← Perimeter is the sum of all side lengths · Perimeter by Tracing Every Side

Perimeter is the sum of all side lengths · 12 practice problems

3.MD.D.85.NF.A.15.NF.A.2

Generated variants — 12

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

Variant 1 easy answer: 2132\frac{1}{3} cm

The perimeter of a rectangle is 99 cm. If its width is 2162\frac{1}{6} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 2162\frac{1}{6} cm. The length of the vertical side is the value to find.

2 1 6 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 99 cm and a width of 2162\frac{1}{6} cm. We need its height.

Givens
  • Perimeter: 99 cm.
  • Width: 2162\frac{1}{6} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 2162\frac{1}{6} cm, and both vertical sides are the height we want.
perimeter=2×216+2×height\text{perimeter} = 2 \times 2\frac{1}{6} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
216+216=4132\frac{1}{6} + 2\frac{1}{6} = 4\frac{1}{3}
4134\frac{1}{3} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
9−413=4239 - 4\frac{1}{3} = 4\frac{2}{3}
4234\frac{2}{3} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
423÷2=2134\frac{2}{3} \div 2 = 2\frac{1}{3}
The height is 2132\frac{1}{3} cm.
Answer: 2132\frac{1}{3} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 2162\frac{1}{6} + 2 x 2132\frac{1}{3} = 99 cm, which is what was given.

Another way: Halve the perimeter first: 99 / 2 = 4124\frac{1}{2} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 2 easy answer: 2142\frac{1}{4} cm

The perimeter of a rectangle is 7567\frac{5}{6} cm. If its width is 1231\frac{2}{3} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 1231\frac{2}{3} cm. The length of the vertical side is the value to find.

1 2 3 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 7567\frac{5}{6} cm and a width of 1231\frac{2}{3} cm. We need its height.

Givens
  • Perimeter: 7567\frac{5}{6} cm.
  • Width: 1231\frac{2}{3} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 1231\frac{2}{3} cm, and both vertical sides are the height we want.
perimeter=2×123+2×height\text{perimeter} = 2 \times 1\frac{2}{3} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
123+123=3131\frac{2}{3} + 1\frac{2}{3} = 3\frac{1}{3}
3133\frac{1}{3} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
756−313=4127\frac{5}{6} - 3\frac{1}{3} = 4\frac{1}{2}
4124\frac{1}{2} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
412÷2=2144\frac{1}{2} \div 2 = 2\frac{1}{4}
The height is 2142\frac{1}{4} cm.
Answer: 2142\frac{1}{4} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 1231\frac{2}{3} + 2 x 2142\frac{1}{4} = 7567\frac{5}{6} cm, which is what was given.

Another way: Halve the perimeter first: 7567\frac{5}{6} / 2 = 311123\frac{11}{12} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 3 easy answer: 1561\frac{5}{6} cm

The perimeter of a rectangle is 8168\frac{1}{6} cm. If its width is 2142\frac{1}{4} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 2142\frac{1}{4} cm. The length of the vertical side is the value to find.

2 1 4 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 8168\frac{1}{6} cm and a width of 2142\frac{1}{4} cm. We need its height.

Givens
  • Perimeter: 8168\frac{1}{6} cm.
  • Width: 2142\frac{1}{4} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 2142\frac{1}{4} cm, and both vertical sides are the height we want.
perimeter=2×214+2×height\text{perimeter} = 2 \times 2\frac{1}{4} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
214+214=4122\frac{1}{4} + 2\frac{1}{4} = 4\frac{1}{2}
4124\frac{1}{2} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
816−412=3238\frac{1}{6} - 4\frac{1}{2} = 3\frac{2}{3}
3233\frac{2}{3} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
323÷2=1563\frac{2}{3} \div 2 = 1\frac{5}{6}
The height is 1561\frac{5}{6} cm.
Answer: 1561\frac{5}{6} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 2142\frac{1}{4} + 2 x 1561\frac{5}{6} = 8168\frac{1}{6} cm, which is what was given.

Another way: Halve the perimeter first: 8168\frac{1}{6} / 2 = 41124\frac{1}{12} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 4 easy answer: 1141\frac{1}{4} cm

The perimeter of a rectangle is 9129\frac{1}{2} cm. If its width is 3123\frac{1}{2} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 3123\frac{1}{2} cm. The length of the vertical side is the value to find.

3 1 2 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 9129\frac{1}{2} cm and a width of 3123\frac{1}{2} cm. We need its height.

