Many of us teach area and perimeter but I’m guessing that most of us have not spent a lot of time thinking about the relationship between the two. This meeting began a investigation that was totally new to me. Tyler started Friday’s meeting by showing us a coordinate plane with an x axis labeled Perimeter and y axis labeled Area with 2 plotted points. Tyler asked us: What do you notice? What questions do you have?
Some of our questions:
- What does the dot represent? Does it represent a rectangle?
- What is the relationship between area and perimeter?
- Are both of the figures represented by these dots possible? Which is possible and which is impossible?
- Could the point describe a triangle? Pentagon? Other shapes?
Tyler then gave us graph paper and the following prompt:
“Draw a shape on squared paper and plot a point to show its perimeter and area. Which points on the grid represents squares, rectangles, etc? Draw a shape that may be represented by the point (4, 12) or (12, 4). Find all the “impossible” points.”
(Desmos animation of different rectangles with a perimeter of 12.)
Over the next hour, we drew figures and calculated area and perimeter for rectangles, triangles, circles, etc. Some of us plotted rectangles on the coordinate plane and noticed patterns about where squares showed, for example, and determined how to know whether a point was possible or impossible.
Click to see possible plotted points [spoiler].
Click to see a curve of all squares [spoiler].
Questions to explore:
- Why are squares interesting when we think about the relationship between area and perimeter?
- What would the graph of all possible triangles look like?
- The graph of all possible circles?
At the end of the meeting, we did some algebraic work to determine the exact dimensions of a rectangle with an area of 4 and a perimeter of 12. We knew from trial and error that 5.3 x .7 was close. The perimeter (5.3 + 5.3 + .7 + .7) is 12, but the area (5.3 x .7) is 3.71, not 4.
We started with the following two equations and solved a system of equations:
x * y = 4
2x + 2y = 12
(Click to see a solution of this system. Sorry for the coffee stain.)
We came to 5.24 x .76, which is closer but not exact due to rounding. 5.24 + 5.24 + .76 + .76 = 12 and 5.24 x .76 = 3.9824 (not 4).
What do you notice about the solution to the following system of equations, from the other point (4, 12) on the coordinate plane)?
x * y = 12
2x * 2y = 4
On the way out, Tyler shared the Gold Rush problem with us and told us about Math Memos, the new section on CollectEdNY.org which features teacher write-ups of rich problems, along with samples of student work and discussion of multiple solution methods. Tyler’s write-up of the Gold Rush problem is highly recommended. If you missed this meeting, I recommend that you try the Gold Rush problem on your own, then think about connections to the area vs. perimeter puzzle above. Share your thoughts about the problems or teaching questions below.
(By the way, Tyler is looking for teachers who are interested in writing up problems for Math Memos. Is anyone interested in teaching a problem from a CAMI meeting, collecting student work and writing it up for the site?)
Question for the group: Would anyone use the Area vs. Perimeter prompt at the beginning with an HSE class? How could we scaffold and prepare students for the open-ended nature and abstraction of the task? Is the Gold Rush problem adequate preparation?
Extension:
An integer-sided rectangle with area A is called a rectangular personality of A. Which integer from 1 through 100 has the most rectangular personalities?
In other words, which integer area value between 1 and 100 has the most possible integer-only perimeters?
In attendance: Avril, Chaim, Deneise, Eric, Jane, Nikko, Solange, Tyler
Programs represented: Bard Prison Initiative, BMCC, Brooklyn Public Library, CUNY Adult Literacy, Fifth Avenue Committee
Location: The Brooklyn Public Library, Pacific Street Branch at 25 4th Avenue
One thing that I think I’m noticing is that:
x * y = 12
2x * 2y = 4
is Not possible, e.g. if x =1 and y=12, then 2(1) times 2(12) cannot and never will be a number less than 12, but can only be some number more than 12, in this case 2 times 24 + 48.
In other words it’s not posssible for 2 time a number to be less than the original number before multiplying it twice
Secondly, I think one way to approach solving the problem will be to compare the formulars for area for a square, i.e. L=2 times W=6 = 12
triangle1/2B (12)=6 H=2 = 12
circle pi times r squared = 12
I don’t know if I’m correct but I got r squared=3.82
Therefore 3.14 times 3.8256 = 12 when 3.8216 rounds to 3.8256
Although I am still unable to solve the problem someone may have an easier time of by gleaning info from what’s similar & what’s different when calculating the same area of these 3 different
figures.
“An integer-sided rectangle with area A is called a rectangular personality of A. Which integer from 1 through 100 has the most rectangular personalities? In other words, which integer area value between 1 and 100 has the most possible integer-only perimeters?”
Tyler,
I’m trying to understand the extension. Am I on the right track?
I’m looking at all the numbers between 1 and 100. Which numbers (as areas of different rectangles) have the most different associated rectangles with whole number sides? For example, a rectangle with an area of 3 has only one possible perimeter with integers (1×3), so there is only one rectangle with the personality of 3. An area of 4 though gives me two different rectangles with the same personality (1×4 and 2×2).
I’m looking for area values that have the most different rectangles with that same personality (area)?
Eric