Honors Geometry, Spring Exam Review
page 1
Honors Geometry
Review Exercises for the May Exam
C
1.
Given: CA  CB < 1  < 2
<3  <4
34
CAM  CBM
Prove:
Proof:
1
A
2
B
M
1.
2.
3.
4.
<1  <2
< 1 is supp to < CAM
< 2 is supp to < CBM
< CAM  < CBM
1.
2.
3.
4.
given
Linear pairs are supp.
Linear pairs are supp.
Supps of congruent angles are
5.
6.
<3  <4
CAM  CBM
5.
Given
6.
ASA
congruent
D
2.
Given: AB || CD
C
AB  CD
ABD  CDB
Prove:
Proof:
A
1.
AB  CD
2.
AB || CD
3.
4.
5.
< ABD
 <CDB
DB  BD
ABD  CDB
16
B
1.
given
2.
3.
4.
5.
Given
P AIC
Reflexive for 
SAS
Honors Geometry, Spring Exam Review
3.
page 2
Given: BC  DF AD = CE
< DCB  < CDF
B
F
ACB  EDF
Prove:
Proof:
A
1.
2.
3.
4.
5.
6.
7.
8
9.
BC  DF
< DCB  < CDF
AD = CE
AD + CD = CE + CD
AD + CD = AC
CE + CD = DE
AC = DE
AC  DE
ACB  EDF
1.
2.
3.
4.
5.
6.
7.
8.
9.
D
C
Given
Given
Given
addition property of =
Segment addition postulate
Segment addition postulate
Substitution
Defn of congruent
SAS
D
4.
Given: AB || CD
Prove:
E
C
AB  CD
AD || CB
Proof:
A
1.
AB  CD
2.
AB || CD
3.
4.
5.
6.
< ABD
7.
 <CDB
DB  BD
ABD  CDB
<ADB < CBD
AD || CB
B
1.
given
2.
3.
4.
5.
6
Given
P AIC
Reflexive for 
SAS
CPCPT
7.
AIC  P
A newer version of the proof (which we can do at the end of the semester is …)
1.
AB  CD
2.
AB || CD
3.
ABCD is a parallelogram
1.
given
2.
3.
Given
If one set of sides is both parallel
and congruent, then the quad is a
parallelogram.
Honors Geometry, Spring Exam Review
4.
5.
AD || CB
Given: BA  BC
Prove:
page 3
4.
Defn parallelogram
DB bisects < ABC
< AED
 < CED
Proof:
1.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Given
DB bisects < ABC
2.
Given
< ABE  < CBE
BE  BE
ABE  CBE
< BEA  <BEC
< BEA is supp to < AED
< BEC is supp to < CED
< AED  < CED
3.
4.
5.
6.
7.
8.
9.
Defn bisect
Reflexive for 
SAS
CPCTC
Linear paris are supp.
Linear paris are supp.
Congruent supps. theorem
B
6.
Given:
BA  BC
AB  CB , BD = BE
Prove: AF  CF
Honors Geometry, Spring Exam Review
page 4
Proof:
D
E
F
A
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
AB  CB
1.

<B <B
2.
BD = BE
3.
4.
BD  BE
ABE  CBD
5.

<A <C
6.
AB = CB
7.
AD + BD = AB
8.
CE + BE = BC
9.
AD + BD = CE + BE 10.
AD = CE
11.
AD  CE
12.
< AFD < CFE
13.
AFD  CFE
14.
AF  CF
15.
C
Given
Reflexive for congruence
Given
Defn congruent
SAS
CPCTC
Defn congruent
Segment addition postulate
Segment addition postulate
Substitution
Subtraction property of =
Defn congruent
Vertical angles are congruent
AAS
CPCTC
Honors Geometry, Spring Exam Review
page 5
Solve for the missing parts of each triangle. Give all possible solutions.
15
7.
58
63
8.
40
22
68
Third < =
900
x
x
Solve for x.
Solve or x
x2= 402 + 582 – 2(40)(58)cos150
sin 680 =
= 482.104166
x=
x = 21.9569  22
63
x
63
sin 68
= 67.948  68
A
9.
220
Solve for all angle measures.
180
a2 = b2 + c2- 2bc*cosA
C
253
B
2532 = 2202 + 1802 – 2(220)(180)cosA
2532 -2202 - 1802
-2(220)(180)
=0.2120
m< A = 77.75997  780
cosA =
220
253

