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NAME
-c
_ DATE
Practice with Examples
For use with pages 527-534
Solve problems involving similar right triangles formed by the altitude
drawn to the hypotenuse of a right triangle and use a geometric mean to
solve problems
Theorem 9.1
If the altitude is drawn to the hypotenuse of a right triangle, then the two
triangles formed are similar to the original triangle and to each other.
Theorem 9.2
In a right triangle, the altitude from the right angle to the hypotenuse
divides the hypotenuse into two segments. The length of the altitude
is the geometric mean of the lengths of the two segments.
Theorem 9.3
In a right triangle, the altitude from the right angle to the hypotenuse
divides the hypotenuse into two segments. The length-of each leg of the
right triangle is a geometric mean of the lengths of the hypotenuse and
the segment of the hypotenuse that is adjacent to the leg.
Consider the right triangle shown.
a. Identify the similar triangles.
b. Find the heighte of MBC.
6
SOILUT~ON
a. MBC- ACBD- MCD
A
10
Sketch the three similar right triangles so that the corresponding
angles and sides have the same orientation.
B
C
~
C
B
C
D
A
A
b. Use the fact that MBC- ACED to write a proportion.
CD
AC
CB
AB
h
6
10
·11.6
11.6h
=
6(10)
h = 5.2
Geometry
Practice Workbook
vvith Examples
Corresponding side lengths are in proportion.
Substitute.
Cross product property
Solve for h.
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LESSON
NAME
_
DATE
Practice with Examples
CONTINUED
For use- with pages 527-534
.~~If!!.I?~~f!.~.
!.f!.~.~l!.'!.I!!.I!.~I!..!.
.
Find the height, h, of the given right triangle.
1.
I
N
M
w
I:
3.
2.
Z
4
!
'.
I
!
~
Using a Geometric Mean
Find the value of each variable.
b.
a.~
10
4
SOl.UTION
a. Apply Theorem 9.3.
10
b. Apply Theorem 9.2.
+4
x
3
l
x
4
y
1
56
y2
=
3
y
=
-J3
X2 =
x =
-J56
= ~
=
2-Jl4
t'
'.",
Copyright © McDougal Littell Inc.
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Geometry
Practice Workbook with Examples
LESSON
NAME
~------------
DATE
Practice with Examples
CONTINUED
For use with pages 527-534
;~~.f!!.~~~I!.~.
!.l!!..
~!!.l!.'!!.I!.~~.?
.
Find the value of each variable to the nearest tenth.
4.
5.
1
~
x
6.
Geometry
Practice Workbook
Copyright © McDougal Littell Inc.
vvith Examples
All rights reserved.
NAME
_
DATE
rectice with Examples
I
For use with pages 535-541
I
I
GZIIa
Find the Length,ot a Hypotenuse'
Find the length of the hypotenuse of the right triangle.
Tell whether the side lengths form a Pythagorean triple.
6
SO,LUTION
(hypotenuse)? = (leg)2
+ (leg)2
Pythagorean
Theorem
Substitute.
+
X2
= 36
X2
= 100
Add.
= 10
Find the positive square root.
x
64
Multiply.
Because the side lengths 6, 8, and 10 are integers, they form a
Pythagorean.triple .
.~~.~!.~
~~~~
..~C?~.
~l!.~I!!.I!.~
~.!
'"
Find the length of the hypotenuse of the right triangle. Tell
whether the side lengths form a Pythagorean triple.
1.
8
2.
3.
7
l'
I
I
.i
i
Copyright © McDougal. Littell Inc .
All rights reserved.
Geometry
Practice
Workbook
with Examples
.
p
LESSON
NAME __ ~
~~
~
_
9~2 Proctice with Examples
DATE
CONTINUED
For use with pages 535-541
_
Finding the Length of a Leg
Find the length of the leg of the right triangle .
. SOU.JTBON
(hypotenuse)? =. (leg)?
122
=
144
=
63 =
.,j63
+
(leg)?
