CI/Math 321 – Final Examination – Spring 2009
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1. (7 points) Prove that the sum of the measures of the interior angles of a convex n-gon is
(n-2)180˚.
ANS: Select one vertex. Sketch the diagonals to all nonadjacent vertices. This creates (n-2) triangles whose angles
taken together cover all the vertex angles of the polygon.
Since the angles of one triangle add up to 180°, we get a sum
of (n-2)(180) degrees for all the angles of the polygon.
2. (7 points) The figure shown is a regular hexagon. What is the measure of angle XYZ?
X
Y
Z
ANS: The sum of the vertex angles in this
hexagon is (6 – 2)(180°), or 4×180°, or 720°.
Since all the angles are congruent, each one is
720/6, or 120°. Since the (interior) vertex
angle at Y + <XYZ = 180°,
<XYZ = 120°
3. (7 points) In the triangle below, it is given that AB ≅ AC. Prove that ∠C ≅ ∠B. (Hint: try
inserting a median OR an angle bisector.)
A
IF we insert median AP, then we know:
AB ≅ AC (given)
BP ≅ PC (because median BISECTS BC)
AP ≅ AP.
So by SSS triangles ABP and ACP are
congruent.
B
Then we are allowed to say that all the other
P
parts are congruent and in particular, <B ≅ <C.
OR: IF we insert angle bisector AP, then we know:
AB ≅ AC (given)
<BAP ≅ <CAP (definition of angle BISECTOR)
AP ≅ AP
So by SAS, triangles ABP and ACP are congruent. Then we are allowed to say that
all the other parts are congruent and in particular, <B ≅ <C.
C
CI/Math 321 – Final Examination – Spring 2009
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4. (7 points) In this figure, triangles ABC and DEF are congruent with BC = EF. What is the
measure of angle EGC?
A
D
ANS: Since triangles ABC and DEF are congruent,
we can say <GEC = 40° and <GCE = 60°. The
sum of the angles in triangle GEC is 180°, so
<EGC must be 80°.
G
B
60º
40º
E
C
F
Explain how we know that segments AB and ED are parallel.
ANS: As noted above, <GEC is 40° and so is <ABC. Considering BC as a transversal for lines AB
and ED, <GEC and <ABC are “corresponding angles.”
When corresponding angles are congruent, the Euclidean Parallel Postulate allows us to say the
lines are parallel. Here, corresponding angles ABC and DEF are congruent because triangles ABC
and DEF are congruen. So we may ssay lines AB and DE are parallel.
5. (8 points) Do EITHER (a) or (b).
a) Imagine that you are teaching multiplication with fractions. To make this meaningful for
1
2
kids, what would be a good model for 1 × 2 =
4
3
Extending models we teach for whole numbers,
the area model for this multiplication would
show a rectangle with dimensions
1 ¼ by 2 2/3. The large black boxes indicate unit
squares. The thick gray subdivisions show the
¼’s (vertically) and 1/3’s (horizon-tally). The
blue region represents the product: 8 x 5 = 40
gray “boxes”. Each gray box is 1/12 so answer
is 40/12 or 31/3 units2.
CI/Math 321 – Final Examination – Spring 2009
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b) For each of the following figures,
I) Sketch a plane of reflectional symmetry if possible. If no plane of symmetry
exists, state NO REFLECTIONAL SYMMETRY.
II) Sketch a line of rotational symmetry, if possible. If no line of rotational
symmetry exists, state NO ROTATIONAL SYMMETRY.
Other correct
answers possible
Square pyramid reflectional symmetry
Other correct
answers possible
Square pyramid rotational symmetry
Center of front to
center of back face
Other correct
answers are
possible
Triangular prism reflectional symmetry
Triangular prism rotational symmetry
6. (7 points) On a quiz show a contestant stands at the entrance to a maze that opens into two
rooms labeled A and B. the master of ceremonies’ assistant is to place a new car in one room
and a donkey in the other. The contestant must walk through the maze into one of the rooms
and will win whatever is in that room. If the contestant makes each decision at random, in
which room should the assistant place the car to give the contestant the best chance to win?
