← Find a missing side, then sum sides as fractions · Perimeter by Tracing Every Side

Find a missing side, then sum sides as fractions · 10 practice problems

4.NF.B.34.MD.A.3

Generated variants — 10

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

Variant 1 easy answer: 181318\frac{1}{3} cm

The width of a rectangle is 3163\dfrac{1}{6} cm, and its height is 2562\dfrac{5}{6} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 3 16\frac{1}{6} cm. Its height is 2 56\frac{5}{6} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 3 16\frac{1}{6} cm.
  • Height = width + 2 56\frac{5}{6} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 6; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 6.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 2 56\frac{5}{6} = 3 16\frac{1}{6} + 2 56\frac{5}{6}. Whole: 3+2 = 5. Fractions: 16\frac{1}{6} + 56\frac{5}{6} = 66\frac{6}{6}. So height = 6 cm.
316+256=63\tfrac{1}{6}+2\tfrac{5}{6}=6
Same-denominator add, then regroup if the fraction part reaches 6/6.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 3 16\frac{1}{6} + 6 = 9 16\frac{1}{6} cm. This is half the perimeter.
316+6=9163\tfrac{1}{6}+6=9\tfrac{1}{6}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 9 16\frac{1}{6} = 18 26\frac{2}{6} = 18 13\frac{1}{3} cm.
2×916=1826=18132\times 9\tfrac{1}{6}=18\tfrac{2}{6}=18\tfrac{1}{3}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 181318\frac{1}{3} cm
4 · Reviewdoes it hold up?

Width is about 3.17 cm and height about 6.00 cm, so perimeter is about 2 x 9.17 = 18.33 cm, matching 18 13\frac{1}{3} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 3 16\frac{1}{6} + 3 16\frac{1}{6} + 6 + 6 = 18 26\frac{2}{6} = 18 13\frac{1}{3} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 6 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 2 easy answer: 135713\frac{5}{7} cm

The width of a rectangle is 2472\dfrac{4}{7} cm, and its height is 1571\dfrac{5}{7} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 2 47\frac{4}{7} cm. Its height is 1 57\frac{5}{7} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 2 47\frac{4}{7} cm.
  • Height = width + 1 57\frac{5}{7} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 7; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 7.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 1 57\frac{5}{7} = 2 47\frac{4}{7} + 1 57\frac{5}{7}. Whole: 2+1 = 3. Fractions: 47\frac{4}{7} + 57\frac{5}{7} = 97\frac{9}{7}. So height = 4 27\frac{2}{7} cm.
247+157=4272\tfrac{4}{7}+1\tfrac{5}{7}=4\tfrac{2}{7}
Same-denominator add, then regroup if the fraction part reaches 7/7.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 2 47\frac{4}{7} + 4 27\frac{2}{7} = 6 67\frac{6}{7} cm. This is half the perimeter.
247+427=6672\tfrac{4}{7}+4\tfrac{2}{7}=6\tfrac{6}{7}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 6 67\frac{6}{7} = 13 57\frac{5}{7} = 13 57\frac{5}{7} cm.
2×667=1357=13572\times 6\tfrac{6}{7}=13\tfrac{5}{7}=13\tfrac{5}{7}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 135713\frac{5}{7} cm
4 · Reviewdoes it hold up?

Width is about 2.57 cm and height about 4.29 cm, so perimeter is about 2 x 6.86 = 13.71 cm, matching 13 57\frac{5}{7} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 2 47\frac{4}{7} + 2 47\frac{4}{7} + 4 27\frac{2}{7} + 4 27\frac{2}{7} = 13 57\frac{5}{7} = 13 57\frac{5}{7} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 7 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 3 easy answer: 123412\frac{3}{4} cm

The width of a rectangle is 2382\dfrac{3}{8} cm, and its height is 1581\dfrac{5}{8} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 2 38\frac{3}{8} cm. Its height is 1 58\frac{5}{8} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 2 38\frac{3}{8} cm.
  • Height = width + 1 58\frac{5}{8} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 8; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 8.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 1 58\frac{5}{8} = 2 38\frac{3}{8} + 1 58\frac{5}{8}. Whole: 2+1 = 3. Fractions: 38\frac{3}{8} + 58\frac{5}{8} = 88\frac{8}{8}. So height = 4 cm.
238+158=42\tfrac{3}{8}+1\tfrac{5}{8}=4
Same-denominator add, then regroup if the fraction part reaches 8/8.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 2 38\frac{3}{8} + 4 = 6 38\frac{3}{8} cm. This is half the perimeter.
238+4=6382\tfrac{3}{8}+4=6\tfrac{3}{8}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 6 38\frac{3}{8} = 12 68\frac{6}{8} = 12 34\frac{3}{4} cm.
2×638=1268=12342\times 6\tfrac{3}{8}=12\tfrac{6}{8}=12\tfrac{3}{4}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 123412\frac{3}{4} cm
4 · Reviewdoes it hold up?