Pythagorean Theorem
+ X2
81 + X2
Multiply.
x2
Subtract 81 from each side .
92
Substitute.
= X
Exercises for Example 2
............................................
Find the positive square root.
.,
-
.
Find the unknown side length. Round to the nearest tenth, if
necessary.
5.
4.
6.
~1.
L~
5
7.1
24
Geometry
Practice Workbook vvith Examples
Copyright © McDougal Littell Inc.
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-
.::~
LESSON
NAME~
_ DATE
Practice with Examples
CONTINUED
For use with pages 535-541
Finding the Area of a Triangle
Find the area of the triangle to the nearest tenth.
4
SOLUTION
In this case, the side of length 4 can be used as the height and
the side of unknown length can be used as the base. To find
the length of the unknown side, use the Pythagorean Theorem.
(hypotenu~e)2
=
(leg)?
152
=
42
-J209
=
b
+
(leg)?
+ b2
Pythagorean Theorem
Substitute.
Solve for b.
Now find the area of the triangle.
A
=
~bh
=
~(-J209)(4)
= 28.9 square units
.~~.~!.~~~~~.!.l!.~.~l!.E!.'!!I!.~~.~
.
Find the area of the triangle to the nearest tenth.
7.
8.
9.
13
~2---l
)
i
j
I
oj
-:1
I
l
J
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Geometry
Practice WorkbOok
with Examples
I
_=='-'. _.
&l:t*m~-~~T·~~£.~1*¥.&j£G!':t::5f~-::---:=-~~i!ff5~
~~~~-..-.-,;,z.:n.
NAME
DATE
~----------~--------------~--
Practice with Examples
For use with pages 543--548
Use the converse of the Pythagorean Theorem to solve problems and use
side lengths to classify triangles by their angle measures
Theorem 9.5
'I
I
Converse of the Pythagorean Theorem
If the square of the length of the longest side of a triangle is equal to the
sum of the squares of the lengths of the other two sides, then the triangle
is a right triangle. .
!
I Theorem 9.6
.
I If the square of the length
.
.
of the longest side of a triangle is less than the
I, sum of the squares of the lengths of the other two sides, then the triangle
is acute.
I
.
i Theorem
9.7
If the square of the length of the longest side of a triangle is greater than
the sum of the squares of the lengths of the other two sides, then the
triangle is obtuse.
Verifying Right Triangles
The triangles below appear to be right triangles. Tell whether they are
right triangles.
a.
,~-
b.
Let c represent the length of the longest side of the triangle (you do not
want to call this the "hypotenuse" because you do not yet know if the
triangle is a right triangle). Check to see whether the side lengths satisfy
the equation c2 = a2 + b2.
?
a. 102 ~ 82
?
100 ~
100
+ 72
64 + 49
* 113
The triangle is not a right triangle.
?
b. 202 ~ 122
?
400 ~
400
+ 162
144 + 256
= 400
The triangle is a right triangle.
".r.·~.<
~,
~
Geometry
Practice Workbook vvith Examples
Copyright © McDougal Littell Inc.
All rights reserved.
',"/
-: .•..
NAME
CONTINUED
_
DATE
Practice with Examples
For use with pages 543-548
.~~.~!.~~~f!.~
..~I!.~.~l!.?I!!.I!.~~.!
.
In Exercises 1-3, determine if the triangles are right triangles.
1.
2.
3.
4
;:
'I
:1
Classifying Triangles
Decide whether the set of numbers can represent the side lengths of a
triangle. If they can, classify the-triangle as right, acute, or obtuse.
a. 58,69,80
b. 11,30,39
,
"';
:!
~,~'{~,\
\'~:~~s:,'"
"
,
Copyright © McDougal Littell Inc.
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Geometry
Practice Workbook with Examples
LESSON
9.3
NAME
_
DATE
Practice with Examples
CONTINUED
For use with pages 543-548
SCU ..UTUON
You can use the Triangle Inequality.to confirm that each set of numbers
can represent the side lengths of a triangle.