1/2
If each decision is made at random, then at each
junction we can show the probabilities of taking each
choice. (Shown on the diagram)
The probability of ending up in room A is:
1/3
(1/3)(1/2)+(1/3)(1/2)(1/3)+(1/3)(1/2)(1/2)+(1/3)(1/2)
= 1/6 + 1/18 + 1/12 + 1/6
=(6+2+3+6)/36
=17/36.
1/3
The probability of ending up in room B is 19/36.
Therefore they should put the car in room B.
1/2
1/3
1/3
1/3
1/3
1/2
1/2
1/2
1/2
A
B
CI/Math 321 – Final Examination – Spring 2009
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7. (7 points) What is the distance between the centers of these circles?
(ANS.) The two triangles formed by
joining the centers of the circles are
similar because their angles are
24/5
congruent. We know this because
the two angles marked with open
dots are congruent by vertical
angles; two right angles are given;
12
the third angles follow because all
36/5
three angles must add up to 180
degrees. “AAA” guarantees
similarity.
Hint: insert the line joining the two centers. Explain why the two triangles are similar. Then
use the similarity to solve this problem.
ANS (cont’d) The ratios of lengths in one triangle to lengths in the other must be 2 to 3 (because
of the “2” and the “3” that are given). Therefore the length of “12” joining the two right angles
must be cut in a 2 to 3 ratio. That’s a total of 5 equal parts so the 12 is subdivided into segments
of length 24/5 and 36/5, marked on the diagram.
In the smaller triangle, the hypotenuse c satisfies:
c2 = 4 + (24/5)2 = (100+576)/25 = 676/25.
c = 26/5
The hypotenuse of the larger triangle must be 3/2 times the smaller hypotenuse, or 39/5.
The total distance between the centers is 26/5 + 39/5 = 65/5 = 13 units.
CI/Math 321 – Final Examination – Spring 2009
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8. (7 points) DO EITHER (a) OR (b)
a) Imagine that you are teaching division with fractions. To make this meaningful for kids,
3 2
what would be a good model for 2 ÷ =
4 3
One way to model this is to ask: How many steps of size 2/3 are there in 2 ¾?
Show 2 ¾:
Now show steps of size 2/3 from the start of the 2 ¾:
The alternating yellow and green colors show steps of size 2/3 – so far there are 4 such
steps. Now we need to know what fraction of a STEP is left to cover the red distance?
By subdividing the unit cells into 1/12 size pieces, we see that the piece left to cover the
red part with steps of size 2/3 is 1/8 OF A STEP.
ANS: Since there 4 ¼ steps of size 2/3 in a length of 2 ¾, we know 2 ¾ ÷ 2/3 = 4 1/8.
b) Use straight edge and compass to construct a line through P that is perpendicular to line
AB.
A
P
B
CI/Math 321 – Final Examination – Spring 2009
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9. (7 points) What is the radius of the circle shown below?
ANS: By the Pythagorean Theorem, (18+r)2 = r2 + 242.
182 + 36r + r2 = r2 + 242
182 + 36r = 242
Divide through by 6x6, noting that 182 is 6x6x3x3 and 242 is 6x6x4x4
9 + r = 16
r=7
CHECK: 242 + 72 = 576 + 49 = 625 = 252. 25 = 18 + 7 so the answer checks.
10. (7 points) The angles of an n-gon have an average measure of 165º.
a) What is n? (Explain.)
ANS: We know that the sum of all the angles is (n-2)x180 and there are n angles altogether, so
the average angle measure is (n-2)×180/n. Therefore:
(n-2)×180/n = 165
(n-2)×180 = 165n
180n – 360 = 165n
15n = 360
n = 24
b) Suppose the polygon has flexible joints at the vertices. As the polygon is flexed to take
on new shapes, what happens to the average measure of the interior angles? Explain your
reasoning.
ANS: The AVERAGE measure will not change because the total of all the angles remains
the same as the joints are flexed.
