-inch wire is bent into a rectangle whose long side is 3 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 88 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 3 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 88 inches (all wire used).
  • Long side = 3 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (3 shorts), and again, giving 8 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 3 shorts, so the perimeter is 1 + 3 + 1 + 3 = 8 short-side lengths.
1+3+1+3=8 short sides1 + 3 + 1 + 3 = 8 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 8 equal pieces.
#6 Guess and Check 3.OA.C.7
The 88-inch wire is divided into 8 equal short-side pieces.
88÷8=11 inches88 \div 8 = 11 \text{ inches}
Sharing the total length equally among the 8 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 3 times the short side, so multiply 11 by 3.
11×3=33 inches11 \times 3 = 33 \text{ inches}
The long side is 3 short pieces joined, so scale the short side by 3.
Answer: 33 inches

Review

Check: short 11, long 33, perimeter = 2 times (11 + 33) = 2 times 44 = 88 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 3s) = 8s = 88, so s = 11 and long side = 3s = 33.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 88-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 88 by 8 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 3 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 8 short sides, so divide, then multiply for the long side!
Variant 2 answer: 28 inches

Mason used a piece of wire 6464 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 77 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 64-inch wire is bent into a rectangle whose long side is 7 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 64 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 7 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 64 inches (all wire used).
  • Long side = 7 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (7 shorts), and again, giving 16 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 7 shorts, so the perimeter is 1 + 7 + 1 + 7 = 16 short-side lengths.
1+7+1+7=16 short sides1 + 7 + 1 + 7 = 16 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 16 equal pieces.
#6 Guess and Check 3.OA.C.7
The 64-inch wire is divided into 16 equal short-side pieces.
64÷16=4 inches64 \div 16 = 4 \text{ inches}
Sharing the total length equally among the 16 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 7 times the short side, so multiply 4 by 7.
4×7=28 inches4 \times 7 = 28 \text{ inches}
The long side is 7 short pieces joined, so scale the short side by 7.
Answer: 28 inches

Review

Check: short 4, long 28, perimeter = 2 times (4 + 28) = 2 times 32 = 64 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 7s) = 16s = 64, so s = 4 and long side = 7s = 28.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 64-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 64 by 16 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 7 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 16 short sides, so divide, then multiply for the long side!
Variant 3 answer: 25 inches

Noah used a piece of wire 6060 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 55 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 60-inch wire is bent into a rectangle whose long side is 5 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 60 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 5 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 60 inches (all wire used).
  • Long side = 5 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (5 shorts), and again, giving 12 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 5 shorts, so the perimeter is 1 + 5 + 1 + 5 = 12 short-side lengths.
1+5+1+5=12 short sides1 + 5 + 1 + 5 = 12 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 12 equal pieces.
#6 Guess and Check 3.OA.C.7
The 60-inch wire is divided into 12 equal short-side pieces.
60÷12=5 inches60 \div 12 = 5 \text{ inches}
Sharing the total length equally among the 12 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 5 times the short side, so multiply 5 by 5.
5×5=25 inches5 \times 5 = 25 \text{ inches}
The long side is 5 short pieces joined, so scale the short side by 5.
Answer: 25 inches

Review

Check: short 5, long 25, perimeter = 2 times (5 + 25) = 2 times 30 = 60 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 5s) = 12s = 60, so s = 5 and long side = 5s = 25.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 60-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 60 by 12 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 5 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 12 short sides, so divide, then multiply for the long side!
Variant 4 answer: 24 inches

Emma used a piece of wire 7272 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 22 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 72-inch wire is bent into a rectangle whose long side is 2 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 72 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 2 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 72 inches (all wire used).
  • Long side = 2 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (2 shorts), and again, giving 6 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 2 shorts, so the perimeter is 1 + 2 + 1 + 2 = 6 short-side lengths.
1+2+1+2=6 short sides1 + 2 + 1 + 2 = 6 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 6 equal pieces.
#6 Guess and Check 3.OA.C.7
The 72-inch wire is divided into 6 equal short-side pieces.
72÷6=12 inches72 \div 6 = 12 \text{ inches}
Sharing the total length equally among the 6 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 2 times the short side, so multiply 12 by 2.
12×2=24 inches12 \times 2 = 24 \text{ inches}
The long side is 2 short pieces joined, so scale the short side by 2.
Answer: 24 inches

Review

Check: short 12, long 24, perimeter = 2 times (12 + 24) = 2 times 36 = 72 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 2s) = 6s = 72, so s = 12 and long side = 2s = 24.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 72-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 72 by 6 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 2 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 6 short sides, so divide, then multiply for the long side!
Variant 5 answer: 36 inches

Ava used a piece of wire 9696 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 33 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 96-inch wire is bent into a rectangle whose long side is 3 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 96 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 3 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 96 inches (all wire used).
  • Long side = 3 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (3 shorts), and again, giving 8 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 3 shorts, so the perimeter is 1 + 3 + 1 + 3 = 8 short-side lengths.
1+3+1+3=8 short sides1 + 3 + 1 + 3 = 8 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 8 equal pieces.
#6 Guess and Check 3.OA.C.7
The 96-inch wire is divided into 8 equal short-side pieces.
96÷8=12 inches96 \div 8 = 12 \text{ inches}
Sharing the total length equally among the 8 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 3 times the short side, so multiply 12 by 3.
12×3=36 inches12 \times 3 = 36 \text{ inches}
The long side is 3 short pieces joined, so scale the short side by 3.
Answer: 36 inches

