Practice · Read a repeated unit length off the picture · Sensim Math

Variant 1 easy

The figure at right shows rectangle ABCD made by joining $6$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $44\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $6$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $5$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 6 identical small rectangles. Its perimeter is 44 in, and we need its area.

Givens

6 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 5 lie sideways, stacked.
The perimeter of ABCD is 44 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 5 sideways rectangles stacked on the right. Stacking 5 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 5 short sides. Call one short side a unit.

\text{long side} = 5 \times \text{short side}

One long side is worth 5 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 5 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 5 units = 6 units.

\text{height} = 5\ \text{units}, \quad \text{width} = 1 + 5 = 6\ \text{units}

A 6-by-5 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (6 + 5) = 22 units. That means the 44-inch perimeter is 22 unit lengths.

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

22 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

22 unit lengths equal 44 inches, so one unit length is 44 divided by 22 = 2 inches. Check it: if 1 unit = 2 in, the sides are width = 12 in and height = 10 in, and 2 x (12 + 10) = 44 in. It matches.

22 \times \square = 44 \;\Rightarrow\; \square = 44 \div 22 = 2\ \text{in}

One unit is 2 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 12 inches wide and 10 inches tall, so its area is width times height.

A = 12 \times 10 = 120\ \text{in}^2

120 square inches.

Answer: 120 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 2 by 10 inches, so it covers 20 square inches, and 6 of them cover 120 square inches -- the same 120.

Another way: Guessing side lengths directly would mean testing pairs that add to 22; only 12 and 10 also fit the 6 : 5 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 2 easy

The figure at right shows rectangle ABCD made by joining $5$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $54\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $5$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $4$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 5 identical small rectangles. Its perimeter is 54 in, and we need its area.

Givens

5 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 4 lie sideways, stacked.
The perimeter of ABCD is 54 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 4 sideways rectangles stacked on the right. Stacking 4 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 4 short sides. Call one short side a unit.

\text{long side} = 4 \times \text{short side}

One long side is worth 4 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 4 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 4 units = 5 units.

\text{height} = 4\ \text{units}, \quad \text{width} = 1 + 4 = 5\ \text{units}

A 5-by-4 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (5 + 4) = 18 units. That means the 54-inch perimeter is 18 unit lengths.

P = 2 \times (5 + 4) = 2 \times 9 = 18\ \text{units}

18 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

18 unit lengths equal 54 inches, so one unit length is 54 divided by 18 = 3 inches. Check it: if 1 unit = 3 in, the sides are width = 15 in and height = 12 in, and 2 x (15 + 12) = 54 in. It matches.

18 \times \square = 54 \;\Rightarrow\; \square = 54 \div 18 = 3\ \text{in}

One unit is 3 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 15 inches wide and 12 inches tall, so its area is width times height.

A = 15 \times 12 = 180\ \text{in}^2

180 square inches.

Answer: 180 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 3 by 12 inches, so it covers 36 square inches, and 5 of them cover 180 square inches -- the same 180.

Another way: Guessing side lengths directly would mean testing pairs that add to 27; only 15 and 12 also fit the 5 : 4 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 3 easy

The figure at right shows rectangle ABCD made by joining $4$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $70\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $4$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $3$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 4 identical small rectangles. Its perimeter is 70 in, and we need its area.

Givens

4 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 3 lie sideways, stacked.
The perimeter of ABCD is 70 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 3 sideways rectangles stacked on the right. Stacking 3 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 3 short sides. Call one short side a unit.

\text{long side} = 3 \times \text{short side}

One long side is worth 3 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 3 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 3 units = 4 units.

\text{height} = 3\ \text{units}, \quad \text{width} = 1 + 3 = 4\ \text{units}

A 4-by-3 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (4 + 3) = 14 units. That means the 70-inch perimeter is 14 unit lengths.

P = 2 \times (4 + 3) = 2 \times 7 = 14\ \text{units}

14 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

14 unit lengths equal 70 inches, so one unit length is 70 divided by 14 = 5 inches. Check it: if 1 unit = 5 in, the sides are width = 20 in and height = 15 in, and 2 x (20 + 15) = 70 in. It matches.