CI/Math 321 – Final Examination – Spring 2009
11. (6 points) Finding the distance from one point to
another in the coordinate plane is an application
of the Pythagorean theorem.
(a) What are the coordinates of the points shown?
A ___(-2, -1)___
B ___(6, 5)___
(b) What is the distance from point A to point B
(the length of line segment AB)?
AB2 = 82 + 62 = 64 + 36 = 100
AB = 10
12. (6 points) In the figure at the right, the original
shape has been shaded and labeled “O”. The
figures labeled “A” and “B” have been
translated from the original. Give (describe) the
rule (or “code”) for translation for A and B.
From original to figure A:
(x, y) Æ __(x+10, y+3)__________
To get from figure O to figure A, everything moves
10 right and 3 up.
From original to figure B:
(x, y) Æ ____(x+2, y-7)_________
To get from figure O to figure B, everything moves 2
right and 7 down.
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CI/Math 321 – Final Examination – Spring 2009
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13. (12 points)
y
(a) Give the coordinates the
vertices of figure ABC.
A ___(2, 4)____
C1
A
B ___(8, 2)____
C ____(3, -4)___
B
(b) Draw the reflection of figure
ABC using the x-axis as the
line of reflection. Label
corresponding points A1, B1,
C1 and give their coordinates.
x
B1
A1 C
A1 __(2, -4)______
B1 ___(8, -2)____
C1 ___(3, 4)____
y
(c) Now draw the reflection of
figure ABC using the line l as
the line of reflection. Label
the corresponding points of
your figure A2, B2, C2 and
give their coordinates.
l
B2
A
C2
A2
A2 ___(4, 2)____
x
B2 ___(2, 8)____
C2 ____(-4, 3)___
B
C
CI/Math 321 – Final Examination – Spring 2009
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(d) Draw the figure produced when
triangle ABC is rotated 90
degrees clockwise about the
origin. Label the corresponding
points of this figure A3, B3, C3
and give their coordinates.
y
A
A3 __(4, -2)_____
B3 ___(2, -8)____
B
C3 ___(-4, -3)___
x
A3
C3
C
B3
14. (8 points) Using the rectangular dot paper at the right, draw squares on the sides of triangle
ABC. Each side of the given triangle will be one side of each of these squares.
What is the area of the square
on side AB?
ANS: 3 by 3, so 9 square units
B
What is the area of the square on side BC?
ANS: 5 by 5, so 25 square units
C
What is the area of the square on side AC?
ANS: Has to be 9 + 25 = 34 square units
by the Pythagorean Theorem.
A
CI/Math 321 – Final Examination – Spring 2009
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15. (7 points) The colored region shown below is formed by circular arcs drawn from two
opposite corners of a 5-by-5 square. What is the area of the region?
ANS: One way is to figure out the area of each light
region and subtract that from the area of the square. ONE
light region is formed by cutting a quarter circle of radius
5 from a square of radius 5. So ONE light region has area
[25 – (.25)π(25)].
Therefore the dark shaded region has area
25 – [25 – (.25)π(25)] - [25 – (.25)π(25)] =
-25 + (.25)π(25) +(.25)π(25) =
-25 + .5 π(25) =
25(-1 + π/2) square units
This is approximately 14.2699 square units.
16. (6 points) Decide whether or not each of the following is an axis of rotational symmetry for
the given figure. If it is, give the order of the rotation. If it is not, say why not.
ANS: YES, this is an axis or rotational
symmetry of order 2.
ANS: NO, this is not an axis of
rotational symmetry. The figure will
not “fit on itself” until it has completed
a full rotation.
CI/Math 321 – Final Examination – Spring 2009
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17. (8 points) Consider the following stem-and-leaf plot for a set of 31 test scores.
9
8
7
6
5
4
1,2,7,8,8
0,2,3,5,5,7,9,9
0,1,1,4,6,8,9
6,7,7,9
0,7,9
2,3,5,6
Find the median, upper quartile, and lower quartile.