Width is about 2.38 cm and height about 4.00 cm, so perimeter is about 2 x 6.38 = 12.75 cm, matching 12 34\frac{3}{4} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 2 38\frac{3}{8} + 2 38\frac{3}{8} + 4 + 4 = 12 68\frac{6}{8} = 12 34\frac{3}{4} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 8 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 4 medium answer: 217921\frac{7}{9} cm

The width of a rectangle is 4294\dfrac{2}{9} cm, and its height is 2492\dfrac{4}{9} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 4 29\frac{2}{9} cm. Its height is 2 49\frac{4}{9} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 4 29\frac{2}{9} cm.
  • Height = width + 2 49\frac{4}{9} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 9; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 9.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 2 49\frac{4}{9} = 4 29\frac{2}{9} + 2 49\frac{4}{9}. Whole: 4+2 = 6. Fractions: 29\frac{2}{9} + 49\frac{4}{9} = 69\frac{6}{9}. So height = 6 69\frac{6}{9} cm.
429+249=6694\tfrac{2}{9}+2\tfrac{4}{9}=6\tfrac{6}{9}
Same-denominator add, then regroup if the fraction part reaches 9/9.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 4 29\frac{2}{9} + 6 69\frac{6}{9} = 10 89\frac{8}{9} cm. This is half the perimeter.
429+669=10894\tfrac{2}{9}+6\tfrac{6}{9}=10\tfrac{8}{9}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 10 89\frac{8}{9} = 21 79\frac{7}{9} = 21 79\frac{7}{9} cm.
2×1089=2179=21792\times 10\tfrac{8}{9}=21\tfrac{7}{9}=21\tfrac{7}{9}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 217921\frac{7}{9} cm
4 · Reviewdoes it hold up?

Width is about 4.22 cm and height about 6.67 cm, so perimeter is about 2 x 10.89 = 21.78 cm, matching 21 79\frac{7}{9} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 4 29\frac{2}{9} + 4 29\frac{2}{9} + 6 69\frac{6}{9} + 6 69\frac{6}{9} = 21 79\frac{7}{9} = 21 79\frac{7}{9} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 9 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 5 medium answer: 243524\frac{3}{5} cm

The width of a rectangle is 53105\dfrac{3}{10} cm, and its height is 17101\dfrac{7}{10} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 5 310\frac{3}{10} cm. Its height is 1 710\frac{7}{10} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 5 310\frac{3}{10} cm.
  • Height = width + 1 710\frac{7}{10} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 10; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 10.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 1 710\frac{7}{10} = 5 310\frac{3}{10} + 1 710\frac{7}{10}. Whole: 5+1 = 6. Fractions: 310\frac{3}{10} + 710\frac{7}{10} = 1010\frac{10}{10}. So height = 7 cm.
5310+1710=75\tfrac{3}{10}+1\tfrac{7}{10}=7
Same-denominator add, then regroup if the fraction part reaches 10/10.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 5 310\frac{3}{10} + 7 = 12 310\frac{3}{10} cm. This is half the perimeter.
5310+7=123105\tfrac{3}{10}+7=12\tfrac{3}{10}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 12 310\frac{3}{10} = 24 610\frac{6}{10} = 24 35\frac{3}{5} cm.
2×12310=24610=24352\times 12\tfrac{3}{10}=24\tfrac{6}{10}=24\tfrac{3}{5}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 243524\frac{3}{5} cm
4 · Reviewdoes it hold up?

Width is about 5.30 cm and height about 7.00 cm, so perimeter is about 2 x 12.30 = 24.60 cm, matching 24 35\frac{3}{5} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 5 310\frac{3}{10} + 5 310\frac{3}{10} + 7 + 7 = 24 610\frac{6}{10} = 24 35\frac{3}{5} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 10 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 6 medium answer: 2171121\frac{7}{11} cm

The width of a rectangle is 46114\dfrac{6}{11} cm, and its height is 18111\dfrac{8}{11} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 4 611\frac{6}{11} cm. Its height is 1 811\frac{8}{11} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 4 611\frac{6}{11} cm.
  • Height = width + 1 811\frac{8}{11} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 11; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 11.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 1 811\frac{8}{11} = 4 611\frac{6}{11} + 1 811\frac{8}{11}. Whole: 4+1 = 5. Fractions: 611\frac{6}{11} + 811\frac{8}{11} = 1411\frac{14}{11}. So height = 6 311\frac{3}{11} cm.
4611+1811=63114\tfrac{6}{11}+1\tfrac{8}{11}=6\tfrac{3}{11}
Same-denominator add, then regroup if the fraction part reaches 11/11.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 4 611\frac{6}{11} + 6 311\frac{3}{11} = 10 911\frac{9}{11} cm. This is half the perimeter.
4611+6311=109114\tfrac{6}{11}+6\tfrac{3}{11}=10\tfrac{9}{11}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 10 911\frac{9}{11} = 21 711\frac{7}{11} = 21 711\frac{7}{11} cm.
2×10911=21711=217112\times 10\tfrac{9}{11}=21\tfrac{7}{11}=21\tfrac{7}{11}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 2171121\frac{7}{11} cm
4 · Reviewdoes it hold up?

