Free math worksheets

- Count grid rows and columns from the picture before writing any expression. - The vertical boundary divides the full width into two smaller widths while keeping the same row count. - Each smaller product should represent one complete section of the rectangle.

1. The model has \(8\) rows and \(8\) columns. The vertical line splits the columns into \(5\) on the left and \(3\) on the right. 2. The whole rectangle is \(8 \times 8\), and the two sections are \(8 \times 5+8 \times 3\). 3. \(8 \times 5+8 \times 3=40+24=64\), so \(8 \times 8=64\). 4. The distributive property lets us split the rectangle into the two smaller rectangles and add their products.

a) \(8\) rows; \(8\) columns; widths \(5\) and \(3\) b) \(8 \times 8=8 \times 5+8 \times 3=64\) c) Distributive property

- Read the rows and columns in each smaller rectangle. - Find the product of each smaller rectangle before finding the whole. - Add the two smaller products, then write the whole multiplication fact.

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

a) \(5\times7=35\) and \(1\times7=7\); \(35+7=42\); whole: \(6\times7=42\) b) \(5\times8=40\) and \(2\times8=16\); \(40+16=56\); whole: \(7\times8=56\) c) \(5\times9=45\) and \(3\times9=27\); \(45+27=72\); whole: \(8\times9=72\)

- Use the picture to find how many columns are in each row. - Find the multiplication fact shown by the first \(5\) rows. - Find how many squares the bottom row adds.

1. The first \(5\) rows each have \(7\) squares, so they represent \(5 \times 7=35\). 2. The additional bottom row adds \(1 \times 7=7\) unit squares. 3. Therefore, \(6 \times 7=5 \times 7+1 \times 7=35+7=42\).

\(6 \times 7=5 \times 7+1 \times 7=35+7=42\). The additional bottom row adds \(7\) unit squares.

- Use the image to count the rows and columns in the first part. - Find how many squares are in the extra rightmost column. - Add the two parts to describe the whole grid.

1. The first \(5\) columns have \(6\) rows, so they represent \(6 \times 5=30\). 2. The extra rightmost column adds \(6 \times 1=6\) unit squares. 3. Therefore, \(6 \times 6=6 \times 5+6 \times 1=30+6=36\). The known fact is \(6 \times 5\).

\(6 \times 6=6 \times 5+6 \times 1=30+6=36\)

- Use the picture to find the total number of rows and columns. - Compare the total rows with the \(5\) rows in the upper section. - Use the number of columns to find the number of squares in the lower rows.

1. The image shows a \(7\)-row by \(8\)-column grid. 2. The upper section has \(5\) rows, so the lower section has \(7-5=2\) rows. 3. The lower section contains \(2\times8=16\) unit squares. 4. The whole grid is \(7\times8=5\times8+2\times8=40+16=56\).

The lower section contains \(16\) unit squares. \(7 \times 8=5 \times 8+2 \times 8=40+16=56\).

- Count the rows and the columns in each section. - Write one multiplication fact for the left section and one for the right section. - Add those two products to find the whole grid.

1. The left section has \(6\) rows and \(5\) columns, so it represents \(6\times5=30\). 2. The right section has \(6\) rows and \(3\) columns, so it represents \(6\times3=18\). 3. Add the two smaller products: \(6\times8=6\times5+6\times3=30+18=48\).

\(6\times8=(6\times5)+(6\times3)=30+18=48\)

- Find the \(7\)-by-\(7\) square made by the first \(7\) columns. - Find how many squares the eighth column adds. - Start with \(7\times7=49\), then add the extra column.

1. The first \(7\) columns form a \(7\)-by-\(7\) square, so they represent \(7 \times 7=49\). 2. The added eighth column contains \(7\) unit squares. 3. Therefore, \(7 \times 8=7 \times 7+7 \times 1=49+7=56\).

The first \(7\) columns show \(7 \times 7=49\). The added eighth column adds \(7\), so \(7 \times 8=49+7=56\).

