7 Geometry PDF
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Fuad_Khafizov_5342Objectives
After studying this section you will be able to:
1. Understand the fundamental geometric concepts of point, line, and plane.
2. Describe the concepts of line segment, ray, and half line.
3. Understand the concepts of plane and space.
4. Describe the relation between two lines.
5. Describe the relation between a line and a plane.
6. Describe the relation between two planes.
A. POINT, LINE, AND PLANE
1. Introduction
In this chapter we will look at the fundamental concepts we need in order to begin our study
of geometry.
Definition geometry
The word geometry comes from two Greek words, ‘geo’ and ‘metric’, which together mean ‘to
measure the earth.’ Geometry is now the branch of mathematics that studies space, shape,
area, and volume.
Nature displays an infinite array of geometric shapes,
from the smallest atom to the biggest galaxy. Snowflakes,
the honeycomb of a bees’ nest, the spirals of seashells,
spiders’ webs, and the basic shapes of many flowers are
just a few of nature’s geometric masterpieces.
The Egyptians and Babylonians studied the area and
volume of shapes and established general formulas.
However, the first real book about geometry was written by
a Greek mathematician, Euclid. Euclid’s book, The
Elements, was published in about 300 BC. It defined the
most basic concepts in geometry and proved some of their
properties.
Geometry as a science has played a great role in the
development of civilization. Throughout history, geometry
has been used in many different areas such as
architecture, art, house design, and agriculture.
8 Geometriy 7
The three most basic concepts of geometry are point, line,
and plane. Early mathematicians tried to define these
terms. In fact, it is not really possible to define them using
any other concepts, because there are no simpler
concepts for us to build on. Therefore, we need to
understand these concepts without a precise definition.
Let us look instead at their general meaning.
2. Point
When you look at the night sky, you see billions of stars, each represented as a small dot of
light in the sky. Each dot of light suggests a point, which is the basic unit of geometry.
Concept point
A point is a position. It has no size, length, width, or thickness, and it is infinitely small.
We use a dot to represent a point. We name a point with a capital letter such as A, B, C, etc.
Nature’s Great Book is
written in mathemati-
cal symbols.
Galileo Galilei
All geometric figures consist of collections of points, and many terms in geometry are defined
using points.
3. Line
Concept line
A line is a straight arrangement of points. It is the second fundamental concept of geometry.
There are infinitely many points in a line. A line has no width or thickness, and extends
without end in both directions.
Geometric Concepts 9
A line is usually named by any two of its points, or by a lower-case letter.
Look at the diagram. The line that passes through points A and B is written AB. We say it is
line AB. The line on the right is simply called line .
Let none unversed in
geometry enter here.
Plato
The arrows at each end of a line show that the line extends to infinity in both directions.
If any point C is on a line AB or a line d, we write C AB, or C d.
d
A C B A B C
C Î AB A, B, C Î d
Property
There exists exactly one line passing through any two distinct points.
By this property, a line is determined by two distinct points. However, remember that a line
consists of more than just two points. There are infinitely many points on a line.
4. Plane
Concept plane
A plane is the third fundamental concept of geometry. A plane has length and width but no
thickness. It is a flat surface that extends without end in all directions.
A plane is suggested by a flat surface such as a table top, a wall, a floor, or the
surface of a lake. We represent a plane with a four-sided figure, like a piece of
paper drawn in perspective. Of course, all of these things are only parts of
planes, since a plane extends forever in length and in width.
We use a capital letter (A, B, C, ...) to name a plane. We write plane P, or (P),
to refer to a plane with name P.
10 Geometriy 7
We can now understand the meaning of the terms point, line, and plane without a formal
definition. We can use these undefined terms to define many new geometric figures and
terms.
5. Collinear Points
Definition collinear points
Points that lie on the same line are called collinear points.
d l
A B C M N
P
For example, in the diagram above, points A, B, and C lie on the same line d. Therefore A, B,
and C are collinear points. However, point P is not on line so M, P, and N are not collinear
points. We say that, M, P, and N are noncollinear points.
EXAMPLE 1 Look at the given figure.
a. Name the lines. A
S
b. Write all the collinear points.
B P
c. Give two examples of noncollinear points. T
a. There are three lines, AC, CN, and SR. C M N
Solution
R
b. The points A, B, C, the points S, T, R, and
the points C, M, N are on the same line, so
they are collinear.
c. The points A, N, C and the points M, T, N A
are not on the same line. They are exam- d
ples of noncollinear points. B
Now consider the three noncollinear points in m
the figure on the right. Since we know that two
C
distinct points determine a straight line, we
can draw the lines AB, AC and BC passing l
through A, B, and C. Therefore, there are three
lines that pass through three noncollinear points.
Geometric Concepts 11
Definition triwise points
If three points are noncollinear then they are also called triwise points.
When we say, n triwise noncollinear points, A
we mean that any three of n points are B
noncollinear.
E
For example, the diagram opposite shows five C D
triwise noncollinear points. Any set of three
points in the diagram is noncollinear.
Theorem
n ( n – 1)
different lines pass through n triwise points.
2
EXAMPLE 2 How many different lines pass through each number of triwise noncollinear points?
a. 4 b. 5 c. 9 d. 22
Solution 4 (4 – 1) 4 3 5 (5 – 1) 5 4
a. = = 6 lines b. = = 10 lines
2 2 2 2
9 (9 – 1) 9 8 22 (22 – 1) 22 21
c. = = 36 lines d. = = 231
2 2 2 2
Check Yourself 1
1. Describe the three undefined terms in geometry.
A E
2. Name the collinear points in the figure.
B
C
D
3. Look at the figure. A C
a. Name the lines.
F E
b. Name all the collinear points.
D B G
c. Give two examples of noncollinear points.
12 Geometriy 7
4. How many different lines can pass through each number of triwise noncollinear points?
a. 8 b. 14 c. 64 d. 120
Answers
1. Point: A point is a position. It has no size, length, width, or thickness, and it is infinitely
small. Line: A line a straight arrangement of points. There are infinitely many points in a
line. A line has no width or thickness, and extends without end in both directions. Plane:
A plane has length and width but no thickness. It is is a flat
surface that extends without end in all directions.
2. The points A, B, C and the points D, B, E are collinear points.
3. a. The lines: AC, AB, DG b. The points A, E, B, the points D, B, G are collinear points c.
The points A, F, G, and the points D, B, C are non collinear points.
4. a. 28 b. 91 c. 2016 d. 7140
B. LINE SEGMENT, RAY, AND HALF LINE
1. Line Segment
Definition line segment
The line segment AB is the set of points
A B
consisting of point A, point B, and all the
points between A and B. A and B are called the line segment
AB or [AB]
endpoints of the segment. We write [AB] to
refer to the line segment AB.
This definition describes one type of line segment: a closed line segment. There
are three types of line segment.
a. Closed Line Segment
A line segment whose endpoints are included in the A B
segment is called a closed line segment.
[AB] in the diagram is a closed line segment.
b. Open Line Segment
Physical model of a line
A B
segment: a piece of string. A line segment whose endpoints are excluded from the
segment is called an open line segment.
The line segment AB in the diagram is an open line segment and denoted by ]AB[.
We use an empty dot ( ) to show that a point is not included in a line segment.
Geometric Concepts 13
c. Half-Open Line Segment
A line segment that includes only one of its endpoints is called a half-o
open line segment.
A B A B
half-open line segment AB half-open line segment AB
[AB[ ]AB]
EXAMPLE 3 Name the closed, open and half-open line seg- A B
ments in the figure on the right.
Solution Closed line segment: [AB]
Half-open line segments: [AC[, [BC[, and [BD[ C D
Open line segment: ]CD[
Property
If C is a point between A and B, then
[AC] [CB]
[AC] + [CB] = [AB].
A C B
Using this property, we can conclude that if
[AB]
three points are collinear, then one of them is
between the other points.
A B C
Point B is between the points A and C.
2. Ray
Definition ray
The ray AB is the part of the line AB that contains point A and all the points on the line
segment that stretches from point A through point B to infinity. The ray AB is denoted by [AB.
In the diagrams, each ray begins at a point and extends to infinity in one direction. A is the
endpoint of [AB, and C is the endpoint of [CD.
A B D C
ray AB, or [AB [CD
14 Geometriy 7
A half line extends to infinity in one direction. A half line is like a ray,
A B
but it begins at an open endpoint.
half line AB
]AB
C. PLANE AND SPACE
1. Plane
We can think of the floor and ceiling of a room as parts of horizontal planes. The walls of a
room are parts of vertical planes.
vertical planes horizontal planes
D A point can be an element of a plane.
A B In the diagram, point A is an element of plane P. We can write
C
P A (P). Similarly, B (P), C (P), D (P), and E (P).
E
plane P: (P)
a. Coplanar Points
Definition coplanar points
Points that are in the same plane are called coplanar points.
In the figure, points A, B, and C are all in the plane P. They B
A
are coplanar points. Points K, L, and M are also coplanar C
P
points. A, K, and M are not coplanar points, because they K M
do not lie in the same plane. L Q
Geometric Concepts 15
b. Coplanar Lines
Definition coplanar lines
Lines that are in the same plane are called coplanar lines.
m n
For example, in the figure, the lines m and n are both in
the plane P. They are coplanar lines. P
Theorem
For any three points, there is at least one plane that contains them. For any three non-
collinear points, there is exactly one plane that contains them.
In the figure, the plane P is determined by the C
A
noncollinear points A, B, and C. B
P
2. Space
Definition space
Space is the set of all points.
We have seen that lines and planes are defined by sets of points.
According to the definition of space, all lines and planes can be considered as subsets of space.
D. RELATION BETWEEN TWO LINES
1. Intersecting Lines
d l
Two lines that intersect each other in a plane are called
A
intersecting lines, or concurrent lines.
In the figure on the left, line d and line l intersect each P
other at point A. They are intersecting lines.
16 Geometriy 7
2. Parallel Lines
d
l
Two lines are parallel if they are in the same plane and
do not have a common point.
In the figure on the left, line d and line l are parallel d l and d Ç l = Æ
lines. We write d l to show that lines d and l are par-
allel.
3. Coincident Lines
Two lines are coincident if each one contains all the l
points of the other. d
In the figure on the left, line d and line l are coincident
d=l
lines. We write d = l to show that lines d and l are
coincident.
4. Skew Lines
l
Two lines are skew if they are non-coplanar and they do d
not intersect. F
In the figure on the left, E and F are two non-parallel
planes. Hence, lines d and l are in different planes, and
E
since they do not intersect, they are skew lines.
EXAMPLE 4 In the figure there are three intersecting lines. Decide whether each statement is true or
false.
m C d
a. point A is the intersection of l and d
l
b. point C is the intersection of d and l B
A
P
c. point B is the intersection of l and m
Solution a. True, since point A is the common point of l and d.
b. False, since point C is not a common point of d and l.
c. True, since point B is the common point of l and m.
Geometric Concepts 17
E. RELATION BETWEEN A LINE AND A PLANE
We have seen the different possibilities for the relation between two lines. Let us look at the
possible relations between a line and a plane.
1. The Intersection of a Line and a Plane
A line can intersect a plane at one point. d
In the diagram on the left, the line d intersects the plane
E at point A. A
E
d Ç (E) = {A}
2. Parallelism of a Line and a Plane
A line can be parallel to a plane. d
In the diagram on the left, there is no common point
between line d and plane E. They are parallel. E
d Ç (E) = Æ
3. A Line Lies in a Plane
If at least two points of a line lie in a plane, then the line d
lies in the plane. We write d (E) to show that line d lies B
in plane E. E A
A, B Î d
In the diagram, points A and B are in plane E, so the line d Î (E)
A, B Î (E)
AB lies in the plane E.
F. RELATION BETWEEN TWO PLANES
1. Parallel Planes
If two planes have no common point, they are called
parallel planes. We write (A) (B) to show that two
planes are parallel. The opposite walls of a room are an
example of parallel planes. E
P Q
F
(P) (Q) (E) (F)
18 Geometriy 7
2. Intersecting Planes
If two planes have only one common line, they are called d
B
intersecting planes.
E A
F
(E) Ç (F) = d
3. Coincident Planes
A
If two planes have three noncollinear points in common,
they are called coincident planes. (P) and (Q) in the C
B
Q
figure are coincident planes. We write (P) = (Q) to show P
that planes P and Q are coincident. A, B, C Î (P)
(P) = (Q)
A, B, C Î (Q)
l
4. Half Planes half plane half plane
(E1) (E2)
A line in a plane separates the plane into two disjoint
E
regions that are called half planes. (E1) and (E2) in the
figure are half planes of (E). boundary of two half planes
(E1) Ç (E2) = Æ
(E1) È (E2) È l = (E)
Geometric Concepts 19
EXERCISES 1 .1
1. Explain why the concepts of point, line, and plane 10. Name all the lines, rays, line segments, and half
cannot be defined in geometry. lines in the given figure.
2. Draw five points on a piece of paper, and make
A B C D
sure that no three are of them collinear. Draw all
H
the lines passing through these points. How many
L
lines can you draw?
E F G
3. Explain the difference between a ray and a half
line.
11. Write the meaning of the following.
4. At least how many points determine a line?
a. [CD] b. [PQ[ c. ]AB[ d. [KL e. ]MN f. EF
5. At least how many noncollinear points determine
a plane? Why? 12. Describe the intersection of the line and the
6. Give examples from daily life to illustrate the plane in each figure.
concepts of point, line, and plane. a. b. m c.
n
7. Write words to complete the sentences. A
C
B
E l F G
a. A point has no __________ and no __________.
b. Two points determine a ___________ .
c. Three noncollinear points determine a _____ .
d. Two lines that lie in different planes and do 13. Write the coplanar points M N
P L
not intersect are called ___________ lines. in the given figure. R K
S
8. D
l
A B C 14. Draw a diagram to show that the intersection of
E two planes can be a line.
Determine whether the statements are true or
false for the given figure. 15. Draw a diagram to show that the intersection of
a. A, B, and C are collinear points three planes can be a point.
b. points D and E are not in the line l
16. Look at the figure. E
c. B l
d. E l a. How many planes are S
there?
e. C, D, and E are noncollinear DR C
T
9. How many different lines can pass through each b. Write the intersection
of the planes. Q
P
number of triwise noncollinear points? A B
a. 5 b. 7 c. 21 d. 101 c. How many lines pass through each point?
20 Geometriy 7
CHAPTER REVIEW TEST 1A
1. Which concept is precisely defined in geometry? 5. According to the figure, l d
B
which statement is A m
A) point B) line C) plane
false? C
D) space E) ____
A) l d = {C} B) l m = {A}
C) l d m = {A, B, C} D) m d = {B}
E) ____
2. A plane has no
A) thickness. B) length. C) width.
D) surface. E) ____
6. ABCD is a rectangle in a plane P. E is a point such
that E (P). How many planes are there that
include point E, with one or more of points A, B,
C, and D?
A) 7 B) 8 C) 9 D) 10 E) ?
3. A ray with an open endpoint is called
A) a line. B) a half line.
C) a line segment. D) an open line segment.
E) ____
7. How many lines do five points determine if no
three of the points are collinear?
A) 15 B) 12 C) 10 D) 9 E) ?
4. According to the figure,
l D d
which statement is true? C
B E
A 8. Space is
A) A, B and E are collinear points A) the intersection of two planes.
B) l d = {B} B) the set of all points.
C) C l C) a subset of a plane.
D) D, B, and E are noncollinear points D) a very large plane.
E) _ E) ?
Chapter Review Test 1A 21
9. According to the figure, l 12. Let [AB] + [BC] = [AC], and [MN] + [NK] = [MK].
which statement is Which points are between two other points?
false? d
C A) A and M B) B and M
A
A) C (E) E B C) C and K D) B and N
B) l (E) = {A} E) ____
C) l d = {B}
D) l and d are skew lines
E) ?
10. Which figure shows ]AB?
A) B)
A B A B
C) D)
A B A B
E) ____
11. According to the figure, m
which statement is false?
l
A) (P) l = l
B) (P) m = m A
n
C) (P) n = n
D) l m n = {A}
P
E) ?
22 Geometriy 7
Objectives
After studying this section you will be able to:
1. Define the concept of angle and the regions an angle forms.
2. Measure angles.
3. Classify angles with respect to their measures.
4. Classify angles with respect to their positions.
5. Classify angles with respect to the sum of their measures.
A. REGIONS OF AN ANGLE
1. Angle
One of the basic figures in geometry is the angle.
A television antenna is a physical model of an angle.
