## Friday, August 8, 2014

### Area model as a reasoning tool

4.NBT.5
Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.

Area models play an important role in the CCSS-M, particularly in relationship to understanding multiplication. For example, 4.NBT.5 (above) suggests that area models may be useful to explain the process of multiplication calculation. I have previously discussed how area models may play a role in developing the whole number multiplication algorithm (Sept. – Oct. 2009) as well as how area models may be used to model division of fractions.

Although the CCSS-M does not necessarily discuss area models as a problem solving tool, I want to illustrate how area models may be used to make sense of and solve problems in certain situations.

Let’s consider the following problem (probably very familiar to most readers except the specific numbers used in the problem may be different).

A farmer has both pigs and chickens on his farm. There are 78 feet and 27 heads. How many pigs and how many chickens are there?

One way to solve this problem is to set up an equation, by letting the number of chickens as x. Then, we have the following equation: 4(27 – x) + 2x = 78. Alternately, we can use an area model to represent this problem situation. The number of feet of the pigs will be 4 × (number of pigs) and the number of feet of the chickens is 2 × (number of chicken). Since area models can be used to represent a product, we can draw two rectangles for the number of feet of the pigs and the number of feet for chickens as shown below.

In these rectangles, the vertical dimension represents the number of feet of an animal while the horizontal dimension represents the number of an animal. The area of each rectangle represents the number of feet.

Since the total number of feet is given as 78, if we draw these rectangles side by side to form an L-shape (see below), we know that the area of this L-shape represents the total number of feet, while the total of the horizontal dimension must be 27.

This diagram suggests a couple of possible solution approaches. One possibility is to note that the “area” of the top left corner of the L-shape (shaded in the figure below) must be 24 (= 78 – 54). Since the vertical dimension of the shaded region is 2 (feet), we can easily tell that the number of pigs must be 12. Then, since the total number of animals is 27, we can find that the number of chickens is 15 (= 27 – 12).

Another approach is to note that the “area” of the top right corner (shaded in the figure below) must be 30 (= 4 × 27 – 78). Once again, since the vertical dimension of this (shaded) rectangle is 2, we can easily find that the number of chickens must be 15. Then, by simple subtraction, we can find the number of pigs.

These solution approaches correspond to common solution strategies for this type of problems. The first approach is parallel to the idea of “let’s pretend all the animals were chicken.” Then, we should have only 54 feet, so, the extra 24 must be from the 2 extra feet each pig will bring to the total. The second approach, on the other hand will pretend if all the animals were pigs. Thus, 30 feet that are short must be due to the presence of chickens each of which contribute 2 fewer feet to the total. Perhaps the area models more clearly illustrate what 24 and 30 represent in the problem context.

Let’s see if you can use the area model to solve a similar problem by making use of area models.

Suppose there is a simple coin-toss game. On each toss, you receive 7 points if you get a Heads and 3 points if you get a Tails. After you played the game 20 times (that is, 20 coin tosses), your score was 104 points. How many times did you get Heads? How many times did you get Tails?

Here is another problem, but this one may require you to think a little more carefully about the situation.

There are some candies and children. If we give each child 4 candies there will be 18 candies left. However, in order to give each child 6 candies, we will need 12 additional candies. How many candies and how many children are there?

The answers to these two questions will be in the next post.