Practice · Perimeter by tracing every side · Sensim Math

Variant 1 answer: 68 cm

Five identical rectangles, each $7\,\text{cm}$ long and $3\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining five identical rectangles (each $7\,\text{cm}$ wide and $3\,\text{cm}$ tall) edge to edge. The top row has $4$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $7\,\text{cm}$ along its horizontal side and $3\,\text{cm}$ along its vertical side.

Show solution

Understand

Five identical rectangles, each 7 cm wide and 3 cm tall, are joined into a staircase shape: four sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 7 cm wide and 3 cm tall.
Four rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 7 cm and 3 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 4 times 7 = 28 cm across and 3 cm down. The extra rectangle sits below the leftmost column, spanning 7 cm across and another 3 cm down, so that column reaches 6 cm tall.

4 \times 7 = 28\text{ cm (top width)}, \quad 2 \times 3 = 6\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (28 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (28 - 7 = 21 cm), and the bottom of the lower rectangle (7 cm).

28 + 21 + 7 = 56

Grade 3 addition: the left-and-right edges of the outline add to 56 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (3 cm), the short drop at the step (3 cm), and the tall left side (6 cm).

3 + 3 + 6 = 12

Grade 3 addition: the up-and-down edges of the outline add to 12 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

56 + 12 = 68

Perimeter is just the whole trip around the edge: 56 cm across plus 12 cm up and down is 68 cm.

Answer: 68 cm

Review

A single 7 by 3 rectangle has perimeter 20 cm; 5 of them total 100 cm, but joining them hides several shared edges. Our 68 cm is comfortably below 100 cm and well above one rectangle's 20 cm, so the magnitude makes sense for a 5-rectangle staircase.

Treat the figure as a 28 by 3 top bar plus a 7 by 3 tab. Their perimeters are 62 cm and 20 cm; subtract twice the 7 cm shared edge (2 times 7 = 14 cm): 62 + 20 - 14 = 68 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 68 cm!

Variant 2 answer: 68 cm

Four identical rectangles, each $8\,\text{cm}$ long and $5\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining four identical rectangles (each $8\,\text{cm}$ wide and $5\,\text{cm}$ tall) edge to edge. The top row has $3$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $8\,\text{cm}$ along its horizontal side and $5\,\text{cm}$ along its vertical side.

Show solution

Understand

Four identical rectangles, each 8 cm wide and 5 cm tall, are joined into a staircase shape: three sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 8 cm wide and 5 cm tall.
Three rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 8 cm and 5 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 3 times 8 = 24 cm across and 5 cm down. The extra rectangle sits below the leftmost column, spanning 8 cm across and another 5 cm down, so that column reaches 10 cm tall.

3 \times 8 = 24\text{ cm (top width)}, \quad 2 \times 5 = 10\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (24 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (24 - 8 = 16 cm), and the bottom of the lower rectangle (8 cm).

24 + 16 + 8 = 48

Grade 3 addition: the left-and-right edges of the outline add to 48 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (5 cm), the short drop at the step (5 cm), and the tall left side (10 cm).

5 + 5 + 10 = 20

Grade 3 addition: the up-and-down edges of the outline add to 20 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

48 + 20 = 68

Perimeter is just the whole trip around the edge: 48 cm across plus 20 cm up and down is 68 cm.

Answer: 68 cm

Review

A single 8 by 5 rectangle has perimeter 26 cm; 4 of them total 104 cm, but joining them hides several shared edges. Our 68 cm is comfortably below 104 cm and well above one rectangle's 26 cm, so the magnitude makes sense for a 4-rectangle staircase.

Treat the figure as a 24 by 5 top bar plus a 8 by 5 tab. Their perimeters are 58 cm and 26 cm; subtract twice the 8 cm shared edge (2 times 8 = 16 cm): 58 + 26 - 16 = 68 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 68 cm!

Variant 3 answer: 98 cm

Four identical rectangles, each $11\,\text{cm}$ long and $8\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining four identical rectangles (each $11\,\text{cm}$ wide and $8\,\text{cm}$ tall) edge to edge. The top row has $3$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $11\,\text{cm}$ along its horizontal side and $8\,\text{cm}$ along its vertical side.