Changing the length of the antenna does not change the
angle. However, moving the two antennae closer
together or further apart changes the angle.
Definition angle
An angle is the union of two rays that have a common endpoint. The rays are called the sides
of the angle. The common endpoint is called the vertex of the angle.
Look at the diagram. [BA and [BC are the sides of the
A
angle. The vertex is the common endpoint B.
e
vertex
sid
The symbol for an angle is . We name the angle in the
side
diagram ABC, or CBA, and say ‘angle ABC’, or ‘angle B C
CBA’. We can also name angles with numbers or
lower-case letters, or just by their vertex.
Note
In three-letter angle names the letter in the middle must always be the vertex.
24 Geometriy 7
EXAMPLE 1 Name the angles in the diagrams. a.
A
b. c. d.
Solution a. AOB or BOA b. A a
O A 1
c. 1 d. a B
Definition interior and exterior region of an angle
The region that lies between the sides (arms) of an angle is called the interior region of the angle.
The region that lies outside an angle is called the exterior region of the angle.
EXAMPLE 2 Answer the questions for the angle ABC on the right.
A
a. Which points are in the interior region of the angle? G
D
b. Which points lie on the angle? B
E H
C
c. Which points are in the exterior region of the angle? F
Solution a. The points D and E are in the interior region of the angle.
b. The points A, B, C, and H lie on the angle.
c. The points G and F are in the exterior region of the angle.
B. MEASURING ANGLES
Angles are measured by an amount of B
rotation. We measure this rotation in units
360°
Babylonian astronomers chose called degrees. One full circle of rotation is
the number 360 to represent one
360 degrees. We write it as 360°.
full rotation of a ray back on to
A
itself. We can show the size of an angle on a diagram
Why this number was chosen? using a curved line between the two rays at the
O 45°
It is because 360 is close to the vertex, with a number. When we write the size
number of days in a year and it is of an angle, we write a lowercase m in front of B
divisible by 2, 3, 4, 5, 6, 8, 9, 10,
the angle symbol.
12, and many other numbers.
For example, mAOB = 45° means that angle AOB measures 45° degrees.
Look at some more examples of angle measures in the diagrams.
10° 90°
150° E
A B
30°
C D 360°
mÐA = 10° mÐB = 30° mÐC = 90° mÐD = 150° mÐE = 360°
Angles 25
Notice that the symbol for a 90° angle is a small square at the vertex. A 90° angle
60°
70°
80° 90° 100°
110°
120°
is also called a right angle in geometry.
50° 130°
40° 110°100°
120°
130°
90° 80° 70°
60°
50°
140° It is important to read angles carefully in geometry problems. For example, an
30°
angle in a problem might look like a right angle (90°). However, if it is not labelled
140° 40° 150°
20° 150° 30°
160°
160° 20°
10°
0°
170°
180°
10°
170°
as a right angle, it may be a different size. We can only use the given information
0° 180°
in a problem. We calculate other information using the theorems in geometry.
0 1 2 3 4 5 6 7 8 9 10
Definition protractor
The geometric tool we use to measure angles on paper is called a protractor.
A protractor has a semi-circular shape and a scale with units from 0 to 180.
To measure angles with a protractor, follow the three steps below.
80° 90° 100° 80° 90° 100°
70° 110° 70° 110°
60° 120° 60° 120°
50° 130° 50° 130°
90° 80° 70° 90° 80° 70°
40° 110°100° 40° 110°100°
120° 60° 140° 120° 60° 140°
130° 50° 130° 50°
30° 150° 30° 150°
140° 40° 140° 40°
20° 150° 30° 20° 150° 30°
160° 160°
160° 20° 160° 20°
10° 170° 10° 170°
170° 10° 170° 10°
0° 180° 0° 180° 0° 180° 0° 180°
0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10
1. Place the central hole (dot) of the protractor on the vertex of the angle.
2. Place the zero measure on the protractor along one side of the angle.
3. Read the measure of the angle where the other side of the angle crosses the protractor’s scale.
Notice that there are two semicircular scales of numbers on the protractor. If the angle
measure is smaller than 90° then we read the angle using the scale with the smaller number.
If the angle measure is greater than 90° then we use the scale with the larger number.
EXAMPLE 3 Read the protractor to find the measure of
D
each angle. C
80° 90° 100°
a. mAOB b. mAOC c. mAOD E 70° 110°
60° 120°
50° 130°
d. mAOE e. mAOF f. mBOC 40° 110°100°
120°
90° 80° 70°
60°
50°
140°
B
130°
30° 150°
140° 40°
g. mCOF h. mDOE 20° 150° 30°
160°
F 10°
160° 20°
170°
170° 10° A
0° 180° 0° 180°
O
0 1 2 3 4 5 6 7 8 9 10
26 Geometriy 7
Solution a. mAOB = 22° b. mAOC = 68°
c. mAOD = 90° d. mAOE = 140°
e. mAOF = 175° f. mBOC = mAOC – mAOB = 68° – 22° = 46°
g. mCOF = mAOF – mAOC = 175° – 68° = 107°
h. mDOE = mAOE – mAOD = 140° – 90° = 50°
A B For example, let us use a protractor to draw an angle of 56°.
1. Draw a ray.
C
A B 2. Place the centre point of the protractor on the endpoint (A) of the ray.
Align the ray with the base line of the protractor.
C 3. Locate 56° on the protractor scale. Make a dot at that point and label it as C.
mÐBAC = 56°
56°
A
B 4. Remove the protractor and draw [AC.
After learning to how use a protractor we can easily draw and measure angles.
Check Yourself 1
1. Name the following angles.
a. A b. c. d.
3
b
O B A
2. Find the following sets of points in the figure.
a. L {X} K Z
X
b. int L {X} T L
M S Y
c. ext L L
d. int L ext L
e. int L {S}
f. L {T, S, }
g. int L {Z, Y}
h. ext L {Z, Y}
Angles 27
3. Measure each angle using a protractor.
a. b. c. d.
A B C D
4. Draw the angles.
a. 45° b. 83° c. 174° d. 180° e. 225°
Answers
1. a. AOB b. A c. 3 d. b
2. a. b. {x} c. d. e. f. {S, K} g. h. {Z, Y}
C. TYPES OF ANGLE WITH RESPECT TO THEIR MEASURES
We can classify angles according to their measures.
1. Acute Angle
An angle that measures less than 90° is called an acute
C
angle. 25° 45° 80°
The angles on the left are all examples of acute angles A
B
because they measure less than 90°.
2. Right Angle
An angle that measures exactly 90° is called a right angle.
O
The angles on the left are all examples of right angles N
because they measure exactly 90°. We use a special
square symbol at the vertex to show a right angle. M
3. Obtuse Angle
An angle that measures between 90° and 180° is called Y
an obtuse angle. 120° 165°
91°
The angles on the left are all obtuse angles. X
O
4. Straight Angle
An angle that measures exactly 180° is called a straight 180°
angle. In the diagram, A is a straight angle. A
28 Geometriy 7
5. Complete Angle
An angle that measures exactly 360° is called a complete E
angle. In the diagram, E is a complete angle. 360°
EXAMPLE 4 Classify the angles according to their a.
180°
b.
360°
c.
measure.
125°
Solution a. 180° is a straight angle.
d. e.
b. 360° is a complete angle.
35°
c. 125° is between 90° and 180°, so
it is an obtuse angle.
d. 90° is a right angle.
e. 35° is less than 90°, so it is an acute angle.
D. TYPES OF ANGLE WITH RESPECT TO THEIR POSITION
1. Adjacent Angles
Definition adjacent angles
Adjacent angles are two angles in the same plane that have a common vertex and a common
side, but do not have any interior points in common.
In the diagram, the angles AOC and BOC have a B
common vertex and a common side ([OC) with C
non-intersecting interior regions.
Therefore, AOC and BOC are adjacent angles. O A
Theorem
If two angles are vertical then they are also congruent, i.e. they have equal measures.
EXAMPLE 5 Determine whether the pairs of angles are vertical or not, l
using the figure. b
c a
a. a, b b. a, c c. d, a d. b, d d
k
Solution The lines l and k intersect at one point. Therefore,
a. a and b are not vertical angles,
b. a and c are vertical angles, because they are in opposite directions,
c. d and a are not vertical angles, and
d. b and d are vertical angles.
Angles 29
E. TYPES OF ANGLE WITH RESPECT TO THE SUM OF
THEIR MEASURES
1. Complementary Angles
Definition complementary angles
If the sum of the measures of two angles is 90°, then the
A
angles are called complementary angles. B C
Each angle is called the complement of the other angle.
For example, in the diagram opposite, ANB and CMD 30°
are complementary angles, because the sum of their 60°
measures is 90°: N M D
mANB + mCMD = 30° + 60° = 90°.
2. Supplementary Angles
Definition supplementary angles
If the sum of the measures of two angles is 180°, then the angles are called supplementary
angles. Each angle is called the supplement of the other angle.
In the diagram, XYZ and MNO are supplementary M
angles because the sum of their measures is 180°: X
40° 140°
mXYZ + mMNO = 40° + 140° = 180°.
Z Y N O
EXAMPLE 6 Find x if the given angles are a. b.
complementary.
3x+30
Solution a. If x and 2x are 2x
x 2x+10
complementary, then
x + 2x = 90°.
Therefore, x = 30°.
b. 2x + 10° + 3x + 30° = 90°
2x + 3x + 10° + 30° = 90°
5x = 50°
x = 10°
30 Geometriy 7
EXAMPLE 7 Find x if the given angles are
a. b.
supplementary.
Solution a. If 2x and 4x are supplemen- 4x 3x+50°
2x 2x+60°
tary, then
2x + 4x = 180°.
Therefore, x = 30°.
b 2x + 60° + 3x + 50° = 180°
2x + 3x + 60° + 50° = 180°
5x + 110° = 180°
5x = 70°
x = 14°
Check Yourself 2
1. Find x if the given angles are complementary.
a. b. c.
15°
3x+
°
30
2x+20° 2x+
4x
x x 20°
2. Find x if the given angles are supplementary.
a. b. c.
6x 4x+40° 5x 12°
3x 2x 10
2x
18°
Answers
1. a. 18° b. 30° c. 9°
2. a. 20° b. 25° c. 30°
Angles 31
Objectives
After studying this section you will be able to:
1. Identify corresponding angles, alternate interior angles, and alternate exterior angles.
2. Identify interior angles on the same side of a transversal.
3. Describe the properties of angles with parallel sides.
4. Define an angle bisector.
A. CORRESPONDING ANGLES AND ALTERNATE ANGLES
Definition supplementary angles
Let m and n be two lines in a plane. A third line l that intersects each of m and n at different
points is called a transversal of m and n.
In the diagram, line AB is a transversal of m and n.
Let us look at the types of angle formed in the figure of two parallel lines with a transversal.
Remember the notation for parallel lines: m n means that m is parallel to n.
1. Corresponding Angles
Definition corresponding angles
In a figure of two parallel lines with a transversal, the m//n l
angles in the same position at each intersection are 2 1
called corresponding angles. m
3 4
In the diagram, 1 and 5 are corresponding angles.
6 5
Also, the angle pairs 2 and 6, 3 and 7, and 4 and n
7
8 are corresponding angles. 8
Property
Corresponding angles are congruent.
l
m//n
Therefore, in the diagram, m1 = m5, 2 1
m
m2 = m6, 3 4
m3 = m7, and
m4 = m8. 6 5
n
7
8
32 Geometriy 7
2. Alternate Interior Angles
Definition alternate interior angles
In a figure of two parallel lines with a transversal, the interior angles between the parallel
lines on opposite sides of the transversal are called alternate interior angles.
l
m//n 2 1
m
In the diagram, the angles 4 and 6 are alternate 3 4
interior angles. Also, 3 and 5 are alternate interior
angles. 6 5
n
7
8
Property
Alternate interior angles are congruent. m//n l
Therefore, in the diagram, m
3 4
m4 = m6, and m3 = m5.
6 5
n
3. Alternate Exterior Angles
Definition alternate exterior angles
In a figure of two parallel lines with a transversal, the angles outside the parallel lines on
opposite sides of the transversal called alternate exterior angles.
l
m//n
2 1
In the diagram, the angles 1 and 7 are alternate m
3
exterior angles. Also, 2 and 8 are alternate exterior 4
angles. 6 5
n
7
8
Property
Alternate exterior angles are congruent. m//n l
Therefore, in the diagram, 2 1
m
m1 = m7, and m2 = m8.
n
7
8
Angles 33
4. Interior Angles on the Same Side of a Transversal
Definition alternate interior angles
In a figure of two parallel lines intersected by a m//n l
transversal, interior angles on the same side of the trans- m
versal are supplementary. x
Therefore, in the diagram, mx + my = 180°.
y
n
A B
EXAMPLE 8 In the digaram, [BA DE. 100°
C
Find mBCD.
30°
D E
Solution A B A B B
100° 100°
F x F
C 80°
y C
30°
C
30° 30°
D E D E D
If [BA [CF then ABC EDC and y are mBCD = mx + my
and x are supplementary. alternate interior = 80° + 30°
mABC + mx = 180° angles. = 110°
100° + mx = 180° mEDC = my
mx = 80° my = 30°
5. Angles with Parallel Sides
Theorem
The measures of two angles with parallel sides in the same direction are equal.
Proof Consider the diagram on the right. [OA [LK K
1. AOB and LTB are corresponding angles. [OB [LM
A
AOB LTB
S L
2. LTB KLM (corresponding angles) M
AOB KLM
O T B
mAOB = mKLM
34 Geometriy 7
EXAMPLE 9 In the figure, [AC [DF, [AB [DE, mCAB = 2x + 40°, C F
and mFDE = 6x – 20°. Find mCAB. 6x 20°
D
E
Solution mCAB = mFDE
2x + 40°
2x + 40° = 6x – 20° A B
40° + 20° = 6x – 2x
60° = 4x
15° = x
So mCAB = 70°.
Theorem
The measures of two angles with parallel sides in opposite directions are equal.
Proof 1. KLM BPL (corresponding angles) [OA // [LM A
[OB // [LK
2. AOB BPL AOB KLM
P
O
B
K
L R
Property
In the figure, if d k and B is the intersection of [BA and
[BC, then A
d
a
mb = ma + mc.
b B
c
k
C
EXAMPLE 10 In the figure, [AE [BF, mA = 40°, and mB = 30°. A
40°
E
Find mAOB.
Solution mAOB = mOAE + mOBF ? O
mAOB = 40° + 30° 30°
B
F
= 70°
Angles 35
Property
In the figure, if d k and B is the intersection of [BA and A
d
[BC, then a
ma + mb + mc = 360°. b B
c
k
C
EXAMPLE 11 In the figure, AB CD, mAEF = 5x, mEFH = 3x, and A E B
mFHD = 2x. Find x. 5x
3x F
Solution mFHC + mFHD = 180° (supplementary angles)
2x
mFHC = 180° – 2x C H D
mAEF + mEFH + mFHC = 360°
5x + 3x + 180° – 2x = 360°
6x + 180° = 360°
6x = 180°
x= 30°
Property
In a figure such as the figure opposite, the sum of the A E B
a
measures of the angles in one direction is equal to the
x F
sum of the measures of the angles in the other direction. H b
mx + my + mz = ma + mb + mc + md y G
K c
z L
d
C M D
AB CD
EXAMPLE 12 In the figure, [AE [DF, mEAB = 35°, mBCD = 25°, A
35°
E
and mABC = 4 mCDF. Find mABC.
4x B
Solution 35° + 25° = 4x + x C 25°
60° = 5x x
D
F
x = 12°
Therefore, mABC = 48°.
36 Geometriy 7
Two lines are called perpendicular lines if they intersect
A
at right angles. We write AB CD to show that two lines
AB and CD are perpendicular. C D
O
Property
In the diagram, if [OA [LK and [OB [LB then
K A
mAOB + mNLB = 180°.
N
L
O B
EXAMPLE 13 In the figure, [BA [FD, [BC [FE, [BA [FG, A
mABC = 60°, and mGFE = x. Find x.
D G
Solution mABC + mDFE = 180° F x
60° + mDFE = 180°
60°
mDFE = 120°
mGFD = 90° B E C
mGFD + mGFE + mDFE = 360°
90° + mGFE + 120° = 360°
90° + x + 120° = 360°
x + 210° = 360°
x = 150°
EXAMPLE 14 In the figure, [AC [ED, mABE = 23°, A C
and mBED = 118°, Find mBAC. D
E
23° 118°
Solution Let us draw a line [BF parallel to [ED.
mDEB + mEBF = 180° (interior angles on the same B
118° + mEBF = 180° side of a transversal)
mEBF = 62° A C
mBAC + mABF = 180°
E D
mBAC + 85° = 180°
mBAC = 95° 23° 118°
62° F
B
Angles 37
EXAMPLE 15 In the figure, [AD [CE, mDAB = 112°, C E
and mBCE = 120°. D A 120°
Find mABC. 112°
?
