← An L-shaped outline costs the same walk as its box · Perimeter by Tracing Every Side

An L-shaped outline costs the same walk as its box · 12 practice problems

3.MD.C.73.MD.D.8

0 / 12 solved

Show all answers (parent / teacher)

Generated variants — 12

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

Variant 1 easy

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $6\ \text{cm}$ has a rectangle whose height is $4\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $2\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 6 cm square and a rectangle of the same area, 4 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 6 cm.
The rectangle is 4 cm tall and has the same area as the square.
The square sticks 2 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 6 x 6 = 36 cm2. The rectangle has the same area and a height of 4 cm, so its width is 36 divided by 4.

6 \times 6 = 36,\quad 36 \div 4 = 9

The rectangle is 9 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 2 cm to the left of the rectangle, and the rectangle (9 cm wide) runs from there to the right.

2 + 9 = 11

The whole figure is 11 cm across and 6 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 11, the rectangle's right side is 4, the ledge along the top of the rectangle is 11 - 6 = 5, the riser up the square's right side is 6 - 4 = 2, the square's top is 6, and its left side is 6. Every one of those slides out to the 11 by 6 box without changing length.

(11 + 6) \times 2 = 17 \times 2 = 34

The way around is 34 cm.

Answer: 34 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 11 + 4 + 5 + 2 + 6 + 6 = 34 cm, the same as the box's 34 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 2 easy

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $8\ \text{cm}$ has a rectangle whose height is $4\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $3\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 8 cm square and a rectangle of the same area, 4 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 8 cm.
The rectangle is 4 cm tall and has the same area as the square.
The square sticks 3 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 8 x 8 = 64 cm2. The rectangle has the same area and a height of 4 cm, so its width is 64 divided by 4.

8 \times 8 = 64,\quad 64 \div 4 = 16

The rectangle is 16 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 3 cm to the left of the rectangle, and the rectangle (16 cm wide) runs from there to the right.

3 + 16 = 19

The whole figure is 19 cm across and 8 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 19, the rectangle's right side is 4, the ledge along the top of the rectangle is 19 - 8 = 11, the riser up the square's right side is 8 - 4 = 4, the square's top is 8, and its left side is 8. Every one of those slides out to the 19 by 8 box without changing length.

(19 + 8) \times 2 = 27 \times 2 = 54

The way around is 54 cm.

Answer: 54 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 19 + 4 + 11 + 4 + 8 + 8 = 54 cm, the same as the box's 54 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 3 easy

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $9\ \text{cm}$ has a rectangle whose height is $3\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $4\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 9 cm square and a rectangle of the same area, 3 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 9 cm.
The rectangle is 3 cm tall and has the same area as the square.
The square sticks 4 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 9 x 9 = 81 cm2. The rectangle has the same area and a height of 3 cm, so its width is 81 divided by 3.

9 \times 9 = 81,\quad 81 \div 3 = 27

The rectangle is 27 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 4 cm to the left of the rectangle, and the rectangle (27 cm wide) runs from there to the right.

4 + 27 = 31

The whole figure is 31 cm across and 9 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 31, the rectangle's right side is 3, the ledge along the top of the rectangle is 31 - 9 = 22, the riser up the square's right side is 9 - 3 = 6, the square's top is 9, and its left side is 9. Every one of those slides out to the 31 by 9 box without changing length.

(31 + 9) \times 2 = 40 \times 2 = 80

The way around is 80 cm.

Answer: 80 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 31 + 3 + 22 + 6 + 9 + 9 = 80 cm, the same as the box's 80 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 4 easy

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $10\ \text{cm}$ has a rectangle whose height is $5\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $4\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 10 cm square and a rectangle of the same area, 5 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 10 cm.
The rectangle is 5 cm tall and has the same area as the square.
The square sticks 4 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 10 x 10 = 100 cm2. The rectangle has the same area and a height of 5 cm, so its width is 100 divided by 5.

10 \times 10 = 100,\quad 100 \div 5 = 20

The rectangle is 20 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 4 cm to the left of the rectangle, and the rectangle (20 cm wide) runs from there to the right.

4 + 20 = 24

The whole figure is 24 cm across and 10 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 24, the rectangle's right side is 5, the ledge along the top of the rectangle is 24 - 10 = 14, the riser up the square's right side is 10 - 5 = 5, the square's top is 10, and its left side is 10. Every one of those slides out to the 24 by 10 box without changing length.

(24 + 10) \times 2 = 34 \times 2 = 68

The way around is 68 cm.

Answer: 68 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 24 + 5 + 14 + 5 + 10 + 10 = 68 cm, the same as the box's 68 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 5 medium

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $12\ \text{cm}$ has a rectangle whose height is $9\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $3\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 12 cm square and a rectangle of the same area, 9 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 12 cm.
The rectangle is 9 cm tall and has the same area as the square.
The square sticks 3 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 12 x 12 = 144 cm2. The rectangle has the same area and a height of 9 cm, so its width is 144 divided by 9.

12 \times 12 = 144,\quad 144 \div 9 = 16

The rectangle is 16 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 3 cm to the left of the rectangle, and the rectangle (16 cm wide) runs from there to the right.

3 + 16 = 19

The whole figure is 19 cm across and 12 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 19, the rectangle's right side is 9, the ledge along the top of the rectangle is 19 - 12 = 7, the riser up the square's right side is 12 - 9 = 3, the square's top is 12, and its left side is 12. Every one of those slides out to the 19 by 12 box without changing length.

(19 + 12) \times 2 = 31 \times 2 = 62

The way around is 62 cm.

Answer: 62 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 19 + 9 + 7 + 3 + 12 + 12 = 62 cm, the same as the box's 62 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 6 medium

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $14\ \text{cm}$ has a rectangle whose height is $7\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $8\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 14 cm square and a rectangle of the same area, 7 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 14 cm.
The rectangle is 7 cm tall and has the same area as the square.
The square sticks 8 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 14 x 14 = 196 cm2. The rectangle has the same area and a height of 7 cm, so its width is 196 divided by 7.

14 \times 14 = 196,\quad 196 \div 7 = 28

The rectangle is 28 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 8 cm to the left of the rectangle, and the rectangle (28 cm wide) runs from there to the right.

8 + 28 = 36

The whole figure is 36 cm across and 14 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 36, the rectangle's right side is 7, the ledge along the top of the rectangle is 36 - 14 = 22, the riser up the square's right side is 14 - 7 = 7, the square's top is 14, and its left side is 14. Every one of those slides out to the 36 by 14 box without changing length.

(36 + 14) \times 2 = 50 \times 2 = 100

The way around is 100 cm.

Answer: 100 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 36 + 7 + 22 + 7 + 14 + 14 = 100 cm, the same as the box's 100 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 7 medium

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $15\ \text{cm}$ has a rectangle whose height is $9\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $6\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 15 cm square and a rectangle of the same area, 9 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 15 cm.
The rectangle is 9 cm tall and has the same area as the square.
The square sticks 6 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 15 x 15 = 225 cm2. The rectangle has the same area and a height of 9 cm, so its width is 225 divided by 9.

15 \times 15 = 225,\quad 225 \div 9 = 25

The rectangle is 25 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 6 cm to the left of the rectangle, and the rectangle (25 cm wide) runs from there to the right.

6 + 25 = 31

The whole figure is 31 cm across and 15 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 31, the rectangle's right side is 9, the ledge along the top of the rectangle is 31 - 15 = 16, the riser up the square's right side is 15 - 9 = 6, the square's top is 15, and its left side is 15. Every one of those slides out to the 31 by 15 box without changing length.

(31 + 15) \times 2 = 46 \times 2 = 92

The way around is 92 cm.

Answer: 92 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 31 + 9 + 16 + 6 + 15 + 15 = 92 cm, the same as the box's 92 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 8 medium

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $16\ \text{cm}$ has a rectangle whose height is $8\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $5\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 16 cm square and a rectangle of the same area, 8 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 16 cm.
The rectangle is 8 cm tall and has the same area as the square.
The square sticks 5 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 16 x 16 = 256 cm2. The rectangle has the same area and a height of 8 cm, so its width is 256 divided by 8.

16 \times 16 = 256,\quad 256 \div 8 = 32

The rectangle is 32 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 5 cm to the left of the rectangle, and the rectangle (32 cm wide) runs from there to the right.

5 + 32 = 37

The whole figure is 37 cm across and 16 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 37, the rectangle's right side is 8, the ledge along the top of the rectangle is 37 - 16 = 21, the riser up the square's right side is 16 - 8 = 8, the square's top is 16, and its left side is 16. Every one of those slides out to the 37 by 16 box without changing length.

(37 + 16) \times 2 = 53 \times 2 = 106

The way around is 106 cm.

Answer: 106 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 37 + 8 + 21 + 8 + 16 + 16 = 106 cm, the same as the box's 106 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 9 hard

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $18\ \text{cm}$ has a rectangle whose height is $12\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $7\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 18 cm square and a rectangle of the same area, 12 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 18 cm.
The rectangle is 12 cm tall and has the same area as the square.
The square sticks 7 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 18 x 18 = 324 cm2. The rectangle has the same area and a height of 12 cm, so its width is 324 divided by 12.

18 \times 18 = 324,\quad 324 \div 12 = 27

The rectangle is 27 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 7 cm to the left of the rectangle, and the rectangle (27 cm wide) runs from there to the right.

7 + 27 = 34

The whole figure is 34 cm across and 18 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 34, the rectangle's right side is 12, the ledge along the top of the rectangle is 34 - 18 = 16, the riser up the square's right side is 18 - 12 = 6, the square's top is 18, and its left side is 18. Every one of those slides out to the 34 by 18 box without changing length.

(34 + 18) \times 2 = 52 \times 2 = 104

The way around is 104 cm.

