← 4-2 · Perimeter of overlapped squares and rectangles · Perimeter by Tracing Every Side

Perimeter of overlapped squares and rectangles · 10 practice problems

4.MD.A.34.G.A.2

Generated variants — 10

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

Variant 1 answer: 38 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $3\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $8\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 3 cm and the left side AB is 8 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 3 cm.
Left side AB = 8 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 8 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 8 cm, and because MDCN is a square, every side of it is 8 cm, so MD = 8 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 8 cm, and the dividing segment MN is the same height as the rectangle, so MN = 8 cm.

MN = AB = 8 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 8 cm, so MD = 8 cm.

MD = MN = 8 \text{ cm}

A square's four sides are all the same length, so once one side is 8 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 3 + 8 = 11 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 11 cm wide and 8 cm tall, so the perimeter is two widths plus two heights.

2 \times (11 + 8) = 2 \times 19 = 38 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 38 cm

Review

The rectangle is 11 cm by 8 cm, so the perimeter 38 cm equals 2 x (11 + 8). The width 11 cm is more than the 8 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 11 + 8 + 11 + 8 one side at a time to confirm 38 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 8 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 11 cm and the perimeter 2(11+8) = 38 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 2 answer: 36 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $4\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $7\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 4 cm and the left side AB is 7 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 4 cm.
Left side AB = 7 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 7 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 7 cm, and because MDCN is a square, every side of it is 7 cm, so MD = 7 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 7 cm, and the dividing segment MN is the same height as the rectangle, so MN = 7 cm.

MN = AB = 7 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 7 cm, so MD = 7 cm.

MD = MN = 7 \text{ cm}

A square's four sides are all the same length, so once one side is 7 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 4 + 7 = 11 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 11 cm wide and 7 cm tall, so the perimeter is two widths plus two heights.

2 \times (11 + 7) = 2 \times 18 = 36 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 36 cm

Review

The rectangle is 11 cm by 7 cm, so the perimeter 36 cm equals 2 x (11 + 7). The width 11 cm is more than the 7 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 11 + 7 + 11 + 7 one side at a time to confirm 36 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 7 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 11 cm and the perimeter 2(11+7) = 36 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 3 answer: 50 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $7\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $9\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 7 cm and the left side AB is 9 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 7 cm.
Left side AB = 9 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 9 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 9 cm, and because MDCN is a square, every side of it is 9 cm, so MD = 9 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 9 cm, and the dividing segment MN is the same height as the rectangle, so MN = 9 cm.

MN = AB = 9 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 9 cm, so MD = 9 cm.

MD = MN = 9 \text{ cm}

A square's four sides are all the same length, so once one side is 9 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 7 + 9 = 16 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 16 cm wide and 9 cm tall, so the perimeter is two widths plus two heights.

2 \times (16 + 9) = 2 \times 25 = 50 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 50 cm

Review

The rectangle is 16 cm by 9 cm, so the perimeter 50 cm equals 2 x (16 + 9). The width 16 cm is more than the 9 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 16 + 9 + 16 + 9 one side at a time to confirm 50 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 9 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 16 cm and the perimeter 2(16+9) = 50 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 4 answer: 52 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $6\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $10\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 6 cm and the left side AB is 10 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 6 cm.
Left side AB = 10 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 10 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 10 cm, and because MDCN is a square, every side of it is 10 cm, so MD = 10 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 10 cm, and the dividing segment MN is the same height as the rectangle, so MN = 10 cm.

MN = AB = 10 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 10 cm, so MD = 10 cm.

MD = MN = 10 \text{ cm}

A square's four sides are all the same length, so once one side is 10 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 6 + 10 = 16 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 16 cm wide and 10 cm tall, so the perimeter is two widths plus two heights.