14 \times \square = 70 \;\Rightarrow\; \square = 70 \div 14 = 5\ \text{in}

One unit is 5 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 20 inches wide and 15 inches tall, so its area is width times height.

A = 20 \times 15 = 300\ \text{in}^2

300 square inches.

Answer: 300 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 5 by 15 inches, so it covers 75 square inches, and 4 of them cover 300 square inches -- the same 300.

Another way: Guessing side lengths directly would mean testing pairs that add to 35; only 20 and 15 also fit the 4 : 3 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 4 easy

The figure at right shows rectangle ABCD made by joining $3$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $70\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $3$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $2$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 3 identical small rectangles. Its perimeter is 70 in, and we need its area.

Givens

3 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 2 lie sideways, stacked.
The perimeter of ABCD is 70 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 2 sideways rectangles stacked on the right. Stacking 2 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 2 short sides. Call one short side a unit.

\text{long side} = 2 \times \text{short side}

One long side is worth 2 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 2 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 2 units = 3 units.

\text{height} = 2\ \text{units}, \quad \text{width} = 1 + 2 = 3\ \text{units}

A 3-by-2 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (3 + 2) = 10 units. That means the 70-inch perimeter is 10 unit lengths.

P = 2 \times (3 + 2) = 2 \times 5 = 10\ \text{units}

10 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

10 unit lengths equal 70 inches, so one unit length is 70 divided by 10 = 7 inches. Check it: if 1 unit = 7 in, the sides are width = 21 in and height = 14 in, and 2 x (21 + 14) = 70 in. It matches.

10 \times \square = 70 \;\Rightarrow\; \square = 70 \div 10 = 7\ \text{in}

One unit is 7 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 21 inches wide and 14 inches tall, so its area is width times height.

A = 21 \times 14 = 294\ \text{in}^2

294 square inches.

Answer: 294 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 7 by 14 inches, so it covers 98 square inches, and 3 of them cover 294 square inches -- the same 294.

Another way: Guessing side lengths directly would mean testing pairs that add to 35; only 21 and 14 also fit the 3 : 2 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 5 medium

The figure at right shows rectangle ABCD made by joining $8$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $90\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $8$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $7$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 8 identical small rectangles. Its perimeter is 90 in, and we need its area.

Givens

8 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 7 lie sideways, stacked.
The perimeter of ABCD is 90 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 7 sideways rectangles stacked on the right. Stacking 7 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 7 short sides. Call one short side a unit.

\text{long side} = 7 \times \text{short side}

One long side is worth 7 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 7 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 7 units = 8 units.

\text{height} = 7\ \text{units}, \quad \text{width} = 1 + 7 = 8\ \text{units}

A 8-by-7 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (8 + 7) = 30 units. That means the 90-inch perimeter is 30 unit lengths.

P = 2 \times (8 + 7) = 2 \times 15 = 30\ \text{units}

30 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

30 unit lengths equal 90 inches, so one unit length is 90 divided by 30 = 3 inches. Check it: if 1 unit = 3 in, the sides are width = 24 in and height = 21 in, and 2 x (24 + 21) = 90 in. It matches.

30 \times \square = 90 \;\Rightarrow\; \square = 90 \div 30 = 3\ \text{in}

One unit is 3 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 24 inches wide and 21 inches tall, so its area is width times height.

A = 24 \times 21 = 504\ \text{in}^2

504 square inches.

Answer: 504 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 3 by 21 inches, so it covers 63 square inches, and 8 of them cover 504 square inches -- the same 504.

Another way: Guessing side lengths directly would mean testing pairs that add to 45; only 24 and 21 also fit the 8 : 7 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 6 medium

The figure at right shows rectangle ABCD made by joining $7$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $104\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $7$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $6$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 7 identical small rectangles. Its perimeter is 104 in, and we need its area.

Givens

7 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 6 lie sideways, stacked.
The perimeter of ABCD is 104 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 6 sideways rectangles stacked on the right. Stacking 6 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 6 short sides. Call one short side a unit.

\text{long side} = 6 \times \text{short side}

One long side is worth 6 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 6 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 6 units = 7 units.

\text{height} = 6\ \text{units}, \quad \text{width} = 1 + 6 = 7\ \text{units}

A 7-by-6 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (7 + 6) = 26 units. That means the 104-inch perimeter is 26 unit lengths.

P = 2 \times (7 + 6) = 2 \times 13 = 26\ \text{units}

26 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

26 unit lengths equal 104 inches, so one unit length is 104 divided by 26 = 4 inches. Check it: if 1 unit = 4 in, the sides are width = 28 in and height = 24 in, and 2 x (28 + 24) = 104 in. It matches.

26 \times \square = 104 \;\Rightarrow\; \square = 104 \div 26 = 4\ \text{in}

One unit is 4 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 28 inches wide and 24 inches tall, so its area is width times height.

A = 28 \times 24 = 672\ \text{in}^2

672 square inches.

Answer: 672 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 4 by 24 inches, so it covers 96 square inches, and 7 of them cover 672 square inches -- the same 672.

Another way: Guessing side lengths directly would mean testing pairs that add to 52; only 28 and 24 also fit the 7 : 6 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 7 medium

The figure at right shows rectangle ABCD made by joining $5$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $108\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $5$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $4$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 5 identical small rectangles. Its perimeter is 108 in, and we need its area.

Givens

5 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 4 lie sideways, stacked.
The perimeter of ABCD is 108 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 4 sideways rectangles stacked on the right. Stacking 4 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 4 short sides. Call one short side a unit.

\text{long side} = 4 \times \text{short side}

One long side is worth 4 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 4 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 4 units = 5 units.

\text{height} = 4\ \text{units}, \quad \text{width} = 1 + 4 = 5\ \text{units}

A 5-by-4 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (5 + 4) = 18 units. That means the 108-inch perimeter is 18 unit lengths.

P = 2 \times (5 + 4) = 2 \times 9 = 18\ \text{units}

18 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

18 unit lengths equal 108 inches, so one unit length is 108 divided by 18 = 6 inches. Check it: if 1 unit = 6 in, the sides are width = 30 in and height = 24 in, and 2 x (30 + 24) = 108 in. It matches.

18 \times \square = 108 \;\Rightarrow\; \square = 108 \div 18 = 6\ \text{in}

One unit is 6 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 30 inches wide and 24 inches tall, so its area is width times height.

A = 30 \times 24 = 720\ \text{in}^2

720 square inches.

Answer: 720 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 6 by 24 inches, so it covers 144 square inches, and 5 of them cover 720 square inches -- the same 720.

Another way: Guessing side lengths directly would mean testing pairs that add to 54; only 30 and 24 also fit the 5 : 4 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 8 medium

The figure at right shows rectangle ABCD made by joining $4$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $112\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $4$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $3$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 4 identical small rectangles. Its perimeter is 112 in, and we need its area.

Givens

4 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 3 lie sideways, stacked.
The perimeter of ABCD is 112 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 3 sideways rectangles stacked on the right. Stacking 3 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 3 short sides. Call one short side a unit.

\text{long side} = 3 \times \text{short side}

One long side is worth 3 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 3 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 3 units = 4 units.

\text{height} = 3\ \text{units}, \quad \text{width} = 1 + 3 = 4\ \text{units}

A 4-by-3 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (4 + 3) = 14 units. That means the 112-inch perimeter is 14 unit lengths.

P = 2 \times (4 + 3) = 2 \times 7 = 14\ \text{units}

14 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

14 unit lengths equal 112 inches, so one unit length is 112 divided by 14 = 8 inches. Check it: if 1 unit = 8 in, the sides are width = 32 in and height = 24 in, and 2 x (32 + 24) = 112 in. It matches.

14 \times \square = 112 \;\Rightarrow\; \square = 112 \div 14 = 8\ \text{in}

One unit is 8 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 32 inches wide and 24 inches tall, so its area is width times height.

A = 32 \times 24 = 768\ \text{in}^2

768 square inches.

Answer: 768 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 8 by 24 inches, so it covers 192 square inches, and 4 of them cover 768 square inches -- the same 768.

Another way: Guessing side lengths directly would mean testing pairs that add to 56; only 32 and 24 also fit the 4 : 3 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 9 hard

The figure at right shows rectangle ABCD made by joining $3$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $120\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $3$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $2$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 3 identical small rectangles. Its perimeter is 120 in, and we need its area.

