6.2: Area Model for Multiplication
When people refer to "measurement", often they think of using a ruler, yardstick, or tape measure. That is because one of the first forms of measurement we learn are based on a linear measurement model. In mathematics, we see this in the fact that often the first way we model relationships between numbers abstractly is with the number line.
We know that multiplication and area are related. But it is important to explicitly note what area actually is: a measurement describing two-dimensional space. It is also important that we understand why the area perspective of multiplication makes sense.
Suppose our basic unit is one square, as shown below.
So, you could describe 3 units using three squares. Using the repeated addition perspective of multiplication, we can picture 4 × 3 as 4 groups of 3 squares each, as shown below.
But if we stack up the groups, we would have 4 rows, with 3 squares in each row.
So we can think about 4 × 3 as a rectangle that has length 3 and width 4. The product, 12, is the total number of squares in that rectangle. Since each unit was actually a square, we refer to the area using "square units". That is, the area of the 4 x 3 rectangle is 12 square units.
The Distributive Property
In addition to many other benefits to thinking of multiplication in terms of area, one thing area makes very clear is that the Distributive Property makes sense.
For any three real numbers, the distributive property states that the product of one and the sum of the other two is equal to the sum of the products of the one with each of the other two. Symbolically, we write that if \(a\), \(b\), and \(c\) are any real numbers, then \(a \cdot (b+c) = a\cdot b + a\cdot c\). Sometimes we summarize this by stating that multiplication distributes over addition .
Use a rectangle to show that \(8 \cdot (2+5) = 8\cdot 2 + 8\cdot 5\). Then calculate the value of the expression.
Solution
First, we will draw an 8 x 7 rectangle. The 7 is because \(2+5 = 7\).
Now, we will split the side that is 7 units long into two pieces: one that is 2 units long and one that is 5 units long.
Since it's the same rectangle, the area is the same. The area of the first version is \(8\cdot (2+5)\). The area of the second version is the same value, but we can think of it as the sum of the areas of the two smaller rectangles that form it: \(8\cdot 2 + 8\cdot 5\).
Thus, \(8 \cdot (2+5) = 8\cdot 2 + 8\cdot 5 = 16+40 = 56\).
If you already know that \(8\times 7 = 56\), then it is easy to see that the answer above for \(8 \cdot (2+5) = 56\) is correct. What if both dimensions of a rectangle are split into pieces?
Try the following problem on your own and share it with a classmate, friend, or tutor.
Vera drew this picture as a model for 15 × 17. Use her picture to help you compute 15 × 17. Explain your work.
The problem above is an example of the following fact: distribution can be applied multiple times in one expression!
Use the distributive property twice to write \((a+b)\cdot (c+d)\) as a sum of products.
Solution
First, we use the distributive property to distribute the \((a+b)\).
\((a+b)\cdot (c+d) = (a+b)\cdot c + (a+b)\cdot d\)
Now, although we have not explcitily stated so before, we can use the commutativity of multiplication to rewirte the producst with the \((a+b)\) factors second.
\((a+b)\cdot (c+d) = (a+b)\cdot c + (a+b)\cdot d = c \cdot (a+b) + d\cdot (a+b)\)
This means the distributive property can be used again, but this time we distribute \(c\) and \(d\) in their respective terms.
\((a+b)\cdot (c+d) = (a+b)\cdot c + (a+b)\cdot d = c \cdot (a+b) + d\cdot (a+b)\)
\(=c \cdot a + c\cdot b + d\cdot a + d\cdot b\)
We use the pictures and the distrubitive property to make many calculations easier. Try the following exercises, keeping in mind that it is easy to multiply number that end in 0. For the third product, it may be useful to think about the fact that you can split a rectangle into as many small rectangles are you want!
Draw pictures like Vera’s for each of these multiplication exercises. Use your pictures to find the products without using a calculator or the standard American multiplication algorithm.
- \(23 \times 37\)
- \(8 \times 43\)
- \(371 \times 42\)
The Standard Algorithm for Multiplication
How were you taught to compute 83 × 27 in school? Were you taught to write something like the following?
Or maybe you were taught to leave in the zeros rather than leaving them out?
This is really no different than drawing the rectangle and using Vera’s picture for calculating! Try the following and discuss your answers with a classmate, friend, or tutor.
- Use \(83\times 27\) to explain why Vera’s rectangle method and the standard algorithm are really the same.
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Calculate the products below using both methods. Explain where you’re computing the same pieces in each algorithm.
- \(23 \times 14\)
- \(106 \times 21\)
- \(213 \times 31\)
Other Algorithms: Lines and Intersections
Here’s an unusual way to perform multiplication. To compute 22 × 13, for example, draw two sets of vertical lines, the left set containing two lines and the right set two lines (for the digits in 22) and two sets of horizontal lines, the upper set containing one line and the lower set three (for the digits in 13).
There are four sets of intersection points. Count the number of intersections in each and add the results diagonally as shown:
The answer 286 appears!
Use an area model like Vera's to calculate the same product. Can you see where the values 2, 2, 6, and 6 are in the area model?
There is one possible glitch in this lines & intersections algorithm, as illustrated by the computation 246 × 32 below:
Although the answer 6 thousands, 16 hundreds, 26 tens, and 12 ones is absolutely correct, one needs to carry digits and translate this as 7,872. Once again, use an area model to calculate the same product and see if you can find where the numbers of intersections appear in the area model. Try the following now.
- Compute 131 × 122 using the lines and intersections algorithm. Check your answer using another method.
- Compute 15 × 1332 using the lines and intersections algorithm. Check your answer using another method.
- Can you adapt the method to compute 102 × 3054?
- Why does the method work in general?
Other Algorithms: Lattice Multiplication
Throughout history, many people have used what is now commonly known as the lattice method .
To multiply 43 and 218, for example, draw a 2 × 3 grid. Write the digits of one number along the right side of the grid and the digits of the other number along the top.
Divide each cell of the grid diagonally and write in the product of the column digit and row digit of that cell, separating the tens from the units across the diagonal of that cell. If the product is a one digit answer, place a 0 in the tens place.
To get the answer, add the entries in each diagonal, carrying tens digits over to the next diagonal if necessary. In our example, we have
\(218 \times 43 = 9374\)
Now try the following.
- Compute 5763 × 345 using the lattice method.
- Explain why the lattice method is really the standard American multiplication algorithm in disguise.
- What is the specific function of the diagonal lines in the grid?