Show solution

Understand

Four identical rectangles, each 11 cm wide and 8 cm tall, are joined into a staircase shape: three sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 11 cm wide and 8 cm tall.
Three rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 11 cm and 8 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 3 times 11 = 33 cm across and 8 cm down. The extra rectangle sits below the leftmost column, spanning 11 cm across and another 8 cm down, so that column reaches 16 cm tall.

3 \times 11 = 33\text{ cm (top width)}, \quad 2 \times 8 = 16\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (33 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (33 - 11 = 22 cm), and the bottom of the lower rectangle (11 cm).

33 + 22 + 11 = 66

Grade 3 addition: the left-and-right edges of the outline add to 66 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (8 cm), the short drop at the step (8 cm), and the tall left side (16 cm).

8 + 8 + 16 = 32

Grade 3 addition: the up-and-down edges of the outline add to 32 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

66 + 32 = 98

Perimeter is just the whole trip around the edge: 66 cm across plus 32 cm up and down is 98 cm.

Answer: 98 cm

Review

A single 11 by 8 rectangle has perimeter 38 cm; 4 of them total 152 cm, but joining them hides several shared edges. Our 98 cm is comfortably below 152 cm and well above one rectangle's 38 cm, so the magnitude makes sense for a 4-rectangle staircase.

Treat the figure as a 33 by 8 top bar plus a 11 by 8 tab. Their perimeters are 82 cm and 38 cm; subtract twice the 11 cm shared edge (2 times 11 = 22 cm): 82 + 38 - 22 = 98 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 98 cm!

Variant 4 answer: 124 cm

Five identical rectangles, each $12\,\text{cm}$ long and $7\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining five identical rectangles (each $12\,\text{cm}$ wide and $7\,\text{cm}$ tall) edge to edge. The top row has $4$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $12\,\text{cm}$ along its horizontal side and $7\,\text{cm}$ along its vertical side.

Show solution

Understand

Five identical rectangles, each 12 cm wide and 7 cm tall, are joined into a staircase shape: four sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 12 cm wide and 7 cm tall.
Four rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 12 cm and 7 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 4 times 12 = 48 cm across and 7 cm down. The extra rectangle sits below the leftmost column, spanning 12 cm across and another 7 cm down, so that column reaches 14 cm tall.

4 \times 12 = 48\text{ cm (top width)}, \quad 2 \times 7 = 14\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (48 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (48 - 12 = 36 cm), and the bottom of the lower rectangle (12 cm).

48 + 36 + 12 = 96

Grade 3 addition: the left-and-right edges of the outline add to 96 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (7 cm), the short drop at the step (7 cm), and the tall left side (14 cm).

7 + 7 + 14 = 28

Grade 3 addition: the up-and-down edges of the outline add to 28 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

96 + 28 = 124

Perimeter is just the whole trip around the edge: 96 cm across plus 28 cm up and down is 124 cm.

Answer: 124 cm

Review

A single 12 by 7 rectangle has perimeter 38 cm; 5 of them total 190 cm, but joining them hides several shared edges. Our 124 cm is comfortably below 190 cm and well above one rectangle's 38 cm, so the magnitude makes sense for a 5-rectangle staircase.

Treat the figure as a 48 by 7 top bar plus a 12 by 7 tab. Their perimeters are 110 cm and 38 cm; subtract twice the 12 cm shared edge (2 times 12 = 24 cm): 110 + 38 - 24 = 124 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 124 cm!

Variant 5 answer: 100 cm

Three identical rectangles, each $15\,\text{cm}$ long and $10\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining three identical rectangles (each $15\,\text{cm}$ wide and $10\,\text{cm}$ tall) edge to edge. The top row has $2$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $15\,\text{cm}$ along its horizontal side and $10\,\text{cm}$ along its vertical side.

Show solution

Understand

Three identical rectangles, each 15 cm wide and 10 cm tall, are joined into a staircase shape: two sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 15 cm wide and 10 cm tall.
Two rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 15 cm and 10 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 2 times 15 = 30 cm across and 10 cm down. The extra rectangle sits below the leftmost column, spanning 15 cm across and another 10 cm down, so that column reaches 20 cm tall.