ANS. Since there are 31 scores, the middle one – the median – will be the 16th score in the list
ordered from smallest to largest (or largest to smallest). Reading carefully from the stem-andleaf plot, the median is 76.
The bottom half of the data, excluding the median, is the lowest 15 scores. The middle one is the
8th score, 66. This is the lower quartile.
The top half of the data, excluding the median, is the top 15 scores. The middle one is the 8th
score, 87. This is the upper quartile.
Make a box-and-whisker representation of this data.
66
76
87
42
0
10
20
30
40
98
50
60
70
80
90
18. (9 points) A jar contains 3 red, 4 blue and 5 green marbles. One marble is selected from the
jar and the color is noted.
(a) What are the possible outcomes of this experiment?
ANS. Since the experiment asks for “color”, the outcomes are Red, Blue and Green.
(b) Are all colors equally probable? Explain how you would help a student understand this.
ANS. No. Green is more likely than blue and blue is more likely than red. A way to help
students see this is to have them take turns to draw a marble from this jar and note the color.
After 3 or 4 turns each, tally the total number of red, blue and green outcomes. They should be
close to the “theoretical” possibilities of 3/12, 4/12 and 5/12.
(c) For this jar, what are the odds against selecting a red marble? Explain.
ANS. The odds against selecting a red marble is the ratio of the number of NONred to the
number of red; in this case 9 to 3.
100
CI/Math 321 – Final Examination – Spring 2009
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19. (8 points) A flask in the shape of a rectangular prism is partially filled with water and then
tilted as shown. The base of the prism is 6″ by 8 inches as shown.
a) Find the total surface area of the 3-dimensional shape formed by the water when the
flask is tilted as shown (do not simplify).
Bottom: 6×8 = 48. Left side: 5×6 = 30. Right side: 7×6 = 42. Front trapezoid (.5)(5+7)(8)
= 48. Back trapezoid = 48 also. Top is 6 × ?? We find ?? from the Pythagorean Theorem.
82 + 22 = ??2. So 68 = ??2 and ?? = 68 . So top is 6 × 68 , which is about 49.4773.
TOTAL surface area: 48 + 30 + 42 + 48 + 48 + 49.4773 square inches. (It’s OK to leave
answer unsimplified!)
b) Find the volume of the water in the flask as shown.
The water forms a trapezoidal prism with volume =
(Area of trapezoid) × (dimension perpendicular to
trapezoid)
= (.5)(5+7)(8) × (6)
= 48 × 6
= 288 cubic inches.
5″
7″
6″
8″
20. (7 points) Find the distances x and y in the cube.
1
First, x is the hypotenuse of a right triangle with legs of 1 and 1,
so x = 2
y
1
x
1
Then y is the hypotenuse of a right triangle with legs 1 and x so
1 + x2 = y2
1 + 2 = y2
3 = y2
y= 3
CI/Math 321 – Final Examination – Spring 2009
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21. (7 points) On isometric dot paper, the triangles formed by dots positioned like these are
equilateral triangles. Assume that the horizontal space between the dots is 1 unit.
Use the Pythagorean Theorem to find the altitude of triangle ABC below. Give an exact answer.
Do not give an estimate or round off any numbers.
Find the area of triangle ABC
Observe that the equilateral triangle “splits” into
two right triangles. The legs of one right triangle
are h and 2, where h is the altitude. The
hypotenuse is 4. So
h2 + 4 = 16
h2 = 12
h = 12
22. (7 points)
For each of the following, determine x if the
lines a and b are parallel.
z
ANS: Extend one or more of the lines and
use the Euclidean Parallel Postulate.
y
Angle z is 30° by "alternate interior angles".
So angle y must be 180 - 20 - 30 = 130
degrees.
OR put a line through the vertex of angle x
parallel to both lines and use alternate
interior angles.
So angle x must be 50°
In the second figure, angle w is 70° by
"corresponding angles", so angle t must be
110° and then angle x must be 180 - 30 - 110
= 40 degrees.
t w
CI/Math 321 – Final Examination – Spring 2009
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23. (8 points) A coin is tossed four times.
a) What is the probability of getting 2 heads and 2 tails in any order?
b) What is the probability of getting 1 head and 3 tails in any order?