Givens

3 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 2 lie sideways, stacked.
The perimeter of ABCD is 120 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 2 sideways rectangles stacked on the right. Stacking 2 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 2 short sides. Call one short side a unit.

\text{long side} = 2 \times \text{short side}

One long side is worth 2 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 2 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 2 units = 3 units.

\text{height} = 2\ \text{units}, \quad \text{width} = 1 + 2 = 3\ \text{units}

A 3-by-2 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (3 + 2) = 10 units. That means the 120-inch perimeter is 10 unit lengths.

P = 2 \times (3 + 2) = 2 \times 5 = 10\ \text{units}

10 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

10 unit lengths equal 120 inches, so one unit length is 120 divided by 10 = 12 inches. Check it: if 1 unit = 12 in, the sides are width = 36 in and height = 24 in, and 2 x (36 + 24) = 120 in. It matches.

10 \times \square = 120 \;\Rightarrow\; \square = 120 \div 10 = 12\ \text{in}

One unit is 12 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 36 inches wide and 24 inches tall, so its area is width times height.

A = 36 \times 24 = 864\ \text{in}^2

864 square inches.

Answer: 864 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 12 by 24 inches, so it covers 288 square inches, and 3 of them cover 864 square inches -- the same 864.

Another way: Guessing side lengths directly would mean testing pairs that add to 60; only 36 and 24 also fit the 3 : 2 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 10 hard

The figure at right shows rectangle ABCD made by joining $4$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $154\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $4$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $3$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 4 identical small rectangles. Its perimeter is 154 in, and we need its area.

Givens

4 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 3 lie sideways, stacked.
The perimeter of ABCD is 154 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 3 sideways rectangles stacked on the right. Stacking 3 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 3 short sides. Call one short side a unit.

\text{long side} = 3 \times \text{short side}

One long side is worth 3 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 3 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 3 units = 4 units.

\text{height} = 3\ \text{units}, \quad \text{width} = 1 + 3 = 4\ \text{units}

A 4-by-3 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (4 + 3) = 14 units. That means the 154-inch perimeter is 14 unit lengths.

P = 2 \times (4 + 3) = 2 \times 7 = 14\ \text{units}

14 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

14 unit lengths equal 154 inches, so one unit length is 154 divided by 14 = 11 inches. Check it: if 1 unit = 11 in, the sides are width = 44 in and height = 33 in, and 2 x (44 + 33) = 154 in. It matches.

14 \times \square = 154 \;\Rightarrow\; \square = 154 \div 14 = 11\ \text{in}

One unit is 11 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 44 inches wide and 33 inches tall, so its area is width times height.

A = 44 \times 33 = 1452\ \text{in}^2

1452 square inches.

Answer: 1452 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 11 by 33 inches, so it covers 363 square inches, and 4 of them cover 1452 square inches -- the same 1452.

Another way: Guessing side lengths directly would mean testing pairs that add to 77; only 44 and 33 also fit the 4 : 3 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 11 hard

The figure at right shows rectangle ABCD made by joining $6$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $198\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $6$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $5$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 6 identical small rectangles. Its perimeter is 198 in, and we need its area.

Givens

6 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 5 lie sideways, stacked.
The perimeter of ABCD is 198 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 5 sideways rectangles stacked on the right. Stacking 5 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 5 short sides. Call one short side a unit.

\text{long side} = 5 \times \text{short side}

One long side is worth 5 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 5 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 5 units = 6 units.

\text{height} = 5\ \text{units}, \quad \text{width} = 1 + 5 = 6\ \text{units}

A 6-by-5 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (6 + 5) = 22 units. That means the 198-inch perimeter is 22 unit lengths.

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

22 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

22 unit lengths equal 198 inches, so one unit length is 198 divided by 22 = 9 inches. Check it: if 1 unit = 9 in, the sides are width = 54 in and height = 45 in, and 2 x (54 + 45) = 198 in. It matches.

22 \times \square = 198 \;\Rightarrow\; \square = 198 \div 22 = 9\ \text{in}

One unit is 9 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 54 inches wide and 45 inches tall, so its area is width times height.

