# Measurement Relationships Part A: Area and Perimeter (45 minutes)

## Session 9, Part A

### Constant Perimeter

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.

### Problem A1

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.

1.  What are the dimensions of the possible rectangular pens?
2. What are the areas of these pens?

### Problem A2

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?

### Problem A3

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.

### Problem A4

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.

### Problem A5

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.

### Problem A6

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.

### Constant Area

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

1. What is the smallest perimeter possible using 12 square tiles?
2. What is the largest possible perimeter?

Problem A9

1. Make figures with an area of 12 square units with perimeters of 14 through 26 units. Keep in mind that it is not possible to create all of these perimeters. Sketch the shapes and record their areas and perimeters.
2. Choose one perimeter between 14 and 26 that you could not make and explain why it is impossible.

Problem A10

Under what circumstances might you want the smallest perimeter for a set area?

### Solutions

Problem A1

1. There are many possibilities, but the length and width must add up to 36 ft., since the perimeter (made up of two lengths and two widths) is 72 ft.
2. Possible areas go from 35 ft2 (35 by 1) to 324 ft2 (18 by 18).
 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 = 2 r, 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 ft2. 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 ft2. The largest rectangle would be 18 by 36, with an area of 848 ft2, 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 ft2. 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

1. The smallest possible perimeter is 14 units and is found by using a 3-by-4 rectangle of tiles.
2. The largest possible perimeter is 26 units. Many different shapes have this perimeter, but the simplest example is a 1-by-12 rectangle. A large L-shaped piece and a snaking piece where the squares are laid end to end will also have this maximum perimeter.

Problem A9

1. The possible perimeters are all even numbers between 14 and 26. Some possibilities:
14 units: a 3-by-4 rectangle
16 units: a 2-by-6 rectangle
18 units: a 2-by-5 rectangle with two extra pieces added, which don’t touch one another: 20 units: a 2-by-4 rectangle with four extra pieces added, none of which touch one another: 22 units: a 2-by-3 rectangle with a row of six extra pieces added: 24 units: a 2-by-2 square with a row of eight extra pieces added: 26 units: a row of 12 pieces
2. When pieces border, two segments of perimeter are removed at once. This means that each time the pieces touch, the potential total perimeter remains an even number at all times. No odd number may be the perimeter.

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.