Solution Let us draw a line [BK parallel to [AD.
C E
mDAB + mABK = 180°
D A 120°
112° + mABK = 180°
112°
mABK = 68° ?
68°
mBCE = mABC + mABK K B
120° = mABC + 68° mABC = 52°
6. Bisector of an Angle
Definition angle bisector
A ray that divides an angle into two congruent angles is called the bisector of the angle.
In the figure, [OB is the bisector of AOC: A
mAOB = mBOC B
a
1 O a
= AOC.
2 C
EXAMPLE 16 In the figure, [BE and [BF are the bisectors of ABD and D
CBD respectively. Find mEBF. E
F
Solution mABD + mCBD = 180°
mABD m CBD A B C
mEBF = + = 90°
2 2
Definition
The bisectors of two adjacent supplementary angles are perpendicular to each other.
38 Geometriy 7
EXAMPLE 17 In the figure, [OE is the bisector of FOD. G
c
mBAD = a, mEBC = b, and mFCG = c. C F
Show that a + c = 2 b. E
b
B
a
O A
D
Solution Let [OH AG.
H
mHOC = mGCF = c G
c
mHOB = mCBE = b C F
mHOD = mBAD = a E
Let mCOB = mBOA = x. b
c B
c+x=b x=b–c b+x=a a
b+b–c=a O A
D
Therefore, a + c = 2 b.
Angles 39
EXERCISES 2
1. Using the given fig- 4. Draw the angles.
O P
ure, find each set of
M a. 20° b. 35° c. 75° d. 120°
points. K
N e. 175° f. 210° g. 240° h. 330°
a. O {P}
b. O {N} c. O {K, O, M}
5. Classify the types of angle.
d. int O {P} e. int O {N}
a. b. c.
f. int O {K, O, M} g. ext O {N}
135°
45°
h. ext O {P} i. O int O
j. O ext O k. int O ext O O d. e.
360°
2. Find each set of B
E
points for the given C
A 6. Find x in each figure if the angles are
D F
figure. complementary.
H
G
a. ABC ACD a. b. c.
2x+25°
3x
b. int ABC CAD
°
4x+30°
15
35
x+
°
3x5°
c. ABC int CAD x
d. ext ABC CAD
e. ABC ext CAD 7. Find x in each figure if the angles are
supplementary.
a. b. c.
3. Measure the angles using a protractor.
3x 2x+20° 3x+15° 4x+25° 125° + 2x
45° x
a. b. c.
8. In the figure, m n, l
d. e. f. l is a transversal and 3 2
m
m7 = 115°. 4 1
Find the measures.
a. m1 b. m2 7 6
n
c. m3 d. m4 8
5
g. h. i.
e. m5 f. m6
g. m8
40 Geometriy 7
9. Given [BA [DE, find A B 16. In the figure, B A
30° x
mx. [BA [EF,
C
x mBCD = 100°,
100°
C E F
80° mCDE = 25°, and
D 105°
E mFED = 105°. 25°
10. In the figure, Find mABC.
A D D
[BA [ED, F K 4x70°
d
[BC [EF, E 30°
mABC = 3x – 30°, and 3x
3x30°
17. In the figure, d l.
mDEF = 4x – 70°. B C 40°
Find x. x
Find x.
2x
l d l
11. In the figure,
d l. 115° C
18. In the figure,
Find mx. x
[BA [DE, A B
[BC [DF and
d [BC [BD], and
12. In the figure, E F
50° mGDE = 40°.
x l 40° D
d l.
100° Find mABC. G
Find mx.
19. In the figure, A
13. In the figure, A B [BC [DF, G
[BA [DE, x
E
[BA [DG, and
D
mBCD = 40°, and [ED] is the angle B 140° D
40° 120°
bisector of mGDF. E
mCDE = 120°. C
F
Find mABC. C Find mABC.
A B
14. In the figure, E
2x
[BA [FG, G 120° 20 . In the figure, 60°
E
mEFG = 120°, and F x D [AB [CD.
mABC = 130°. A B Find mAEC. 2x
4x
Find mx. 130°
C D
C
D A B
15. In the figure, F
[BA [ED] and 21 . In the figure,
z
[CD] [EF. AB CD. E
F 70°
Find the relation
E B y Find mBFC.
x
between x, y, and z. A
C
D C
Angles 41
CHAPTER REVIEW TEST 2
1. The complement of an angle x is 10° more than 7. In the figure, K A m
three times mx. Find the measure of the bigger m n, 130°
x B
angle. mKAB = 130°, and
L
mLCD = 40°. n
A) 50° B) 60° C) 70° D) 80° 40° C
Find mABC. D
A) 100° B) 90° C) 80° D) 70°
2. The sum of the measures of the supplementary
and complementary angles of an angle x is 250°.
8. In the figure, d
Find mx. 3x
d l. Find x.
6x10°
A) 10° B) 20° C) 30° D) 40°
4x20°
l
3. What is the measure of the angle between the A) 40° B) 30° C) 20° D) 10°
bisectors of two adjacent supplementary angles?
9. In the figure, A x B
A) 45° B) 60° C) 75° D) 90° [AB] [BE.
70° C
Find mx. D 60°
4. In the figure, [OA [OB,
B 130° E
mAOC = a, C
mCOB = b, and b A) 30° B) 40° C) 50° D) 60°
a
a 2 . Find b. O
= A
10. In the figure,
b 3 D
A) 30° B) 36° C) 54° D) 60° [BC [DE and x A
[BA [DF. F
What is the relation
y
5. The ratio of two complementary angles is 2 . Find between mx and B
C
7 my? E
the measure of the supplementary angle
of the smaller angle. A) mx + my = 90° B) mx + my = 180°
C) mx = my D) mx – my = 30°
A) 170° B) 160° C) 150° D) 110°
11. In the figure, x
A
6. In the figure, l k. k
[BC is the angle
85°
115° bisector of ABC.
Find mx. l
x 35° Find mx. B 125°
C
A) 110° B) 100° C) 90° D) 80° A) 65° B) 55° C) 50° D) 45°
42 Geometriy 7
Objectives
After studying this section you will be able to:
1. Define a triangle.
2. Name the elements of a triangle.
3. Describe the types of triangle accordin to sides.
4. Describe the types of triangle according to angles.
A. THE TRIANGLE AND ITS ELEMENTS
The roofs of many buildings have a triangular cross-section. A triangle makes a simple musi-
cal instrument, and many traffic signs have a triangular shape. These are just some exam-
ples of how triangles are used in the world around us.
Activity Making a Poster - Triangles
Challenge!
Without lifting your pencil, Make a poster to show how triangles are used in everyday life. You can take photographs,
join the following four
points with three segments make drawings or collect pictures from magazines or newspapers to show buildings,
to form a closed figure. designs, signs and artwork which use triangles.
In this section we will consider the main features of triangles and how we can use them to
solve numerical problems.
1. Definition
The word triangle means ‘three angles’. Every triangle has three angles and three sides.
Definition triangle, vertex, side
A triangle is a plane figure which is formed by three line segments joining three noncollinear
points. Each of the three points is called a vertex of the triangle. The segments are called the
sides of the triangle.
A
We name a triangle with the symbol
followed by three capital letters, each
The plural of vertex is corresponding to a vertex of the triangle. We
vertices.
can give the letters in any order, moving
clockwise or counterclockwise around the
B C
triangle.
44 Geometriy 7
For example, we can refer to the triangle shown at the right as ABC. We can also call it BCA,
CAB, ACB, BAC or CBA. The vertices of ABC are the points A, B and C. The sides of ABC
are the segments AB, BC and CA.
Definition interior and exterior angles of a triangle
In a triangle ABC, the angles BAC, ABC and
ACB are called the interior angles of the
A¢
Notice that a triangle is triangle. They are written as A, B and C,
A
defined as the union of
three line segments. Since
respectively. The adjacent supplementary
an angle lies between two angles of these interior angles are called the B¢ B C
rays (not two line segments),
a triangle has no angles by exterior angles of the triangle. They are
C¢
this definition. However, written as A, B and C, respectively.
we can talk about the
angles of a triangle by
assuming the existence
of rays: for example, the
We can refer to the sides of a triangle ABC by
A
rays AB and AC form using the line segments AB, BC and AC, or by
angle A of a triangle ABC.
A
using the lower-case form of the vertex
c b
opposite each side.
a
For instance, in ABC at the right:
a is the side opposite vertex A, B a C
B C
b is the side opposite vertex B, and
c is the side opposite vertex C.
EXAMPLE 1 Look at the figure.
N
a. Name all the triangles in the figure.
b. Name all the interior angles of MNE.
c. Name all the vertices of NEP.
d. Name all the sides of MNP. M E P
e. Name all the exterior angles of ENM.
Solution a. MNE, NEP and MNP d. segment MP, segment PN and segment
b. M (or NME), MNE and MEN. NM
c. points N, E and P e. E, N and M
Triangles and Construction 45
Definition perimeter of a triangle
The sum of the lengths of the three sides of a triangle is called the perimeter of the triangle.
We write P(ABC) to mean the perimeter of a triangle ABC.
For instance, the perimeter of the triangle A
Challenge!
Move exactly three ABC in the figure is
toothpicks in the following P(ABC) = BC + CA + AB = a + b + c. b
arrangement to make five c
triangles.
B a C
EXAMPLE 2 In the figure, P(ABC) = P(DEF). Find x. A D
x+2 10 16 14
Solution P(ABC) = P(DEF)
x + 2 + x +10 = 16 + 14 + x (given)
B x C E x F
2x + 12 = x + 30
x = 18
Check Yourself 1
1. Three distinct points K, M and N lie on a line m, and a fourth point T is not on the line
m. Point T is joined to each of the other points. Find how many triangles are formed and
name each one. D F C
2. Find and name all the triangles in the figure at the right.
G K L
A E B
3. Polygon ABCDE is a regular polygon and its diagonals are D
shown in the figure. Name
Triangles in the world N M
around us
a. all the triangles whose three vertices lie on the polygon. E C
b. all the triangles which have exactly one vertex on the P L
polygon. K
A regular polygon is a
c. all the triangles which have two sides on the polygon.
polygon in which all sides A B
have the same length and d. the sides of all the triangles which do not have a side on the polygon.
all angles are equal.
4. The side AC of a triangle ABC measures 12.6 cm, and the two non-congruent sides AB and
BC are each 1 cm longer or shorter than AC. Find P(ABC).
46 Geometriy 7
5. Point X is on the side KN of a triangle KMN. Find the length
of MX if the perimeters of the triangles KXM, XMN and KMN
are 24, 18, and 30, respectively.
Answers
1. Three triangles are formed: KMT, MNT and TKN.
2. AEL, LEB, LBC, AKL, AGK, ALB, ABC, AFC,
ADF, AGL, ADC The picture shows the ‘food
triangle’ of different types of food.
3. a. ABC, BCD, CDE, DEA, EAB Can you see what the different
regions mean?
b. BKL, CLM, DMN, ENP, APK
c. ABC, BCD, CDE, DEA, EAB
d. sides of BKL: BK, KL, BL; sides of CLM: CL, ML, CM; sides of DMN: DM, MN, DN;
sides of ENP: EN, NP, EP; sides of APK: AP, KP, AK
4. 37.8 cm 5. 6
2. Regions of a Triangle
Any given triangle ABC separates the plane which contains it into three distinct regions:
1. The points on the sides of the triangle form the triangle itself.
2. The set of points which lie inside the triangle form the interior of the triangle, denoted
int ABC.
3. The set of points which lie outside the
triangle form the exterior of the triangle, A exterior
denoted ext ABC.
interior
The union of a triangle with its interior and
B C
exterior region forms a plane. In the figure E
opposite, the plane is called E. We can write
E = int ABC ABC ext ABC.
EXAMPLE 3 Write whether each statement is true or false
according to the figure opposite. D B
A
a. Point T is in the interior of DFE. K
M
b. M ext BDE F E
T
c. ADF BED = P C
d. ext FDE int FCE = FCE
e. Points T and K are in the exterior of DFE.
Solution a. false b. true c. false d. false e. true
Triangles and Construction 47
Check Yourself 2
A
Answer according to the figure.
L
a. Name five points which are on the triangle. J
N S
b. Name three points which are not on the triangle.
c. Name two points which are in the exterior of the triangle. B T C
d. What is the intersection of the line ST and the triangle ABC?
e. What is the intersection of the segment NS and the exterior of the triangle ABC?
A physical model of a
triangle with its interior Answers
region a. points A, B, C, T and S b. points J, L and N c. points J and L d. points S and T
e.
3. Auxiliary Elements of a Triangle
Three special line segments in a triangle can often help us to solve triangle problems. These
segments are the median, the altitude and the bisector of a triangle.
a. Median
Definition median
In a triangle, a line segment whose endpoints are a vertex and the midpoint of the side
opposite the vertex is called a median of the triangle.
In the figure, the median to side BC is the A
line segment AD. It includes the vertex A and
the midpoint of BC.
Auxiliary elements are Va
extra or additional
elements.
B D C
We usually use the capital letter V to indicate A
the length of a median. Accordingly, the
lengths of the medians from the vertices of a Va
triangle ABC to each side a, b and c are Vb Vc
written as Va, Vb and Vc, respectively. As we
can see, every triangle has three medians. B D C
48 Geometriy 7
EXAMPLE 4 Name the median indicated in each triangle and indicate its length.
a. b. c.
K S P
D E
L M T V
R F N
Solution a. median MD, length Vm
b. median TE, length Vt
c. median PF, length Vp
Activity Paper Folding - Medians
1. Follow the steps to construct the median of a triangle by paper folding.
Take a triangular piece of paper Fold the paper again from the DM is the median of EF.
and fold one vertex to another midpoint to the opposite vertex.
vertex. This locates the
midpoint of a side.
2. Cut out three different triangles. Fold the triangles carefully to construct the three
medians of each triangle. Do you notice anything about how the medians of a triangle
intersect each other?
Definition centroid of a triangle
The medians of a triangle are concurrent. Their common point is called the centroid of the
triangle.
Triangles and Construction 49
The centroid of a triangle is the center of gravity of the triangle. In other words, a triangular
model of uniform thickness and density will balance on a support placed at the centroid of
Concurrent lines are lines
which all pass through a
the triangle. The two figures below show a triangular model which balances on the tip of a
common point. pencil placed at its centroid.
b. Angle bisector
Definition triangle angle bisector
An angle bisector of a triangle is a line segment which bisects an angle of the triangle and
which has an endpoint on the side opposite the angle.
In the figure, AN is the angle bisector which A
divides BAC into two congruent parts.
a a
We call this the bisector of angle A
because it extends from the vertex A. nA
Since AN is an angle bisector, we can write
m(BAN) = m(NAC).
B N C
We usually use the letter n to indicate the A
length of an angle bisector in a triangle.
Hence the lengths of the angle bisectors of a nA
triangle ABC from vertices A, B and C are
written nA, nB and nC, respectively. As we can nB nC
see, every triangle has three angle bisectors.
B N C
50 Geometriy 7
Activity Paper Folding - Angle Bisectors
Follow the steps to explore angle bisectors in a triangle.
1. Cut out three different triangles.
2. Fold the three angle bisectors of each triangle as shown below.
3. What can you say about the intersection of the angle bisectors in a triangle?
Folding the angle bisector of A. AN is the angle bisector of A. BM is the angle bisector of B.
Definition incenter of a triangle
The angle bisectors in a triangle are concurrent A
and their intersection point is called the
incenter of the triangle. The incenter of a L K
O
triangle is the center of the inscribed circle of
The inscribed circle of a
triangle is a circle which the triangle.
is tangent to all sides of
the triangle. B N C
O is the incenter of DABC
As an exercise, try drawing a circle centered at the incenter of each of your triangles from
the previous activity. Are your circles inscribed circles?
We have seen that nA, nB and nC are the bisectors of the interior angles of a triangle ABC. We
can call these bisectors interior angle bisectors. Additionally, the lengths of the bisectors of
the exterior angles A, B and C are
written as nA, nB and nC respectively. These
K
bisectors are called the exterior angle
bisectors of the triangle. nK¢
In the figure at the right, segment KN is the
exterior angle bisector of the angle K in
N M T
KMT and its length is nK.