A = 54 \times 45 = 2430\ \text{in}^2

2430 square inches.

Answer: 2430 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 9 by 45 inches, so it covers 405 square inches, and 6 of them cover 2430 square inches -- the same 2430.

Another way: Guessing side lengths directly would mean testing pairs that add to 99; only 54 and 45 also fit the 6 : 5 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Variant 12 hard

The figure at right shows rectangle ABCD made by joining $7$ identical rectangles edge to edge without overlapping. When the perimeter of rectangle ABCD is $260\,\text{in}$ , what is the area of rectangle ABCD, in square inches?

Figure description: A large rectangle ABCD is formed by joining $7$ congruent rectangles edge to edge without overlapping. The top-left vertex is A, the bottom-left is B, the bottom-right is C, and the top-right is D. One rectangle stands upright on the left; the other $6$ lie sideways, stacked on top of one another to fill the right-hand part. Interior lines show the boundaries between the smaller rectangles.

Your answer

Show solution

1 · Understandwhat's really being asked

A big rectangle is tiled by 7 identical small rectangles. Its perimeter is 260 in, and we need its area.

Givens

7 congruent rectangles tile ABCD with no gaps or overlaps.
One stands upright; the other 6 lie sideways, stacked.
The perimeter of ABCD is 260 in.

Unknowns

The area of rectangle ABCD.

Constraints

No side length is given directly -- only the way the pieces fit.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#6 Guess and Check

The picture fixes the shape of one small rectangle. Call its short side one unit, write every side of the big rectangle in units, and the given perimeter turns into a division.

3 · Execute5 carry out the plan

1Name a unit length from the picture

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

Look at the 6 sideways rectangles stacked on the right. Stacking 6 of their short sides reaches from the top to the bottom of the big rectangle, and that same height is exactly the long side of the upright rectangle on the left. So the long side is 6 short sides. Call one short side a unit.

\text{long side} = 6 \times \text{short side}

One long side is worth 6 units.

2Write each side of the big rectangle in units

#7 Identify Subproblems 3.MD.D.8

The height of ABCD is one long side = 6 units. The width of ABCD is the upright rectangle's short side plus the sideways rectangles' long side, which is 1 unit + 6 units = 7 units.

\text{height} = 6\ \text{units}, \quad \text{width} = 1 + 6 = 7\ \text{units}

A 7-by-6 rectangle, measured in units.

3Express the perimeter as a multiple of the unit length

#7 Identify Subproblems 3.MD.D.8

Perimeter is twice the width plus twice the height, so it is 2 x (7 + 6) = 26 units. That means the 260-inch perimeter is 26 unit lengths.

P = 2 \times (7 + 6) = 2 \times 13 = 26\ \text{units}

26 units of fence, all the same length.

4Find the unit length

#6 Guess and Check 4.MD.A.3

26 unit lengths equal 260 inches, so one unit length is 260 divided by 26 = 10 inches. Check it: if 1 unit = 10 in, the sides are width = 70 in and height = 60 in, and 2 x (70 + 60) = 260 in. It matches.

26 \times \square = 260 \;\Rightarrow\; \square = 260 \div 26 = 10\ \text{in}

One unit is 10 inches.

5Compute the area

#7 Identify Subproblems 4.MD.A.3

The big rectangle is 70 inches wide and 60 inches tall, so its area is width times height.

A = 70 \times 60 = 4200\ \text{in}^2

4200 square inches.

Answer: 4200 square inches

4 · Reviewdoes it hold up?

Count it in pieces instead: each small rectangle is 10 by 60 inches, so it covers 600 square inches, and 7 of them cover 4200 square inches -- the same 4200.

Another way: Guessing side lengths directly would mean testing pairs that add to 130; only 70 and 60 also fit the 7 : 6 shape the picture forces.

Standardsmin grade 4

3.MD.D.8 Solve real-world problems involving perimeters of polygons — Writing the perimeter as a count of equal unit lengths.
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — Turning the unit length into real side lengths and an area.

💡Takeaway. When no length is given, name the repeated one as a unit -- the picture will tell you how many of them each side is worth.

Read a repeated unit length off the picture · 12 practice problems

Generated variants — 12

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