Answer: 104 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 34 + 12 + 16 + 6 + 18 + 18 = 104 cm, the same as the box's 104 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 10 hard

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $20\ \text{cm}$ has a rectangle whose height is $16\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $5\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 20 cm square and a rectangle of the same area, 16 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 20 cm.
The rectangle is 16 cm tall and has the same area as the square.
The square sticks 5 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 20 x 20 = 400 cm2. The rectangle has the same area and a height of 16 cm, so its width is 400 divided by 16.

20 \times 20 = 400,\quad 400 \div 16 = 25

The rectangle is 25 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 5 cm to the left of the rectangle, and the rectangle (25 cm wide) runs from there to the right.

5 + 25 = 30

The whole figure is 30 cm across and 20 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 30, the rectangle's right side is 16, the ledge along the top of the rectangle is 30 - 20 = 10, the riser up the square's right side is 20 - 16 = 4, the square's top is 20, and its left side is 20. Every one of those slides out to the 30 by 20 box without changing length.

(30 + 20) \times 2 = 50 \times 2 = 100

The way around is 100 cm.

Answer: 100 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 30 + 16 + 10 + 4 + 20 + 20 = 100 cm, the same as the box's 100 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 11 hard

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $21\ \text{cm}$ has a rectangle whose height is $9\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $10\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 21 cm square and a rectangle of the same area, 9 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 21 cm.
The rectangle is 9 cm tall and has the same area as the square.
The square sticks 10 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 21 x 21 = 441 cm2. The rectangle has the same area and a height of 9 cm, so its width is 441 divided by 9.

21 \times 21 = 441,\quad 441 \div 9 = 49

The rectangle is 49 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 10 cm to the left of the rectangle, and the rectangle (49 cm wide) runs from there to the right.

10 + 49 = 59

The whole figure is 59 cm across and 21 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 59, the rectangle's right side is 9, the ledge along the top of the rectangle is 59 - 21 = 38, the riser up the square's right side is 21 - 9 = 12, the square's top is 21, and its left side is 21. Every one of those slides out to the 59 by 21 box without changing length.

(59 + 21) \times 2 = 80 \times 2 = 160

The way around is 160 cm.

Answer: 160 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 59 + 9 + 38 + 12 + 21 + 21 = 160 cm, the same as the box's 160 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Variant 12 hard

The figure on the right is made by placing a rectangle on top of a square. When the square and the rectangle have the same area, what is the perimeter of the whole figure, in $\text{cm}$ ?

Figure description: A square with a side of $24\ \text{cm}$ has a rectangle whose height is $18\ \text{cm}$ laid over it, shifted to the right so the two shapes overlap. The square rises higher than the rectangle, and the rectangle sticks out farther to the right than the square, so the combined outline forms a staircase (L-shaped) figure. Along the bottom, the left part of the square that the rectangle does not cover is $9\ \text{cm}$ long.

Your answer

Show solution

1 · Understandwhat's really being asked

A 24 cm square and a rectangle of the same area, 18 cm tall, overlap to make a staircase outline. We need the distance around it.

Givens

The square's side is 24 cm.
The rectangle is 18 cm tall and has the same area as the square.
The square sticks 9 cm out to the left of the rectangle.

Unknowns

The perimeter of the combined outline.

Constraints

Only the outside of the combined figure counts; the edges buried in the overlap do not.

2 · Planchoose the strategy

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

Equal areas is the only thing that pins the rectangle's width down, so start there. Then draw the outline and walk it.

3 · Execute3 carry out the plan

1Find the rectangle's width from equal area

#7 Identify Subproblems 3.MD.C.7

The square's area is 24 x 24 = 576 cm2. The rectangle has the same area and a height of 18 cm, so its width is 576 divided by 18.

24 \times 24 = 576,\quad 576 \div 18 = 32

The rectangle is 32 cm wide.

2Sketch the outline and find the total width

#1 Draw a Diagram 3.MD.D.8

Drawing both shapes on the same baseline: the square's left edge sticks out 9 cm to the left of the rectangle, and the rectangle (32 cm wide) runs from there to the right.

9 + 32 = 41

The whole figure is 41 cm across and 24 cm tall.

3Trace the outside edges and add them

#7 Identify Subproblems 3.MD.D.8

Going around the staircase: the bottom is 41, the rectangle's right side is 18, the ledge along the top of the rectangle is 41 - 24 = 17, the riser up the square's right side is 24 - 18 = 6, the square's top is 24, and its left side is 24. Every one of those slides out to the 41 by 24 box without changing length.

(41 + 24) \times 2 = 65 \times 2 = 130

The way around is 130 cm.

Answer: 130 cm

4 · Reviewdoes it hold up?

Add the six edges one by one: 41 + 18 + 17 + 6 + 24 + 24 = 130 cm, the same as the box's 130 cm.

Another way: The step's exact position does not matter. Sliding the rectangle further right would change where the corner sits but not the total, as long as the outline stays a plain staircase.

Standardsmin grade 3

3.MD.C.7 Relate area to the operations of multiplication and addition — Getting the rectangle's width out of the equal-area condition.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Walking the staircase outline and adding its edges.

💡Takeaway. A staircase outline costs exactly the same walk as the box that just fits around it.

Where next?

← Back to the problem: An L-shaped outline costs the same walk as its box