Triangles and Construction 51
Definition excenter of a triangle
The bisectors of any two exterior angles of a
B
triangle are concurrent. Their intersection is
called an excenter of the triangle.
In the figure, ABC is a triangle and the bisectors A V C
of the exterior angles A and C intersect at
T S
the point O. So O is an excenter of ABC. In O
addition, O is the center of a circle which is
tangent to side AC of the triangle and the
extensions of sides AB and BC of the triangle.
An escribed circle of a This circle is called an escribed circle of
triangle is a circle which is
tangent to one side of the ABC.
triangle and the extensions
of the other two sides. As we can see, a triangle has three excenters and three corresponding escribed circles.
EXAMPLE 5 Find all the excenters of KMN in the figure K
by construction.
N
M
Solution To find the excenters, we first construct the
bisector of each exterior angle using the
E2
method we learned in Chapter 1. Then we
K
use a straightedge to extend the bisectors
until they intersect each other. E1
The intersection points E1, E2 and E3 are the
N
excenters of KMN. M
E3
52 Geometriy 7
c. Altitude
Definition altitude of a triangle
An altitude of a triangle is a perpendicular line segment from a vertex of the triangle to the
line containing the opposite side of the triangle.
In the figure, AH is the altitude to side A
BC because AH is perpendicular to BC.
ha
B H C
In a triangle, the length of an altitude is called a height of the triangle.
The heights from sides a, b and c of a triangle
A
ABC are usually written as ha, hb and hc,
respectively. As we can see, every triangle has
three altitudes. ha
hb hc
B H C
EXAMPLE 6 Name all the drawn altitudes of all the B
triangles in the figure.
A C
K
Solution There are eight triangles in the figure. Let us look at them one by one and name the drawn
altitudes in each.
Triangles and Construction 53
B B
B B
C A K
C A K K
A
K
D altitudes
altitude BK altitude AK altitudes BK and CK AK and BK
B
K
K
K C A
A C K C
D D
D D altitudes
altitude DK altitude CK altitudes CK and DK AK and DK
Activity Paper Folding - Altitudes
To fold an altitude, we fold a triangle so that a side matches up with itself and the fold
contains the vertex opposite the side.
Cut out three different triangles. Fold them carefully to construct the three altitudes of
each triangle. What can you say about how the altitudes intersect?
54 Geometriy 7
Definition orthocenter of a triangle
The altitudes of a triangle are concurrent. Their common point is called orthocenter of the
triangle.
Since the position of the altitudes of a A
triangle depends on the type of triangle, the
position of the orthocenter relative to the
K
triangle changes. In the figure opposite, the ha
hb hc
orthocenter K is in the interior region of the
triangle. Later in this chapter we will look
B H C
at two other possible positions for the K is the orthocenter of DABC
orthocenter.
Once we know how to draw an altitude of a triangle, we can use it to find the area of the
triangle.
Definition area of a triangle
The area of a triangle is half the product of the length of a side (called the base of the
triangle) and the height of the altitude drawn to that base. We write A(ABC) to mean the
area of ABC.
For example, the area of ABC in the figure A
BC AH a h
is A( ABC ) = = . Area is usually
2 2
h
expressed in terms of a square unit.
B H C
a
EXAMPLE 7 Find the area of each triangle.
a. b. 2 cm
D c. K
A T
12 cm
6 cm
4 cm 5 cm
B 3 cm H 7 cm C E F M 8 cm N
BC AH
Solution a. A( ABC ) = (Definition of the area of a triangle)
2
10 4
= (Substitute)
2
= 20 cm2 (Simplify)
Triangles and Construction 55
FT DE
b. A( DEF ) = (Definition of the area of a triangle)
2
5 14
= (Substitute)
2
= 35 cm2 (Simplify)
KM MN
c. A( KMN ) = (Definition of the area of a triangle)
2
6 8
= (Substitute)
2
= 24 cm2 (Simplify)
Definition perpendicular bisector of a triangle
In a triangle, a line that is perpendicular to a side of the triangle at its midpoint is called a
perpendicular bisector of the triangle.
A
In the figure, HN, DN and EN are the
The picture below hangs perpendicular bisectors of triangle ABC. E D
straight when the hook
lies on the perpendicular
Perpendicular bisectors in a triangle are N
bisector of the picture’s always concurrent. B C
top edge. H
Definition circumcenter of a triangle
The intersection point of the perpendicular bisectors of a triangle is called the circumcenter
of the triangle. The circumcenter of a triangle is the center of the circumscribed circle of the
triangle.
The circumscribed circle
of a triangle is a circle
which passes through all
the vertices of the triangle.
EXAMPLE 8 Find the circumcenter of each triangle by construction.
a. b. c.
56 Geometriy 7
Solution First we construct the perpendicular bisector of each side of the triangle. Their intersection
point is the circumcenter of the triangle.
a. b. c.
Activity Perpendicular Bisector of a Triangle
There are three main faculties on
a university campus. The university
wants to build a library on the
campus so that it is the same
distance from each faculty.
1. Make a geometric model of the
problem.
2. Find the location of the library in
the picture opposite.
As an exercise, draw three more triangles on a piece of paper and construct their
circumcenters. Check that each circumcenter is the center of the inscribed circle.
Check Yourself 3
1. Name the auxiliary element shown in each triangle using a letter (n, h or V) and a vertex
or side.
a. b. c. d. e. f.
A M X K N M P J
N S H M
B C N P Y W Z L M N K L
Triangles and Construction 57
2. In a triangle MNP, the altitude NT of side MP and the median MK of side NP intersect at
the point R.
a. Name all the triangles in the figure formed. b. Name two altitudes of MTN.
3. In a triangle DEF, EM is the median of side DF. If DE = 11.4, MF = 4.6 and the perimeter
of DEF is 27, find the length of side EF.
4. In a triangle KLM, LN is the altitude of the side KM. We draw the angle bisectors LE and
LF of angles KLN and MLN respectively. If the angles between the angle bisectors and the
altitude are 22° and 16° respectively, find m(KLM).
A
5. In the figure, A(ABH) = A(AHC). Find x.
10
8
x
4
B H C
6. Write one word or letter in each gap to D A
make true statements about the figures.
R
a. Point O is a(n) __________ . O
b. Segment ________ is a median.
E P F B C
c. Point _______ is an excenter.
Y
d. Segment ________ is an altitude. N
e. Point B is a(n) _____________. G
V
f. Segment ER is a(n) __________ X Z M
J
___________.
S T L
g. Point _________ is a circumcenter.
K
h. Line TM is a(n) __________
__________.
i. Point ________ is a centroid.
Answers
1. a. nB b. hp c. Vx d. Vl e. hn f. nL
2. a. MNK, MKP, MNT, NTP, MRT, MNR, RNK, MNP b. NT, TM
3. 6.4 4. 76° 5. 5
6. a. incenter b. ET c. K d. AB (or BC) e. orthocenter (or vertex) f. angle bisector
g. M h. perpendicular bisector i. G
58 Geometriy 7
B. TYPES OF TRIANGLE
Some triangles are given special names according to the lengths of their sides or the
measures of their angles.
1. Types of Triangle According to Sides
A triangle can be called scalene, isosceles or equilateral, depending on the lengths of its sides.
Definition scalene triangle
A triangle is called scalene if all of its sides A
have different lengths. In other words, a
scalene triangle has no congruent sides. c b
B a C
a ¹ b ¹ c, so DABC is a scalene triangle
Activity Euler Lines
The Euler line of a triangle is the line which passes through
the orthocenter, circumcenter and centroid of the triangle. centroid orthocenter
Draw a scalene triangle and find its Euler line using circumcenter
Euler line
a. a ruler and set square. b. a compass and straightedge. c. dynamic geometry software.
Dynamic geometry
software is a powerful tool
for studying geometric
concepts. Geometry
programs allow us to
change and manipulate
figures, so that we can
explore and experiment
with geometrical
concepts instead of just
memorizing them.
Which method was easier?
Definition isosceles triangle
A triangle is called isosceles if it has at least A
two congruent sides.
c b
B a C
b = c, so DABC is isosceles
Triangles and Construction 59
In an isosceles triangle, the congruent sides vertex
A
are called the legs of the triangle. The third angle
side is called the base of the triangle. legs
The two angles between the base and the c b
legs of the triangle are congruent. They are
base
called the base angles of the triangle. base angles
The angle opposite the base is called the B a C
vertex angle.
Activity Golden Triangles
The head of this knee
hammer forms an isosceles A golden triangle is a triangle in which the
triangle.
ratio of the length of the legs to the length 36°
of the base is the golden ratio. The angle
between the two legs of a golden triangle is golden
always 36°. triangle
72° 72° The sides of the Great Pyramid
of Giza are golden triangles.
To construct a golden triangle, first draw a
F
A a B b C square ABCD and mark the midpoint E of G
AB. Find the point F on the extension of AB B C
Line segments AB and
BC are in the golden by making EF = EC. Then find G by making
ratio if
AF = AG and BA = BG. Finally, draw FG E
a+ b a
= , and AG. Then BGF is a golden triangle.
a b
A D
1+ 5
= 1.6180339...
2 1. Construct a golden triangle using a straightedge and compass.
2. Repeat the construction using dynamic geometry software.
3. In both constructions, check the measures of the interior angles.
EXAMPLE 9 Segment EM is a median of an isosceles E
triangle DEF with base DF. Find the length of
EM if the perimeter of EMF is 65 and the
b
perimeter of DEF is 100. b
Solution Let us draw an appropriate figure.
D a M a F
In the figure opposite,
in DEM, a + b + x = 65, (1)
in DEF, 2(a + b) = 100. So a + b = 50. (2)
Substituting (2) into (1) gives us 50 + x = 65; x = 15. So EM = 15.
60 Geometriy 7
EXAMPLE 10 In KMN, K . Given that KN is 4 cm K
less than MN and MK is 2 cm more than
three times KN, find the perimeter of KMN. x
x4
Solution We begin by drawing the figure opposite.
If MK = x then KN = x – 4. Also, MK = MN
N x M
because K N.
Also, we are given MN = 3KN + 2
x = 3(x – 4) + 2
10 = 2x
5 = x.
Since P(KMN) = 3x – 4, P(KMN) = (3 5) – 4 = 11 cm.
EXAMPLE 11 In ABC opposite, O is the intersection point A
of the bisectors of the interior angles of the 4
triangle. Given that OE BC, OD AB, D
AD = 4 cm, DE = 5 cm and EC = 6 cm, find
5
P(EOD).
O
E
6
B C
Solution Let us join points A and C to O. We know A
from the question that OA and OC are the
4
bisectors of A and C, respectively. D
Since OD AB,
4 5
m(OAB) = m(AOD). (Alternate Interior 6
O E
Angles Theorem)
6
So ODA is an isosceles triangle and
AD = OD = 4 cm. (1) B C
Similarly, since OE BC,
m(EOC) = m(OCB). (Alternate Interior Angles Theorem)
So EOC is also an isosceles triangle and OE = EC = 6 cm. (2)
By (1) and (2), P(EOD) = OE + OD + DE = 6 + 4 + 5 = 15 cm.
Triangles and Construction 61
Definition equilateral triangle
A triangle is called equilateral if it has three A
congruent sides.
a a
How many equilateral
triangles can you see in
the figure below? B a C
In an equilateral triangle, all of the interior A
angles are congruent and measure 60°.
60°
Notice that an equilateral triangle is also an
isosceles triangle, but an isosceles triangle is
not always equilateral.
60° 60°
You will probably ‘see’ two
triangles, one on top of the B C
other. This is actually an
optical illusion, though,
as the white triangle is
not actually drawn.
Activity Toothpick Triangles
Find six toothpicks and try to do each thing below. Some things may not be possible.
Can you explain why?
1. Make one equilateral triangle with six toothpicks.
2. Make two equilateral triangles with six toothpicks.
3. Make three equilateral triangles with six toothpicks.
4. Make four equilateral triangles with six toothpicks.
EXAMPLE 12 The three sides of a triangle measure 5n + 8, n+12 and 3n+10 with n N. Which value of
n makes this triangle equilateral?
Solution If the triangle is equilateral, all the sides must be congruent.
So 5n + 8 = n + 12 = 3n + 10. Let us solve the first equality to find n:
5n + 8 = n + 12
4n = 4
n = 1.
If we substitute 1 for n, the side lengths become (5 1) + 8 = 13, 1 + 12 = 13 and
(3 1) + 10 = 13. So the triangle is equilateral when n = 1.
62 Geometriy 7
Check Yourself 4
A
1. In ABC opposite, DE BC and point O is the incenter of the
triangle. If BD = 6 and EC = 4, find DE.
O
D E
6 4
B C
2. The perimeter of an isosceles triangle is 18.4 and its base measures 4 units more than the
length of one leg. Find the length of a leg of this triangle.
3. The sides of an isosceles triangle have lengths in the ratio 4 : 5 : 5. Find the length of the
base of the triangle if its perimeter is 28.
4. The perimeter of an isosceles triangle is 22.8. An equilateral triangle is drawn such that one
side is congruent to the base of the isosceles triangle. If the perimeter of the equilateral
triangle is 24.6, find the length of one leg of the isosceles triangle.
5. In an isosceles triangle NTM, MN = NT, MN = 35, TN = 4x +15 and MT = 40 – x2. Find
MT.
B
6. In the figure, all triangles are equilateral, F
AG = 24.12 cm and AC = 3CE = 2EG. Find
The picture shows a puzzle the perimeter of each triangle. C
called the Three Companions A E G
Puzzle. Get your own and
try to free one of the triangles D
from the string. Can you do 7. The three sides of a triangle measure 3a,
it? a+10 and 6a – 15. Which value of a makes the triangle equilateral?
8. Construct an isosceles and an equilateral triangle.
Answers
1. 10 2. 4.8 3. 8 4. 7.3 5. 15 6. 12.6 cm, 6.3 cm, 4.2 cm 7. 5
2. Types of Triangle According to Angles
A triangle can be called acute, right or obtuse, depending on the measures of its angles.
Definition acute triangle, right triangle, obtuse triangle
A triangle is called an acute triangle if all its angles are acute.
A triangle is called a right triangle if it has a right angle.
A triangle is called an obtuse triangle if it has an obtuse angle.
Triangles and Construction 63
In a right triangle, the sides adjacent to the
A
Challenge!
right angle are called the legs of the triangle. legs hypotenuse
Try to change the The side opposite the right angle is called the
equilateral triangle in the
figure so that it points hypotenuse of the triangle.
upwards by moving only
three balls. B C
Then try to make the
triangle in this figure Note
point downwards by using
the least number of balls Notice that a triangle can be only one of obtuse, acute or right.
possible.
Triangles
Acute Obtuse Right
Scalene Isosceles Equilateral Scalene Isosceles Scalene Isosceles
EXAMPLE 13 Name all the right triangles in the figure. D C
K
Solution There are four smaller right triangles (ABK,
BKC, CKD and DKA) and four larger
triangles (ABC, BCD, CDA and DAB). A B
EXAMPLE 14 Classify each triangle according to its side lengths and angle measures.
a. b. c.
80° 80°
45° 45° 70° 30° 50° 50°
Solution a. isosceles right triangle b. scalene acute triangle c. isosceles acute triangle
64 Geometriy 7
Activity Tangram
‘Tangram’ is a fun puzzle and a good way to exercise your brain. The name comes from
tan, which means ‘Chinese’, and gram, which means ‘diagram’ or ‘arrangement’. The
puzzle first appeared in China thousands of years ago, and it is now known all over the
world. There are seven pieces in a tangram set: five triangles, one square and one
parallelogram. The challenge of the puzzle is to use the seven pieces together to make
different shapes. You must use all the pieces, and they must all touch but not overlap.
All seven tangram pieces are made up of right triangles with this shape:
The first tangram challenge is to make a square with all seven pieces. The solution is
shown below.
1
5
2
3
7
Find a tangram set, or copy the figure above to make your own.
1. Make one right triangle using all of the pieces.
2. Can you make an obtuse triangle by using all of the pieces?
3. Can you make an acute triangle by using all of the pieces?
EXAMPLE 15 Draw a right triangle and divide it using
a. two parallel lines which are perpendicular to one of the legs.
b. two parallel lines which are not perpendicular to legs.
c. two perpendicular lines to create two more right triangles.
d. two intersecting lines which are not perpendicular to each other to create two more right
triangles.