2 \times 15 = 30\text{ cm (top width)}, \quad 2 \times 10 = 20\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (30 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (30 - 15 = 15 cm), and the bottom of the lower rectangle (15 cm).

30 + 15 + 15 = 60

Grade 3 addition: the left-and-right edges of the outline add to 60 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (10 cm), the short drop at the step (10 cm), and the tall left side (20 cm).

10 + 10 + 20 = 40

Grade 3 addition: the up-and-down edges of the outline add to 40 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

60 + 40 = 100

Perimeter is just the whole trip around the edge: 60 cm across plus 40 cm up and down is 100 cm.

Answer: 100 cm

Review

A single 15 by 10 rectangle has perimeter 50 cm; 3 of them total 150 cm, but joining them hides several shared edges. Our 100 cm is comfortably below 150 cm and well above one rectangle's 50 cm, so the magnitude makes sense for a 3-rectangle staircase.

Treat the figure as a 30 by 10 top bar plus a 15 by 10 tab. Their perimeters are 80 cm and 50 cm; subtract twice the 15 cm shared edge (2 times 15 = 30 cm): 80 + 50 - 30 = 100 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 100 cm!

Variant 6 answer: 78 cm

Four identical rectangles, each $9\,\text{cm}$ long and $6\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining four identical rectangles (each $9\,\text{cm}$ wide and $6\,\text{cm}$ tall) edge to edge. The top row has $3$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $9\,\text{cm}$ along its horizontal side and $6\,\text{cm}$ along its vertical side.

Show solution

Understand

Four identical rectangles, each 9 cm wide and 6 cm tall, are joined into a staircase shape: three sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 9 cm wide and 6 cm tall.
Three rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 9 cm and 6 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 3 times 9 = 27 cm across and 6 cm down. The extra rectangle sits below the leftmost column, spanning 9 cm across and another 6 cm down, so that column reaches 12 cm tall.

3 \times 9 = 27\text{ cm (top width)}, \quad 2 \times 6 = 12\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (27 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (27 - 9 = 18 cm), and the bottom of the lower rectangle (9 cm).

27 + 18 + 9 = 54

Grade 3 addition: the left-and-right edges of the outline add to 54 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (6 cm), the short drop at the step (6 cm), and the tall left side (12 cm).

6 + 6 + 12 = 24

Grade 3 addition: the up-and-down edges of the outline add to 24 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

54 + 24 = 78

Perimeter is just the whole trip around the edge: 54 cm across plus 24 cm up and down is 78 cm.

Answer: 78 cm

Review

A single 9 by 6 rectangle has perimeter 30 cm; 4 of them total 120 cm, but joining them hides several shared edges. Our 78 cm is comfortably below 120 cm and well above one rectangle's 30 cm, so the magnitude makes sense for a 4-rectangle staircase.

Treat the figure as a 27 by 6 top bar plus a 9 by 6 tab. Their perimeters are 66 cm and 30 cm; subtract twice the 9 cm shared edge (2 times 9 = 18 cm): 66 + 30 - 18 = 78 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 78 cm!

Variant 7 answer: 56 cm

Three identical rectangles, each $10\,\text{cm}$ long and $4\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining three identical rectangles (each $10\,\text{cm}$ wide and $4\,\text{cm}$ tall) edge to edge. The top row has $2$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $10\,\text{cm}$ along its horizontal side and $4\,\text{cm}$ along its vertical side.

Show solution

Understand

Three identical rectangles, each 10 cm wide and 4 cm tall, are joined into a staircase shape: two sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 10 cm wide and 4 cm tall.
Two rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 10 cm and 4 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 2 times 10 = 20 cm across and 4 cm down. The extra rectangle sits below the leftmost column, spanning 10 cm across and another 4 cm down, so that column reaches 8 cm tall.

2 \times 10 = 20\text{ cm (top width)}, \quad 2 \times 4 = 8\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (20 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (20 - 10 = 10 cm), and the bottom of the lower rectangle (10 cm).

20 + 10 + 10 = 40

Grade 3 addition: the left-and-right edges of the outline add to 40 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (4 cm), the short drop at the step (4 cm), and the tall left side (8 cm).

