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Distributive property with area models

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5199823
A sports club orders equipment for \(8\) teams. Each team receives \(5\) soccer balls and \(3\) basketballs. a) Use an area model split into an \(8 \times 5\) rectangle and an \(8 \times 3\) rectangle. Write two expressions for the total number of balls. b) Name the property that explains why the expressions are equivalent.

Hints

- Represent each team with one row of the array. - Split the columns into groups of \(5\) and \(3\). - Compare the area of the whole rectangle with the sum of the two smaller areas.

Solution

1. One expression counts the total per team first: \(8 \times (5 + 3) = 8 \times 8 = 64\). 2. The split area model gives \(8 \times 5 + 8 \times 3 = 40 + 24 = 64\). 3. The distributive property explains why the two expressions are equivalent.

Answer

a) \(8 \times (5 + 3)\) and \(8 \times 5 + 8 \times 3\) b) Distributive property
5203873
Use a split area model to find each product. Split one factor into \(5\) and a remaining part, as in \(7 \times 8 = (5 + 2) \times 8\). a) \(6 \times 7\) b) \(7 \times 8\) c) \(8 \times 9\)

Hints

- Draw each product as a rectangle. - Split one side into \(5\) and the remaining amount. - Add the areas of the two smaller rectangles.

Solution

1. For a), split a \(6 \times 7\) rectangle into \(5 \times 7\) and \(1 \times 7\): \(35 + 7 = 42\). 2. For b), split a \(7 \times 8\) rectangle into \(5 \times 8\) and \(2 \times 8\): \(40 + 16 = 56\). 3. For c), split an \(8 \times 9\) rectangle into \(5 \times 9\) and \(3 \times 9\): \(45 + 27 = 72\).

Answer

a) \((5 + 1) \times 7 = 35 + 7 = 42\) b) \((5 + 2) \times 8 = 40 + 16 = 56\) c) \((5 + 3) \times 9 = 45 + 27 = 72\)
5372643
In the rectangular grid, \(5\) complete rows are blue and one additional row is orange. Use the known fact \(5 \times 7\) to find \(6 \times 7\). Write the decomposition and explain what the orange row represents.
Figure for problem 537264

Hints

- Find the number of unit squares in the blue part first. - Use the number of columns to find the unit squares in the orange row. - Add the two parts.

Solution

1. The blue part contains \(5 \times 7 = 35\) unit squares. 2. The orange row contains \(1 \times 7 = 7\) more unit squares. 3. Therefore, \(6 \times 7 = 5 \times 7 + 1 \times 7 = 35 + 7 = 42\).

Answer

\(6 \times 7 = 5 \times 7 + 1 \times 7 = 35 + 7 = 42\). The orange row represents \(7\) additional unit squares.
5372743
The green part shows \(6 \times 5\). One extra purple column is attached on the right. Use the array to find \(6 \times 6\). Which neighboring fact is used?
Figure for problem 537274

Hints

- Find the product represented by the green part. - Count the unit squares in the additional column. - One extra column adds one unit square to each row.

Solution

1. The green part contains \(6 \times 5 = 30\) unit squares. 2. The purple column contains \(6 \times 1 = 6\) more unit squares. 3. Therefore, \(6 \times 6 = 6 \times 5 + 6 \times 1 = 30 + 6 = 36\). The neighboring fact is \(6 \times 5\).

Answer

\(6 \times 6 = 6 \times 5 + 6 \times 1 = 30 + 6 = 36\). The neighboring fact is \(6 \times 5\).
5373463
Read the rectangular grid as \(7 \times 8\). The blue section shows \(5\) complete rows. How many unit squares are in the gray section that completes the grid? Write a decomposition equation.
Figure for problem 537346

Hints

- First determine how many rows are missing. - Each missing row contains \(8\) unit squares.

Solution

1. The grid needs \(7\) rows, and \(5\) rows are blue, so \(7 - 5 = 2\) rows are gray. 2. Each gray row contains \(8\) unit squares, so there are \(2 \times 8 = 16\) gray unit squares. 3. The complete grid is \(7 \times 8 = 5 \times 8 + 2 \times 8 = 40 + 16 = 56\).

Answer

There are \(16\) unit squares in the gray section. \(7 \times 8 = 5 \times 8 + 2 \times 8 = 40 + 16 = 56\).
5373483
The vertical color boundary splits the array into two smaller arrays. Use the split to find \(6 \times 8\) without recalling the \(8\)s fact directly.
Figure for problem 537348

Hints

- Read the width of each colored section. - Find each partial product, then add them.

