String of Related Problems

Students will be encouraged to explore the area model to apply the use of partial products made possible with the distributive property through the following string of related problems.

Since we discovered in the previous lesson that the area of any circle can be found by decomposing the circle into a rectangle or parallelogram, we will leverage a context involving area of a square and area of a rectangle to build fluency and flexibility approximating the area of a circle.

Throughout the string, you’ll notice that each “pair” of problems in the string involves first finding the area of a rectangle with a length that is 3 times the width (representing the radius multiplied by half of the circumference) and the second represented by tripling the area of a square (representing the area of a square with side lengths equivalent to the length of the radius, tripled). Each “pair” of problems should produce the same resulting area.

To state the context, you might consider a script similar to the following:

A builder is trying to determine how many square units of flooring they will need for different rooms of a home. Help them out by determining the area for each room.

Room 1 is rectangular with dimensions of 5 units by 15 units

Room 2 was measured in 3 rectangular chunks with dimensions of 5 units by 5 units.

Room 3 is rectangular with dimensions of 6 units by 18 units.

Room 4 was measured in 3 rectangular chunks with dimensions of 6 units by 6 units.

Room 5 is rectangular with dimensions of 10 units by 30 units.

Room 6 was measured in 3 rectangular chunks with dimensions of 10 units by 10 units.

Encourage or create opportunities to represent student thinking using the area model by pausing with questions such as “would it be ok if I show your thinking using this area model” if students do not initiate the use of a model. If students want to model their thinking another way however, this should be honoured not discouraged.

Be sure to draw out the equivalence between each pair of problems. For example, from problem 1 and 2:

5 x 15
= 5(5 + 5 + 5)
= (5 x 5) + (5 x 5) + (5 x 5)
= 3(5 x 5)
= 3(52)

Note that we have explicitly drawn out the idea that the area of a circle can be found by “a little more” than tripling a square with side lengths equivalent to the radius. So while most textbooks would suggest that the area of a circle formula is \(πr^2\), what we are seeing is that we can approach this idea by multiplying the radius by “a little more” than triple the radius.

From here, you might consider asking students to think backwards now and ask:

For each equivalent pair of problems in the string, what would the approximate diameter be of a circle if we wanted to cover the same area?

If students are building flexibility with their understanding of area of a circle, they may be able to quickly identify that in each problem from the string, the the square side length or the shorter side length from the rectangle is the radius. Therefore, doubling that value would be the approximate diameter.

Facilitator Note:

Note that as we have done so previously, we have approximated the multiplicative relationship between the circumference and the diameter, Pi (π), to be 3 instead of approximately \(\frac{22}{7}\) or 3.14.