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If you have 72 ft. of fencing and you want to use it to make a rectangular pen for your Highland terrier, you must consider both the perimeter of the pen and its area. What relationships exist between these two measures? Do shapes with the same perimeter have the same area? Let’s investigate this situation.
Imagine that you want to use all 72 ft. of fencing to make the rectangular pen, that the dimensions of the pen in feet will be whole-number values, and that you want the maximum area for your puppy.
All of the pens have a perimeter of 72 ft., yet the areas of the pens differ. What do you notice about the shapes of the pens with small areas as opposed to those with large areas? What are the characteristics of a shape with the greatest area?
In the above problems, you’ve seen that when you form the fencing into a long, skinny rectangle, the area is small. But the area increases as the rectangle becomes more square-like, and the greatest area occurs when the fencing is in the shape of a square or square-like rectangle. This leads us to consider shapes other than rectangles. For example, if the perimeter remained the same, would an equilateral triangle or a regular pentagon or a regular hexagon have the same area or more or less area than the square?
Imagine the pen were in the shape of an equilateral triangle. What is the area of this triangular pen?
Draw a picture of a triangle. How long is each side? Draw the height of the triangle, and use the Pythagorean theorem to find the height.
Imagine the pen were in the shape of a regular hexagon. What is the area of this hexagonal pen?
Draw a picture of a hexagon. What is the length of each side? You can divide the hexagon into six equilateral triangles and calculate the area of a single triangle using the method from Problem A3.
Would other shapes give the puppy even more square footage? Imagine building a circular pen. Find the area when the circumference is exactly 72 ft.
In the activities above, you’ve seen that when the perimeter is fixed, shapes that have many sides have a greater area. In fact, the shape with the greatest area when the perimeter remains constant is a circle.
Imagine you have a barn that is 70 ft. long on your property. You plan to use a part of the existing barn wall as one side of the fence. What are some options for the shape of the pen? What shape of the pen will give the greatest area under these conditions?
Consider several different shapes for the pen — square, rectangle, semicircle — and collect data for each of them.
Video Segment
In this video segment, David Cellucci and David Russell examine how to maximize the area of the puppy pen built against a barn. They try several different shapes, including rectangles and semicircles. Watch this segment after you’ve completed Problem A6. Were your findings similar? You can find this segment on the session video approximately 4 minutes and 56 seconds after the Annenberg Media logo. |
There are other situations that involve both area and perimeter. Consider this one: Joel, a student in a sixth-grade class, completed an exercise similar to one in Session 6. After tracing his hand on grid paper, he was asked to approximate its area. Instead of counting squares, he took a piece of string and traced the perimeter of his hand. He then took the length of string that represented the perimeter of his hand, reshaped it into a rectangle, and found the area of the rectangle. Joel concluded that the area of his hand and the area of the rectangle were the same.
Problem A7
Think about Joel’s strategy and conclusion. Will his strategy work in all situations? Do you agree or disagree with his conclusions?
Think about what you discovered in Problems A1 and A2 and how that information might be used to analyze Joel’s strategy.
Looking at this situation from another direction, do figures with the same area always have the same perimeter? Why or why not? And if not, which perimeters are possible, and which are impossible?
Use this Interactive Activity to investigate this question. You can arrange 12 square “tiles” to make plane figures like the ones below. Each square must share at least one side with another square.
For this activity, use 12 one-inch square plastic or ceramic tiles (or Scrabble tiles).
Problem A8
Problem A9
Problem A10
Under what circumstances might you want the smallest perimeter for a set area?
Problem A1
Length |
Width |
Area |
1 | 35 | 35 |
2 | 34 | 68 |
3 | 33 | 99 |
4 | 32 | 128 |
5 | 31 | 155 |
6 | 30 | 180 |
7 | 29 | 203 |
8 | 28 | 224 |
9 | 27 | 243 |
10 | 26 | 260 |
11 | 25 | 275 |
12 | 24 | 288 |
13 | 23 | 299 |
14 | 22 | 308 |
15 | 21 | 315 |
16 | 20 | 320 |
17 | 19 | 323 |
18 | 18 | 324 |
Length |
Width |
Area |
19 | 17 | 323 |
20 | 16 | 320 |
21 | 15 | 315 |
22 | 14 | 308 |
23 | 13 | 299 |
24 | 12 | 288 |
25 | 11 | 275 |
26 | 10 | 260 |
27 | 9 | 243 |
28 | 8 | 224 |
29 | 7 | 203 |
30 | 6 | 180 |
31 | 5 | 155 |
32 | 4 | 128 |
33 | 3 | 99 |
34 | 2 | 68 |
35 | 1 | 35 |
Problem A2
The shape with the largest area for this particular perimeter is the square 18 by 18 ft. In general, figures whose length and width are close to one another have larger areas than figures whose length and width are very different.
Problem A3
If the pen were triangular, and its perimeter equal to 72 ft., the length of each side would be 72 ÷ 3 = 24 ft. So the area is A = (1/2)bh. Using the Pythagorean theorem to calculate the height (h), we’d get the following:
.
So the area is This is significantly less than the areas of some of the rectangular pens in Problem A1.
Problem A4
If the pen were hexagonal, and its perimeter equal to 72 ft., the length of each side would be 72 ÷ 6 = 12 ft. The area will be six times the area of the equilateral triangle inside the hexagon (see picture).
So the area is as follows:
As you see, the area of the hexagon is larger than that of any of the rectangular shapes in Problem A1.
Problem A5
Since we know that the circumference of the pen is C = 72 ft., and that C = 2r, we can calculate the radius, r = C ÷ 2π = 72 ÷ 2π, or approximately 11.46 ft. Its area is π • (11.46)^{2}, or about 412.53 ft^{2}. This is quite a bit roomier than the largest rectangular pen that could be built.
Problem A6
In this situation, the fencing completes three sides of a rectangular pen, and the existing barn wall completes the fourth side. So the largest square you could enclose would have side length 24 ft. and area 576 ft^{2}. The largest rectangle would be 18 by 36, with an area of 848 ft^{2}, larger in area than the square. (Note that because the fencing is not enclosing all four sides, a square does not enclose the largest are in this situation — in fact, the largest area for a rectangular pen occurs when the fencing makes three sides of “half a square,” with length equal to twice the width).
A semicircle would have radius 72 ÷ π, or approximately 22.92 ft. (since C = 2πr and half the circumference is equal to πr). The area of the semicircle is (1/2) • π • (22.92)^{2}, or about 825 ft^{2}. This is larger in area than any square or rectangular pen.
Problem A7
No, this strategy does not work. As you saw in Problems A1-A6, figures with a constant perimeter can have varied areas depending on the shape of the figure. Because the shapes of Joel’s hand and the rectangle are inherently different, it is unlikely that the areas will be the same, even though he may be able to construct the rectangle that approximates the same area. The method he used for determining the perimeter would also have been affected by whether or not he had his fingers spread open.
Problem A8
Problem A9
Problem A10
Some examples include a room that can be painted as efficiently as possible, or a packing box that can be made out of the least amount of cardboard. You may also want to build the pen for your puppy with the least amount of fencing.