- Use the picture to find the number of rows and total lights in each row. - The right section shows the same number of broken lights in every row. - Subtract the broken rectangular section from the whole array.

1. The image shows \(9\) rows with \(7\) lights in each row. 2. Since \(2\) lights in each row are broken, \(7 - 2 = 5\) lights work in each row. 3. Thus, \(9 \times 5 = 45\) lights are on. 4. The same result is \(9 \times 7 - 9 \times 2 = 63 - 18 = 45\).

\(9 \times 7 - 9 \times 2 = 63 - 18 = 45\). There are \(45\) working lights.

- Use the image to determine the whole rectangle's dimensions and the width of each colored section. - Write one multiplication expression for each colored section. - Combine the two partial products to recover the whole product.

1. The image shows \(7\) rows and \(12\) columns, split into \(10\) columns and \(2\) columns. 2. The partial products are \(7 \times 10 = 70\) and \(7 \times 2 = 14\). 3. Add the adjacent sections: \(70 + 14 = 84\). Therefore, \(7 \times 12 = 84\). 4. The partial products are added because the two sections together make the whole rectangle.

\(7 \times 12 = 7 \times 10 + 7 \times 2 = 70 + 14 = 84\). The two partial products are added because the sections together form the whole array.

- Use the two images to compare the numbers of rows. - The number of spaces in each row stays the same. - Write the enlarged rectangle as the original rectangle plus only the new rows.

1. Diagram a) shows \(6\) rows of \(9\), so the original bed has \(6\times9=54\) spaces. 2. Diagram b) shows \(8\) rows of \(9\), so the enlarged bed has \(8\times9=72\) spaces. 3. The added rectangle has \(2\) rows of \(9\), or \(18\) spaces. 4. The relationship is \(8\times9=6\times9+2\times9=54+18=72\).

Original bed: \(6\times9=54\) Enlarged bed: \(8\times9=72\) \(8\times9=6\times9+2\times9\), so \(18\) new spaces were added.

- Read the row count and total column count from the grid. - The vertical boundary splits only the columns; both sections keep all the rows. - Write one smaller product for each visible section.

1. The rectangle has \(4\) rows and \(7\) columns, so the whole is \(4 \times 7\). 2. The vertical split divides the \(7\) columns into \(5\) and \(2\). 3. The two smaller rectangles are \(4 \times 5\) and \(4 \times 2\). 4. Therefore, \(4 \times 7=4 \times 5+4 \times 2=20+8=28\).

a) \(4 \times 7=28\) b) \(4 \times 7=4 \times 5+4 \times 2\) c) \(28\)

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

1. The vertical boundary splits the \(9\) columns into a left section of \(5\) columns and a right section of \(4\) columns. 2. Use \(6 \times (5+4)=6 \times 5+6 \times 4\). 3. Calculate \(30+24=54\). This is easier because it begins with the familiar fact \(6 \times 5\).

One way is \(6 \times (5+4)=6 \times 5+6 \times 4=30+24=54\). It uses the familiar fact \(6 \times 5\).

- Compare the numbers of columns in the two images. - Find the product represented by the larger \(\times 10\) grid. - Remove exactly the rightmost column that was added.

1. Grid A has \(8\) rows and \(9\) columns. Grid B has the same \(8\) rows and \(10\) columns. 2. Grid B contains \(8 \times 10=80\) unit squares, and the added rightmost column contains \(8\) unit squares. 3. Subtract the added column: \(80-8=72\). Thus, \(8 \times 9=72\). 4. This works well because the product \(8 \times 10\) is easy to find.

\(8 \times 9=8 \times 10-8=80-8=72\). The strategy uses the easy \(\times 10\) fact and removes the added rightmost column.

- Evaluate each pair of partial products separately. - Compare the two ways of decomposing the factor \(12\). - A valid distributive decomposition must keep the same total factor.