2 \times (16 + 10) = 2 \times 26 = 52 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 52 cm

Review

The rectangle is 16 cm by 10 cm, so the perimeter 52 cm equals 2 x (16 + 10). The width 16 cm is more than the 10 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 16 + 10 + 16 + 10 one side at a time to confirm 52 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 10 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 16 cm and the perimeter 2(16+10) = 52 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 5 answer: 76 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $10\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $14\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 10 cm and the left side AB is 14 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 10 cm.
Left side AB = 14 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 14 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 14 cm, and because MDCN is a square, every side of it is 14 cm, so MD = 14 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 14 cm, and the dividing segment MN is the same height as the rectangle, so MN = 14 cm.

MN = AB = 14 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 14 cm, so MD = 14 cm.

MD = MN = 14 \text{ cm}

A square's four sides are all the same length, so once one side is 14 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 10 + 14 = 24 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 24 cm wide and 14 cm tall, so the perimeter is two widths plus two heights.

2 \times (24 + 14) = 2 \times 38 = 76 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 76 cm

Review

The rectangle is 24 cm by 14 cm, so the perimeter 76 cm equals 2 x (24 + 14). The width 24 cm is more than the 14 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 24 + 14 + 24 + 14 one side at a time to confirm 76 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 14 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 24 cm and the perimeter 2(24+14) = 76 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 6 answer: 58 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $5\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $12\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 5 cm and the left side AB is 12 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 5 cm.
Left side AB = 12 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 12 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 12 cm, and because MDCN is a square, every side of it is 12 cm, so MD = 12 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 12 cm, and the dividing segment MN is the same height as the rectangle, so MN = 12 cm.

MN = AB = 12 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 12 cm, so MD = 12 cm.

MD = MN = 12 \text{ cm}

A square's four sides are all the same length, so once one side is 12 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 5 + 12 = 17 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 17 cm wide and 12 cm tall, so the perimeter is two widths plus two heights.

2 \times (17 + 12) = 2 \times 29 = 58 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 58 cm

Review

The rectangle is 17 cm by 12 cm, so the perimeter 58 cm equals 2 x (17 + 12). The width 17 cm is more than the 12 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 17 + 12 + 17 + 12 one side at a time to confirm 58 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 12 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 17 cm and the perimeter 2(17+12) = 58 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 7 answer: 22 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $1\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $5\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 1 cm and the left side AB is 5 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 1 cm.
Left side AB = 5 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 5 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 5 cm, and because MDCN is a square, every side of it is 5 cm, so MD = 5 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 5 cm, and the dividing segment MN is the same height as the rectangle, so MN = 5 cm.

MN = AB = 5 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 5 cm, so MD = 5 cm.

MD = MN = 5 \text{ cm}

A square's four sides are all the same length, so once one side is 5 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 1 + 5 = 6 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 6 cm wide and 5 cm tall, so the perimeter is two widths plus two heights.

2 \times (6 + 5) = 2 \times 11 = 22 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 22 cm

Review

The rectangle is 6 cm by 5 cm, so the perimeter 22 cm equals 2 x (6 + 5). The width 6 cm is more than the 5 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 6 + 5 + 6 + 5 one side at a time to confirm 22 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 5 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 6 cm and the perimeter 2(6+5) = 22 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 8 answer: 48 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $4\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $10\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 4 cm and the left side AB is 10 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 4 cm.
Left side AB = 10 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 10 cm.

Plan

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

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 10 cm, and the dividing segment MN is the same height as the rectangle, so MN = 10 cm.

MN = AB = 10 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 10 cm, so MD = 10 cm.

MD = MN = 10 \text{ cm}

A square's four sides are all the same length, so once one side is 10 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 4 + 10 = 14 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 14 cm wide and 10 cm tall, so the perimeter is two widths plus two heights.

2 \times (14 + 10) = 2 \times 24 = 48 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 48 cm

Review

The rectangle is 14 cm by 10 cm, so the perimeter 48 cm equals 2 x (14 + 10). The width 14 cm is more than the 10 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 14 + 10 + 14 + 10 one side at a time to confirm 48 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 10 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 14 cm and the perimeter 2(14+10) = 48 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 9 answer: 28 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $2\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $6\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 2 cm and the left side AB is 6 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 2 cm.
Left side AB = 6 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 6 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 6 cm, and because MDCN is a square, every side of it is 6 cm, so MD = 6 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 6 cm, and the dividing segment MN is the same height as the rectangle, so MN = 6 cm.