Givens
  • Perimeter: 9129\frac{1}{2} cm.
  • Width: 3123\frac{1}{2} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 3123\frac{1}{2} cm, and both vertical sides are the height we want.
perimeter=2×312+2×height\text{perimeter} = 2 \times 3\frac{1}{2} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
312+312=73\frac{1}{2} + 3\frac{1}{2} = 7
77 cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
912−7=2129\frac{1}{2} - 7 = 2\frac{1}{2}
2122\frac{1}{2} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
212÷2=1142\frac{1}{2} \div 2 = 1\frac{1}{4}
The height is 1141\frac{1}{4} cm.
Answer: 1141\frac{1}{4} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 3123\frac{1}{2} + 2 x 1141\frac{1}{4} = 9129\frac{1}{2} cm, which is what was given.

Another way: Halve the perimeter first: 9129\frac{1}{2} / 2 = 4344\frac{3}{4} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 5 medium answer: 1161\frac{1}{6} cm

The perimeter of a rectangle is 9569\frac{5}{6} cm. If its width is 3343\frac{3}{4} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 3343\frac{3}{4} cm. The length of the vertical side is the value to find.

3 3 4 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 9569\frac{5}{6} cm and a width of 3343\frac{3}{4} cm. We need its height.

Givens
  • Perimeter: 9569\frac{5}{6} cm.
  • Width: 3343\frac{3}{4} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 3343\frac{3}{4} cm, and both vertical sides are the height we want.
perimeter=2×334+2×height\text{perimeter} = 2 \times 3\frac{3}{4} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
334+334=7123\frac{3}{4} + 3\frac{3}{4} = 7\frac{1}{2}
7127\frac{1}{2} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
956−712=2139\frac{5}{6} - 7\frac{1}{2} = 2\frac{1}{3}
2132\frac{1}{3} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
213÷2=1162\frac{1}{3} \div 2 = 1\frac{1}{6}
The height is 1161\frac{1}{6} cm.
Answer: 1161\frac{1}{6} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 3343\frac{3}{4} + 2 x 1161\frac{1}{6} = 9569\frac{5}{6} cm, which is what was given.

Another way: Halve the perimeter first: 9569\frac{5}{6} / 2 = 411124\frac{11}{12} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 6 medium answer: 2142\frac{1}{4} cm

The perimeter of a rectangle is 105610\frac{5}{6} cm. If its width is 3163\frac{1}{6} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 3163\frac{1}{6} cm. The length of the vertical side is the value to find.

3 1 6 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 105610\frac{5}{6} cm and a width of 3163\frac{1}{6} cm. We need its height.

Givens
  • Perimeter: 105610\frac{5}{6} cm.
  • Width: 3163\frac{1}{6} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 3163\frac{1}{6} cm, and both vertical sides are the height we want.
perimeter=2×316+2×height\text{perimeter} = 2 \times 3\frac{1}{6} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
316+316=6133\frac{1}{6} + 3\frac{1}{6} = 6\frac{1}{3}
6136\frac{1}{3} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
1056−613=41210\frac{5}{6} - 6\frac{1}{3} = 4\frac{1}{2}
4124\frac{1}{2} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
412÷2=2144\frac{1}{2} \div 2 = 2\frac{1}{4}
The height is 2142\frac{1}{4} cm.
Answer: 2142\frac{1}{4} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 3163\frac{1}{6} + 2 x 2142\frac{1}{4} = 105610\frac{5}{6} cm, which is what was given.

Another way: Halve the perimeter first: 105610\frac{5}{6} / 2 = 55125\frac{5}{12} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 7 medium answer: 1351\frac{3}{5} cm

The perimeter of a rectangle is 87108\frac{7}{10} cm. If its width is 2342\frac{3}{4} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 2342\frac{3}{4} cm. The length of the vertical side is the value to find.

2 3 4 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 87108\frac{7}{10} cm and a width of 2342\frac{3}{4} cm. We need its height.

Givens
  • Perimeter: 87108\frac{7}{10} cm.
  • Width: 2342\frac{3}{4} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 2342\frac{3}{4} cm, and both vertical sides are the height we want.
perimeter=2×234+2×height\text{perimeter} = 2 \times 2\frac{3}{4} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
234+234=5122\frac{3}{4} + 2\frac{3}{4} = 5\frac{1}{2}
5125\frac{1}{2} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
8710−512=3158\frac{7}{10} - 5\frac{1}{2} = 3\frac{1}{5}
3153\frac{1}{5} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
315÷2=1353\frac{1}{5} \div 2 = 1\frac{3}{5}
The height is 1351\frac{3}{5} cm.
Answer: 1351\frac{3}{5} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 2342\frac{3}{4} + 2 x 1351\frac{3}{5} = 87108\frac{7}{10} cm, which is what was given.