Triangles and Construction 65
Solution a. b. c. d.
EXAMPLE 16 Draw each triangle and use a set square to find its orthocenter. Write the orthocenter as an
intersection of lines or line segments.
a. acute scalene b. right scalene c. obtuse scalene
Solution Remember that the orthocenter of a triangle is the intersection point of its altitudes. We draw
the altitudes in each triangle by using a set square.
A
a. orthocenter K,
E
K = AF BD EC K D
30°
B F C
60°
b. orthocenter A, B
A = AB CA AD D
a 30°-60° set square
A C
c. orthocenter T, A
45°
T = AT BT TC
45°
B C
a 45° set square T
Ealier in this chapter we said that the position of the orthocenter of a triangle depends on
the type of triangle. One position is in the interior of the triangle. Can you see what the other
two possible positions are, after studying the example above? How do they correspond to the
types of triangle shown?
66 Geometriy 7
Check Yourself 5
1. Classify each triangle according to its angle measures.
a. b. c. S
d. e.
A M X K
20°
80°
40°
50° 70° 60° Z 120°
B 60° 60°
C N P K Y L M
T
D
2. Name all the right triangles in the figure.
C
E
This shark’s fin forms a
right triangle with the
A B
water.
3. At most how many of each type of angle can one triangle have?
a. acute angle b. right angle c. obtuse angle
(Hint: Try to draw a suitable figure for each case using a protractor.)
4. Draw a right triangle and divide it using
a. two intersecting lines which are perpendicular to each other.
b. two intersecting lines which are not perpendicular to each other, to make three more
right triangles.
5. Construct a right isosceles triangle.
Answers
1. a. right triangle b. acute triangle c. acute triangle
d. obtuse triangle e. obtuse triangle
How many triangles?
2. DKB, KAB, KBC, KCD, KDA
3. a. three b. one c. one
4. a. b.
Triangles and Construction 67
EXERCISES 3 .1
A. The Triangle and Its Elements 8. In a triangle ABC, two points different to A and B
A on the side AB are joined to the vertex C by line
1. Find and name all the
segments. Similarly, three points different to B
triangles in the figure. E
D and C on side BC are joined to the vertex A by line
K
segments. How many regions inside the triangle
B F C are formed by the intersection of these segments?
2. How many triangles can be formed by joining any 9. In an isosceles triangle DEF, DF is the base
three points D, E, F and G if no three of the given and FT is a median. Given that P(DEF) = 23 cm
points are collinear? Name each triangle.
and P(EFT) is 1 cm more than the perimeter of
triangle DTF, find DF.
6
3. In a triangle ABC, AB is of AC, AB = BC and
5
AC = 15 cm. Find P(ABC). 10. Draw three triangles ABC A
as in the figure and
18 construct each element
4. In a triangle KMN, KM =cm, MN is 75% of
5 separately, using a compass
KM and KN is 0.1 cm more than KM. Find B C
and straightedge.
P(KMN). a. ha b. Va c. nA
5. The side AC of a triangle ABC measures 12.8 cm,
which is 2.6 cm less than the sum of the lengths 11. Draw four triangles KMN as in N
of the other sides. Find the perimeter of this the figure and find each
triangle. point separately using
only a compass and
K M
6. Answer according to B D straightedge.
the figure. E a. centroid b. incenter
a. Name four F M
c. orthocenter d. circumcenter
collinear points on
ABC. A C
12. Repeat question 11 with a protractor and ruler.
b. Name a point which is in the interior of
ADC. 13. Find the excenters of the triangle in question 11
c. What is the intersection of ABC and ADC? by using a protractor and ruler.
d. What is the intersection of ABC and int MAC?
14. The sides AB, BC and AC of a triangle ABC measure
7. Draw four figures to show how two triangles can 13, 14 and 15 units respectively. Given that the
intersect to form a four-sided, five-sided, length of the altitude to side BC is 12, find the
six-sided and three-sided polygon. lengths of the remaining altitudes.
68 Geometriy 7
B. Types of Triangle 21. Write always, sometimes or never to make true
A
statements.
15. Look at the figure and
a. If a triangle is isosceles then it is ______________
name
F equilateral.
a. an isosceles triangle. 60° b. If a triangle is equilateral then it is ___________
30°
b. three right triangles. isosceles.
B E C
c. an obtuse isosceles triangle. c. If a triangle is scalene then it is ______________
d. an acute triangle. isosceles.
e. an equilateral triangle. d. If a triangle is obtuse then it is _______________
isosceles.
16. Find the circumcenter of a right triangle using a e. An obtuse triangle is __________________ a right
ruler and protractor. triangle.
17. State whether each type of triangle is possible or f. In a triangle DEF, if DE EF then DF is
not. ________________ perpendicular to EF.
a. an isosceles acute triangle g. A scalene triangle ________________________ has
b. a right equilateral triangle an acute angle.
c. a scalene acute triangle h. If a triangle has two complementary angles
d. an obtuse isosceles triangle then it is ____________________ a right triangle.
e. an obtuse equilateral triangle
18. The sides of a triangle measure 2x + 8, 3x – 6, 22. In each case, draw a triangle with the given
and 12 + x. property.
a. Find the value(s) of x that make(s) the triangle a. All three angle bisectors are medians.
isosceles. b. No altitude is a median.
b. Which value(s) of x make(s) the triangle
c. Only one angle bisector is the perpendicular
equilateral?
bisector of a side.
19. The sum of the lengths of the legs of an d. Only one altitude is in the interior region of
isosceles right triangle is 22 cm. Find the area of the triangle.
this triangle. e. The medians, altitudes and angle bisectors
20. Complete the table showing the location (in the coincide.
interior, on the triangle or in the exterior) of the f. Exactly one of the three altitudes is also a
intersection of the segments or lines for each type median.
of triangle.
Perpendicular Angle Line containing
Medians
bisectors bisectors the altitudes
Acute
triangle a. b. c. d. 23. Divide any right triangle using two lines so that
Right
triangle e. f. g. h. the figure contains a total of
Obtuse
triangle i. j. k. l. a. five right triangles. b. six right triangles.
Triangles and Construction 69
Objectives
After studying this section you will be able to:
1. Identify congruent triangles
2. Construct a circle
3. Construct congruent segments
4. Find the midpoint of a segment
5. Construct perpendicular lines and parallel lines
6. Construct congruent angles and an angle bisector
7. Construct atriangle tram given information
8. Desctibe and use the properties of isosceles, equilateral and right triangles.
9. Describe and use the triangle Angle Bisector Theorem.
A. THE CONCEPT OF CONGRUENCE
In the previous section we studied triangles and their features and properties. In this section
we will look at possible relations between two or more triangles.
If we are given two triangles, how can we compare them? We might notice that they are the
same size and shape. This important relation in geometry is called congruence. Let us start
our study of congruence with a general definition of congruence in figures and polygons.
1. Congruent Figures
The world around us is full of objects of various shapes and sizes.
If we tried to compare some of these objects we could put them
in three groups:
objects which have a different shape and size,
objects which are the same shape but a different size, and
objects which are the same shape and size.
The tools in the picture at the right have different shape and size.
The pictures below show tools which have the same shape but different size. In geometry,
figures like this are called similar figures. We will study similar figures in Chapter 3.
Congruence is a basic
geometric relationship.
70 Geometriy 7
The pictures below show objects which are the same size and shape. In this section, we will
study figures which have this property.
Factories often need to
produce many parts
with exactly the same
size and shape.
Definition congruent figures
Figures that have the same size and shape are called congruent figures. We say ‘A is congruent
to B’ (or ‘B is congruent to A’) if A and B are congruent figures.
The pictures at the bottom of the previous page show some examples of congruent objects.
The pictures below show two more examples. In these two examples there is only one piece
left to fit in the puzzle. Therefore, without checking anything, we can say that each piece and
its corresponding place are congruent.
Congruence in nature:
the petals of this flower
are congruent.
Activity Making a Poster - Congruent Figures
Make a poster to show congruent figures in everyday life. You can take photos, draw
pictures or collect pictures from magazines or newspapers that show buildings, designs,
signs and artwork with congruent parts.
EXAMPLE 17 Which piece is congruent to the empty space?
a. b. c. d.
Solution If we compare the vertices and sides, we can easily see that only c. fits into the space.
Triangles and Construction 71
Activity Congruent Dissections
When you learned common fractions, you probably learned them by working with
figures divided into congruent parts. Often the figures are circles and rectangles, as
these are the easiest to divide into any number of congruent parts.
Dividing (also called dissecting) a figure into congruent parts can also be a puzzle. As an
example, can you see how to dissect the first figure below into two congruent pieces?
Answer:
Now try the two puzzles below. The answers are at the back of the book.
1. Dissect each figure into four congruent pieces.
A car has many
congruent parts.
2. The polygon below left can be dissected into four congruent polygons, as shown in the
figure below right. There is also a way to divide this polygon into five congruent
polygons. Can you find it?
72 Geometriy 7
We can think of congruent figures as figures that are exact copies of each other. In other
words, we can put congruent figures one on top of the other so that each side, angle and
vertex coincides (i.e. matches perfectly).
Definition corresponding elements or parts
The points, lines and angles which match perfectly when two congruent figures are placed
one on top of the other are called corresponding elements or corresponding parts of the
congruent figures.
We can see that by definition, corresponding parts of congruent figures are congruent. We
Sometimes we need to can write this in a shorter way as CPCFC.
move or modify a figure
to see that it is congruent You are already familiar with congruent segments (segments that have equal lengths) and
to another figure. The
basic changes that we
congruent angles (angles that have equal measures). In the rest of this section we will look
can make to a figure are at congruent figures which are made up of segments and angles. These figures are polygons
reflection (flipping),
rotation (turning) and and especially triangles.
translation (sliding). We
will study these in
Chapter 3.
2. Congruent Triangles
Definition congruent triangles
Two triangles are congruent if and only if their corresponding sides and angles are congruent.
We write ABC DEF to mean that ABC and DEF are congruent.
Challenge!
Remove five toothpicks
In the figure below, ABC and DEF are congruent because their corresponding parts are
to make five congruent congruent. We can write this as follows:
triangles.
A D AB DE
B E and BC EF
C F AC DF.
We can show this symbolically in a figure as follows:
A D
B C E F
DABC DDEF
Triangles and Construction 73
EXAMPLE 18 Given that MNP STK, state the congruent angles and sides in the two triangles without
drawing them.
Solution The figure at the right shows how the DMNP @ DSTK
vertices of each triangle correspond to each
M corresponds to S
A short history of the symbol:
other. Because MNP STK and CPCTC
Gottfried Wilhelm N corresponds to T
Leibniz (corresponding parts of congruent triangles P corresponds to K
(1640-1716) are congruent), we can write
introduced for congruence
in an unpublished manuscript in M S MN ST
1679.
In 1777,
N T and NP TK
Johann Friedrich
P K PM KS.
Häseler
(1372-1797) As we can see, the order of the vertices in congruent triangles is important when we are
used (with the tilde reversed).
considering corresponding elements. Any mistake in the ordering affects the correspondence
In 1824,
Carl Brandan between the triangles.
Mollweide
(1774-1825)
If two triangles are congruent then we can write this congruence in six different ways. For
used the modern symbol for instance, if ABC is congruent to DEF, the following statements are all true:
congruence in Euclid’s Elements.
ABC DEF
ACB DFE
BAC EDF
BCA EFD
CAB FDE
CBA FED.
EXAMPLE 19 Complete each statement, given that PRS KLM.
a. PR _____ b. _____ K c. _____ SP
d. S _____ e. ML _____ f. L _____
Solution a. PR KL b. P K c. MK SP
d. S M e. ML SR f. L R
EXAMPLE 20 Decide whether or not the two triangles in A K 4 N
the figure are congruent and give a reason for 60°
your answer. 30°
8 8
B 4 C M
74 Geometriy 7
Solution Let us calculate the missing angles:
m(C) = 60° (Triangle Angle-Sum Theorem in ABC)
m(M) = 30° (Triangle Angle-Sum Theorem in KMN)
Now we can write the congruence of corresponding parts:
AB KM (Given)
BC KN (BC = KN = 4)
AC MN (AC = MN = 8)
A M (m(A) = m(M) = 30°)
B K (m(B) = m(K) = 90°)
C N (m(C) = m(N) = 60°)
Therefore, ABC MKN by the definition of congruent triangles.
EXAMPLE 21 ABC EFD is given with AB = 11 cm, BC = 10 cm and EF + ED = 19 cm. Find the
perimeter of EFD.
Solution Since ABC EFD, AB = EF, BC = FD and A E
AC = ED by the definition of congruence.
So by substituting the given values we get 11 11 8
11 = EF, 10 = FD and AC = ED.
Since we are given that EF + ED = 19 cm, B 10 C F 10 D
we have 11 + ED = 19 cm; ED = 8 cm.
So P(EFD) = EF + ED + FD = 11 + 8 + 10 = 29 cm.
Check Yourself 6
1. KLM XYZ is given. State the corresponding congruent angles and sides of the
triangles.
2. State the congruence JKM SLX in six different ways.
3. Triangles KLM and DEF are congruent. P(KLM) = 46 cm, the shortest side of KLM
measures 14 cm, and the longest side of the DEF measures 17 cm. Find the lengths of
all the sides of one of the triangles.
What would happen if
the blades of this ship’s 4. Triangles DEF and KLM are congruent. If DE = 12.5 cm, EF = 14.4 cm and the perimeter
propellor or these wheels
were not congruent? of the triangle KLM is 34.6 cm, find the length of the side DF.
Triangles and Construction 75
5. Two line segments KL and AB bisect each other at a point T. If AL = 7 and the lengths of
the segments KL and AB are 22 and 18 respectively, find the perimeter of KTB.
Answers
1. KL XY K X 2. PKM SLN, KMP LNS, MPK NSL,
LM YZ L Y PMK SNL, KPM LSN, MKP NLS
KM XZ M Z
3. 14 cm, 15 cm, 17 cm 4. 7.7 cm 5. 27
B. CONSTRUCTIONS
In this section we will construct geometric figures using only two instruments,
a straightedge and a compass.
straightedge
1. Basic Constructions
We use a straightedge to construct a line,
compass ray, or segment when two points are given.
A straightedge is like a ruler without num-
bers. r
r O
We use a compass to construct an arc or a
pencil circle, given a point O and a length r
compass point
point (a radius).
Construction 1
Constructing a congruent segment.
Given [AB], A B
construct [CD] such that [CD] [AB].
A B A B
C A B C D
Use a straightedge to draw Set your compass Using C as the center,
a line. to the length of [AB]. draw an arc intersecting
Mark a point C on the line. line [CD]. Label the point of
intersection D.
Result: [CD] [AB].
76 Geometriy 7
Construction 2
Finding the midpoint of a given segment.
Given [AB],
A B
construct M such that such that [AM] [MB].
X
X
A M B
A B A B
Y
Y
Using any radius greater Draw an arc with the same Draw [XY].
radius and center B. Mark and name
than 1 |AB| , draw an
2 Label the points of the intersection point M.
arc with center A. intersection of the arcs
X and Y.
Result: M is the midpoint of [AB].
Construction 3
Constructing a perpendicular to a line at a given point on the line.
Given point M on the line l,
l
M
construct [MN] l.
N N
l l l
A M B A M B A M B
Using M as the center and Using centers A and B Draw [MN].
any radius, use a compass to and a radius greater
draw arcs intersecting than |MA|, draw two arcs
l at A and B. and find the intersection point N.
Result: [MN] l.
Triangles and Construction 77
Construction 4
Constructing a perpendicular to a given line through a point outside the given line.
N
Given line l and a point N outside the line,
construct [MN] l.
l
N N N
l l l
A B A B A B
M M
Using N as a center, Use A and B as centers to draw arcs
draw an arc that intersects with the same radius that intersect Draw [MN].
l at two points A and B. at a point M.
Result: [MN] l.
Construction 5
Constructing a congruent angle.
C
Given A, construct A such that A A.
A B
C C
A B A B
C¢ C¢
A¢ B¢ A¢ B¢ A¢ B¢
Use a straightedge to draw a ray. Use |BC| as a radius and Draw [A¢C¢.
Name its initial point A¢. center B¢ to draw an
Using a compass at center A, arc which intersects the
draw BïC. Keep the same first arc at point C¢.
radius and draw an arc which
intersects the ray from A¢ at point B¢.
Result: AA.
78 Geometriy 7
Construction 6
Constructing a parallel to a line through a point outside the line.