1. Student A: \(6\times10+6\times2=60+12=72\). 2. Student B: \(6\times6+6\times6=36+36=72\). 3. Student A may be more efficient because multiplying by \(10\) is especially simple. Student B is also valid because \(6+6=12\). 4. Both represent \(6\times12\) because the factor \(12\) is decomposed into addends whose sum is still \(12\).

a) Both decompositions equal \(72\). b) Sample answer: Student A is more efficient because \(6\times10\) is easy to find. c) Both are valid because \(10+2=12\) and \(6+6=12\).

- Use the picture to determine the width of each colored section. - Write a multiplication expression for each section before adding. - Check that each partial product counts every row in that section.

1. The large section has \(10\) columns, so it represents \(7 \times 10 = 70\). 2. The smaller section has \(2\) columns, so it represents \(7 \times 2 = 14\), not \(2\). 3. Add the partial products: \(70 + 14 = 84\). Therefore, \(7 \times 12 = 84\). 4. One check is to reverse the factors and view the same array as \(12 \times 7 = 84\).

a) The student used \(2\) instead of \(7 \times 2\) for the smaller partial product. b) \(7 \times 12 = 7 \times 10 + 7 \times 2 = 70 + 14 = 84\) c) One possible check is \(12 \times 7 = 84\).

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

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 split.

Array A, because each row is split into \(4\) unit squares and \(3\) unit squares.

- Compare both side lengths in the image with \(10\). - Use a nearby \(\times 10\) product to calculate the exact value. - Compare the exact result with the estimate.

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 \(10\), so the exact product is only \(1\) less than \(100\).

The estimate is about \(100\) dots. The exact number is \(99\) dots.

- In Grid B, compare the \(10\)-column rectangle with the \(9\)-column target. - In Grid A, split the \(8\) rows into two equal groups of \(4\). - Check that both ways give the same product.

1. Grid B shows \(8 \times 10=80\). Remove its added rightmost column of \(8\) squares: \(80-8=72\). 2. Grid A can be split into two equal groups of \(4\) rows. Each half is \(4 \times 9=36\), so \(2 \times 36=72\). 3. Both methods give \(8 \times 9=72\).

a) \(8\times9=80-8=72\) b) \(8\times9=2\times36=72\)

- Read the whole rectangle’s rows and columns from the grid. - The marked strip is exactly one column wide. - Your subtraction equation should show the whole product and the strip product. - For the final equation, express the whole rectangle as the sum of its two rectangular parts.

1. The whole grid has \(7\) rows and \(10\) columns, so its area is \(7 \times 10 = 70\) square units. 2. The marked strip has \(7\) rows and \(1\) column, so its area is \(7 \times 1 = 7\) square units. 3. Subtract the strip from the whole: \(7 \times 10 - 7 \times 1 = 70 - 7 = 63\). 4. The same partition can be written \(7 \times 10 = 7 \times 9 + 7 \times 1\). Thus the unmarked rectangle has area \(7 \times 9 = 63\) square units.

a) \(7 \times 10 = 70\) b) \(7 \times 1 = 7\) c) \(7 \times 10 - 7 \times 1 = 70 - 7 = 63\) d) \(7 \times 10 = 7 \times 9 + 7 \times 1\)

- Read the width of each section from the vertical partition. - Both smaller rectangles have the same \(6\) rows. - Add the two smaller products to find the whole rectangle.

1. The left section is \(5\) columns wide, so it represents \(6 \times 5=30\). 2. The right section is \(3\) columns wide, so it represents \(6 \times 3=18\). 3. Together the widths make \(8\) columns, so \(6 \times 8=(6 \times 5)+(6 \times 3)=30+18=48\).

\(6 \times 8=(6 \times 5)+(6 \times 3)=48\)

- Look at the number of rows in both parts of the grid. - The second part needs a multiplication fact, not just its width. - Check that the two smaller products add to the whole product.

1. Both smaller rectangles have \(8\) rows. 2. The second rectangle is \(8\) rows by \(2\) columns, so it is \(8\times2\), not just \(2\). 3. The correct equation is \(8\times7=8\times5+8\times2=40+16=56\).

The missing factor is \(8\). The correct equation is \(8\times7=8\times5+8\times2=56\).