Width is about 4.55 cm and height about 6.27 cm, so perimeter is about 2 x 10.82 = 21.64 cm, matching 21 711\frac{7}{11} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 4 611\frac{6}{11} + 4 611\frac{6}{11} + 6 311\frac{3}{11} + 6 311\frac{3}{11} = 21 711\frac{7}{11} = 21 711\frac{7}{11} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 11 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 7 medium answer: 135613\frac{5}{6} cm

The width of a rectangle is 17121\dfrac{7}{12} cm, and its height is 39123\dfrac{9}{12} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 1 712\frac{7}{12} cm. Its height is 3 912\frac{9}{12} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 1 712\frac{7}{12} cm.
  • Height = width + 3 912\frac{9}{12} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 12; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 12.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 3 912\frac{9}{12} = 1 712\frac{7}{12} + 3 912\frac{9}{12}. Whole: 1+3 = 4. Fractions: 712\frac{7}{12} + 912\frac{9}{12} = 1612\frac{16}{12}. So height = 5 412\frac{4}{12} cm.
1712+3912=54121\tfrac{7}{12}+3\tfrac{9}{12}=5\tfrac{4}{12}
Same-denominator add, then regroup if the fraction part reaches 12/12.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 1 712\frac{7}{12} + 5 412\frac{4}{12} = 6 1112\frac{11}{12} cm. This is half the perimeter.
1712+5412=611121\tfrac{7}{12}+5\tfrac{4}{12}=6\tfrac{11}{12}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 6 1112\frac{11}{12} = 13 1012\frac{10}{12} = 13 56\frac{5}{6} cm.
2×61112=131012=13562\times 6\tfrac{11}{12}=13\tfrac{10}{12}=13\tfrac{5}{6}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 135613\frac{5}{6} cm
4 · Reviewdoes it hold up?

Width is about 1.58 cm and height about 5.33 cm, so perimeter is about 2 x 6.92 = 13.83 cm, matching 13 56\frac{5}{6} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 1 712\frac{7}{12} + 1 712\frac{7}{12} + 5 412\frac{4}{12} + 5 412\frac{4}{12} = 13 1012\frac{10}{12} = 13 56\frac{5}{6} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 12 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 8 hard answer: 1717 cm

The width of a rectangle is 25142\dfrac{5}{14} cm, and its height is 311143\dfrac{11}{14} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 2 514\frac{5}{14} cm. Its height is 3 1114\frac{11}{14} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 2 514\frac{5}{14} cm.
  • Height = width + 3 1114\frac{11}{14} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 14; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 14.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 3 1114\frac{11}{14} = 2 514\frac{5}{14} + 3 1114\frac{11}{14}. Whole: 2+3 = 5. Fractions: 514\frac{5}{14} + 1114\frac{11}{14} = 1614\frac{16}{14}. So height = 6 214\frac{2}{14} cm.
2514+31114=62142\tfrac{5}{14}+3\tfrac{11}{14}=6\tfrac{2}{14}
Same-denominator add, then regroup if the fraction part reaches 14/14.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 2 514\frac{5}{14} + 6 214\frac{2}{14} = 8 714\frac{7}{14} cm. This is half the perimeter.
2514+6214=87142\tfrac{5}{14}+6\tfrac{2}{14}=8\tfrac{7}{14}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 8 714\frac{7}{14} = 17 = 17 cm.
2×8714=17=172\times 8\tfrac{7}{14}=17=17
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 1717 cm
4 · Reviewdoes it hold up?

Width is about 2.36 cm and height about 6.14 cm, so perimeter is about 2 x 8.50 = 17.00 cm, matching 17 cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 2 514\frac{5}{14} + 2 514\frac{5}{14} + 6 214\frac{2}{14} + 6 214\frac{2}{14} = 17 = 17 cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 14 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 9 hard answer: 164516\frac{4}{5} cm