Solution

1. The blue section has \(6\) rows and \(5\) columns, so it represents \(6 \times 5 = 30\). 2. The green section has \(6\) rows and \(3\) columns, so it represents \(6 \times 3 = 18\). 3. Add the partial products: \(6 \times 8 = 6 \times 5 + 6 \times 3 = 30 + 18 = 48\).

Answer

\(6 \times 8 = 6 \times 5 + 6 \times 3 = 30 + 18 = 48\).
5373493
A student knows \(7 \times 7 = 49\) but does not know \(7 \times 8\). Use the array to explain how the unknown fact can be built from the known fact.
Figure for problem 537349

Hints

- Locate the blue square first. - What does the additional column add?

Solution

1. The blue square shows \(7 \times 7 = 49\). 2. The gray additional column contains \(7\) unit squares. 3. Therefore, \(7 \times 8 = 7 \times 7 + 7 \times 1 = 49 + 7 = 56\).

Answer

The blue square shows \(7 \times 7 = 49\). The gray column adds \(7\) unit squares, so \(7 \times 8 = 49 + 7 = 56\).
5373743
A light wall has \(9\) rows with \(7\) lights in each row. The last \(2\) lights in every row are broken. Use the array to write a decomposition equation and find how many lights are still on.
Figure for problem 537374

Hints

- First determine how many lights work in each row. - Use the two sections of the array to write a subtraction equation.

Solution

1. Each row has \(7 - 2 = 5\) working lights. 2. With \(9\) rows, \(9 \times 5 = 45\) lights are on. 3. The array also shows \(9 \times 7 - 9 \times 2 = 63 - 18 = 45\).

Answer

\(9 \times 7 - 9 \times 2 = 63 - 18 = 45\). There are \(45\) working lights.
5373983
Use the array to find \(7 \times 12\). Split the \(12\) columns into \(10\) columns and \(2\) columns. Explain why the two partial products are added.
Figure for problem 537398

Hints

- The left section is \(10\) columns wide, and the right section is \(2\) columns wide. - Both sections have \(7\) rows.

Solution

The array has \(7\) rows and \(12\) columns. Splitting the columns into \(10\) and \(2\) gives \(7 \times 12 = 7 \times 10 + 7 \times 2 = 70 + 14 = 84\). The two partial products are added because the adjacent sections together make the entire array.

Answer

\(7 \times 12 = 70 + 14 = 84\). The partial products are added because the two sections together make the entire array.
5374333
Diagrams a) and b) show a garden bed before and after it is enlarged from \(6 \times 9\) planting spaces to \(8 \times 9\) planting spaces. How many new spaces are added? Explain the difference as a rectangle.
Figure for problem 537433

Hints

- Compare the numbers of rows. - The number of spaces in each row stays the same.

Solution

1. The original bed has \(6 \times 9 = 54\) planting spaces. 2. The enlarged bed has \(8 \times 9 = 72\) planting spaces. 3. The number of new spaces is \(72 - 54 = 18\). 4. The added rectangle has \(2\) new rows of \(9\) spaces, so \(2 \times 9 = 18\).

Answer

\(18\) new spaces are added. The added rectangle has \(2\) rows of \(9\) spaces, so \(2 \times 9 = 18\).
5372993
Use the vertical split in the rectangular grid to find an efficient way to calculate \(6 \times 9\). Write the calculation and explain why your method is efficient.
Figure for problem 537299

Hints

- Use the vertical boundary to determine the two widths. - Look for a familiar multiplication fact in one colored section. - Explain both your calculation and why the decomposition helps.

Solution

1. The vertical color boundary splits the \(9\) columns into \(5\) columns and \(4\) columns. 2. Use the decomposition \(6 \times (5 + 4) = 6 \times 5 + 6 \times 4\). 3. Calculate \(30 + 24 = 54\). This method is efficient because it begins with the familiar fact \(6 \times 5\).

Answer

One efficient method is \(6 \times (5 + 4) = 6 \times 5 + 6 \times 4 = 30 + 24 = 54\). It uses the familiar fact \(6 \times 5\).
5373013
Grid A shows \(8 \times 9\). In grid B, one gray column has been added on the right to make \(10\) columns. Use the “multiply by \(10\), then subtract one column” strategy to find \(8 \times 9\). Explain why the strategy is efficient.
Figure for problem 537301

Hints

- Compare the number of columns in grids A and B. - Find the value of the complete \(8 \times 10\) grid. - Subtract the added column.

Solution

1. Grid B contains \(8 \times 10\) unit squares. The added gray column contains \(8 \times 1\) unit squares. 2. Subtract the added column: \(8 \times (10 - 1) = 8 \times 10 - 8 \times 1\). 3. Therefore, \(80 - 8 = 72\). The method is efficient because multiplying by \(10\) is simple.

