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Australia
Year 5

3.07 Distributive property of multiplication

Are you ready?

We have already practiced breaking apart numbers to help us  solve multiplication problems  . Let's try this problem to review.

Examples

Example 1

Let's use an area model to find 77 \times 3.

a

Fill in the areas of each rectangle.

A rectangle with a length of 77 and a height of 3 divided into 4 rectangles. Ask your teacher for more information.
Worked Solution
Create a strategy

For each rectangle, use the formula \text{Area} = \text{length} \times \text{width}.

Apply the idea

Area of top left rectangle: 70 \times 2= 140

Area of top right rectangle: 7 \times 2= 14

Area of bottom left rectangle: 70 \times 1= 70

Area of bottom right rectangle: 7 \times 1= 7

A rectangle with a length of 77 and a height of 3 divided into 4 rectangles. Ask your teacher for more information.
b

What is the total area of all four rectangles?

Worked Solution
Create a strategy

Add the areas from part (a) together using a place value table.

Apply the idea

Put the numbers in a place value table:

HundredsTensOnes
140
70
14
7

Now we can add the numbers down each column and regroup where needed:

HundredsTensOnes
140
70
14
+117
=231
  • 4+7 = 11 so we put a 1 in the ones column and carry the 1 to the tens column.

  • Then 4+7+1+1=13 so we put a 3 in the tens column and carry the 1 to the hundreds column.

So the total area is 231.

c

Find 77 \times 3.

Worked Solution
Create a strategy

Use the answer from part (b).

Apply the idea

The rectangle from part (a) has a length of 77 and a width of 3. So we can find its area using 77\times 3. But since we already found its area in part (b) to be 231, we get: 77\times 3 = 231

Idea summary

We have different ways we can solve multiplication problems, such as the area method and arrays. But as our numbers get larger, the algorithm method can be more useful.

Distributive property of multiplication

This video introduces the distributive property.

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Examples

Example 2

We want to use the distributive property to rewrite 2\times 19 as easier multiplications.

a

This diagram shows how 2 groups of 19 objects can be split up.

2 rectangles divided into 19 equal parts. 10 parts are shaded green and 9 parts are shaded blue.

Use the diagram to complete the blank to make the statement true.2\times19=2\times(10+⬚)

Worked Solution
Create a strategy

We can use the diagram to split up the number.

Apply the idea

The diagram shown below that one group of 19, objects can be split up into 10 and 9.

A rectangle divided into 19 equal parts. 10 parts are shaded green and 9 parts are shaded blue.

This means: 19=10+9

For 2 groups of 19, we need to double this sum.

So 2 groups of 19 objects is the same as 2 \times 19 = 2 \times (10 + 9)

b

Complete the blanks to show how 2 groups of (10+9) can be split up into smaller multiplications.

2\times(10+9)=2\times10 + 2 \times ⬚

Worked Solution
Create a strategy

Count the groups of 10 and 9 in the diagram.

2 rectangles divided into 19 equal parts. 10 parts are shaded green and 9 parts are shaded blue.
Apply the idea

The above images shows 2 groups of 19 squares.

We can think of it as 2 groups of 10 squares on the left, and 2 groups of 9 squares on the right.

So 2 \times 19 is also equal to 2 \times 10 + 2 \times 9.But we already showed that 2 \times 19 = 2\times(10+9).So now we have:

2\times \left(10+9\right)=2\times 10+2\times 9

Idea summary

The distributive property of multiplication states that multiplying a sum is the same as multiplying each part and then adding. \begin{aligned} 5 \times 13 &= 5 \times (3+10) \\ &= 5 \times 3+5 \times 10 \end{aligned}

Distributive property to solve problems

This video shows us how to use the distributive property to solve problems.

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Examples

Example 3

We want to find 2 \times 45.

Use the area model to complete the following:

A rectangle with a length of 45 and a height of 2, divided into 2 smaller rectangles with lengths 40 and 5.
\displaystyle 2\times 45\displaystyle =\displaystyle 2\times \left(40+5\right)
\displaystyle =\displaystyle 2\times ⬚ + 2\times5
\displaystyle =\displaystyle ⬚ + ⬚
\displaystyle =\displaystyle ⬚
Worked Solution
Create a strategy

We can find the area of the two smaller rectangles and add the two areas.

Apply the idea

We can think of 2\times 45 as representing the area of this rectangle:

A rectangle with a length of 45 and a height of 2.

which is equal to the sum of the two smaller rectangles.

A rectangle with a length of 45 and a height of 2, divided into 2 smaller rectangles with lengths 40 and 5.

The rectangle on the left has an area of 2 \times 40, and the rectangle on the right has an area of 2 \times 5.

\displaystyle 2\times 45\displaystyle =\displaystyle 2\times \left(40+5\right)
\displaystyle =\displaystyle 2\times 40 + 2\times5Add the areas of the smaller rectangles
\displaystyle =\displaystyle 80 + 10Multiply the numbers
\displaystyle =\displaystyle 90Add 80 and 10
Idea summary

If we have a number sentence such as 5\times 12 it can be rewritten as 5\times 10 + 5\times 2.

Solve problems with 3 digit numbers

This video has another example of using it to solve problems.

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Examples

Example 4

We want to find 6 \times 795.

Use the area model to complete the following:

A rectangle with a length of 795 and a height of 6, divided into 3 smaller rectangles with lengths 700, 90, and 5.
\displaystyle 6\times 795\displaystyle =\displaystyle 6\times \left(700+900+5\right)
\displaystyle =\displaystyle 6\times ⬚ + 6\times ⬚ + 6\times ⬚
\displaystyle =\displaystyle ⬚ + ⬚ + ⬚
\displaystyle =\displaystyle ⬚
Worked Solution
Create a strategy

We can find the area of the three smaller rectangles and add the three areas.

Apply the idea

We can think of 6\times 795 as representing the area of this rectangle:

A rectangle with a length of 795 and a height of 6.

which is equal to the sum of the three smaller rectangles.

A rectangle with a length of 795 and a height of 6, divided into 3 smaller rectangles with lengths 700, 90 and 5.

The rectangle on the left has an area of 6 \times 700, the rectangle on the middle has an area of 6 \times 90, and the rectangle on the right has an area of 6 \times 5.

\displaystyle 6 \times 795\displaystyle =\displaystyle 6\times \left(700+90+5\right)
\displaystyle =\displaystyle 6\times 700 + 6 \times 90 + 6\times 5Add the areas of the smaller rectangles
\displaystyle =\displaystyle 4200 + 540 + 30Multiply the numbers
\displaystyle =\displaystyle 4770Add the results
Idea summary

We can represent the distributive property as splitting a rectangle into three parts, finding the area of each part and then adding them together.

Outcomes

AC9M5N09

use mathematical modelling to solve practical problems involving additive and multiplicative situations including financial contexts; formulate the problems, choosing operations and efficient calculation strategies, using digital tools where appropriate; interpret and communicate solutions in terms of the situation

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