 sinB = 0.849798  m<B  580. So m<C  440.
sin B sin(77.7599)
10.
Solve for all missing lengths and angles.
A
Honors Geometry, Spring Exam Review
52
page 6
19.8
B
C
sin 52 =
BC
19.8
m<B = 380
cos 52 =
BC  15.6
11.
AC
19.8
AC  12.2
Solve for all missing parts.
A
26
24
48
B
C
24
26
26sin 48
 sinB =
 sin B = 0.8051  m<B =53.60 or m< B =

sin 48 sin B
24
126.40
So m< A = 78.40 or m<A = 5.60
24
a
24 sin 78.4
 a=
 a = 31.635  a  32

sin 48 sin 78.4
sin 48
24
a
24sin 5.6
 a=
 a = 3.15

sin 48 sin 5.6
sin 48
12.
Robin sits at the top of a 35 foot tree looking down at an earthworm crawling along the
ground towards the base of the tree. He measures the angle of depression to the worm
and finds it to be 54. How far is the earthworm from the base of the tree?
tan 54 =
35
35
 e=
e
tan 54
Honors Geometry
y, Spring Ex
xam Review
w
page 7
e  25.4 ft.
13.
A ship is saailing east at 6 knots ( 6 nautical
n
milees per hour) when the caaptain sightss a
small island
d at an anglee of 15 to th
he north of thhe ship’s couurse. After 110 minutes, tthe
angle is 28. How far away
a
is the island at the time of the ssecond sightting?
6 knots *
1
hours = 1 nautical mii = distance
6
1
x
sin15
 x=
 x = 1. 15 nau
ut. miles

sin13 sin
sin13
n15
15
14.
15.
m < 1 = 20
m < 2 = 60
m < 3 = 100
0
m < 4 = 20
16.
28
m < 1 = 36
m < 2 = 144
Describe ho
ow you woulld construct the centroidd of a triangle.
The centro
oid of a trian
ngle is a thee point wherre the mediaans concur.
17.
Describe wh
here the orth
hocenter of an
a obtuse triaangle is.
Honors Geometry, Spring Exam Review
page 8
The orthocenter is on the outside of a triangle (where the altitudes concur).
18.
a)
What is the point of concurrence of the medians called?
centroid
b)
If the three medians are constructed in a triangle, and one of the medians has
length 18, into what lengths is it divided by the point named in part (a)?
12 and 6 (12 is from the vertex to the centroid, and the 6 is from the centroid
to the midpoint of the opposite side.)
19.
Calculate ALL of the angles in the figure on the next page. The outside polygon is a
regular pentagon. Mark the angles on the drawing
Is ACDE a trapezoid? Give evidence for your conclusion.
Since m < ACD = 72 and m < EDC = 108, then the
same side interior angles are supplementary. Thus
AC / / DE but AE is not parallel to DC . So ACDE is
a trapezoid.
D
C
E
F
A
20.
Name the largest and smallest angle of the triangle.
B
21
29
Largest angle = < C
Smallest angle = < A
A
23
C
B
Honors Geometry, Spring Exam Review
21.
page 9
Name the longest side in the triangle
B
m< A = 350
70
Longest side = AB
75
A
22.
C
Complete the chart by checking the figures for which each property is true.
Honors Geometry, Spring Exam Review
23.
Given:
page 10
ABCD
DA  DE
Prove: EDBC is a
Proof:
24.
Given:
1.
2.
Parallelogram ABCD
AD  BC
1.
2.
Given
Opp sides of
parallelograms
are congruent
defn parallelogram
3.
AD || BC
3.
4.
DA  DE
5.
6.
DE  BC
EDBC is a parallelogram
4.
5.
6.
Given
Transitive for congruence
If one set of opp sides of a
quad are congruent and
parallel, then the quad is
a parallelogram
1.
2.
Given
given
3.
Through 2 points, there is
exactly one line
Def midsegments
MNOP
A, B, C, and are the midpoints
of MN , NO, OP, and PM
respectively
Prove: ABCD is a parallelogram
Proof:
1.
2.
3.
4.
5.
Parallelogram MNOP
A, B, C, and are the midpoints
of MN , NO, OP, and PM
respectively
Draw MO
DC and AB are midsegments
of  MOP and  MON
DC || MO and AB || MO
4.
5.
Midsegments are parallel
to the third side of the
triangle
Honors Geometry, Spring Exam Review
6.
7.
8.
9.
10.
25.
DC 
1
1
MO and AB  MO
2
2
DC || AB
DC = AB
DC  AB
ABCD is a parallelogram
page 11
6.
Midsegments measure
7.
8.
9.
10.
one-half of the third side
of the triangle
Transitive for ||
Transitive for =
Def congruent
If one set of opp sides of a
quad are congruent and
parallel, then the quad is
a parallelogram
A(-4 , 0) B(0, 3) C(12,0) D(12, -4)
a) Show that ABCD is a trapezoid.
mBC = mDA= 
1
so BC || DA
4
b) Analyze the diagonals of ABCD.
The diagonals are not perpendicular (check slopes of AC and BD ), nor are they
congruent, using the distance formula.
26.
Rhombus PQRS
PQ = 13 SQ = 24
PR = 10
27.
TUVW
m < 1 = 160
m < 2 = 300
m < 3 = 640
Honors Geometry, Spring Exam Review
28.
page 12
Kite KITE
IE = 6
IT = 5 KS = 3
3
3
KT = 4 + 3
m< SKI = 300
3
m < ITS = tan-1 4 = 36.80
29.
LM = 10
MF = 102  62  100  36  2 34
Area (LMQF) =
30.
1
6(10  18)  84
2
What does “divided proportionally” mean?
Divided proportionally means that two segments are each dvidied into the same
ratio. We often use this with a triangle where one segment is drawn parallel to a
side of the triangle.
31.
10 12
3
20

( )    x 
x
8
2
3
Solve for x and y.
10
x
12
12 20
5
48

( )    y 
y 16
4
5
y
8
16
Honors Geometry, Spring Exam Review
32.
page 13
Solve for each variable
6
x
20
6 y
 y2 = 120  y = 2 30
y 20
y
6 x
x2 = 6*26 = 4*39  y = 2 39

x 26
z
20 z
z2 = 20*26 =4*130 

z 26
z = 2 130
33.
3x 2 x


18 4 x
34.
x 1
x=3

6 2
Honors Geometry, Spring Exam Review
RI IV
RI 15  3 



    RI = 27
36 20  4 
EB VE
35.
Find the geometric mean between 5 and 19.
5 x
 x2 = 95 x =

x 19
36.
95
Two similar polygons are shown. Find the values of x, y, and z for each.
A.
B.
page 14
Honors Geometry, Spring Exam Review
page 15
Find the values of x and y
37.
38.
39.
40.
12
18  3 

    3y + 36=48  y = 4
y  12 24  4 
41.
x
7
49
 18x = (7)(14)  x =

14 18
9
Bobbie Mo Smith, 5’ 8” tall, wants to estimate the height of her favorite tulip tree on
Melrose Ave. She asks Ms. Wilson for a mirror from the science department and heads
down the Gordon field. She faces the tree and places the mirror between herself and the
tree, on the ground, in such a way that she can see the top of the tree in the mirror. She is
4 feet from the mirror and the mirror is 12 feet from the tree. Her eyes are 4” from the top
of her head. About how tall is the tree?
16
t
3

4 1
 
12  3 
t
5’4”
t = 16’
4’
12’
Honors Geometry, Spring Exam Review
page 16
Find the value of the variable in each example below.
42.
43.
44.
9
10
19

  x 
10 x  9
9
4 y

   x  2 17
y 17
z
12

   z 2  7 z  144    z  16, z  9
12 z  7
Find the lengths of all of the sides of the right triangles.
45.
46.
3-4-5  12-16-20
47.
48.
49.
50.
Honors Geometry, Spring Exam Review
51.
page 17
52.
x2 + (3x)2 = 202
x2 + 9x2= 400
10x2 = 400
X2 = 40
X = 40  2 10
Find the measure of the numbered angles, or the measure of x (in #56)
53.
m<1 =
55.
54.
1
(68 + 21) = 44.50
2
m<2 =
56.
m<1 = 660
1
(188-56) = 660
2
Honors Geometry, Spring Exam Review
page 18
1
(310-x-x)
2
82 = 310 – 2x
2x = 228
x = 114
41 =
Fill in the blanks with a correct answer.
57.
When you solve a triangle numerically that is given by SSA, then it is possible to
find ____0, 1, or 2____ (give a number or numbers) different solutions.
58.
When an angle of a triangle is bisected, then there is a proportion of the two
parts of the cut side to the ratio of ___the adjacent sides____. This could also be
correctly stated as the ratio of one part of the cut side is to its ____adjacent side____
as the other part of the cut side is to its _______ adjacent side __________.
59.
The Triangle Inequality says that _____the sum of the lengths of any two sides of a
triangle is greater than the length of the third side______.
60.
The geometric mean of two numbers is the ______ square root ____ of their product.
61.
When the perpendicular bisectors of the three sides of a triangle concur on one side of the
triangle, then the triangle is a __right_ triangle.
62.
In a 30-60-90 triangle, then you calculate the length of the short leg by
__dividing___ the longest leg by a factor of ___
3 ____.
The Power Theorems for Circles (which is an application of similar triangles)
63.
Given: Plane figure with a circle
Fill in the result: (EC) ( DC ) = ( AC ) ( BC )
Honors Geometry, Spring Exam Review
page 19
Defn: The power of C with respect to the circle is the product (EC)(DC), which is the same
value as (AC)(BC)
64.
Given: Chords SR and QP intersect in a circle at L
Fill in the result: (PL) ( LQ ) = (SL ) ( LR )
Defn: The power of L with respect to the circle is the product (PL)(LQ), which is the same
value as (SL)(LR)
Solve for x.
65.
66.
10x = (12)(4)
x = 4.8 or
24
5
(x)(15) = (22)(12)
88
x=
5
Honors Geometry, Spring Exam Review
page 20
67.
x2 = (21)(12)
x=
68.
Solve for x.
What is the power of C?
(13)(7) = (x+5)(5)
91 = 5x + 25
66 = 5x
66
x=
5
Power of C = (13)(7) which is 91
69.
Solve for RW.
What is the power of W?
(RW)(4)=(5)(8)
RW = 10
Power of W = (5)(8) = 40
(7)(3)(4)(3)  6 7
Honors Geometry, Spring Exam Review
page 21
Calculate the area of each figure. Label your answer (with “A =”) and write the formula you
used in each case.
20
70.
71.
2
13
13
8
30
10
10
1
1
h(b1  b2 ) = 12(10  20)
2
2
= 180
1
1
h(b1  b2 ) = 4(2  10)
2
2
= 24
A=
72.
Rhombus with a diagonal
of length 10 and a side of
length 13.
A=
73.
A=
74.
1
1
d1d 2 = (10(24)
2
2
= 120
An equilateral triangle whose sides
measure 12 meters.
1
1
h(b1  b2 ) = 9(6  15)
2
2
189
=
2
A=
A=
75.
s 2 3 12 2 3
=
= 36 3 sq. meters
4
4
Find the area of the kite KITE.
A=
1
1
d1d 2 = (12)(23) =138
2
2
Honors Geometry, Spring Semester Exam Review
76.
Find the area of the triangle.
1
1
A= bc sin A = (17)(26.3)sin 58
2
2
= 189.581
77.
Calculate h.
A = bh = (10)(16) = 24h
h=
20
3
page 22