N
Given line l with point N which is not on l,
construct a line through N which is parallel to l. l
N t
N Q
l l
M M P
Draw a line k which intersects At N, construct ÐRNQ
line l at point M, and passes congruent to ÐNMP.
through point N.
Result: l t.
Construction 7
Constructing an angle bisector.
C
Given CAB, construct the bisector of CAB.
A B
C C
C D D
A B A B A B
Draw an arc BïC with With B and C as centers and a Draw [AD.
center A. radius greater than 1 |BC|,
2
draw arcs intersecting at D.
Result: [AD bisects A
Triangles and Construction 79
EXAMPLE 22 a. Construct two congruent line segments. d. Construct an isosceles triangle.
b. Construct an obtuse angle and bisect it. e. Construct an equilateral triangle.
c. Construct two parallel line segments.
Solution a.
l l
A B A¢ A¢ B¢
Use a straightedge to Draw another line l. Use the radius |AB| and set
draw [AB]. Choose any point on the compass point at A. Draw
Set the compass at the line l and label it A. an arc intersecting l.
points A and B. Label the point of intersection
B. Now [AB] [AB]
b. D
A A
B C B C
Draw any obtuse angle ABC. Use B as the Label the point D where the two
center, and draw an arc AïC. Next, draw two arcs, arcs intersect. Draw [BD.
one with center A and the other with center C. [BD is the angle bisector of ABC.
c. Look at construction 6.
d. C C
l l
A B A B
Draw a line segment [AB]. Use any radius greater Draw the triangle ABC.
1
than 2 | AB| and draw two arcs with centers A |AC| = |BC|, so the triangle is
and B. Name the intersection point C. isosceles.
e. C
l l
A B A B
Use a straightedge to draw [AB]. Next, open Label the intersection point C. Draw [AC]
the compass to |AB| and draw two arcs, one and [BC]. All the sides have equal length,
with center A and the other with center B. so ABC is an equilateral triangle.
80 Geometriy 7
Practice Problems 7
1. Construct a 30° angle. (Hint: construct a 60° angle and bisect it.)
2. Construct a right triangle with legs which are congruent to [AB] A B
and [CD] in the figure.
C D
3. Construct a right triangle whose legs are in the ratio 2:1.
4. Construct a line segment and divide it into four equal parts.
Answers
C D
1. By constructing equilateral triangles:
30°
A B
2. D
C
A B A B A B
M N
Use a straightedge Open your compass Open your compass
to draw [AB] more than [MA] and to [CD] and draw an
draw two arcs, one arc with center A.
with center M, te Label the point of
other with center N. intersection D. Draw
Draw a line from A [AD] and [DB.
to the point of
intersection.
3. D
A C B A C B A C B
Find the midpoint of Draw a perpendicular Open your compass
[AD] (see construction 2) line to [AB] from A. to as [AC] then draw
(see construction 3) [AD] and [DB].
4.
C B A D C E A D C E B
A B
Find the midpoint of [AB]. Find the midpoint of [AC] Label the new points D and
and the midpoint of [CB]. E.
Triangles and Construction 81
2. Constructing Triangles
We can construct basic geometric figures using only a straightedge and a compass. However,
to construct triangles we need a compass, a ruler and a protractor. We use the ruler to
measure the sides of triangle, and the protractor to draw the angles.
We have seen that a triangle has six basic elements: three angles and three sides. To
construct a triangle, we need to know at least three of these elements, and one of these three
elements must be the length of a side. Let us look at the possible cases.
Note
In any triangle, the sum of any two given angles is less than 180° and the sides satisfy the
triangle inequality.
a. Constructing a Triangle from Three Known Sides
Let us construct ABC, B a C
where |AB| = c, |BC| = a, and |AC| = b, given that
A b C
a < b < c.
A c B
Construction 1
Draw a line d. d
Construction 2
Locate point A on d. d
A
Construction 3
Open the compass as much as length c and put the d
A c B
sharp point of the compass on A. Then draw an arc.
Name the intersection point B.
Construction 4
Again open the compass as much as length b and put the
sharp point on A. Then draw an arc on the upper side of d
A c B
d.
82 Geometriy 7
Construction 5
Finally, open the compass as much as length a and put C
the sharp point on B. Then draw an arc which intersects
d
the other arc drawn before. Name the intersection point A c B
C.
Construction 5
After determining the point C, draw [AC] and [BC]. The C
result is the constructed triangle. b a
d
A c B
Note
Remember that in a triangle, side a is opposite A, side b is opposite B, and side c is
opposite C. When we talk about ‘side b’ we mean the side opposite B, or the length of this
side.
EXAMPLE 23 Construct ABC given |AB| = 10 cm, |BC| = 8 cm, and |AC| = 6 cm.
Solution C C
6 cm 8 cm 6 cm 8 cm
A 10 cm B A 10 cm B A 10 cm B
Draw a line and C is the intersection point Connect the vertices.
locate points A and B. of the two arcs.
b. Constructing a Triangle from Two Known Angles and a Known Side
Let us construct the triangle ABC, where A, B, and the side c are given.
Construction 1
Draw a line d. d
Construction 2
Locate point A on the line. A
Triangles and Construction 83
Construction 3
Using a protractor, take the point A as a vertex and draw X
a ray [AX to construct A.
d
A
Construction 4
Using a compass, locate the point B on d such that X
|AB| = c.
d
A c B
Construction 5
Using a protractor, take the point B as vertex and draw a Y X
C
ray [BY to construct B. Label the intersection point of
[AX and [BY as C. The construction is complete. d
A c B
EXAMPLE 24 Construct ABC given mB = 40°, mC = 70°, and |BC| = 12 cm.
Solution X
X X Y
A
40° 40° 40° 70° C
B B 12 cm C B 12 cm
Mark the point Using a compass Draw ÐC to find
B and draw ÐB. and ruler, find C. the point A on [BX.
3. Constructing a Triangle from Two Known Sides and a Known
Angle
Finally, let us construct ABC given A c B
|AB| = c, |BC| = a and the known angle B. a
B C
Construction 1
Draw a line d and use a compass to locate the points B
d
and C such that |BC| = a.
84 Geometriy 7
Construction 2
Use a protractor to construct B and the ray [BX. X
d
B a C
Construction 3
Use a compass or ruler to locate the point A on [BX such X
A
that |AB| = c. c
d
B a C
Construction 4
Join the points A and C. The result is the constructed X
triangle.
d
A c B
EXAMPLE 25 Construct ABC given |BC| = 5 cm, |AB| = 10 cm, and mB = 70o.
Solution
X X X
A A
10 cm 10 cm
70° 70° 70°
B 5 cm C B 5 cm C B 5 cm C
Locate the points B and Locate A on [BX. Join A and C.
C and draw ÐB.
Triangles and Construction 85
Practice Problems 8
1. State the things you need to know in order to construct a triangle.
2. Draw an equaliteral triangle with sides 6 cm long.
3. Construct ABC given a = 5 cm, b = 4 cm and c = 2 cm.
4. Construct ABC given a = 7 cm, b = 6 cm and c = 8 cm.
5. Construct DEF given d = 6 cm, e = 8 cm and f = 10 cm.
6. Construct ABC given mA = 40o, mB = 65o and |AB| = 10 cm.
7. Construct KLM given mM = 45o, mL = 70o and |ML| = 7 cm.
8. Construct PQR given mR = 40o, mQ = 60o and |RQ| = 4 cm.
9. Construct MNP given mM = 30o, mN = 65o and |MN| = 15 cm.
10.Construct ABC given mB = 90o, |AB| = 5 cm and |BC| = 12 cm.
11.Construct PQR given mQ = 80o, |PQ| = 7 cm and |QR| = 4 cm.
12.Construct GHK given mH = 50o, |GH| = 6 cm and |HK| = 9 cm.
13.Construct XYZ given mY = 110o, |XY| = 3 cm and |YZ| = 5 cm.
14.Can you draw a triangle from only three given angles?
C. ISOSCELES, EQUILATERAL AND RIGHT TRIANGLES
Isosceles, equilateral and right triangles are useful triangles because they have many special
properties. If we can identify one or more of these triangles in a figure then we can often use its
properties to solve a geometric problem. In this section we will look at some fundamental
theorems about isosceles, equilateral and right triangles, and some useful additional properties.
1. Properties of Isosceles and Equilateral Triangles
a. Basic Properties
Theorem Isosceles Triangle Theorem
If two sides of a triangle are congruent then the angles opposite these sides are also
congruent.
A
Proof Let us draw an appropriate figure.
Given: AB = AC
Prove: B C
B N C
86 Geometriy 7
Let AN be the bisector of A.
Statements Reasons
1. AB AC 1. Given
2. BAN CAN 2. Definition of an angle bisector
3. AN AN 3. Reflexive property of congruence
4. ABN ACN 4. SAS Congruence Postulate
5. B C 5. CPCTC
Theorem Converse of the Isosceles Triangle Theorem
If two angles in a triangle are congruent then the sides opposite these angles are also
congruent.
Proof Let us draw an appropriate figure. A
Given: B C
Prove: AB AC
We begin by drawing the bisector AN, and
continue with a paragraph proof.
B N C
Since AN is the angle bisector,
BAN CAN.
It is given that B C.
By the reflexive property of congruence, AN AN.
So ABN ACN by the AAS Congruence Theorem.
Since CPCTC, we have AB AC.
EXAMPLE 26 In a triangle DEF, T DF such that DT = DE. Given m(EDT) = 40° and m(DEF) = 85°,
find m(TEF).
Solution Let us draw an appropriate figure. D
Since DE = DT, DET is an isosceles triangle.
40°
So by the Isosceles Triangle Theorem,
m(DET) = m(DTE).
So by the Triangle Angle-Sum Theorem in
DET, E T
m(EDT) + m(DET) + m(DTE) = 180° F
40° + 2m(DET) = 180°
m(DET) = 70°.
So m(TEF) = m(DEF) – m(DET) = 85° – 70° = 15°.
Triangles and Construction 87
Corollary Corollary of the Isosceles Triangle Theorem
If a triangle ABC is equilateral then it is also equiangular. In other words,
if a = b = c then m(A) = m(B) = m(C).
Corollary Corollary of the Converse of the Isosceles Triangle Theorem
If a triangle ABC is equiangular then it is also equilateral, i.e. if m(A) = m(B) = m(C)
then a = b = c.
EXAMPLE 27 In the figure, ABC and DEF are equilateral A
D
triangles. If BF = 17 cm and EC = 3 cm, find
AB + AH + DH + DF.
H
B E C F
Solution In the figure,
m(HCE) = 60° and m(HEC) = 60°. (ABC and DEF are equilateral)
So in HEC,
m(H) + m(E) + m(C) = 180° (Triangle Angle-Sum Theorem)
m(H) = 60°.
So HEC is equiangular. (m(C) = 60°, m(E) = 60°, m(H) = 60°)
Therefore HEC is equilateral. (By the previous Corollary)
So HE = HC = EC = 3 cm.
Let a and b be the lengths of the sides of ABC and DEF, respectively.
In ABC, AB = a, BE = a – 3 and AH = a – 3. (EC = 3 cm, given)
In DEF, DF = b, CF = b – 3 and DH = b – 3. (EC = 3 cm, given)
So
A
AB + AH + DH + DF = a + a – 3 + b – 3 + b D
60°
= 2(a + b) – 6. (1)
a
60°
3
a
b
Moreover, BF = a – 3 + 3 + b – 3 (Segment H b
Addition Postulate) 3
3 60°
60° 60°
17 = a + b – 3 60° 60°
B a3 E 3 C b3 F
20 = a + b. (2)
Substituting (2) into (1) gives us AB + AH + DH + DF = 34 cm.
88 Geometriy 7
Practice Problems 9
1. In a triangle ABC, the interior angle bisector at the vertex A makes an angle of 92° with
the side opposite A and has the same length as one of the remaining sides. Find all the
angles in ABC.
A
2. In the figure, CE is the angle bisector of C, HD BC and
HD = 5 cm. Find the length of AC. E 5 D
H
B C
3. In the figure, DCE is an equilateral triangle and DC = BC.
?
m( A ) 1
= is given. Find m(A).
m( B) 13
E
Answers D
1. 8°, 84° and 88° 2. 10 cm 3. 5°
B C
b. Further properties
Properties 6
1. For any isosceles triangle, the following statements are true.
a. The median to the base is also the angle bisector of the vertex and the altitude to the
base.
b. The altitude to the base is also the angle bisector of the vertex and the median to the
base.
c. The angle bisector of the vertex is also the altitude and the median to the base.
In other words, if AB = AC in any triangle ABC then nA = Va = ha.
2. In an equilateral triangle, the medians, angle bisectors and altitudes from the same
vertex are all the same, i.e., ha = nA = Va, hb = nB = Vb, and hc = nC = Vc.
Moreover, all of these lines are the same length:
ha = hb = hc = nA = nB = nC = Va = Vb = Vc.
In other words, if AB = BC = AC in ABC then ha = hb = hc = nA = nB = nC = Va = Vb = Vc.
3. If AB = AC in any triangle ABC then nB = nC, Vb = Vc and hb = hc.
4. If ha = nA or ha = Va or nA = Va in ABC then ABC is isosceles.
Triangles and Construction 89
5. a. If ABC is an isosceles triangle with A
AB = AC, P BC, E AC, D AB,
PE AB and PD AC,
then PE + PD = b = c. E
D
B P C
b. If P, E and D are any three points on A
the sides of an equilateral triangle ABC D
such that PE and PD are parallel to
two distinct sides of ABC, then E
PE + PD = AB = BC = AC.
B P C
6. a. In any isosceles triangle ABC with A
AB = AC, the sum of the lengths of two
lines drawn from any point on the base
perpendicular to the legs is equal to
the height of the triangle from the H
D
vertex B or C. In other words, if
AB = AC, P BC, H AC, D AB, B P C
PH AC and PD AB, then
PH + PD = hc = hb.
b. In any equilateral triangle ABC, the sum of the lengths of two lines drawn from any
point on any side perpendicular to the other sides is equal to the height of the triangle
from any vertex. In other words, if AB = BC = AC, P BC, H AC, D AB, PH AC
and PD AB, then PH + PD = ha = hb = hc.
7. In any equilateral triangle ABC, A
if P int ABC and points D, E and F
lie on the sides of ABC such that D
PE AB, PD AC and PF BC, then
P F
PE + PF + PD = AB.
B E C
90 Geometriy 7
8. In any equilateral triangle ABC, A
if P int ABC and points D, E and F lie
on the sides of ABC such that PE AB, D
PD AC and PF BC, E
then PD + PE + PF = AH where P
AH BC.
B F C
EXAMPLE 28 An isosceles triangle TMS has base TS which measures 8 cm and perimeter 32 cm. The
perpendicular bisector of leg TM intersects the legs TM and MS at the points F and K,
respectively. Find the perimeter of TKS.
Solution Let us draw an appropriate figure. M
Since hk = Vk in TKM, TKM is isosceles
6
by Property 6.4.
F
So TK = KM. (1)
6 K
By the Segment Addition Postulate,
MK + KS = MS. (2)
T 8 S
Since TMS is isosceles and
P(TMS) = 32 cm,
TM = MS = 12 cm. (3)
So P(TKS) = TK + KS + ST
= KM + KS + ST (By (1))
= MS + ST (By (2))
= 12 + 8 = 20 cm. (By (3))
EXAMPLE 29 In the figure, A
AB = AC,
PD AC and E
PE AB.
D
Given AB + AC = 32 cm, find PD + PE.
Solution Since AB + AC = 32 cm and B P C
AB = AC, we have AB = AC = 16 cm.
So by Property 6.5b, PD + PE = AB = 16 cm.
Triangles and Construction 91
EXAMPLE 30 In the figure opposite, H, A and B are C
collinear points with CH HA,
PE AC, PD AB and AB = AC.
P
PE = 6 cm and HC = 10 cm are given. 10 6
Find the length of PD. E
?
Solution By Property 6.6a, PE + PD = CH. H A D B
So 6 + PD = 10 and so PD = 4 cm.
EXAMPLE 31 In the equilateral triangle ABC shown A
opposite, PD BC, PE AB and PF AC.
E
PE = 6 cm, PF = 5 cm and
6
P(ABC) = 42 cm are given.
D P
Find the length of PD.
5
B F C
42
Solution Since ABC is equilateral and its perimeter is 42 cm, AB ==14 cm.
3
By Property 6.7, PD + PE + PF = AB. So PD + 6 + 5 = 14 and so PD = 3 cm.