Compare the square of the length of the longest side with the sum of the
squares of the lengths of the two shorter sides.
a. c27 a2
+ b2
Compare c2 with a2
802 7 582
+ 692
3364 + 4761
Substitute.
64007
Multiply.
c2 is less than a2
6400 < 8125
Because c2 < a2
b. c2 7 a2
Compare c2 with a2
392 7 112+ 3()2
+ b2.
Substitute.
+ 900
Multiply.
c2 is greater than a2
1521 > 1021
Because c2 > a2
+ b2.
+ b2, the triangle is acute.
+ b2
1521 7 121
+ b2.
+
b2.
+ b2, the triangle is obtuse.
Exercises for Example 2
••••••••.••••••••••••
a •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Decide whether the set of numbers can represent the side
lengths of a triangle. if they can, classify the triangle as right,
acute, or obtuse.
4. 5,
.)56,
9
Geometry
Practice Workbook
vvith Examples
5. 23,44,70
6. 12,80,87
7. 4,7, 10
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NAME
_ DATE
Practice with Examples
For use with pages 551-556
-
Find the side lengths of special rjqht triangles
VOCA.BULARY
.. ±.
Right triangles whose angle measures are 45°- 45°- 90° or 30°- 60°- 90°
are called special right triangles.
Theorem 9.8 The 45°· 45°· 90° Triangle Theorem
In a 45°- 45°- 90° triangle, the hypotenuse is -fi times as long as each leg.
Theorem 9.9 The 30°· 60°· 90° Triangle Theorem
In a 30°- 60°- 90° triangle, the hypotenuse is twice as long as the shorter
leg, and the longer leg is .J3 times as long as the shorter leg.
.
Finding
Side~~.=
lengths in a 45°- 45°-90° Triangle
'V""
v
r
;t2~"'~~'~
__
~l~
Find the value of x.
7
x
1
:1'.',
By the Triangle Sum Theorem, the measure of the third angle is 45°. The
triangle is a 45°_ 45°- 90° right triangle, so the length x of the hypotenuse
is -fi times the length of a leg.
,:1
-fi . leg
= -fi . 7
= 7 -fi
Hypotenuse =
x
x
Copyright ©McDougal
All rights reserved.
45° - 45° - 90° Triangle Theorem
Substitute.
Simplify.
Littell Inc.
Geometry
Practice Workbook
with Examples
t:-
LESSON
NAME
_
L
DATE
_
Practice with' Examples
CONTINUED
For use with pages 551-556
.~~.~!.~~~l!.~.
!.~!..
~l!.'!.'!!P..~f!..!.;
~
;
.
Find the value of each variable.
3.
2.
1.
y
y
Finding Side Lengths in a 30° - 60°-90° Triangle
Find the value of x.
SOLUTION
Because the triangle is a 30°- 60°- 90° triangle, the longer leg is
times the length x of the shorter leg.
= fi . shorter leg
22 = J3 . x
Longer leg
22
-=x
fi
fi 22
fifi
-'-=x
22fi'
--=x
3
GeOmletrry
Practice Workbook vvith Examples
fi
30° - 60° - 90° Triangle Theorem
Substitute.
Divide each side by
.)3.
Multiply numerator and denominator by .)3.
Simplify.
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LiiE.SSON
!\lAME
_
DATE
Practice with Examples
CONTINUED
For use with pages 551-555
.~~.~.~~~~~~.
!.~.~.~l!.~'!!.I!.~i!,.?
.
Find the value of each variable.
4.
,\00
Y, 30°
14~X
5.
,I
6.
x
-~f1;~~
-\~'"
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Geometry
Practice Workbook with Examples
NAME
_
DATE
Practice with Examples
For use with pages 558-565
Find the sine, the cosine, and the tangent of an acute angle and use
triqonornetrlc ratios to solve real-life problems
I
VOCABULARY
A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. The three basic trigonometric ratios are sine, cosine, and tangent,
which are abbreviated as sin, cos, and tan, respectively ..