4 + 4 + 8 = 16

Grade 3 addition: the up-and-down edges of the outline add to 16 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

40 + 16 = 56

Perimeter is just the whole trip around the edge: 40 cm across plus 16 cm up and down is 56 cm.

Answer: 56 cm

Review

A single 10 by 4 rectangle has perimeter 28 cm; 3 of them total 84 cm, but joining them hides several shared edges. Our 56 cm is comfortably below 84 cm and well above one rectangle's 28 cm, so the magnitude makes sense for a 3-rectangle staircase.

Treat the figure as a 20 by 4 top bar plus a 10 by 4 tab. Their perimeters are 48 cm and 28 cm; subtract twice the 10 cm shared edge (2 times 10 = 20 cm): 48 + 28 - 20 = 56 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 56 cm!

Variant 8 answer: 60 cm

Four identical rectangles, each $6\,\text{cm}$ long and $6\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining four identical rectangles (each $6\,\text{cm}$ wide and $6\,\text{cm}$ tall) edge to edge. The top row has $3$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $6\,\text{cm}$ along its horizontal side and $6\,\text{cm}$ along its vertical side.

Show solution

Understand

Four identical rectangles, each 6 cm wide and 6 cm tall, are joined into a staircase shape: three sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 6 cm wide and 6 cm tall.
Three rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 6 cm and 6 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 3 times 6 = 18 cm across and 6 cm down. The extra rectangle sits below the leftmost column, spanning 6 cm across and another 6 cm down, so that column reaches 12 cm tall.

3 \times 6 = 18\text{ cm (top width)}, \quad 2 \times 6 = 12\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (18 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (18 - 6 = 12 cm), and the bottom of the lower rectangle (6 cm).

18 + 12 + 6 = 36

Grade 3 addition: the left-and-right edges of the outline add to 36 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (6 cm), the short drop at the step (6 cm), and the tall left side (12 cm).

6 + 6 + 12 = 24

Grade 3 addition: the up-and-down edges of the outline add to 24 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

36 + 24 = 60

Perimeter is just the whole trip around the edge: 36 cm across plus 24 cm up and down is 60 cm.

Answer: 60 cm

Review

A single 6 by 6 rectangle has perimeter 24 cm; 4 of them total 96 cm, but joining them hides several shared edges. Our 60 cm is comfortably below 96 cm and well above one rectangle's 24 cm, so the magnitude makes sense for a 4-rectangle staircase.

Treat the figure as a 18 by 6 top bar plus a 6 by 6 tab. Their perimeters are 48 cm and 24 cm; subtract twice the 6 cm shared edge (2 times 6 = 12 cm): 48 + 24 - 12 = 60 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 60 cm!

Variant 9 answer: 56 cm

Three identical rectangles, each $5\,\text{cm}$ long and $9\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining three identical rectangles (each $5\,\text{cm}$ wide and $9\,\text{cm}$ tall) edge to edge. The top row has $2$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $5\,\text{cm}$ along its horizontal side and $9\,\text{cm}$ along its vertical side.

Show solution

Understand

Three identical rectangles, each 5 cm wide and 9 cm tall, are joined into a staircase shape: two sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 5 cm wide and 9 cm tall.
Two rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 5 cm and 9 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 2 times 5 = 10 cm across and 9 cm down. The extra rectangle sits below the leftmost column, spanning 5 cm across and another 9 cm down, so that column reaches 18 cm tall.

2 \times 5 = 10\text{ cm (top width)}, \quad 2 \times 9 = 18\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (10 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (10 - 5 = 5 cm), and the bottom of the lower rectangle (5 cm).

10 + 5 + 5 = 20

Grade 3 addition: the left-and-right edges of the outline add to 20 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (9 cm), the short drop at the step (9 cm), and the tall left side (18 cm).

9 + 9 + 18 = 36

Grade 3 addition: the up-and-down edges of the outline add to 36 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

20 + 36 = 56

Perimeter is just the whole trip around the edge: 20 cm across plus 36 cm up and down is 56 cm.

Answer: 56 cm

Review

A single 5 by 9 rectangle has perimeter 28 cm; 3 of them total 84 cm, but joining them hides several shared edges. Our 56 cm is comfortably below 84 cm and well above one rectangle's 28 cm, so the magnitude makes sense for a 3-rectangle staircase.