Answer

\(8 \times (10 - 1) = 8 \times 10 - 8 \times 1 = 80 - 8 = 72\). The strategy uses the easy product \(8 \times 10\), then removes one column of \(8\) unit squares.
5373293
Area models A and B both represent \(6 \times 12\), but they decompose \(12\) in different ways. a) Write a calculation that matches each area model. b) Which decomposition is more efficient for you? Explain your choice. c) Find the product.
Figure for problem 537329

Hints

- Read the width of each colored section in both models. - Multiplying by \(10\) or splitting a factor into equal parts can create useful facts. - Different decompositions can represent the same product.

Solution

1. For a), Model A shows \(6 \times (10 + 2) = 6 \times 10 + 6 \times 2 = 60 + 12 = 72\). Model B shows \(6 \times (6 + 6) = 6 \times 6 + 6 \times 6 = 36 + 36 = 72\). 2. For b), Model A may be more efficient because multiplying by \(10\) is especially simple. Model B is also acceptable with a clear explanation, such as using two equal halves. 3. Both decompositions give \(6 \times 12 = 72\).

Answer

a) A: \(6 \times 10 + 6 \times 2 = 60 + 12 = 72\) B: \(6 \times 6 + 6 \times 6 = 36 + 36 = 72\) b) Sample answer: Model A is more efficient because \(6 \times 10\) is easy to find. Model B is also valid with a reasonable explanation. c) \(6 \times 12 = 72\)
5373393
The array represents \(7\times 12\) and is split into a \(10\)-column section and a \(2\)-column section. A student writes \(7\times 12=70+2=72\). a) Find the error. b) Correct the calculation using the two sections of the array. c) Give one way to check the product using the array.
Figure for problem 537339

Hints

- Write a multiplication equation for each section of the array. - Check both partial products before adding them. - Reversing the factors gives another way to view the same array.

Solution

1. In the second partial product, the student used \(2\) instead of calculating \(7\times 2\). 2. The \(10\)-column section has \(7\times 10=70\) dots. The \(2\)-column section has \(7\times 2=14\) dots. 3. Add the partial products: \(70+14=84\). Therefore, \(7\times 12=84\). 4. One check is to reverse the factors and view the array as \(12\times 7\). Another is to count the dots in each section separately.

Answer

a) The student used \(2\) instead of \(7\times 2\) for the second partial product. b) \(7\times 12=7\times 10+7\times 2=70+14=84\) c) One possible check is \(12\times 7=84\).
5373573
Which colored array matches the decomposition \(5 \times 7 = 5 \times 4 + 5 \times 3\)? Explain how the position of the color boundary supports your choice.
Figure for problem 537357

Hints

- Determine whether each boundary separates rows or columns. - In both partial products, \(5\) is the number of rows.

Solution

1. In array A, the boundary is vertical. Each of the \(5\) rows is split into \(4\) unit squares and \(3\) unit squares. 2. Therefore, array A represents \(5 \times 4 + 5 \times 3\). 3. In array B, the boundary separates complete rows, so it represents a different decomposition.

Answer

Array A, because each row is split into \(4\) unit squares and \(3\) unit squares.
5373973
Estimate the number of dots in the array by comparing both side lengths with \(10\). Then find the exact number of dots and explain why the estimate is especially close.
Figure for problem 537397

Hints

- Compare \(9\) and \(11\) with \(10\). - Use \(9 \times (10 + 1)\) to calculate the exact product.

Solution

1. The array has \(9\) rows and \(11\) columns. Since both dimensions are close to \(10\), estimate with \(10 \times 10 = 100\). 2. The exact number is \(9 \times 11 = 99\). Using the distributive property, \(9 \times (10 + 1) = 90 + 9 = 99\). 3. One dimension is \(1\) less than \(10\), and the other is \(1\) greater than \(10\). Their product is only \(1\) less than \(100\), so the estimate is especially close.

Answer

The estimate is about \(100\) dots. The exact number is \(99\) dots. The estimate is especially close because the dimensions are \(9\) and \(11\), one below and one above \(10\).
5374033
Use the array to show two clearly different ways to calculate \(8 \times 9\): one method using \(8 \times 10\), and one method using two equal halves.
Figure for problem 537403

Hints

- For the first method, imagine adding a tenth column. - For the second method, split the \(8\) rows into two groups of \(4\).

Solution

1. Using \(8 \times 10\): Add a tenth column mentally, then subtract one column. \(8 \times 9 = 8 \times 10 - 8 = 80 - 8 = 72\). 2. Using equal halves: Half of the array is \(4 \times 9 = 36\). Therefore, \(8 \times 9 = 2 \times 36 = 72\).

Answer

\(8 \times 9 = 80 - 8 = 72\) and \(8 \times 9 = 2 \times 36 = 72\).

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