Solution manual for college geometry a problem solving approach with applications 2nd edition by musser

One solution is shown.. b One possible solution is shown.. To expose the minimum number of faces, stack the cubes as shown: To expose the maximum number of faces is to stack the 18 block

Chapter 1 •••• 1

Chapter 1

GEOMETRY INVESTIGATION

7 Letters

Section 1.1

2 14 squares

4 12 possible pentominos

6 One solution is shown Many solutions are possible

8 (a) 4 (b) 3 (c) 2

10 (a) Yes it is possible (b) One possible

solution is shown

12 To expose the minimum number of faces, stack the cubes as shown:

To expose the maximum number of faces is to stack the 18 blocks vertically or horizontally

14 One solution is shown Many solutions are possible

16 82 people, 202 people

2 • Chapter 1 •

20 No, there are four points where three lines

meet For at least three of these points, your

tracing will take you to the point, away from

the point, and then back to the point There is

no way to leave any of those points without

retracing a line

22 Lines are labeled in the order they can be

traced Many tracings are possible One

solution is shown

24 167 m by 668 m

26 200 inches

28 Use 4 stacks One stack can contain the 9-inch

and a 1-inch box Another stack can contain

the 7-inch and the 3-inch box The third stack

can contain the 5-inch, 4-inch, and a 1-inch

box The remaining stack can contain the rest

of the boxes Other arrangements are possible

30

32 Make the following cut, as shown on the left,

and arrange the pieces as shown on the right

34

36 Four possible solutions are shown

40 31 rectangles

42 2520 ft

44 (a) One answer is shown Other answers are

possible

(b) One answer is shown Other answers are possible

Chapter 1 •••• 3

Selected Extended Problems

46

Section 1.2

2 (a) 3, 5, 7, 9, 11, 13, 15, 17

(b) 20 triangles

(c) 2n + 1 toothpicks

4 6, 10, 14, 4n + 2

6 4, 9, 14, 5n − 1

2

n n−

10 n(3n − 2)

12 (a) iii (b) ii (c) i (d) iii

16 (a) 19, 22, 37, 3n + 1

(b) 35, 48, 143, n2 − 1

(c) 32, 64, 2048, 2n − 1

(d) 6

7

−

, 7

8

−

, 12 13

− , 1

n n

−

+

18 15 paths

20 14 paths

22 (a) 19, 31, 50

(b) 6, 13, 33

(c) 8, 11, 30

(d) 13, 23, 36, 59

24 35 triangles

26 19 squares, 2n − 1

28 52 squares, 5n + 2

30 44 cubes, 4n + 4

32 1330 cubes

34 (a) 1, 6, 12, 8, 0 (b) 8, 24, 24, 8, 0; 27, 54, 36, 8, 0

(c) (n − 2)3, 6(n − 2)2, 12(n − 2), 8, 0

36 No, it cannot be done Notice the number of black squares and the number of white squares

38 (144)(233) = 33,552

40 There are 6 distinct arrangements The greatest area available for the garden, 42 ft2, occurs with the following arrangement

Selected Extended Problems

42 The sum of the first n odd-numbered terms of the Fibonacci sequence is equal to the nth even numbered term The sum of the first n

even-numbered terms of the Fibonacci sequence is

equal to one less than the n+1st odd-numbered term The sum of any 10 Fibonacci numbers is equal to the product of 11 and the 7th number

in the sum Ratios of consecutive pairs of Fibonacci numbers, that is, ratios of the

(n+1)st divided by the nth Fibonacci number

get closer to 1.618, which is an approximation

to the golden ratio

43

Number

of Bricks

Wall Patterns

1

2

3

4

4 • Chapter 1 •

5

The sequence in the number of possible wall

patterns generated is 1, 2, 3, 5, 8 After the

first two terms, each new term is created by

adding the previous two terms A similar

sequence is not generated if the wall is 3 units

tall and 1-unit by 3-unit bricks are used The

sequence generated is 1, 1, 2, 3, 4, and 4 is not

the sum of the previous two terms