Treat the figure as a 10 by 9 top bar plus a 5 by 9 tab. Their perimeters are 38 cm and 28 cm; subtract twice the 5 cm shared edge (2 times 5 = 10 cm): 38 + 28 - 10 = 56 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 56 cm!

Variant 10 answer: 168 cm

Four identical rectangles, each $20\,\text{cm}$ long and $12\,\text{cm}$ tall, are joined together to make the figure shown below. What is the perimeter of this figure, in $\text{cm}$ ?

Figure description: A staircase-shaped figure is made by joining four identical rectangles (each $20\,\text{cm}$ wide and $12\,\text{cm}$ tall) edge to edge. The top row has $3$ rectangles placed side by side in a horizontal line, and one more rectangle is attached directly below the leftmost rectangle of that row. One rectangle is labeled $20\,\text{cm}$ along its horizontal side and $12\,\text{cm}$ along its vertical side.

Show solution

Understand

Four identical rectangles, each 20 cm wide and 12 cm tall, are joined into a staircase shape: three sit in a top row, and one hangs directly below the leftmost. We need the perimeter of the whole figure in centimeters.

Givens

Each rectangle is 20 cm wide and 12 cm tall.
Three rectangles form the top row, placed side by side.
One more rectangle is attached below the leftmost top rectangle.

Unknowns

The perimeter of the combined staircase figure.

Constraints

Rectangles are joined edge to edge with no gaps or overlaps.
All measurements are multiples of the 20 cm and 12 cm side lengths.

Plan

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

Sketching the figure on a grid lets us trace every outer edge in order. Breaking the boundary into separate horizontal and vertical pieces (subproblems) keeps the additions organized so no side is missed or double counted.

Execute

#1 Draw a Diagram 4.MD.A.3

Place the figure with the top-left corner at the origin. The top row spans 3 times 20 = 60 cm across and 12 cm down. The extra rectangle sits below the leftmost column, spanning 20 cm across and another 12 cm down, so that column reaches 24 cm tall.

3 \times 20 = 60\text{ cm (top width)}, \quad 2 \times 12 = 24\text{ cm (left height)}

Putting the picture on a grid turns 'staircase' into clear straight edges I can measure.

#7 Identify Subproblems 3.NBT.A.2

Going around, the horizontal pieces are: the full top (60 cm), the step in the middle where the top row's bottom juts out past the lower rectangle (60 - 20 = 40 cm), and the bottom of the lower rectangle (20 cm).

60 + 40 + 20 = 120

Grade 3 addition: the left-and-right edges of the outline add to 120 cm.

#7 Identify Subproblems 3.NBT.A.2

The vertical pieces are: the right side of the top row (12 cm), the short drop at the step (12 cm), and the tall left side (24 cm).

12 + 12 + 24 = 48

Grade 3 addition: the up-and-down edges of the outline add to 48 cm.

#7 Identify Subproblems 3.MD.D.8

Add the horizontal total and the vertical total to get the full distance around the figure.

120 + 48 = 168

Perimeter is just the whole trip around the edge: 120 cm across plus 48 cm up and down is 168 cm.

Answer: 168 cm

Review

A single 20 by 12 rectangle has perimeter 64 cm; 4 of them total 256 cm, but joining them hides several shared edges. Our 168 cm is comfortably below 256 cm and well above one rectangle's 64 cm, so the magnitude makes sense for a 4-rectangle staircase.

Treat the figure as a 60 by 12 top bar plus a 20 by 12 tab. Their perimeters are 144 cm and 64 cm; subtract twice the 20 cm shared edge (2 times 20 = 40 cm): 144 + 64 - 40 = 168 cm.

Standards · min grade 4

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Reasoning about the rectangle dimensions and the composite outline on a grid.
3.NBT.A.2 Fluently add and subtract within 1000 — Adding the horizontal and vertical edge lengths.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Combining all outer edges into the total perimeter.

💡 Trace the outline once, adding every edge as you go, and the staircase perimeter is just 168 cm!

Perimeter by tracing every side · 10 practice problems

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