EXAMPLE 32 Prove that in any isosceles triangle, the median to the base is also the angle bisector of the
vertex and the altitude to the base.
A
Solution Look at the figure.
Given: AT is a median and AB = AC.
Prove: AT bisects A and is an altitude of
ABC.
We will write a two-column proof.
Proof: B T C
Statements Reasons
1. AC AC 1. Given
2. ABC ACB 2. Isosceles Triangle Theorem
3. BT TC 3. AT is a median.
4. ABT ACT 4. By 1, 2, 3 and the SAS Congruence Postulate
5. m(TAB) = m(TAC) 5. By 4 (CPCTC)
6. AT bisects A 6. Definition of angle bisector
7. m(ATB) = m(ATC) 7. By 4 (CPCTC)
8. m(ATB) = m(ATC) = 90° 8. Linear Pair Postulate
9. AT is also an altitude of ABC 9. Definition of altitude
92 Geometriy 7
Practice Problems 10
A
1. In the figure, AD and BE are the interior angle
bisectors of A and B, respectively. E
AC = BC and BE + AD = 12 cm are given.
Find the value of BE AD.
B D C
2. In an equilateral triangle ABC, H BC, N AC, BN is the interior angle bisector of B,
and AH is the altitude to the side BC. Given BN = 9 cm, find AH.
A
3. In the figure, BH = HC, 20°
m(HAC) = 20° and
m(BCD) = 30°.
Find m(BDC). D
E 30°
?
B H C
4. In the figure, AB = AC = 13 cm, A
PE AC, PF AB and PE = 4 cm.
Find the length of PF. F
E
B P C
5. In the figure, AB = AC, PE HB, H
A
PF AC and BH HC.
F 12
Given CH = 12 cm, E
find the value of PE + PF.
B P C
6. In the figure, AB = BC, A
PD = x + 3, PH = 3x – 1, and
D
AE = 14. Find the value of x.
P
B E H C
Answers
1. 36 2. 9 cm 3. 80° 4. 9 cm 5. 12 cm 6. 3
Triangles and Construction 93
2. Properties of Right Triangles
a. The Pythagorean Theorem
The Pythagorean Theorem is one of the most famous theorems in Euclidean geometry, and
almost everyone with a high school education can remember it.
Theorem Pythagorean Theorem
In a right triangle ABC with m(C) = 90°,
A
the square of the length of the hypotenuse
is equal to the sum of the squares of the
lengths of the legs, i.e. b c
2 2 2
c =b +a.
C a B
Proof There are many proofs of the Pythagorean
A
Theorem. The proof we will give here uses
the dissection of a square. It proves the
Pythagorean Theorem for the right triangle c
b
ABC shown opposite.
C a B
Imagine that a large square with side length b a
a + b is dissected into four congruent right
a
triangles and a smaller square, as shown in c c b
the figure. The legs of the triangles are a and
b, and their hypotenuse is c. So the smaller b c c
a
square has side length c.
a b
We can now write two expressions for the area S of the larger square:
ab 2 2
S= 4 + c and S = (a + b) .
2
Since these expressions are equal, we can write
ab
4 2
+ c = ( a + b)
2
2
2ab + c2 = a2 + 2ab + b 2
c2 = a2 + b 2 .
This concludes the proof of the Pythagorean Theorem.
94 Geometriy 7
Try the following activity to discover two more proofs of the Pythagorean Theorem.
Activity The Pythagorean Theorem
1. Cut out a square with side length c, and cut out four identical right triangles with
hypotenuse c and legs a and b. Place the four triangles over the square, matching the
hypotenuses to the sides and leaving a smaller square uncovered in the centre. Try
to obtain the Pythagorean Theorem by considering the area of the original square
and the areas of the parts that dissect it.
c c c c c
b b b b
c a a a a
2. Cut out or construct two squares with sides a and b. Try
to dissect these squares and rearrange their pieces to
b
form a new square. Then use the areas of the original
squares and the new square to write the Pythagorean
Theorem. (Hint: Start by cutting out two right triangles b a
with legs a and b.)
EXAMPLE 33 In the figure, ST SQ. Find x and y. S
13
6
x
Solution First we will use the Pythagorean Theorem
in SKT to find x, then we can use it in y
T 12 K Q
SKQ to find y.
SK 2 + KT 2 = ST2 (Pythagorean Theorem in SKT)
x2 + 122 = 132 (Substitute)
2
x + 144 = 169
x2 = 25 (Simplify)
x = –5 is not an answer x=5 (Positive length)
because the length of a
segment cannot be negative. SK2 + KQ2 = SQ2 (Pythagorean Theorem in SKQ)
So the answer is x = 5. 2 2 2
From now on we will 5 +y =6 (Substitute)
always consider only
y2 =36 – 25 (Simplify)
positive values for lengths.
y = ò11
Triangles and Construction 95
EXAMPLE 34 In the figure,
K
PT = TS = KS,
PM = 4 cm and KM = 3 cm. Find ST. S
P T M
Solution Let PT = TS = KS = x.
So SM = KM – KS = 3 – x and TM = PM – PT = 4 – x.
In TMS, TS2 = TM2+ MS2 (Pythagorean Theorem)
Quadratic formula 2 2 2
The roots x1 and x2 of
x = (4 – x) + (3 – x) (Substitute)
the quadratic equation 2 2 2
x = 16 – 8x + x + 9 – 6x + x (Simplify)
ax2 + bx + c = 0 are
2
x – 14x + 25 = 0
– b b2 – 4ac
x1,2 = .
2a x1, 2 = (7 ò24) cm (Quadratic formula)
Since 7 + ò24 is greater than 3 and 4 which are the lengths of the sides, the answer is
x = |ST| = 7 – ò24 cm.
EXAMPLE 35 In the figure, K
m + k = 3 n.
Given A(KMN) = 30 cm2, n
m
find the value of n.
M k N
Solution m + k = 3 n (1) (Given)
A(KMN) = 30 cm2 (Given)
km
= 30 (Definition of the area of a triangle)
2
k m = 60 (2)
In KMN, n2 = k2 + m2 (Pythagorean Theorem)
n2 = (k + m)2 – 2km (Binomial expansion: (k+ m)2 = k2 + 2km + m2)
n2 = (3n)2 – 2 60 (Substitute (1) and (2))
2
8n = 120 (Simplify)
2
n = 15
n = ò15 cm.
Theorem Converse of the Pythagorean Theorem
If the square of one side of a triangle equals the sum of the squares of the other two sides,
then the angle opposite this side is a right angle.
96 Geometriy 7
Proof We will give a proof by contradiction. A A
Suppose the triangle is not a right triangle,
and label the vertices A, B and C. Then there c
b c b
are two possibilities for the measure of angle
C: either it is less than 90° (figure 1), or it is
C a B C a B
greater than 90° (figure 2). figure 1 figure 2
Let us draw a segment DC CB such that
DC = AC.
D D
By the Pythagorean Theorem in BCD, A
A
BD2 = a2 + b2 = c2, and so BD = c. c c
c b
So ACD is isosceles (since DC = AC) and b b c
b
ABD is also isosceles (AB = BD = c). As a
result, CDA CAD and BDA DAB. C a B C B
figure 3 figure 4
However, in figure 3 we have
m(BDA) < m(CDA) and m(CAD) < m(DAB), which gives m(BDA) < m(DAB) if
CDA and CAD are congruent. This is a contradiction of BDA DAB. Also,
in figure 4 we have m(DAB) < m(CAD) and m(CDA) < m(BDA), which gives
m(DAB) < m(BDA) if CAD and CDA are congruent. This is also a contradiction.
So our original assumption must be wrong, and so ABC is a right triangle.
EXAMPLE 36 In the triangle ABC opposite, K AC and
H
B
AN is the interior angle bisector of A.
AB = 16 cm, AN = 13 cm, AK = 12 cm and 16
N
NK = 5 cm are given. Find the area of
ABN. 13 5
A K C
Solution Let us draw an altitude NH from the vertex 12
N to the side AB.
NH AB
To find the area of ABN we need to find NH, because A( ABN ) = and AB is given
2
as 16 cm.
132 = 122 + 52, so m(NKA) = 90°. (Converse of the Pythagorean Theorem)
Also, NH = NK = 5 cm. (Angle Bisector Theorem)
NH AB 5 16
So A( ABN ) = = = 40 cm 2 . (Substitution)
2 2
Triangles and Construction 97
We can use the Pythagorean Theorem to prove other theorems. Here is one example.
Theorem Carnot’s Theorem
If a triangle ABC with P int ABC has
A
perpendiculars PK, PN, and PT drawn to the
sides BC, AC and AB respectively, then N
T
AT 2 + BK 2 + CN 2 = AN2 + BT 2 + CK2. P
B K C
Proof Let us join the point P to the vertices A, B A
and C and use the Pythagorean Theorem in
N
the six right triangles that are created. T
P
B K C
In ATP, AT2 + TP2 = AP2. (Pythagorean Theorem)
In ANP, AN + NP = AP ,
2 2 2
(Pythagorean Theorem)
AT2 + TP2 = AN2 + NP2. (1) (Transitive property of equality)
In BKP, BK2 + KP2 = BP2. (Pythagorean Theorem)
In BTP, BT + TP = BP ,
2 2 2
(Pythagorean Theorem)
2 2 2 2
BK + KP = BT + TP . (2) (Transitive property of equality)
In CNP, CN2 + NP2 = CP2. (Pythagorean Theorem)
In CKP, CK2 + KP2 = CP2, (Pythagorean Theorem)
2 2 2 2
CN + NP = CK + KP . (3) (Transitive property of equality)
From (1), (2) and (3) we get
AT2 + TP2 + BK2 + KP2 + CN2 + NP2 = AN2 + NP2 + BT2 + TP2 + CK2 + KP2 (Addition property
of equality)
AT 2 + BK 2 + CN 2 = AN2 + BT2 + CK2. (Simplify)
98 Geometriy 7
37
A
EXAMPLE Find the length x in the figure. 2
x N
T
P 2x
4
B 4 K 6 C
Solution AT2 + BK2 + CN2 = AN2 + BT2 + CK2 (Carnot’s Theorem)
x2 + 42 + (2x)2 = 22 + 42 + 62 (Substitute)
5x2 = 40 (Simplify)
x2 = 8
x = 2ñ2
Practice Problems 11
1. Find the length x in each figure.
a. D b. A c. P
A Pythagorean triple is a
set of three integers a, b x x 17
and c which satisfy the 8 7 10
Pythagorean Theorem.
The smallest and E 17 F B 24 C M S x N
6
best-known Pythagorean
triple is (a, b, c) = (3, 4, 5). C
K
d. e. f. M
8
7
6 N x
E 15
15 x
L 5
9
x
M A B N ò19 K 12 J
2. In the figure, TN = NK, ST = 12 cm S
and SN = ò69 cm. Find the length of TK.
12
ò69
T N K
3. In a right triangle ABC, m(A) = 90°, AB = x, AC = x – 7 and BC = x + 1. Find AC.
Answers
1. a. 15 b. 25 c. 9 d. 5ñ3 e. 20 f. 10 2. 10 cm 3. 5 cm
Triangles and Construction 99
b. Further properties
Activity Paper Folding - Median to the Hypotenuse
Fold a corner of a sheet of paper, and cut along the fold to get a right triangle.
Label the vertices A, B and C so that m(B) = 90°.
Fold each of the other two vertices to match point B.
Label the point T on the hypotenuse where the folds intersect.
What can you say about the lengths TA, TB and TC? Repeat the steps with a different
right triangle, and see if your conclusions are the same.
Properties 7
1. The length of the median to the hypotenuse of a right triangle is equal to half of the length
of the hypotenuse.
2. a. In any isosceles right triangle, the length of the hypotenuse is ñ2 times the length of
a leg. (This property is also called the 45°-4
45°-9
90° Triangle Theorem.)
b. In any right triangle, if the hypotenuse is ñ2 times any of the legs then the triangle is
a 45°-45°-90° triangle. (This property is also called the Converse of the 45°-4 45°-9
90°
Triangle Theorem).
3. In any 30°-60°-90° right triangle,
a. the length of the hypotenuse is twice the length of the leg opposite the 30° angle.
b. the length of the leg opposite the 60° angle is ñ3 times the length of the leg opposite
the 30° angle. (These properties are also called the 30°-6
60°-9
90° Triangle Theorem.)
4. In any right triangle,
a. if one of the legs is half the length of the hypotenuse then the angle opposite this leg
is 30°.
b. if one of the legs is ñ3 times the length of the other leg then the angle opposite this
first leg is 60°. (These properties are also called the Converse of the 30°-6 60°-990°
Triangle Theorem.) A
5. The center of the circumscribed circle of
r
any right triangle is the midpoint of the
B C
hypotenuse of the triangle. r O r
100 Geometriy 7
EXAMPLE 38 In the figure, C
m(BAC) = 90°, 60°
m(C) = 60° and D
6x 1
BD = DC. 2x + 3
Find BC if AD = 2x + 3 and AC = 6x – 1.
A B
Solution Since AD is a median and the length of the median to the hypotenuse of a right triangle
1
is equal to half the length of the hypotenuse, AD = BC.
2
By the Triangle Angle-Sum Theorem in ABC, m(B) = 30°.
1
By the 30°-60°-90° Triangle Theorem, AC = BC because m(B) = 30° and BC is the
2
hypotenuse.
So by the transitive property of equality, AC = AD, i.e. 6x – 1 = 2x + 3 and so x = 1.
Finally, BC = 2 AC = 2 AD = 2 (2x + 3) = 10.
EXAMPLE 39 In the figure at the right, find m(ADC) if A
m(BAC) = 90°,
m(BAD) = 2x, 2x
m(ACB) = 3x and
3x
BD = DC.
B D C
1
Solution BC.
Since AD is a median, by Property 7.1 we have AD =
2
So AD = BD = DC. Hence DCA and BDA are isosceles triangles.
Since DCA is isosceles, m(DAC) = m(ACD) = 3x.
Additionally, m(BAC) = m(BAD) + m(DAC) by the Angle Addition Postulate.
So 2x + 3x = 90° and x = 18°.
By the Triangle Angle-Sum Theorem in DCA, m(ADC) + 3x + 3x = 180°.
So m(ADC) = 180° – (6 18)°, i.e. m(ADC) = 72°.
Triangles and Construction 101
EXAMPLE 40 One of the acute angles in a right triangle measures 16°. Find the angle between the
bisector of the right angle and the median drawn from the same vertex.
Solution Let us draw an appropriate figure. We need A
to find m(NAT).
45°
According to the figure,
AN is the angle bisector, AT is the 16°
B N T C
median, and m(BAC) = 90°.
m(ACB) = 16° by Property 5.3.
Property 5.3: Since AT is median to hypotenuse, AT = CT = BT.
In any triangle ABC, if
m(B) > m(C) or So ATC is isosceles.
m(B) < m(C) then
Therefore, by the Isosceles Triangle Theorem, m(TAC) = m(ACT) = 16°.
ha < na < Va.
Since AN is an angle bisector and m(BAC) = 90°, m(NAC) = 45°.
So m(NAT) = m(NAC) – m(TAC)= 45° – 16° = 29°.
EXAMPLE 41 In the figure, AB AC and AH BC. A
Given m(C) = 30° and BH = 2 cm, find the
length of HC.
30°
B 2 H ? C
Solution In ABC, since m(C) = 30°,
A
m(B) = 60°.
In ABH, since m(B) = 60°, 60°
4 30°
m(BAH) = 30°.
60° 30°
In ABH, by Property 7.3,
B 2 H C
AB = 2 BH = 2 2 = 4 cm.
8
This set square is in the
form of a 30°-60°-90° In ABC, again by Property 7.3,
triangle.
BC = 2 AB = 2 4 = 8 cm.
So HC = BC – BH = 8 – 2 = 6 cm.
102 Geometriy 7
EXAMPLE 42 Find the value of x in the figure. A
60°
10
x
Solution Let us draw an altitude from C to AB. 45°
In BHC, B 6ñ2 C
BC = ñ2 BH (45°-45°-90° Triangle
Theorem)
A
6ñ2 = ñ2 BH (Substitute)
BH = 6. (Simplify) H 60°
This set square is in the
form of 45°-45°-90° right
AB = AH + HB (Segment Addition
triangle.
Postulate) 45°
10 = AH + 6 (Substitute) B 6ñ2 C
AH = 4. (Simplify)
In AHC,
AC = 2 AH (30°-60°-90° Triangle Theorem)
AC = 2 4 (Substitute)
AC = x = 8. (Simplify)
EXAMPLE 43 Construct a 30° angle by using the Converse of the 30°-60°-90° Triangle Theorem.
