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

Chapter 1 • 1

Section 1.1

Chapter 1
GEOMETRY INVESTIGATION
1. Cat
2. Runner

3.

Boat

2.

14 squares

4.

12 possible pentominos

6.

One solution is shown. Many solutions are
possible.

8.

(a) 4

4. Parallelogram

(b) 3

(c) 2

10. (a) Yes it is possible. (b) One possible
solution is shown.
(
5.

Swan

6. Woman

12. To expose the minimum number of faces,
stack the cubes as shown:
7.

Letters

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
18. (a)

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(b)

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.

34.

36. Four possible solutions are shown.

38. Move 1:

Move 2:

24. 167 m by 668 m
26. 200 inches

Move 3:

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.

40. 31 rectangles

30.

42. 2520 ft

Finished:

44. (a) One answer is shown. Other answers are
possible.

32. Make the following cut, as shown on the left,
and arrange the pieces as shown on the right.

(b) One answer is shown. Other answers are
possible.

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Chapter 1 • 3
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

Selected Extended Problems
46.

36. No, it cannot be done. Notice the number of
black squares and the number of white squares.
38. (144)(233) = 33,552

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

8.

n(3n −1)
2

10. n(3n − 2)
12. (a) iii

(b) ii

14. (a)

(b)

(d)

(e)

(c) i

(d) iii
(c)

16. (a) 19, 22, 37, 3n + 1
(b) 35, 48, 143, n2 − 1
(c) 32, 64, 2048, 2n − 1
−6 −7 −12 −n
,
,
,
(d)
7
8
13 n + 1
18. 15 paths

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 evennumbered 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

20. 14 paths
22. (a)
(b)
(c)
(d)

19, 31, 50
6, 13, 33
8, 11, 30

13, 23, 36, 59

2

3

24. 35 triangles
26. 19 squares, 2n − 1
4
28. 52 squares, 5n + 2
30. 44 cubes, 4n + 4
32. 1330 cubes

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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.

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall