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Written by: Larry E. Collins
Geometry:A Complete Course
(with Trigonometry)
Module E – Instructor's Guidewith Detailed Solutions for
Progress Tests
ERRATA
4/2010
© 2009 VideoTextInteractive Geometry: A Complete Course2
NameUnit V, Part A, Lessons 1, Quiz Form A—Continued—
2. Match each statement in column I with a phrase in column II.
Column I Column II
Rectangle _________
Diagonal of a polygon _________
Polygon _________
Convex Polygon _________
Square _________
Parallelogram _________
Trapezoid _________
Vertex of a polygon _________
Quadrilateral _________
Rhombus _________
3. A list of properties found in the group of seven special quadrilaterals is given below. Write the name of the special quadrilateral(s) beside the given property for which that property is always present.
a. Both pairs of opposite sides are parallel. _________________________________________
b. Exactly one pair of opposite sides are parallel. _________________________________________
c. Both pairs of opposite sides are congruent. _________________________________________
d. Exactly one pair of oppsite sides are congruent. _________________________________________
e. All sides are congruent. _________________________________________
f. All angles are congruent. _________________________________________
a) An equilateral parallelogram
b) A parallelogram that has one right angle
c) A closed “path” of four segments that doesnot cross itself
d) A quadrilateral that has exactly one pair ofparallel sides
e) An end point of a side of a polygon
f) A polygon in which any diagonal liesinside the polygon
g) A quadrilateral with opposite sides parallel
h) A simple closed curve made up entirelyof line segments
i) A segment whose endpoints are two non-consecutive vertices of a polygon
j) An equilangular equilateral quadrilateral
b
i
h
f
j
g
d
e
c
a
Parallelogram, Rectangle, Rhombus, Square
Trapezoid, Isosceles Trapezoid
Parallelogram, Rectangle, Rhombus, Square
Isosceles Trapezoid
Square, Rhombus
Rectangle, Square
© 2009 VideoTextInteractive Geometry: A Complete Course 3
NameUnit V, Part A, Lessons 1, Quiz Form A—Continued—
4. Indicate whether each of the following is true or false.
a) Every square is a rhombus. _________
b) Every rhombus is a square. _________
c) Every square is a kite. _________
d) Every rhombus is a kite. _________
e) If a quadrilateral has three sides of equallength, then it is a kite. _________
f) Every property of every square is a propertyof every rectangle. _________
g) Every property of every trapezoid is a propertyof every parallelogram. _________
h) Every property of a parallelogram is a property of every rhombus. _________
True
False
False
False
False
False
False
True
Name
Class Date Score
Quiz Form A
Unit V - Other PolygonsPart A - Properties of PolygonsLesson 6 -Midsegments
Use the figure to the right for problems 1 and 2.
1. Point H is the midpoint of GJ. GH = ____________
Point L is the midpoint of GK. HJ = _____________
If GJ = 12, and HL = 9, Find GH, HJ, and KJ. KJ = _____________
Problems 1 and 2
2. Point H is the midpoint of GJ.
Point L is the midpoint of GK. m/K = ____________
If m/GLH is 21 degrees and KJ = 14 ,
find m/K and HL. HL = ____________
3. Using the figure to the right, find DE, BC, m/A, m/B, and m/C. DE = ____________
BC = ____________m/A = ____________m/B = ____________m/C = ____________
G
JK
L H
© 2009 VideoTextInteractive Geometry: A Complete Course 33
A
CB
F
D E
G
8
G10
54
54
40O
GH =1
2GJ
GH =1
212
GH = 6
⋅
⋅
HJ = GH
HJ = 6
KJ = 2 HL
KJ = 2 9
KJ = 18
⋅⋅
m K = m GLH
m K = 21
∠ ∠∠
HL =1
2KJ
HL =1
214
1
2
HL =1
2
29
2
HL =29
4= 7
1
4
⋅
⋅
⋅
DE =1
2FG
DE =1
26
DE = 3
⋅
⋅
BC = 2 FG
BC = 2 6
BC = 12
⋅⋅
m A = 180 - 90 - 40
m A = 50
m B = 90
m C = 40
∠∠
∠
∠
12
6
6
18
21O
71
4
3
12
50O
90O
40O
© 2009 VideoTextInteractive Geometry: A Complete Course34
NameUnit V, Part A, Lessons 6, Quiz Form A—Continued—
4. In the figure to the right, W, T, and S are midpoints
of the sides of triangle DEF. If WT = 5, ST = 8, and
SW = 7, What is the perimeter of nDEF?
Permimeter of nDEF = ____________
5. Which of the following named quadrilaterals are parallelograms?
a) b) c)
__________________ __________________ ____________________________________ __________________ ____________________________________ __________________ __________________
6. In the figure to the right, ABCD is a trapezoid with median MN as shown.a) If BC = 10t and MN = 15t, find AD. AD = ____________
b) If AD = 35x and MN = 28x, find BC. BC = ____________
c) If AD = and BC = , find MN. MN = ____________
AF
ED
W T
S
A5 3
3D
4
4 C 4
3
B
3
Z 224W
4
3X3
5
Y
5
G44
3
H
3
2I2
2J
2
ABCD is not a parallelogram
Since point A and D are not
midpoints.
XYZW is a parallelogram GHIJ is a parallelogram
DE = 2 WT
DE = 2 5
DE = 10
⋅⋅
EF = 2 SW
EF = 2 7
EF = 14
⋅⋅
DF = 2 ST
DF = 2 8
DF = 16
⋅⋅
B C
NM
DAMN =
BC + AD
2
15t =10t + AD
230t = 10t + AD
20t = AD
MN =BC + AD
2
28x =BC +35x
256x = BC +35x
21x = BC
MN =BC + AD
2
MN =9 3 +5 6
2
9 3 5 6
40
20t
21x
9 3 +5 6
2
© 2009 VideoTextInteractive Geometry: A Complete Course36
NameUnit V, Part A, Lessons 6, Quiz Form B—Continued—
4. In the figure to the right, point D is the midpoint of AC, and point E is the midpoint of BC.AD = x + 5, DC = 2y + 6, DE = 2x – 5, and AB = y + 8. Find DE and AB.
AB = ____________
DE = ____________
5. Which of the following named quadrilaterals are parallelograms?
a) b) c)
__________________ __________________ __________________
6. In the figure to the right, ABCD is a trapezoid with median MN as shown.
a) If BC = 2x + 5 and MN = 10x – 1.2, find AD. b) If BC = and AD = , find MN.AD = ____________ MN = ____________
c) If BC = 6.7 and AD = 14.4, find MN. MN = ____________
C
BA
D E
W7 7
4
X
4
3Z3
5 Y 4
3 3
3
3
3 3
3
3
M
NQ
P
3C
3
5
D
6
2E24
F
4
WXYZ is not a parallelogram MNPQ is a parallelogram CDEF is not a parallelogram
B
M
C
A
N
D
MN =BC + AD
2
10x -1.2 =2x +5 + AD
220x - 2.4 = 2x +5 + AD
18x --7.4 = AD
MN =BC + AD
2
MN =3 2 +7 2
2
MN =10 2
2
MN =2 5 2
2
MN = 5 2
⋅
MN =BC + AD
2
MN =6.7 +14.4
2
MN =21.1
2
MN =2 10.55
2MN =
⋅
110.55
3 2 7 2
40
40
18x – 7.4 5 2
AD = DC
x +5 = 2y +6
x - 2y = 1
-8x +2y = -36
-7x = -35
x = 5
x +5 = 2y +6
5 +5 = 2y +6
10 = 2y +6
4 = 2y
2 = y
DE = 2x - 5
DE = 2 5 - 5
DE = 10 - 5
DE = 5
AB = y +8
AB = 2 +8
AB = 1
( )
00
AB = 2 DE
y +8 = 2 2x - 5
y +8 = 4x - 10
-4x + y = -18
2(-4x +
( )( )
yy = -18)
-8x +2y = -36
10.55
© 2009 VideoTextInteractive Geometry: A Complete Course 43
Name
Class Date Score
Unit V - Other PolygonsPart B - Areas of PolygonsLesson 1 -Postulate 14 - AreaLesson 2 - Triangles
For problems 1 – 6, find the area of the given polygon using the appropriate Postulate, Theorem, or Corollary from lesson 1 and 2.
1. A = _______ 2. A = _______
3. A = _______ 4. A = _______
5. A = _______ 6. A = _______
36 3 units2
35
2units2
12
60O
30O
5
7
8
8
68
10 45
9
O
x 3
2where x = 12
12 3
2=
2 6 3
2= 6 3
⋅ ⋅
A =1
2
A =1
25 7
A =35
2units2
⋅ ⋅
⋅ ⋅
l l1 2
A =1
2b h
A =1
26 55
A =6 55
2=
2 3 55
2= 3 55units2
⋅ ⋅
⋅ ⋅
⋅ ⋅
3 55units2 64 units2
25 3 units2
81
2units2
A = s s or s
A = 8
A = 64 units
2
2
2
⋅
Height =10 3
2= 5 3
A =1
2b h
A =1
29 2
9 2
2
A =9 2 9 2
2 2
A =81 2
2 2=
81
2u
⋅ ⋅
⋅ ⋅
⋅⋅
⋅⋅
nnits2
Quiz Form B
Base: Side opposite 60O angle
x
2where x = 12
12
2= 6
Height: Side opposite 30O angle
A = b h
A = 6 3 6
A = 36 3 units2
⋅
⋅
Height is the long leg of the righttriangle with short leg equal to 3and hypotenuse equal to 8.
x +3 = 8
x +9 = 64
x = 55
x = 55
2 2 2
2
2
Base of triangle is 10. Height of triangle is the long leg of a right triangle with a hypotenuse of 10. So, the height is where x = 10.x 3
2
Area =1
2b h
=1
210 5 3
=10 5 3
2=
2 5 5 3
2
= 25 3 uni
⋅ ⋅
⋅ ⋅
⋅ ⋅ ⋅
tts2
x 2 = 9
Altitude of triangle is
x where
x =9
2
x =9 2
2 2=
9 2
2⋅
© 2009 VideoTextInteractive Geometry: A Complete Course46
NameUnit V, Part B, Lessons 3&4, Quiz Form A—Continued—
For problems 7 and 8, find the area of each polygonal region.
7. A = _______ 8. A = _______
For Problems 9 and 10, find the area of the shaded region.
9. A = _______
10. A = _______
153
2units2
102 +9 3
o60
o120
123
3
8
12o60 6
17
6
6
12 6
10
13
13
5
5
o60
o609
5
6
3
40 units2
Divide the region into a trapezoid (with bases equal to 10and 6 and height equal to ), and a parallelogram (withbase equal to 6 and height equal to 17 - )
Divide the region into a trapezoid (with one base equal to15, a second base equal to 12 and an altitude equal to 5)and a 3 x 3 square.
The area of the rectangle is b h or 13 ⋅ ⋅110 or 130 units
Area of each trapezoid is
2
11
2h b +b or
1
25 13+5
=5 18
2=
5 2 9
2
1 2( ) ⋅ ⋅ ( )⋅ ⋅ ⋅
== 45 units
Area of shaded triangles is the
2
aarea of the rectangle minus the area
of eachh trapezoid. 130 - 45 - 45 = 40 units2
A =1
2h b +b =
1
25 15 +12 =
5 27
2=
135
2units1 2
2( ) ⋅ ( ) ⋅
A = s s = 3 3 = 9 units2⋅ ⋅
Total area of the polygonal region is 135
2+99 =
135
2+
18
2=
153
2units2
3 3
3 3
A =1
2h b +b
A =1
23 3 12+6
A =3 3 18
2=
3 3 2 9
2= 2
1 2( )
⋅ ( )⋅ ⋅ ⋅
77 3 units2
6 3 units2
Height of trapezoid and parallelogram is 3 3..
(Use 30 ,60 ,90 right triangle propertieO O O ss)
Area of shaded triangle is area of trapezzoid minus area of parallelogram.
1
2h b +b1⋅ ⋅ 22 - b h =
1
23 3 9 +5 - 5 3 3 =
3 3 14
2-15 3 =
42 3
2-
1( ) ⋅ ⋅ ⋅ ( ) ⋅ ⋅ 55 3
1
=42 3
2-
15 3 2
2=
42 3
2-
30 3
2=
12 3
2= 6 3 units2⋅
A = b h
A = 6 17 - 3 3
A = 102 - 18 3
⋅
⋅ ( )
So, total area equals 27 3 +102 -18 3 = 102 +9 3
© 2009 VideoTextInteractive Geometry: A Complete Course48
NameUnit V, Part B, Lessons 3&4, Quiz Form B—Continued—
For Problems 7 and 8, find the area of each polygonal region.
7. A = _______ 8. A = _______
For Problems 9 and 10, find the area of the shaded region.
9. A = _______
10. A = _______
9
21+ 2 units2( ) 45 units2
3
3
3
3 2
3
2
8
91 1
1 11 1
1 11 1
1 1
o
o30 o30
4 4
o30
93
8
6
6 3 units2
39
2units2
Area = b h
= 3 2 = 6
= 4 1 = 4
= 5 1 = 5
= 6 1 = 6
= 7 1 = 7
= 8 1 =
⋅⋅⋅⋅⋅⋅⋅ 88
= 9 1 = 9
Total Area = 45 units2
⋅
Area =1
2b h where h = 2, and b = 2 3.
(Use 30 ,60O
⋅ ⋅
OO O,90 right triangle properties)
Total Area of the polygonal regions is 31
2b h⋅ ⋅ ⋅
⋅ ⋅ ⋅
.
3
1
1
2
2 3
1
2
1= 6 3 units2
For the triangle on the left
A =1
23 3
A =9
2
⋅ ⋅
Area of the shaded region is the area of thee triangle minus
the area of the trapezoid.. Height of the trapezoid is 6
2or 3.
=1
2b⋅ ⋅⋅ ⋅ ( )
⋅ ⋅ ⋅ ( )
h -1
2h b +b
=1
2
8
1
9
1-
1
2
3
18 +3
(Use 30
1 2
O ,,60 ,90 right triangle properties)
=1
2
2 4
O O
⋅ ⋅11
9
1-
1
2
3
1
11
2
= 36 -33
2=
72
2-
33
2=
39
2units2
⋅ ⋅ ⋅
For the triangle on the right
A =1
23 3 2
(Use
⋅ ⋅
45 ,45 ,90
right triangle properties)
A =3
O O O
⋅⋅ 3 2
2=
9 2
2
Total Area =9
2+
9 2
2
=9 +9 2
2
or9 1+ 2
2
or9
21+ 2 un
( )
( ) iits2
(working from the top down.)
© 2009 VideoTextInteractive Geometry: A Complete Course50
NameUnit V, Part B, Lesson 5, Quiz Form A—Continued—
8. Find the area of an equilateral triangle inscribed in a circle, with a radius of units. Area = ______________
9. Find the area of a square with an apothem of 8 inches and a side of length 16 inches. Area = ______________
10. Find the area of a regular hexagon with an apothem of meters and a side of length 22 meters. Area = ______________
A =1
2s a n or A = s = 16 = 256 inches
A =1
2
2 2 2⋅ ⋅ ⋅
⋅116 8 4
A =16 8 2 2
2A = 16 16
A = 256 inches2
⋅ ⋅
⋅ ⋅ ⋅
⋅
256 inches2
A =1
2s a n
A =1
222 11 3 6
A = 1452 meters2
⋅ ⋅ ⋅
⋅ ⋅ ⋅
1452 meters2
4 3
Apothem =1
24 3
=4 3
2=
2 2 3
2= 2 3
s = 2 2 3 3
s = 4 3 3
s = 4
⋅
⋅
⋅ ⋅
⋅⋅⋅ 3
s = 12
A =1
2s a n
A =1
212 2 3 3
A =12 2 3 3
2= 36 3 units2
⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅
36 3 units2
4 3
2 3
11 3
© 2009 VideoTextInteractive Geometry: A Complete Course68
NameUnit V, Unit Test Form A—Continued—
22. Find the area of each of the following labeled polygonal regions using the appropriate postulate, theorem, or corollary. (Note: figures which appear to be regular are regular)
a) Area = _________ b) Area = _________
c) Area = _________ d) Area = _________
e) Area = _________ f) Area = _________
g) Area = _________ h) Area = _________
(B-2) (B-4)
(B-3) (B-3)
(B-1) (B-5)
5 6
7
11
5
9
5
4
o60
11
65
8
4 10
3
5 4
Area of Triangle =1
2b h
=1
27 5
=35
2units2
⋅ ⋅
⋅ ⋅
Area of Trapezoid =1
2h b +b
=1
25 9 +5
=1
1 2⋅ ⋅ ( )
⋅ ⋅ ( )
⋅⋅ ⋅ ⋅ ⋅ ⋅5 14
2=
1 5 2 7
2= 35 units2
Area of Rhombus (Parallelogram) = b h⋅ Area of Parallelogram = b h
= 11 5 = 55 uni
⋅
⋅ tts2
Area of Rectangle = b h or l w
= 8 4 = 32 u
⋅ ⋅
⋅ nnits2
A =1
2s a n
A =1
2
A =2
units2
⋅ ⋅ ⋅
⋅ ⋅ ⋅5 3 5
75
Where x = 4
2 2 + 2 2
4 2
A =1
2s a n
A =1
24 2 2 2 4
A =4 2 2 2 4
2= 4 2 2 4
A = 3
⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅
22 units2
(B-5) (B-5)
35
2units2
35 units2
55 units28 3 units2
32 units2 25 3 units2
32 units2
75
2units2
(Triangle)(Trapezoid)
(Rhombus)(Paralleloram)
(Rectangle) (Regular Triangle)
(Regular Pentagon)
(Square)
h = long leg of a 30 - 60 - 90
=x 3
2=
4 3
V
22 3=
A b h= ⋅
⋅= 4 2 3
= 8 3 units2
Draw line segments from the center of the triangle to the vertices, considering the apothem to be A. We now have two, 30-60-90 right triangles. The long leg of either is x 10 or 5. So, or . 1
2A =
5
35 3
3
A =1
2s a n
A =1
2
A = uni
⋅ ⋅ ⋅
⋅ ⋅ ⋅
= ⋅ ⋅ ⋅⋅
10 5 3 3
2 5 5 3 3
2 3
25 3
A
tts2
© 2009 VideoTextInteractive Geometry: A Complete Course88
NameUnit VI, Part B, Lessons 1,2&3, Quiz Form A—Continued—
2. Use the figure to the right to complete the following statements. In the figure, JT is tangent to (Q at point T.
a) If QT = 6 and JQ = 10, then JT = ____________________
b) If QT = 8 and JT = 15, then JQ = ____________________
c) If m/JQT = 60 and QT = 6, then JQ = ____________________
d) If JK = 9 and KQ = 8, then JT = ____________________
QT
J
K
Since QT JT by Corollary 68a, nQTJ is a right triangle. a +b = c
QT + JT = JQ
6 + JT = 10
36 + JT
2 2 2
2 2 2
2 2 2
( ) ( ) ( )( )( ))( )
2
2
= 100
JT = 64
JT = 8
8
17
12
15
Since QT JT by Corollary 68a, nQTJ is a right triangle. a +b = c
QT + JT = JQ
8 +15 = JQ
64 +225
2 2 2
2 2 2
2 2 2
( ) ( ) ( )( )
== JQ
289 = JQ
17 = JQ
2
2
( )( )
Since QT JT, nQTJ is a right triangle. If m/JQT = 60,nQTJ is a 30-60-90 triangle, and QT is the short leg ofthe triangle opposite the 30 degree angle. QT = 1/2 JQ in a 30-60-90 triangle.
QT =1
2JQ
6 =1
2JQ
12 = JQ
Since QT JT, nQTJ is a right triangle. a2 + =
( ) + ( ) = ( )+ ( ) = +( )
b c
QT JT JQ
JT JK KQ
2 2
2 2 2
2 2 28
644 9 8
64 17
64 289
2 2
2 2
2
+ ( ) = +( )+ ( ) = ( )+ ( ) =
JT
JT
JT
JT(( ) ==
2225
15JT
© 2009 VideoTextInteractive Geometry: A Complete Course104
NameUnit VI, Part C, Lessons 1,2&3, Quiz Form A—Continued—
5. Find EQ in (Q, EQ = _________ 6. Find mCB in (Q, mCB = _________if CD = 10 and DQ = 9. if mCD = 96O
7. Find DC in (Q. DC = _________ 8. Find BD and AC in (Q. BD = _______AC = ________
9. Find CB in (Q, given CB = _________ 10. Find CE in (Q. CE = _______that CD = 16, AQ = 9 and EQ = 5
Q
B D
E
C
A
Q E
C
B
D
A
Q
A
D
E4x
x4
B
C
Q
AB
C
D
6
Q
B
D
E
C
A
Q
A
C
D
B
6
E3
4
x 2
2 14 48O
8 3 1212
4 5 8
DE = 5 from Theorem 73and AB CD, So AB bisects CD
DQ DE EQ
EQ
EQ
( ) = ( ) + ( )( ) = ( ) + ( )
= + ( )
2 2 2
2 2 2
2
9 5
81 25
556
56
4 14
2
2= ( )± =
⋅ =
EQ
EQ EQ cannot be negative
EQ
( )
114 = EQ
( ( (mCB = 48O from Corollary 73aand AB CD, So AB bisects CD
(
AE BE = DE EC
12 4 = x x
48 = x2
⋅ ⋅⋅ ⋅
± =48 x
x cannot be nega( ttive
x
x
so DC
)
,
16 3
4 3
8 3
⋅ =
=
=
AQ QC = BQ QD (call
=
⋅ ⋅⋅ ⋅
⋅ = ⋅
the radius r
r r r
r r
r
r
" ")
6
66
6
r
r =
BD BQ DQ
BD
BD
= += +=
6 6
12
AC AQ CQ
AC
AC
= += +=
6 6
12
AE BE = DE EC
=
⋅ ⋅
⋅ ⋅
=
=
± =
6 4 3
24 3
8
8
2
2
2
x
x
x
x
x cannot be( nnegative
x
x
)
4 2
2 2
⋅ =
=
CE x=
= ( )=
2
22 2
8
Theorem 73 - AB bisects CD, So CE = 8
CB CE BE
CB
CB
( ) = ( ) + ( )( ) = ( ) + −( )( ) = +
2 2 2
2 2 2
2
8 9 5
64 44
64 16
80
80
2
2
2
( )( ) = +
( ) =
= ±
CB
CB
CB
CB cannot be ne( ggative
CB
CB
)
= ⋅
=
16 5
4 5
(
(AC and BD are diameters)
© 2009 VideoTextInteractive Geometry: A Complete Course108
NameUnit VI, Part C, Lessons 4,5,6&7, Quiz Form A—Continued—
3. Find x in (Q. x = _________ 4. Find mAD in (Q. mAD = _________
5. Find CD in (Q. CD = _________ 6. Find AC and AE in (Q. AC = _________AE = _________
7. Find AB and BC in (Q. AB = _________ 8. AB and AC are m/BAE = _________BC = _________ tangents to (Q, and
m/BAC = 42O.Find m/BAE.
Q
B
A
C
12
x -x2
Q
C
D
A
B
94
86
Q
A
B
C
D
12
Q
A
B
CE
D
x+2 x+6
x+1 x
Q
A
B
D
C
1.2
0.4x
0.5x
Q E
B
C
D
A
12
12
( (4 86
AB AC
x x
x x
x x
x or x
=
= −
= − −= −( ) +( )= − =
12
0 12
0 4 3
0 4 0
2
2
++= − =
=
3
4 3
4
x or x
x cannot be negative
x
( )
mAD mAB mBC mCD
mAD
mAD
» » » »
»= − − −
= − − −
360
360 94 86 94
»» = 86
12
CD AB
CD AB
CD
≅== 12
12
9
AB AC AD AE
x x x x x x
⋅ = ⋅
+ +( ) = +( ) +( ) + +( ) 6 1 1 2 +( ) = +( ) +( )+ = + +
= +
x x x x
x x x x
x x
2 6 1 2 3
2 6 2 5 3
6 5 3
2 2
xx = 3
AC x x
AC
AC
= + += ( ) + ( ) +=
6
3 3 6
12
AE x x
AE
AE
AE
= +( ) + +( )= +( ) + +( )= +=
1 2
3 1 3 2
4 5
9
AC
AD
AD
ABx
xx
=
=
( )( ) = ( )
0 9
1 2
1 2
0 41 2 1 2 0 9 0
.
.
.
.. . . .44
1 44 0 361 44
0 364
4
2
2
2
x
x
x
x
x x cannot
( )=
=
=± =
. ..
.
( bbe negative
xx
)
42
==
AB x
AB
AB
= ( ) ⋅= ( )=
0 4
0 4 2
0 8
.
.
.
0.8
1
BC x
BC
BC
= ( ) ⋅= ( )=
0 5
0 5 2
1
.
.
AE bisects /BAC byCorollary 80a
m BAE m BAC∠ = ⋅ ∠
= ⋅
=
1
21
242
21
21O
If we draw radii QA, QB, QC, andQD, we can prove triangles congruent.We can then use Theorem 73 and prove AB > CD.
© 2009 VideoTextInteractive Geometry: A Complete Course 115
Name
Class Date Score
Unit VI - CirclesPart D - Circle ConcurrencyLesson 1 - Theorem 83 - “If you have a triangle, then that triangle is cyclic.”
Lesson 2 - Theorem 84 - “If the opposite angles of a quadrilateral aresupplementary, then the quadrilateral is cyclic.”
1. Quadrilateral ABCD is cyclic. Find x and y. x = _________y = _________
2. Quadrilateral (Kite) ABCD is cyclic. Find mAB. mAB = _________
Quiz Form A
A
B
C
D
110
75
y
x
160
Q
X Y
W
Y
A
Q
X
Z
CB
Z
Q
A B
C
D
E F
G36
84
Q
D
A
B
C
134
A
B
Q
D
A
B
C100
100
C
D
110
75
y
x
Q
X Y
W
X
ZA B
Q
C W
Z
Q
B
A
D
C
160
Q
X Y
W
Y
A
Q
X
Z
CB
Z
Q
A B
C
D
E F
G36
84
Q
D
A
B
C
134
/ABC and /ADC are supplementary. (Corollary 67b)
m ABC m ADC
y
y
m BAD mBD Th
∠ + ∠ =+ =
=
∠ = ⋅
180
90 180
90
1
2» ( eeorem
mBD
mBD or mBCD
67
751
2
150
)
( )
= ⋅
=
»
» ¼
mAB mBCD mDA
x
x
x
» ¼ »+ + =+ + =
+ ==
360
150 110 360
260 360
1100
100
90
46
Quadrilateral ABCD is a kite, so CD > CB,BC > DC (Theorem 81), and mBC = 134.
mAB mBC ABC is a semicircle
mAB
» » ¼
»+ =
+ =
180
134 18
( )
00
46mAB» =
(
( ( (
(
© 2009 VideoTextInteractive Geometry: A Complete Course120
NameUnitVI, Unit Test Form A—Continued—
Determine whether each of the following is always, sometimes, or never true.
________ 13. Congruent chords of different circles intercept congruent arcs.
________ 14. An angle inscribed in a semicircle is a right angle.
________ 15. Two circles are congruent if their radii are congruent.
________ 16. Two externally tangent circles have only two common tangents.
________ 17. A radius is a segment that joins two points on a circle.
________ 18. A polygon inscribed in a circle is a regular polygon.
________ 19. A secant is a line that lies in the plane of a circle, and contains a chord of the circle.
________ 20. The opposite angles of an inscribed quadrilateral are supplementary.
________ 21. If point X is on AB, then mAX + mXB = mAXB.
________ 22. The common tangent segments of two circles of unequal radii are congruent.
________ 23. Tangent segments from an external point to two different circles are congruent.
________ 24. Cyclic quadrilaterals are congruent.
________ 25. If two circles are internally tangent, then the circles have three common tangents.
sometimes
always
always
never
never
sometimes
always
always
always
always
sometimes
sometimes
never
( ( ( (
(C-7)
(B-2)
(A-1)
(A-3)
(A-1)
(A-3)
(A-1)
(B-2)
(Postulate 8 - p198)
(A-3)
(A-3)
(D-2)
(A-3)
© 2009 VideoTextInteractive Geometry: A Complete Course 123
NameUnitVI, Unit Test Form A—Continued—
38. Find m/C m/C = ________ 39. x = ________
40. Find m/D m/D = ________ 41. x = ________
42. Find QE QE = ________ 43. Find BD. BD = ________
95 2 10
80 17
3 10
m A m C Corollary b
m C
m C
∠ + ∠ = ( )+ ∠ =
∠ =
180 67
85 180
95
AC
AD
AD
ABTheorem
x
x
x x
x
= ( )+ =
⋅ = ( ) +( )=
78
4 6
44 4 6
42 ⋅⋅
= ± ⋅
= ⋅
=
10
4 10
4 10
2 10
x
x cannot be negative
x
x
( )
m D m B
m D
∠ + ∠ =
∠ + ==
180
100 180
80
A t r a p e z o i d i n s c r i b e d i n a c i r c l eis cyclic. (Corollary 67b)
x = 17; AB > AC (Theorem 80)
BD = 10, since AD = CD = BD(Corollary 83a)
QF FC QC
QF
QF
( ) + ( ) = ( )( ) + ( ) = ( )
( ) + =
2 2 2
2 2 2
2
4 5
16 25
QQF
QF
QF cannot be negative
QF
QF
( ) =
= ±
==
29
9
9
3
( )
If DC then FC
Theorem
= =8 4
73
,
( )
so, QE = 3. Congruent Chords in the same circle are equidistant from the center of the circle.(See Unit VI, Part C, Lesson 7, Exercise 3)
Q
F
B
EA
C
G
D
Q
ED A
B
C
x
2
B
C
E
D
A
8
12
x
4
4
7
Q Q
A
B
C
D
85
Q
D
A
B
C
6
4
x
B
CD
A100
Q
Q
H
C
DI
E
GA
F
B
B
A
C
17
x
Q
AE B
A
B
CD
l1
l3
l2
C
F
D
Q
M
L
P
RT U
N
5 43
2
Q
T
RC
B
A
D
E
6
x
7
68 17
Q
Qx
6
40
y
C
Q
B
Q
C
B
A
D
Ex
x
4
5
3
AxD
A
B
D
C
A G
D
F
C
E
B
10
8
10
Q
5
Q
F
B
EA
C
G
D
Q
ED A
B
C
x
2
B
C
E
D
A
8
12
x
4
4
7
Q Q
A
B
C
D
85
Q
D
A
B
C
6
4
x
B
CD
A100
Q
Q
H
C
DI
E
GA
F
B
B
A
C
17
x
Q
AE B
A
B
CD
l1
l3
l2
C
F
D
Q
M
L
P
RT U
N
5 43
2
Q
T
RC
B
