Geometry: A Complete Course€¦ · Lesson 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

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


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


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



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


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


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⋅


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


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.)


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


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


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


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)


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.


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» =

(

( ( (

(


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)



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

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.



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

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

QQ

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.

Documents

Geometry: A Complete Course€¦ · Lesson 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