H
H
T
H
H
T
T
H
H
H
T
T
a) There are 16 outcomes altogether.
6 of them fit the description “2
heads and 2 tails” so the Probability
of 2 heads and 2 tails is 6/16, or 3/8.
H
T
T
H
b) 4 of the sixteen outcomes fit the
description “1 head and 3 tails”, so
the probability of 1 head and 3 tails
is 4/16 or ¼.
H
T
H
H
T
T
T
H
H
T
T
H
T
T
CI/Math 321 – Final Examination – Spring 2009
24. (6 points) Find the area of the shaded region (a). OR
(a)
One way: The two white regions are
parallelograms, each with an area of 2x9 sq
cm. So the shaded part left after removing
the white regions is 9×13 – 2×2×9 = 9×9 =
81 sq cm.
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Find the distance x in figure (b):
(b)
Let b be the base. 72 + b2 = 122 so b2 =
144 – 49 = 95.
95 + (7+x)2 = 162 = 256
(7+x)2 = 256 – 95 = 161
7+x = 161 = 12.6886
x =- 5.6886
25. (6 points) A box contains 11 letters, shown as follows. What is the probability of the outcome
MISSISSIPPI in that order if 11 letters are drawn one by one
MIIIIPPSSSS
a) with replacement?
The probability of MISSISSIPPI in that order is:
1 4 4 4 4 4 4 4 2 2 4
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
11 11 11 11 11 11 11 11 11 11 11
b) without replacement?
Without replacement, the probability of MISSISSIPPI in that order is:
1 4 4 3 3 2 1 2 2 1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅1
11 10 9 8 7 6 5 4 3 2
(The probability of getting “M” for the first letter is 1/11. Once the M is gone, there are
10 left and 4 of them are I’s, so the probability of getting I next is 4/10. Once M and one
I are gone, there are 9 left, of which 4 are S’s, so the probability of getting S next is 4/9,
and so on.)
CI/Math 321 – Final Examination – Spring 2009
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26. (6 points) Two angles are supplementary and the ratio of their measure is 8:10. What are the
angle measures?
ANS: the two angles add up to 180 and the 8:10 ratio means there are 18 parts altogether, so
each part is 10 degrees. So the two angles are 80 and 100 degrees.
27. Consider the following test scores for a small class:
47, 64, 68, 73, 75, 79, 81, 82, 86, 89.
a) Find the median (3 points).
Since there are 10 scores the median is the average of the middle two (the 75 & the 79): 77.
b) Find the mean (3 points).
The mean is the sum of all, divided by 10. 744/10 = 74.4
c) Find the variance (4 points).
score 47
64
68
73
75
79
81
82
86
89
score‐
mean squared stdev ‐27.4 750.76 SQRT(Variance)
‐10.4 108.16 11.71495
‐6.4 40.96 ‐1.4 1.96 0.6 0.36 4.6 21.16 6.6 43.56 7.6 57.76 11.6 134.56 14.6 213.16 SUM 1372.4 VARIANCE 137.24 d) Find the standard deviation (2 points).
The standard deviation is the square root of the variance – see table above.
CI/Math 321 – Final Examination – Spring 2009
28. BONUS POINTS. (0.5 points each)
1
9÷
SHOW WORK! DO NOT USE CALCULATOR.
1
=______9 × 2 = 18_______
2
2
3.8 – 5.2 = ___- 1.4_______
3
_25____ % of 80 = 20
4
80 % of __200________ = 160
5
8 ÷ __1/3____ = 24
6
3
6 - 1 =_4 4/7_______
7
7
9.5 ÷ .5 = ___19_______
8
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If you paid $20 for a garment on sale for 33
1
% off the original price, what was the
3
original price? ___$30_______
9
The area of a square is 169 sq. in. What is the length of one side of the square?
_____13_____
10 11.36 ÷ 0.8 = _____14.2______________