Another way: Halve the perimeter first: 87108\frac{7}{10} / 2 = 47204\frac{7}{20} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 8 medium answer: 2122\frac{1}{2} cm

The perimeter of a rectangle is 102310\frac{2}{3} cm. If its width is 2562\frac{5}{6} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 2562\frac{5}{6} cm. The length of the vertical side is the value to find.

2 5 6 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 102310\frac{2}{3} cm and a width of 2562\frac{5}{6} cm. We need its height.

Givens
  • Perimeter: 102310\frac{2}{3} cm.
  • Width: 2562\frac{5}{6} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 2562\frac{5}{6} cm, and both vertical sides are the height we want.
perimeter=2×256+2×height\text{perimeter} = 2 \times 2\frac{5}{6} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
256+256=5232\frac{5}{6} + 2\frac{5}{6} = 5\frac{2}{3}
5235\frac{2}{3} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
1023−523=510\frac{2}{3} - 5\frac{2}{3} = 5
55 cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
5÷2=2125 \div 2 = 2\frac{1}{2}
The height is 2122\frac{1}{2} cm.
Answer: 2122\frac{1}{2} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 2562\frac{5}{6} + 2 x 2122\frac{1}{2} = 102310\frac{2}{3} cm, which is what was given.

Another way: Halve the perimeter first: 102310\frac{2}{3} / 2 = 5135\frac{1}{3} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 9 hard answer: 2232\frac{2}{3} cm

The perimeter of a rectangle is 1021510\frac{2}{15} cm. If its width is 2252\frac{2}{5} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 2252\frac{2}{5} cm. The length of the vertical side is the value to find.

2 2 5 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 1021510\frac{2}{15} cm and a width of 2252\frac{2}{5} cm. We need its height.

Givens
  • Perimeter: 1021510\frac{2}{15} cm.
  • Width: 2252\frac{2}{5} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 2252\frac{2}{5} cm, and both vertical sides are the height we want.
perimeter=2×225+2×height\text{perimeter} = 2 \times 2\frac{2}{5} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
225+225=4452\frac{2}{5} + 2\frac{2}{5} = 4\frac{4}{5}
4454\frac{4}{5} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
10215−445=51310\frac{2}{15} - 4\frac{4}{5} = 5\frac{1}{3}
5135\frac{1}{3} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
513÷2=2235\frac{1}{3} \div 2 = 2\frac{2}{3}
The height is 2232\frac{2}{3} cm.
Answer: 2232\frac{2}{3} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 2252\frac{2}{5} + 2 x 2232\frac{2}{3} = 1021510\frac{2}{15} cm, which is what was given.

Another way: Halve the perimeter first: 1021510\frac{2}{15} / 2 = 51155\frac{1}{15} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 10 hard answer: 2152\frac{1}{5} cm

The perimeter of a rectangle is 911159\frac{11}{15} cm. If its width is 2232\frac{2}{3} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 2232\frac{2}{3} cm. The length of the vertical side is the value to find.

2 2 3 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 911159\frac{11}{15} cm and a width of 2232\frac{2}{3} cm. We need its height.

Givens
  • Perimeter: 911159\frac{11}{15} cm.
  • Width: 2232\frac{2}{3} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 2232\frac{2}{3} cm, and both vertical sides are the height we want.
perimeter=2×223+2×height\text{perimeter} = 2 \times 2\frac{2}{3} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
223+223=5132\frac{2}{3} + 2\frac{2}{3} = 5\frac{1}{3}
5135\frac{1}{3} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
91115−513=4259\frac{11}{15} - 5\frac{1}{3} = 4\frac{2}{5}
4254\frac{2}{5} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
425÷2=2154\frac{2}{5} \div 2 = 2\frac{1}{5}
The height is 2152\frac{1}{5} cm.
Answer: 2152\frac{1}{5} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 2232\frac{2}{3} + 2 x 2152\frac{1}{5} = 911159\frac{11}{15} cm, which is what was given.