The angle that your line of sight makes with a line drawn horizontally is
called the angle of elevation.
TrigOnometric Ratios
be a right triangle. The sine, the cosine, and the tangent of the
I Let MBC
angle LA are defined as follows.
i acute
sin A
.
=
cas A
=
I tan A
=
side opposite LA
hypotenuse'
= ~
side adjacent LA
hypotenuse
=
side opposite LA
side adjacent LA
a
=-
~
B
C
"£
side
a opposite
LA
C
b
~
A
c
b
s_id_e_a_d_ja_C_e_n_tt_o_L__A
~
__ ~
Finding Trigonometric Ratios
Find the sine, the cosine, and the
tangent of the indicated angle.
B
a. LA
~82
b. LB
A
23.1
C
SOLUTiON
a. The length of the hypotenuse is 24.5. For LA, the length of the
opposite side is 8.2, and the length of the adjacent side is 23.l.
'.
sin A
=
A
=
adj ..
hypo
=
23:1 = 09429
24.5
.
tan A
=
opp.
ad].
=
8.2 = 0.3550
23.1
cas
opp.
hypo
=
8.2
24.5 = 0.3347
.Geometiy
Practice Workbook vvith Examples
Copyright © McDougal Littell Inc .
All rights reserved.
NAME~
_
DATE
Practice with Examples
CONTiNUED
for use with pages 558-565
"
b. The length of the hypotenuse is 24.5. ForLB, the length of the
opposite side is 23.1 and the length of the adjacent side is 8.2 .
, J
i
. B
sin
=
opp. = 23.1 = 094' 29
h.<yp. 245.
.
cos B
=
tanB
= -.
adj.
hypo
opp.
ad].
=
8.2
.
24.5 = 0.3347
23.1
= 2.8171
8.2
= -
.~~.~.~~~~l!.~
..~~.~.~l!.I!.I!!.I!.~i!..!
.
find the sine, cosine, and tangent of LA.
1.
2.
A
3
B
4.6
5
c
3.
3.8
6
il
4
c
A
B
A
c
'l
''>1
Copyright © McDougal Littell Inc.
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Geometry
Practice Workbook
vvith Examples
LESSQN
NAME
9.5
CONTINUED
~
DATE
~~----------------------------------
Practice with Example~
for use with pages 558-565
Estimating a Distance '
It is known that a hill frequently used forsled
riding has an angle of elevation of 30° at its
bottom. If the length of a sledder's ride is 52',6 feet,
estimate the height of the hill.
h
SOLUTION
,
0
Use the sine ratio for the 30 angle, because you have the value of the
hypotenuse and you are looking for the value of the side opposite the 30°
angle.
. 0
sin 3
0
"
,h
= 52.6
h = (52.6) . sin 300 = (52.6) . (0.5) = 26.3 feet
.~~~!.~
~~f!.~.
!.C!.~.~1!:'!.r!!l!.{~.?
.
4. In the sled-riding example, find the height of the hill if the angle of
elevation of the hill is 42
0
•
5. If the angle of elevation from your position on the ground to the top
of a building is 67 and you are standing 30 meters from the foot of
the building, approximate the height of the building.
0
Geometry
Practice Vvorkbook vvith Examples
Copyright © McDougal Littelllnc,
All rights reserved,
NAME
_
DATE
Practice with Examples
For use with pages 567-572
Solve a right triangle
IVOCABULARV
I
.
To solve a right triangle means to determine the measures of all six
parts (the right angle, the two acute angles, the hypotenuse, and the
two legs).
.
Solve the right triangle.
A
SOUJTION
Begin by using the Pythagorean Theorem
to find the length of the missing side.
(hypotenuse)? = (leg)"
+ (leg)?
5~
CaB
Pythagorean Theorem.
Substitute:
169
144
12
=
=
a2 + 25
2
a
Multiply.
Subtract 25 from each side.
Find the positive square root.
= a
Then find the measure of LB.
tanB
= opp.
adj.
tan B
I
I'
=
5
12
mLB = 22.6°
Substitute.
