2.10 Multiply decimals

Let's start with a rectangle, we can break it into a series of smaller rectangles, allowing us to work out our problem in smaller steps. Then we can add our answers together at the end.

We can evaluate something like 3.4\times34 by thinking of it as a rectangle whose length and width measure 3.4 units and 34 units respectively. Then we can break it into smaller rectangles and evaluate it in four parts. We finally add the individual solutions together to find our overall answer to our original problem.

The video below shows this process.

Examples

Example 1

In this question, we will use the area model to find the product of a decimal and a whole number.

a

Find the area of each rectangle below as a decimal.

\times	\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,	\,\,\,\,\,\,\,0.6\,\,\,\,\,\,\,	0.07
\,\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,\,\,\,

Worked Solution

Create a strategy

Multiply the sides of each rectangle to get the area.

Apply the idea

\displaystyle \text{Area of 1st rectangle}	\displaystyle =	\displaystyle 4\times9	Multiply the sides
	\displaystyle =	\displaystyle 36	Evaluate

\displaystyle \text{Area of 2nd rectangle}	\displaystyle =	\displaystyle 6\times9	Multiply the sides ignoring the decimal point
	\displaystyle =	\displaystyle 54	Evaluate
	\displaystyle =	\displaystyle 5.4	Insert the decimal point

\displaystyle \text{Area of 3rd rectangle}	\displaystyle =	\displaystyle 7\times9	Multiply the sides ignoring the decimal point
	\displaystyle =	\displaystyle 63	Evaluate
	\displaystyle =	\displaystyle 0.63	Insert the decimal point

Watch question walkthrough

b

Using the answer from part (a), find 4.67\times9 as a decimal.

Worked Solution

Create a strategy

Get the sum of the answers in part (a).

Apply the idea

4.67\times9=36+5.4+0.63

We can work out this addition by computing vertically. If the total sum in any column is bigger than 9, carry the tens into the next column.

\begin{array}{c} &&3&6 \\&&& 5&.&4 \\&+& &0&.&6&3 \\ \hline &&4&2&.&0&3\end{array}

We now have: 4.67\times9=42.03.

Watch question walkthrough

Let's follow through the process for the multiplication 4.83\times5.7 using the standard algorithm.

Before we even begin to calculate the answer it can be a good idea to have an estimation of the answer, especially when dealing with decimals. This will help us confirm our final answer is reasonable.

In this case, we can round both numbers to the nearest whole, giving us the much simpler calculation 5\times6, which we can evaluate to get 30. With this in mind, we would expect our answer to be close to this value, and we can know for sure that it will have digits in the tens and ones columns.

Now, to start the process, we simply ignore the decimal points. In this case we get 483 and 57. We then multiply these together with the method we are used to using for whole numbers:

\begin{array}{c} & & & &4&8&3 \\ &\times & & &  &5&7 \\ \hline & & &3&3&8&1 \\ &+ &2&4&1&5&0 \\ \hline & &2&7&5&3&1 \end{array}

Now we need to account for the decimal point. To do so, we add the total number of decimal places in the original numbers together.

In this case the original numbers are 4.83, which has two decimal places, and 5.7, which has one decimal place. So their product will have 2+1=3 decimal places.

Then to find the final answer, we take the product that we calculated before and insert the decimal point so that there are 3 decimal places:

So we have found that 4.83\times5.7=27.531.

This is very close to our original estimate of 30, and as expected we have a digit in each of the tens column and ones column. If we ended up with a final value of 275.31 we know we have written the decimal point in the wrong place.

The way we evaluate these multiplications is no different to working with whole numbers, and we can use exactly the same methods. We just need to think carefully about where the decimal point goes.

Examples