Another way: Halve the perimeter first: 911159\frac{11}{15} / 2 = 413154\frac{13}{15} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 11 hard answer: 4234\frac{2}{3} cm

The perimeter of a rectangle is 11111511\frac{11}{15} cm. If its width is 1151\frac{1}{5} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 1151\frac{1}{5} cm. The length of the vertical side is the value to find.

1 1 5 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 11111511\frac{11}{15} cm and a width of 1151\frac{1}{5} cm. We need its height.

Givens
  • Perimeter: 11111511\frac{11}{15} cm.
  • Width: 1151\frac{1}{5} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 1151\frac{1}{5} cm, and both vertical sides are the height we want.
perimeter=2×115+2×height\text{perimeter} = 2 \times 1\frac{1}{5} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
115+115=2251\frac{1}{5} + 1\frac{1}{5} = 2\frac{2}{5}
2252\frac{2}{5} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
111115−225=91311\frac{11}{15} - 2\frac{2}{5} = 9\frac{1}{3}
9139\frac{1}{3} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
913÷2=4239\frac{1}{3} \div 2 = 4\frac{2}{3}
The height is 4234\frac{2}{3} cm.
Answer: 4234\frac{2}{3} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 1151\frac{1}{5} + 2 x 4234\frac{2}{3} = 11111511\frac{11}{15} cm, which is what was given.

Another way: Halve the perimeter first: 11111511\frac{11}{15} / 2 = 513155\frac{13}{15} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.
Variant 12 hard answer: 2162\frac{1}{6} cm

The perimeter of a rectangle is 72157\frac{2}{15} cm. If its width is 1251\frac{2}{5} cm, how many centimeters long is its height?

There is a rectangle whose top horizontal side is labeled 1251\frac{2}{5} cm. The length of the vertical side is the value to find.

1 2 5 cm
Show solution
1 · Understandwhat's really being asked

A rectangle has a perimeter of 72157\frac{2}{15} cm and a width of 1251\frac{2}{5} cm. We need its height.

Givens
  • Perimeter: 72157\frac{2}{15} cm.
  • Width: 1251\frac{2}{5} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The height of the rectangle.
Constraints
  • Opposite sides of a rectangle are equal, so each length appears twice in the perimeter.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #11 Work Backwards

Draw the rectangle and mark both widths. The perimeter is all four sides, so taking the two known ones away leaves the two heights -- and halving that gives one.

3 · Execute4 carry out the plan

1Draw and label the rectangle

#1 Draw a Diagram 3.MD.D.8
Both horizontal sides are 1251\frac{2}{5} cm, and both vertical sides are the height we want.
perimeter=2×125+2×height\text{perimeter} = 2 \times 1\frac{2}{5} + 2 \times \text{height}
Four sides, but only two different lengths.

2Add the two widths

#11 Work Backwards 5.NF.A.1
The two horizontal sides together.
125+125=2451\frac{2}{5} + 1\frac{2}{5} = 2\frac{4}{5}
2452\frac{4}{5} cm of the perimeter is accounted for.

3Subtract the widths from the perimeter to leave the two heights

#1 Draw a Diagram 5.NF.A.2
What remains of the perimeter is the two vertical sides.
7215−245=4137\frac{2}{15} - 2\frac{4}{5} = 4\frac{1}{3}
4134\frac{1}{3} cm split between two equal sides.

4Split the two heights into one height

#11 Work Backwards 5.NF.A.1
The two vertical sides are equal, so halve the remainder.
413÷2=2164\frac{1}{3} \div 2 = 2\frac{1}{6}
The height is 2162\frac{1}{6} cm.
Answer: 2162\frac{1}{6} cm
4 · Reviewdoes it hold up?

Rebuild the perimeter: 2 x 1251\frac{2}{5} + 2 x 2162\frac{1}{6} = 72157\frac{2}{15} cm, which is what was given.

Another way: Halve the perimeter first: 72157\frac{2}{15} / 2 = 317303\frac{17}{30} cm is one width plus one height, and taking the width off leaves the height in one step.

Standardsmin grade 5
  • 3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding an unknown side length — Reading the perimeter as the sum of all four sides.
  • 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) — Adding the two widths and halving the leftover.
  • 5.NF.A.2 Solve word problems involving addition and subtraction of fractions referring to the same whole — Subtracting the known sides from the perimeter.
💡Takeaway. Every side of a rectangle appears twice in the perimeter, so take the pair you know away and halve what is left.