The width of a rectangle is 35153\dfrac{5}{15} cm, and its height is 111151\dfrac{11}{15} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 3 515\frac{5}{15} cm. Its height is 1 1115\frac{11}{15} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 3 515\frac{5}{15} cm.
  • Height = width + 1 1115\frac{11}{15} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 15; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 15.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 1 1115\frac{11}{15} = 3 515\frac{5}{15} + 1 1115\frac{11}{15}. Whole: 3+1 = 4. Fractions: 515\frac{5}{15} + 1115\frac{11}{15} = 1615\frac{16}{15}. So height = 5 115\frac{1}{15} cm.
3515+11115=51153\tfrac{5}{15}+1\tfrac{11}{15}=5\tfrac{1}{15}
Same-denominator add, then regroup if the fraction part reaches 15/15.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 3 515\frac{5}{15} + 5 115\frac{1}{15} = 8 615\frac{6}{15} cm. This is half the perimeter.
3515+5115=86153\tfrac{5}{15}+5\tfrac{1}{15}=8\tfrac{6}{15}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 8 615\frac{6}{15} = 16 1215\frac{12}{15} = 16 45\frac{4}{5} cm.
2×8615=161215=16452\times 8\tfrac{6}{15}=16\tfrac{12}{15}=16\tfrac{4}{5}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 164516\frac{4}{5} cm
4 · Reviewdoes it hold up?

Width is about 3.33 cm and height about 5.07 cm, so perimeter is about 2 x 8.40 = 16.80 cm, matching 16 45\frac{4}{5} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 3 515\frac{5}{15} + 3 515\frac{5}{15} + 5 115\frac{1}{15} + 5 115\frac{1}{15} = 16 1215\frac{12}{15} = 16 45\frac{4}{5} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 15 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!
Variant 10 hard answer: 1131011\frac{3}{10} cm

The width of a rectangle is 19201\dfrac{9}{20} cm, and its height is 215202\dfrac{15}{20} cm longer than the width. Find the sum of the lengths of the four sides of this rectangle, in cm.

Show solution
1 · Understandwhat's really being asked

A rectangle has width 1 920\frac{9}{20} cm. Its height is 2 1520\frac{15}{20} cm longer than the width. I must find the perimeter (the sum of all four side lengths) in cm.

Givens
  • Width = 1 920\frac{9}{20} cm.
  • Height = width + 2 1520\frac{15}{20} cm.
  • A rectangle has two widths and two heights.
Unknowns
  • The perimeter (sum of the four sides) of the rectangle.
Constraints
  • Fraction parts share denominator 20; perimeter = 2 x (width + height).
2 · Planchoose the strategy

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

First find the height by adding to the width, then apply the rectangle perimeter formula P = 2 x (width + height). Two clean subproblems, both with denominator 20.

3 · Execute3 carry out the plan

1Find the height

#7 Identify Subproblems 4.NF.B.3
Height = width + 2 1520\frac{15}{20} = 1 920\frac{9}{20} + 2 1520\frac{15}{20}. Whole: 1+2 = 3. Fractions: 920\frac{9}{20} + 1520\frac{15}{20} = 2420\frac{24}{20}. So height = 4 420\frac{4}{20} cm.
1920+21520=44201\tfrac{9}{20}+2\tfrac{15}{20}=4\tfrac{4}{20}
Same-denominator add, then regroup if the fraction part reaches 20/20.

2Add width and height

#7 Identify Subproblems 4.NF.B.3
width + height = 1 920\frac{9}{20} + 4 420\frac{4}{20} = 5 1320\frac{13}{20} cm. This is half the perimeter.
1920+4420=513201\tfrac{9}{20}+4\tfrac{4}{20}=5\tfrac{13}{20}
Adding one width and one height gives half the perimeter.

3Double to get the perimeter

#7 Identify Subproblems 4.MD.A.3
Perimeter = 2 x (width + height) = 2 x 5 1320\frac{13}{20} = 11 620\frac{6}{20} = 11 310\frac{3}{10} cm.
2×51320=11620=113102\times 5\tfrac{13}{20}=11\tfrac{6}{20}=11\tfrac{3}{10}
A rectangle has two equal widths and two equal heights, so doubling the half-perimeter gives the whole.
Answer: 1131011\frac{3}{10} cm
4 · Reviewdoes it hold up?

Width is about 1.45 cm and height about 4.20 cm, so perimeter is about 2 x 5.65 = 11.30 cm, matching 11 310\frac{3}{10} cm. Units stay in cm.

Another way: Add all four sides separately (tool 1): 1 920\frac{9}{20} + 1 920\frac{9}{20} + 4 420\frac{4}{20} + 4 420\frac{4}{20} = 11 620\frac{6}{20} = 11 310\frac{3}{10} cm, the same result.

Standardsmin grade 4
  • 4.NF.B.3 Understand a fraction with numerator greater than one as sum of unit fractions — Adding the mixed-number side lengths with denominator 20 and regrouping improper fractions.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Using perimeter = 2 x (width + height) to total the four sides.
💡Takeaway. This only needs Grade 4 fraction adding plus the rectangle perimeter formula — two widths plus two heights!