Review

Check: short 12, long 36, perimeter = 2 times (12 + 36) = 2 times 48 = 96 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 3s) = 8s = 96, so s = 12 and long side = 3s = 36.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 96-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 96 by 8 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 3 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 8 short sides, so divide, then multiply for the long side!
Variant 6 answer: 18 inches

Lucas used a piece of wire 5454 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 22 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 54-inch wire is bent into a rectangle whose long side is 2 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 54 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 2 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 54 inches (all wire used).
  • Long side = 2 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (2 shorts), and again, giving 6 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 2 shorts, so the perimeter is 1 + 2 + 1 + 2 = 6 short-side lengths.
1+2+1+2=6 short sides1 + 2 + 1 + 2 = 6 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 6 equal pieces.
#6 Guess and Check 3.OA.C.7
The 54-inch wire is divided into 6 equal short-side pieces.
54÷6=9 inches54 \div 6 = 9 \text{ inches}
Sharing the total length equally among the 6 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 2 times the short side, so multiply 9 by 2.
9×2=18 inches9 \times 2 = 18 \text{ inches}
The long side is 2 short pieces joined, so scale the short side by 2.
Answer: 18 inches

Review

Check: short 9, long 18, perimeter = 2 times (9 + 18) = 2 times 27 = 54 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 2s) = 6s = 54, so s = 9 and long side = 2s = 18.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 54-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 54 by 6 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 2 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 6 short sides, so divide, then multiply for the long side!
Variant 7 answer: 16 inches

Olivia used a piece of wire 4040 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 44 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 40-inch wire is bent into a rectangle whose long side is 4 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 40 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 4 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 40 inches (all wire used).
  • Long side = 4 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (4 shorts), and again, giving 10 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 4 shorts, so the perimeter is 1 + 4 + 1 + 4 = 10 short-side lengths.
1+4+1+4=10 short sides1 + 4 + 1 + 4 = 10 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 10 equal pieces.
#6 Guess and Check 3.OA.C.7
The 40-inch wire is divided into 10 equal short-side pieces.
40÷10=4 inches40 \div 10 = 4 \text{ inches}
Sharing the total length equally among the 10 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 4 times the short side, so multiply 4 by 4.
4×4=16 inches4 \times 4 = 16 \text{ inches}
The long side is 4 short pieces joined, so scale the short side by 4.
Answer: 16 inches

Review

Check: short 4, long 16, perimeter = 2 times (4 + 16) = 2 times 20 = 40 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 4s) = 10s = 40, so s = 4 and long side = 4s = 16.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 40-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 40 by 10 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 4 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 10 short sides, so divide, then multiply for the long side!
Variant 8 answer: 35 inches

Ethan used a piece of wire 8484 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 55 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 84-inch wire is bent into a rectangle whose long side is 5 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 84 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 5 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 84 inches (all wire used).
  • Long side = 5 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (5 shorts), and again, giving 12 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 5 shorts, so the perimeter is 1 + 5 + 1 + 5 = 12 short-side lengths.
1+5+1+5=12 short sides1 + 5 + 1 + 5 = 12 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 12 equal pieces.
#6 Guess and Check 3.OA.C.7
The 84-inch wire is divided into 12 equal short-side pieces.
84÷12=7 inches84 \div 12 = 7 \text{ inches}
Sharing the total length equally among the 12 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 5 times the short side, so multiply 7 by 5.
7×5=35 inches7 \times 5 = 35 \text{ inches}
The long side is 5 short pieces joined, so scale the short side by 5.
Answer: 35 inches

Review

Check: short 7, long 35, perimeter = 2 times (7 + 35) = 2 times 42 = 84 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 5s) = 12s = 84, so s = 7 and long side = 5s = 35.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 84-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 84 by 12 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 5 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 12 short sides, so divide, then multiply for the long side!
Variant 9 answer: 20 inches

Sophia used a piece of wire 5050 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 44 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 50-inch wire is bent into a rectangle whose long side is 4 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 50 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 4 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 50 inches (all wire used).
  • Long side = 4 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (4 shorts), and again, giving 10 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 4 shorts, so the perimeter is 1 + 4 + 1 + 4 = 10 short-side lengths.
1+4+1+4=10 short sides1 + 4 + 1 + 4 = 10 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 10 equal pieces.
#6 Guess and Check 3.OA.C.7
The 50-inch wire is divided into 10 equal short-side pieces.
50÷10=5 inches50 \div 10 = 5 \text{ inches}
Sharing the total length equally among the 10 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 4 times the short side, so multiply 5 by 4.
5×4=20 inches5 \times 4 = 20 \text{ inches}
The long side is 4 short pieces joined, so scale the short side by 4.
Answer: 20 inches

