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Learning Math Home
Measurement Session 9: Solutions
 
Session 9 Part A Part B Part C Homework
 
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A B C 
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Solutions for Session 9, Part A

See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6 | A7 | A8 | A9 | A10


Problem A1

a. 

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.

b. 

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

<< back to Problem A1


 

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.

<< back to Problem A2


 

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.

<< back to Problem A3


 

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.

<< back to Problem A4


 

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 2 = 72 2, or approximately 11.46 ft. Its area is • (11.46)2, or about 412.53 ft2. This is quite a bit roomier than the largest rectangular pen that could be built.

<< back to Problem A5


 

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 , or approximately 22.92 ft. (since C = 2r and half the circumference is equal to r). The area of the semicircle is (1/2) • • (22.92)2, or about 825 ft2. This is larger in area than any square or rectangular pen.

<< back to Problem A6


 

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.

<< back to Problem A7


 

Problem A8

a. 

The smallest possible perimeter is 14 units and is found by using a 3-by-4 rectangle of tiles.

b. 

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.

<< back to Problem A8


 

Problem A9

a. 

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

b. 

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.

<< back to 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.

<< back to Problem A10


 

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