Use a calculator.
5
12
Finally, because LA and LB are complements, you can write
mLA =:= 90° - mLB = 90° - 22.6° = 67.4°.
The side lengths of MBC are 5, 12, and 13. MBC has one right angle
and two acute angles whose measures are about 22.6° and 67.4°.
Copyright © McDougal Littell Inc.
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Geometry
Practice Workbook
vvith Examples
I
.
LESSON
=-':'7.oo.:..:~~'f-~":"-':'
~~~~~"~7~~:4£2~
_~;.....;.~
..•_¥_±:_.:;;"S#§:#-~_
'{
:;-.o.~Y~
'90.
,
NAME_'
_ DATE
Practice with Examples
CONTINUED
For use with pages 567-572
~':.f!!.l?,~~f!.~.!.~!..~'!:I!.'!!.I!.~f!..!
.
Solve the right triangle.
1.
y
2.
3.
N
Q
L
x
9.4
fill
Pl...----J....JR
Solve the right triangle.
X
SOLUTiON
v~~
Use trigonometric ratios to find the values of x and y. Z
sinX
. 71 ° =-x
sin
32
=
32(0.9455)=
y
adj.
cos X =-h
yp.
= opp.
hypo
32 sin 71°
x
cos 71° = L
32
32 cos 71°
x
x
32(0.3256)
=
y
=
y
10.4 = y
30.3 = x
Because LX and LY are complements, you can write
I1'LLY
= 90° -
mLx
= 90° - 71° = 19°.
The side lengths of the triangle are about 10.4, 30.3, and 32, The triangle
has one right angle and two acute angles whose measures are 71 ° and 19°.
Geoli111letry
Practice Workbook with Examples
Copyright © McDougal Littell Inc.
All rights reserved.
".,
NAME
DATE
------------------------------------
Practice with Examples
CONTINUED
For use with pages 567-572
.~~~!.~~~~~.
!l!!..~~i!.'!!l!.~'!..?
.
Solve the right triangle.
5.
4. A
6.
M
Q
N
c
Copyright © McDougal Littell Inc.
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41.5
L
Practice Workbook
GeOnletlrY
with EXamples
R
NAME
~
DATE
~----------------------
Practice with Examples
For use with pages 573-579
find the magnitude and the direction of a vector and add vectors
VOCABULARY
~
of a vector AB is the distance from the initial point A
The magnitude
to the terminal pointB andis'written
IAB'I.
The direction of a vector is determined by the angle it makes with a
horizontal line.
Two vectors are equal if they have the same magnitude and direction.
Two vectors are parallel if they have the same or opposite directions.
I Sum of Two Vectors
I ~he ~: of it (aI' bl)
i
=
and
v=
(a2, b2) is
! u + v - (al + a2, bl + b2).
,
--
finding the Magnitude of a Vector
Points P and Q are the Initial and terminal points of the vector
J5Q.
Draw PQ in a coordinate plane. Write the component form of the vector
and find its magnitude.
a. P(l, 2), Q(5, 5)
a. Component form
b. P(O, 4), Q(-2,
=
fQ =
=
(x2
-
Xl'
h - Yl)
!,
I'
(5 - 1,5 - 2)
~---+--
(4,3)
l
Use the Distance Formula to find the
magnitude.
1J5Q1 ~ . .)(5
- 1)2
b. Component form
J5Q
~
I
I
I
I
I
--1----
(-2 - 0, -4 - 4)
=
(-2, -8)
p
4h
I I
! I x
i
I
I
Use the Distance Formula to find the
magnitude.
=
.)(-2 - 0)2 + (-4
- 4)2 =
Q
-J68 = 8.2
1
I
I
Y!
~-----,
I
p
I I
!
.
I
/
I
1
I
2
4
.
1101
I
YIQ
xl' h - YI)
=
I
,
1/
It!
I
;
+ (5 - 2)2 =,$ =5
= (x2
y
c
-4)
1
I
I I
!