A
D
E
6
x
7
68 17
Q
Qx
6
40
y
C
Q
B
Q
C
B
A
D
Ex
x
4
5
3
AxD
A
B
D
C
A G
D
F
C
E
B
10
8
10
Q
5
Q
F
B
EA
C
G
D
Q
ED A
B
C
x
2
B
C
E
D
A
8
12
x
4
4
7
Q Q
A
B
C
D
85
Q
D
A
B
C
6
4
x
B
CD
A100
Q
Q
H
C
DI
E
GA
F
B
B
A
C
17
x
Q
AE B
A
B
CD
l1
l3
l2
C
F
D
Q
M
L
P
RT U
N
5 43
2
Q
T
RC
B
A
D
E
6
x
7
68 17
Q
Qx
6
40
y
C
Q
B
Q
C
B
A
D
Ex
x
4
5
3
AxD
A
B
D
C
A G
D
F
C
E
B
10
8
10
Q
5
Q
F
B
EA
C
G
D
Q
ED A
B
C
x
2
B
C
E
D
A
8
12
x
4
4
7
Q Q
A
B
C
D
85
Q
D
A
B
C
6
4
x
B
CD
A100
Q
Q
H
C
DI
E
GA
F
B
B
A
C
17
x
Q
AE B
A
B
CD
l1
l3
l2
C
F
D
Q
M
L
P
RT U
N
5 43
2
Q
T
RC
B
A
D
E
6
x
7
68 17
Q
Qx
6
40
y
C
Q
B
Q
C
B
A
D
Ex
x
4
5
3
AxD
A
B
D
C
A G
D
F
C
E
B
10
8
10
Q
5
Q
F
B
EA
C
G
D
Q
ED A
B
C
x
2
B
C
E
D
A
8
12
x
4
4
7
Q Q
A
B
C
D
85
Q
D
A
B
C
6
4
x
B
CD
A100
Q
Q
H
C
DI
E
GA
F
B
B
A
C
17
x
Q
AE B
A
B
CD
l1
l3
l2
C
F
D
Q
M
L
P
RT U
N
5 43
2
Q
T
RC
B
A
D
E
6
x
7
68 17
Q
Qx
6
40
y
C
Q
B
Q
C
B
A
D
Ex
x
4
5
3
AxD
A
B
D
C
A G
D
F
C
E
B
10
8
10
Q
5
Q
F
B
EA
C
G
D
Q
ED A
B
C
x
2
B
C
E
D
A
8
12
x
4
4
7
Q Q
A
B
C
D
85
Q
D
A
B
C
6
4
x
B
CD
A100
Q
Q
H
C
DI
E
GA
F
B
B
A
C
17
x
Q
AE B
A
B
CD
l1
l3
l2
C
F
D
Q
M
L
P
RT U
N
5 43
2
Q
T
RC
B
A
D
E
6
x
7
68 17
Q
Qx
6
40
y
C
Q
B
Q
C
B
A
D
Ex
x
4
5
3
AxD
A
B
D
C
A G
D
F
C
E
B
10
8
10
Q
5
(B-2) (C-4)
(B-2) (C-6)
(C-1) (D-1)
Given: l1 and l3 are perpendicularbisectors of the sides of nABC. AC = 20.
Given: AB > CD, DC = 8, the radius of (Q is 5.
© 2009 VideoTextInteractive Geometry: A Complete Course 123
NameUnitVI, Unit Test Form A—Continued—
38. Find m/C m/C = ________ 39. x = ________
40. Find m/D m/D = ________ 41. x = ________
42. Find QE QE = ________ 43. Find BD. BD = ________
95 2 10
80 17
3 10
m A m C Corollary b
m C
m C
∠ + ∠ = ( )+ ∠ =
∠ =
180 67
85 180
95
AC
AD
AD
ABTheorem
x
x
x x
x
= ( )+ =
⋅ = ( ) +( )=
78
4 6
44 4 6
42 ⋅⋅
= ± ⋅
= ⋅
=
10
4 10
4 10
2 10
x
x cannot be negative
x
x
( )
m D m B
m D
∠ + ∠ =
∠ + ==
180
100 180
80
A t r a p e z o i d i n s c r i b e d i n a c i r c l eis cyclic. (Corollary 67b)
x = 17; AB > AC (Theorem 80)
BD = 10, since AD = CD = BD(Corollary 83a)
QF FC QC
QF
QF
( ) + ( ) = ( )( ) + ( ) = ( )
( ) + =
2 2 2
2 2 2
2
4 5
16 25
QQF
QF
QF cannot be negative
QF
QF
( ) =
= ±
==
29
9
9
3
( )
If DC then FC
Theorem
= =8 4
73
,
( )
so, QE = 3. Congruent Chords in the same circle are equidistant from the center of the circle.(See Unit VI, Part C, Lesson 7, Exercise 3)
Q
A
B
C
D
85Q
D
A
B
C
6
4
x
B
CD
A100
B
A
C
17
x
Q
AE B
C
F
D
A
B
CD
l1
l3
l2
(B-2) (C-4)
(B-2) (C-6)
(C-1) (D-1)
Given: l1 and l3 are perpendicularbisectors of the sides of nABC. AC = 20.
Given: AB > CD, DC = 8, the radius of (Q is 5.
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