Review

Check: short 5, long 20, perimeter = 2 times (5 + 20) = 2 times 25 = 50 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 4s) = 10s = 50, so s = 5 and long side = 4s = 20.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 50-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 50 by 10 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 4 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 10 short sides, so divide, then multiply for the long side!
Variant 10 answer: 28 inches

Jackson used a piece of wire 8484 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 22 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 84-inch wire is bent into a rectangle whose long side is 2 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 84 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 2 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 84 inches (all wire used).
  • Long side = 2 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (2 shorts), and again, giving 6 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 2 shorts, so the perimeter is 1 + 2 + 1 + 2 = 6 short-side lengths.
1+2+1+2=6 short sides1 + 2 + 1 + 2 = 6 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 6 equal pieces.
#6 Guess and Check 3.OA.C.7
The 84-inch wire is divided into 6 equal short-side pieces.
84÷6=14 inches84 \div 6 = 14 \text{ inches}
Sharing the total length equally among the 6 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 2 times the short side, so multiply 14 by 2.
14×2=28 inches14 \times 2 = 28 \text{ inches}
The long side is 2 short pieces joined, so scale the short side by 2.
Answer: 28 inches

Review

Check: short 14, long 28, perimeter = 2 times (14 + 28) = 2 times 42 = 84 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 2s) = 6s = 84, so s = 14 and long side = 2s = 28.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 84-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 84 by 6 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 2 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 6 short sides, so divide, then multiply for the long side!
Variant 11 answer: 28 inches

Chloe used a piece of wire 7070 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 44 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 70-inch wire is bent into a rectangle whose long side is 4 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 70 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 4 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 70 inches (all wire used).
  • Long side = 4 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (4 shorts), and again, giving 10 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 4 shorts, so the perimeter is 1 + 4 + 1 + 4 = 10 short-side lengths.
1+4+1+4=10 short sides1 + 4 + 1 + 4 = 10 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 10 equal pieces.
#6 Guess and Check 3.OA.C.7
The 70-inch wire is divided into 10 equal short-side pieces.
70÷10=7 inches70 \div 10 = 7 \text{ inches}
Sharing the total length equally among the 10 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 4 times the short side, so multiply 7 by 4.
7×4=28 inches7 \times 4 = 28 \text{ inches}
The long side is 4 short pieces joined, so scale the short side by 4.
Answer: 28 inches

Review

Check: short 7, long 28, perimeter = 2 times (7 + 28) = 2 times 35 = 70 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 4s) = 10s = 70, so s = 7 and long side = 4s = 28.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 70-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 70 by 10 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 4 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 10 short sides, so divide, then multiply for the long side!
Variant 12 answer: 18 inches

Mia used a piece of wire 4848 inches long, using all of it, to bend a rectangle. The long side of the rectangle is 33 times the length of the short side. How many inches long is the long side?

Show solution

Understand

An 48-inch wire is bent into a rectangle whose long side is 3 times the short side, using all the wire. We need the length of the long side.

Givens
  • The wire is 48 inches long and all of it is used.
  • It forms a rectangle (the wire is the perimeter).
  • The long side is 3 times the short side.
Unknowns
  • The length of the long side in inches.
Constraints
  • The perimeter equals 48 inches (all wire used).
  • Long side = 3 times short side.

Plan

#1 Draw a Diagram · also uses: #6 Guess and Check

Sketch the rectangle and count the short-side units around the perimeter: a short, a long (3 shorts), and again, giving 8 equal short-side pieces; divide the wire among them.

Execute

#1 Draw a Diagram 3.MD.D.8
Going around: short + long + short + long. Each long is 3 shorts, so the perimeter is 1 + 3 + 1 + 3 = 8 short-side lengths.
1+3+1+3=8 short sides1 + 3 + 1 + 3 = 8 \text{ short sides}
Measuring the whole border in short-side units turns the wire into 8 equal pieces.
#6 Guess and Check 3.OA.C.7
The 48-inch wire is divided into 8 equal short-side pieces.
48÷8=6 inches48 \div 8 = 6 \text{ inches}
Sharing the total length equally among the 8 units gives one short side.
#1 Draw a Diagram 3.OA.A.3
The long side is 3 times the short side, so multiply 6 by 3.
6×3=18 inches6 \times 3 = 18 \text{ inches}
The long side is 3 short pieces joined, so scale the short side by 3.
Answer: 18 inches

Review

Check: short 6, long 18, perimeter = 2 times (6 + 18) = 2 times 24 = 48 inches, matching the wire exactly. The long side being longer than the short fits the picture.

Convert to algebra (tool 13): with short side s, perimeter 2(s + 3s) = 8s = 48, so s = 6 and long side = 3s = 18.

Standards · min grade 3

  • 3.MD.D.8 Solve real-world problems involving perimeters of polygons — Expressing the 48-inch perimeter as a sum of short-side units.
  • 3.OA.C.7 Fluently multiply and divide within 100 — Dividing 48 by 8 to find the short side.
  • 3.OA.A.3 Solve multiplication and division word problems within 100 — Scaling the short side by 3 to get the long side.
💡 This only needs Grade 3 perimeter sense: the border is 8 short sides, so divide, then multiply for the long side!