3.12 Applications of multiplication and division

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

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

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

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.

Partition one of the numbers to make the multiplication easier.

97=90+7 so we can write 65\times 97 as 65\times 90+65\times 7. To find each of these products we will use the vertical algorithm.

For 65\times7, set up the vertical algorithm: \begin{array}{c} &&6&5 \\ &\times &&7 \\ \hline &&& \\ \hline \end{array}

7 \times 5 = 35 so we put the 5 in the ones place and carry the 3 to the tens place.

\begin{array}{c} &&{}^36&5 \\ &\times &&7 \\ \hline &&& 5 \\ \hline \end{array}

7 \times 6=42 then add the carried over 3 to get 45. Put the 5 in the tens place and the 4 in the hundreds place.

\begin{array}{c} &&&6&5 \\ &\times &&&7 \\ \hline &&4&5& 5 \\ \hline \end{array}

For 65\times 90:

Since we are multiplying by 90, we will put 0 at the ones place as a place holder:

\begin{array}{c} &6&5 \\ \times &9&0 \\ \hline &&0 \\\hline \end{array}

Now we will multiply 65 by 9.

9\times 5=45. Put the 5 in the tens place and carry the 4 to the tens place.

\begin{array}{c} &{}^46&5 \\ \times &9&0 \\ \hline &5&0 \\\hline \end{array}

9 \times 6=54 then add the carried 4 to get 58. Put the 8 in the hundreds place and the 5 in the thousands place.

\begin{array}{c} &&&&6&5 \\ &\times &&&9&0 \\ \hline &&5&8&5&0 \\\hline \end{array}

Add the two products:

\begin{array}{c} &{}^15&{}^18&5&0 \\ +& &4&5&5 \\ \hline &6&3&0&5 \\\hline \end{array}

65\times 97 = 6305

Use short division.

The working and steps are shown below:

• 3 goes into 1 zero times with remainder 1 which we carry to the hundreds column.

• 3 goes into 12 four times.

• 3 goes into 0 zero times.

• 3 goes into 9 three times.

1209 \div 3=403