MN = AB = 6 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 6 cm, so MD = 6 cm.

MD = MN = 6 \text{ cm}

A square's four sides are all the same length, so once one side is 6 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 2 + 6 = 8 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 8 cm wide and 6 cm tall, so the perimeter is two widths plus two heights.

2 \times (8 + 6) = 2 \times 14 = 28 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 28 cm

Review

The rectangle is 8 cm by 6 cm, so the perimeter 28 cm equals 2 x (8 + 6). The width 8 cm is more than the 6 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 8 + 6 + 8 + 6 one side at a time to confirm 28 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 6 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 8 cm and the perimeter 2(8+6) = 28 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!

Variant 10 answer: 76 cm

The figure below is made by joining a rectangle and a square edge to edge with no overlap. Find the sum of the lengths of the four sides (the perimeter) of rectangle ABCD, in cm.

Figure description: A wide rectangle ABCD (A top-left, D top-right, B bottom-left, C bottom-right) is divided into two parts by a vertical segment MN joining a point M on the top edge to a point N on the bottom edge. The left part AMNB is a rectangle whose top side AM measures $8\text{ cm}$ , and the right part MDCN is a square. The left side AB is marked $15\text{ cm}$ .

Show solution

Understand

A big rectangle ABCD is split by a vertical segment MN into a left rectangle AMNB and a right square MDCN. The top side AM of the left rectangle is 8 cm and the left side AB is 15 cm. I need the perimeter of the big rectangle ABCD.

Givens

ABCD is a rectangle with A top-left, D top-right, B bottom-left, C bottom-right.
A vertical segment MN splits ABCD into left part AMNB and right part MDCN.
The left part AMNB is a rectangle; the right part MDCN is a square.
Top side AM = 8 cm.
Left side AB = 15 cm.

Unknowns

The perimeter of rectangle ABCD, in cm.

Constraints

A square has four equal sides.
In a rectangle, opposite sides are equal.
The height of the whole figure equals AB = MN = 15 cm.

Plan

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

Use the square to find the missing length. The dividing segment MN equals the height AB = 15 cm, and because MDCN is a square, every side of it is 15 cm, so MD = 15 cm. Then the full top AD = AM + MD, and the rectangle's perimeter is twice the length plus twice the width.

Execute

#1 Draw a Diagram 4.G.A.1

The left side AB is 15 cm, and the dividing segment MN is the same height as the rectangle, so MN = 15 cm.

MN = AB = 15 \text{ cm}

MN runs from the top edge to the bottom edge, so it is exactly as tall as the rectangle.

#7 Identify Subproblems 4.G.A.2

The right part MDCN is a square, so all its sides are equal. Its side MN is 15 cm, so MD = 15 cm.

MD = MN = 15 \text{ cm}

A square's four sides are all the same length, so once one side is 15 cm they all are.

#7 Identify Subproblems 4.MD.A.3

The top of the big rectangle is AM plus MD.

AD = AM + MD = 8 + 15 = 23 \text{ cm}

The whole top edge is just the two top pieces laid end to end.

#7 Identify Subproblems 4.MD.A.3

The rectangle is 23 cm wide and 15 cm tall, so the perimeter is two widths plus two heights.

2 \times (23 + 15) = 2 \times 38 = 76 \text{ cm}

Going all the way around a rectangle covers each of the two lengths and two widths exactly once.

Answer: 76 cm

Review

The rectangle is 23 cm by 15 cm, so the perimeter 76 cm equals 2 x (23 + 15). The width 23 cm is more than the 15 cm height, matching a 'wide' rectangle.

Use Draw a Diagram (tool 1): walk the boundary adding 23 + 15 + 23 + 15 one side at a time to confirm 76 cm.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Identifying that MN spans the full height and equals AB.
4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Using the square's equal sides to get MD = MN = 15 cm.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Computing AD = 23 cm and the perimeter 2(23+15) = 76 cm.

💡 Spot the square to fill in the missing length, then go around the rectangle: it only needs Grade 4 perimeter sense!