I
x
I
I
I
I I
I
.~!~..a:
:.....•...•'.
Geometry
Practice VVork:book: with
Examples
Copyright © McDougal Littell Inc.
All rights reserved.
LESSON
NAME
_
DATE
Practice with Examples
CONTINUED
"j
For use with pages 573-579
Exercises for Example 1
•••••••
I ••••••••••••••
, ••
~ ••••••••••••••••••••••••••••••••••••••••••••••••.
0 •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
---'"
Draw PO in a coordinate plane. Write the component
the vector and find its magnitude.
2. P(-2, 1); Q(O,-5)
1. P(3, 2), Q(1, 9)
I
II
I
I
I
I I
! I
I
I
i
! I
'I I
I
I
I
I
iI
!
I
I
I
I
I
,
I
I
I
I
I
I
I
!
I
-n
!
__
I
t
!
I I
I
---+--,--+-r
I !
I, I,
I
I
I
i
I
I
I
I
! I
I !
I
I
iI
!
i
I
I
I i
I
I
I,
I
II
I
I
i
!
I
I
!
I
Describing the Direction of a Vector
The vector AB depicts the velocity of a moving vehicle.
The scale on each axis is in kilometers per hour, Find
the (a) speed of the vehicle and (b) direction it is traveling
relative to east.
N
y
,A
!
I
r-,I
rt
5
w
-----'"
a. The magnitude of the vector AB represents the vehicle's
'speed. Use the Distance Formula.
\ABl =
':
I
I
i
I
I
I
,
i
I
r
!
I I
!
I I
i
I
:
I
!
i I
i i I
I
i
I
I
I
I
I
t
r
1
I
I
I
,
i
i
-+--+-'-1-+-1---
,
1---1-
"
I
I
I
j
I
!
Ii
p( -4, -11), Q(O,2)
I
I
I
I
I
I
I
i
i
!;
I I
!
i
4.
I
I
I
I
I
I
!
I
I
I
I I
I
I---
I
I
I
3. P(3, 8), Q( -1, 10)
I
-I- J,
!
I
I
I !I
I I
I
I
I
!
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form of
=
-,I
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5
I
J152 + 152'
The speed of the vehicle is about 21.2 kilometers per hour.
Copyright © McDougal Littell lnc.
-l
xi
I
i
I
= 21.2
All rights reserved,
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s
J(20 - 5)2 +. (0 - 15)2
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Geometry
Practice Workbook with Examples
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,LESSON
NAME
9.7
DATE
~----------------------------~
Practice with Examples
CONTINUED
For use with pages 573-580
b. The tangent of the angle formed by the vector and a line drawn
,parailel to the x-axis at point A is - ~~. =-l. Use a c~1culator.
i
to find the angle measure.
i,r~f\ll
- 1
=
- 45°
The vehicle is traveling in a direction 45° south of east.
..~?:.~!.I?~!!.l!.~.
!.l!.~.~1!.'!.I!!.I!.~~.?
"
.
in Exercises 5-7, find the vehicle's magnitude and direction if
points A and B are as given.
6. A( -2,4), B(3, - 1)
5. A(O, 0), S(6, 7)
7. A(2, 4), B( - 3, -1)
Finding the Sum of Two Vectors
u = (- 4, 2) and v = (3, 1). Write
Let
the component form of the sum
u + v.
SOB..UTU:)N
To find the sum vector
vertical components of
u + v, add the horizontal
u and
components and add the
V.
u +. v '= (- 4 + 3, 2 + 1) = (-
1, 3)
.~~.~!.I?~!!,i!,~.
!.l!.~.~1!.'!.'!!l!.~~.~
For the given vectors
the sum
+
u e.
8.
10.
u and
.
V, find
u = (0, 8) and v = (- 3, 5)
u=
(3,12) and
the component form of
9.
u = (- 2, -
7) and
v = (2,
10)
v = (- 3, -12)
t;.'
Geome'itt'y
Practice Workbook
vvith Examples
Copyright © McDougal Littell lnc.
All rights reserved.