Solution We will construct a right triangle in which one leg is half of the hypotenuse. Then by the
Converse of the 30°-60°-90° Triangle Theorem, the angle opposite this leg will measure 30°.
1. 2. 3. 4.
m m m m
B B r C B K C
n n n k
5. A 6. A 7. A
30°
r r
B
m m m
r r Kr Cr r K r C r K r C
B B
2 2 2 2 2 2 2 2
n n
n
Triangles and Construction 103
1. Draw a line m.
2. Construct a perpendicular to the line at any point. Name the line n and label the
intersection point of these lines B.
3. Draw an arc with center B and any radius r. Label the intersection point C of this arc and
the line m.
4. Construct the midpoint K of the segment BC and draw the line k perpendicular to m
through this point.
5. Draw an arc with center K and radius BC. Label the intersection point A of the arc and line n.
6. Join A and K to make AK = 2 KB.
7. By the Converse of the 30°-60°-90° Triangle Theorem, m(BAK) = 30°.
EXAMPLE 44 Construct the circumscribed circle of a given right triangle.
Solution 1. 2. 3.
A A A
M
B C
B C B C M
1. Label ABC with m(A) = 90° and m(B) > m(C).
2. Construct the perpendicular bisector of each side and label their point of intersection M.
3. Draw a circle with center M and radius MB. This is the circumscribed circle.
Activity Unsolved Problem - Kobon Triangles
The Kobon triangle problem is an unsolved problem in
geometry which was first stated by Kobon Fujimura. The
problem asks: What is the largest number of non-overlapping
triangles that can be produced by n straight line segments?
This might seem like a simple question, but nobody has yet
found a general solution to the problem.
The mathematician Saburo Tamura proved that the largest
integer below n(n – 2) / 3 is an upper bound for the number of
non-overlapping triangles which can be produced by n lines. If
the number of triangles is exactly equal to this upper bound,
the solution is called a perfect solution. Perfect solutions are A perfect Kobon solution with
known for n = 3, 4, 5, 7, 9, 13 and 15, but for other n-values 15 lines and 65 triangles
(T. Suzuki, Oct. 2, 2005)
perfect Kobon triangle solutions have not been found.
The perfect Kobon solution for five lines creates 5 (5 – 2) / 3 = 5 triangles. Can you find
this solution?
Make Kobon patterns with seven lines. Can you find the perfect solution for seven lines?
104 Geometriy 7
Check Yourself 12
1. In an isosceles right triangle, the sum of the lengths of the hypotenuse and the altitude
drawn to the hypotenuse is 27.3. Find the length of the hypotenuse.
2. In the figure, ABC is a right triangle with C
m(ABC) = 90° and CF = FE, and CE is the D
angle bisector of C. If m(ADB) = 102°, find 102° F
the measure of CAB.
A E B
3. One of the acute angles in a right triangle measures 48°. Find the angle between the
median and the altitude which are drawn from the vertex at the right angle.
4. In a triangle ABC, m(B) = 135°, AC = 17 cm and BC = 8ñ2 cm. Find the length of AB.
5. In a right triangle, the sum of the lengths of the hypotenuse and the shorter leg is 2.4.
Find the length of the hypotenuse if the biggest acute angle measures 60°.
6. In the figure, A
m(C) = 60°, D
x
HC = 4 cm and
2ñ3
DH = 2ñ3 cm. Find the length AC = x. 60°
B H 4 C
7. ABC in the figure is an equilateral triangle with A
x
DH BC, D
BH = 5 cm and
HC = 3 cm.
Find the length AD = x.
B 5 H 3 C
8. The distance from a point to a line k is 10 cm. Two segments non-perpendicular to k are
drawn from this point. Their lengths have the ratio 2 : 3. Find the length of the longer
segment if the shorter segment makes a 30° angle with k.
9. CAB is a right triangle with m(A) = 90° and m(C) = 60°, and D is the midpoint of
hypotenuse. Find the length of the hypotenuse if AD = 3x + 1 and AC = 5x – 3.
10.The hypotenuse of an isosceles right triangle measures 18 cm. Find the distance from the
vertex at the right angle to the hypotenuse.
Answers
1. 18.2 2. 22° 3. 6° 4. 7 cm 5. 1.6 6. 5 cm 7. 2 cm 8. 30 cm 9. 14 10. 9 cm
Triangles and Construction 105
D. THE TRIANGLE ANGLE BISECTOR THEOREM
Theorem Triangle Angle Bisector Theorem
1. The bisector of an interior angle of a A
triangle divides the opposite side in the
same ratio as the sides adjacent to the
angle. In other words, for a triangle ABC
and angle bisector AN, B N C
AB BN
= .
AC CN
2. In any triangle ABC, if the bisector AN of
A
the exterior angle A cuts the extension
of side BC at a point N, then
B C N
AB BN
.
AC CN
A
Proof of 1 We begin by drawing two perpendiculars NK
and NL from N to the sides AB and AC L
respectively, then we draw the altitude K
AH BC.
B H N C
AH BN
A( ABN ) 2 BN
= = (1) (Definition of the area of a triangle and simplify)
A( ANC ) AH NC CN
2
Now let us find the same ratio by using the sides AB and AC and the altitudes NK and NL.
Since N is the point on the angle bisector, by the Angle Bisector Theorem we have NK = NL.
NK AB
A( ABN ) 2 AB
= = (2) (Definition of the area of a triangle and simplify)
A( ANC ) NL AC AC
2
AB BN
= (By (1), (2) and the transitive property of equality)
AC CN
106 Geometriy 7
Proof of 2 We begin by drawing two perpendiculars NK K
and NL from N to the extension of the sides
A
AB and AC respectively, then we draw the
altitude AH BN.
B H C N
L
AH BN
A( ABN ) 2 BN
= = (1) (Definition of the area of a triangle and simplify)
A( ACN ) AH CN CN
2
Now let us find the same ratio by using the sides AB and AC, and the altitudes NK and NL.
Since N is the point on the angle bisector, by the Angle Bisector Theorem we have NK = NL.
NK AB
A( ABN ) 2 AB
= = (2) (Definition of the area of a triangle and simplify)
A( ACN ) NL AC AC
2
AB BN
= (By (1), (2) and the transitive property of equality)
AC CN
EXAMPLE 45 Find the length x in the figure. A
12 6
Solution = (Triangle Angle Bisector 12
8 x 8
Theorem)
3 6
=
2 x
B 6 D x C
x=4
Triangles and Construction 107
EXAMPLE 46 Find the length x in the figure.
A
x
12 4 4
Solution = (Triangle Angle Bisector
12+ x x D 12 B x C
Theorem)
3 1
= ; 3x =12+ x ; x = 6.
12+ x x
Properties 8
1. In any triangle ABC, if AN is an interior A
angle bisector then
AN2 = (AB AC) – (BN NC).
B N C
2. In any triangle ABC, if AN is an
exterior angle bisector then
A
AN2 = (BN CN) – (AB AC).
B C N
EXAMPLE 47 Find the length x in the figure. A
Solution By the Triangle Angle Bisector Theorem,
6 3 6 y
= x
y 2
y = 4.
By Property 8.1, B 3 D 2 C
x2 = 6 4 – 3 2 = 18
x = 3ñ2.
108 Geometriy 7
EXAMPLE 48 In the figure, m(CAB) = 2 m(ABC). C
Given that AC = 4 cm and AB = 5 cm, find
the length of BC. 4
A 5 B
Solution Let AD be the bisector of angle A.
Then m(B) = m(DAB) = m(DAC), since m(CAB) = 2 m(ABC).
So DAB is an isosceles triangle. Let AD = DB = x. If BC = a then CD = a – x.
By the Triangle Angle Bisector Theorem in BAC,
5 x C
= ax a
4 a–x
D
5(a – x) = 4x 4
x x
5a
x= . (1)
9
A 5 B
Now we can use Property 8.1:
x2 = 5 4 – x(a – x)
x2 = 20 – ax + x2
ax = 20. (2)
Substituting (1) into (2) gives
5a
a = 20 ; a2 = 36 ; a = 6 cm.
9
Check Yourself 13
1. In a triangle ABC, N is on side BC and AN is the angle bisector of A. If AB = 8 cm,
AC = 12 cm and BC = 10 cm, find AN.
2. In a triangle KMN, points M, Z, N and T are collinear, KZ is the angle bisector of the
interior angle K, and KT is the angle bisector of the exterior angle K. If MZ = 5 cm and
ZN = 3 cm, find NT.
3. MP is the angle bisector of the interior angle M of a triangle KMN. Find MN if
KP : PN = 3 : 4 and KM = 15 cm.
4. In a triangle ABC, point D is on side BC and AD is the bisector of the angle A. Find BD if
BA : AC = 5 : 3 and BD + DC = 8 cm.
5. ET is an angle bisector in a triangle DEF. Find the length of ET if DE = 14, EF = 21 and
DF = 15.
Answers
1. 6ñ2 cm 2. 12 cm 3. 20 cm 4. 5 cm 5. 4ò15
Triangles and Construction 109
EXERCISES 3. 2
A. The Concept of Congruence 5. Two triangles ABC and CMN are congruent to
1. Find two pairs of congruent figures in each each other with AB = 8 cm, CN = 3 cm, and
picture. Draw each pair. CM = AC = 6 cm. Find BC and MN.
a. b.
B. Constructions
6. Construct each angle.
a. 15° b. 105° c. 75° d. 37.5° e. 97.5°
7. a. Construct an isosceles triangle with base 5 cm
and perimeter 19 cm.
b. Construct an equilateral triangle with perimeter
2. B D M 18 cm.
8 L
C 10 b a+2
M 7 c. Try to construct a triangle with sides of length
70° (x+10°) (n+10°) 80° 2 cm, 3 cm and 6 cm. What do you notice?
A 20y 3 E K 10y+7 P Can you explain why this is so?
In the figure, polygon ABCDE is congruent to
d. Construct a triangle ABC with side lengths
polygon KLMNP. Find each value, using the
a = 5 cm and c = 7 cm, and m(B) = 165°.
information given.
e. Construct a right triangle ABC in which
a. x b. y c. n d. a e. b
m(C) = 90°, b = 5 cm and the hypotenuse
measures 7 cm.
3. ADE KLN is given. List the congruent f. Construct ABC with angles m(C) = 120°
corresponding segments and angles of these and m(B) = 45°, and side b = 5 cm.
triangles. g. Construct an isosceles triangle PQR with
vertex angle m(Q) = 40° and side QP = 3 cm.
4. A triangle ABC is congruent to a second triangle
KMN. Find the unknown value in each statement,
using the properties of congruence. 8. In each case, construct a triangle using only the
a. AC = 5m – 25, KN = 11 – m three elements given.
b. m(BCA) = 45° – v, m(MNK) = 2v – 15°, a. a, b, Vc b. a, b, hc
c. m(B) = 18°, m(M) = u – 4°, c. a, b, nC d. ha, hb, hc
d. BA = 22x – 30, MK = 3 – 2x e. A, a, ha f. Va, Vb, Vc
110 Geometriy 7
C. Isosceles, Equilateral and Right 14. In the figure, A
Triangles ABC is an equilateral
E
9. KMN is an isosceles triangle with base KN. The triangle, PE AC,
PD BC, and PF AB. P D
perpendicular bisector of the side MK intersects
the sides MK and MN at the points T and F, Given P(ABC) = 45,
respectively. Find the length of side KN if find the value of B F C
P(KFN) = 36 cm and KM = 26 cm. PD + PE + PF.
15. In the figure, A
D
BH = HC, ?
10 . In the figure, A AB = DC and
AB = BC, m(B) = 54°. 54°
AD = DB D Find m(BAC). B H C
and BE = EC.
If DC = 3x + 1 and
B E C 16. In the figure, A
AE = 4x – 1,
BE = EC and
find the length x.
AD bisects the interior 12 7
angle A. D
Given AB = 12 and B E C
11. In a triangle DEF, m(E) = 65° and m(F) = 15°, AC = 7, find the
and DK is an angle bisector. Prove that DEK is length of ED.
isosceles.
17. In the figure, A
O is the center of the
12. In a triangle KLM, m(LKM) = 30°, inscribed circle of D
E
m(LMK) = 70° and m(KLM) = 80°.
ABC, OB = CD, O ? 20°
O int KLM and KO = LO = MO are given.
AB = AC and
Find m(OKM), m(OML) and m(OLK).
m(ECD) = 20°.
B C
Find the measure of
BEC.
13. In the figure, A
ABC is an equilateral 18. In a right triangle ETF, the perpendicular
triangle. P 3 E bisector of the leg ET intersects the hypotenuse at
PE BC, PD AC, the point M. Find m(MTF) if m(E) = 52°.
5
PE = 3 and
PD = 5 are given. B D C
Find the length of one side of the equilateral 19. In triangle DEF, DE = EF, m(DEF) = 90° and
triangle. the distance from E to DF is 4.8 cm. Find DF.
Triangles and Construction 111
20 . TF is the hypotenuse of a right triangle TMF, and 25. In the figure, C
the perpendicular bisector of the hypotenuse AB = AC, 8
intersects the leg MF at the point K. Find AD = DB and D
m(KTF) given m(MTF) = 70°. CD = 8 cm. ?
Find the length of DB. A B
21. P
H
26. In a triangle ABC, BC = 7ñ2, m(BAC) = 30° and
m(BCA) = 45°. Find the length of the side AB.
15°
M N
27. The larger acute angle A in an obtuse triangle
In the figure, PMN is a right triangle, MH PN
ABC measures 45°. The altitude drawn from the
MH
and m(N) = 15°. Find . obtuse angle B divides the opposite side into two
PN
(Hint: Draw the median to the hypotenuse.) segments of length 9 and 12. Find the length of
the side BC.
22 . In the figure, D
?
BC = 12, A 28. In the figure, K
m(BAC) = 90°, m(M) = 150°,
m(ADC) = 90° and LM = 2 and
?
m(ABC) = 60°. 60° MN = 3ñ3. N
If AC is the angle B 12 C Find the length of KL. 150° 3ñ3
bisector of C, find the length of AD. L 2 M
23. In the figure, A
29. In the figure, A
m(A) = 30° and m(BAC) = 90°,
30°
AB = AC = 16 cm.
m(AHC) = 90°,
Find the value of 30°
H m(B) = 30° and
PH + PD. D B ? H 1 C
HC = 1 cm.
B P C Find the length of BH.
24. In the figure, D
? 30. In the figure, D 4 A
m(A) = m(B) = 60°,
C m(DBA) = 30°,
AD = 7 and 7
4 m(ABC) = 60° and 30°
BC = 4. 60°
60° 60° AD = 4. Find the
Find DC. B ? C
A B length of BC.
112 Geometriy 7
31. ABC in the figure is A 36. In a right triangle KLM, m(KLM) = 90° and the
x
an equilateral triangle. D perpendicular bisector of the leg LM cuts the
BH = 5 cm and hypotenuse at the point T. If m(LMK) = 20°,
HC = 3 cm are given. find m(TLK).
Find the length
AD = x. B H C
5 3 37. In the figure, A
ABC is an isosceles
30°
32. In the figure, A
triangle, AB = AC,
m(C) = 90°, m(BAC) = 30° and
m(BAD) = m(DAC), 2PE = PD = 4.
D
Find AC. E 2 4
BD = DA and
DC = 3 cm. Find the B ? D 3 C B P C
length of BD.
38. M
33. In the figure, A K
ABC is an equilateral L
6
triangle, BD = 4 cm
? 24°
and AE = 6 cm. Find E
N S P
the length of EC. ?
4
In the figure, NK is the bisector of the interior
B D C
angle N and NL = LK. m(NMP) = 90° and
m(P) = 24° are given. Find m(PSM).
34. In the figure, A
?
m(A) = 90°, D
60°
m(ADC) = 60°, 4 39. In the figure, A
D
m(B) = 30° and 30° m(BAE) = m(DAE) = 60°, 60° 60°
B C
BD = 4 cm. CB = 6 cm and B
Find the length of AD. CE = 3 cm. 6
Find the length of CD.
C 3 E
35. In the figure, A
ABC is an equilateral
triangle, PE AC and E
3
H 40. In the figure, D
?
PD AB. ? AB = BC C 30°
P
PD = 5 cm, m(ADB) = 30° and
5
PE = 3 cm and AC = 6. 6
P(ABC) = 48 cm are B D C Find the length of CD.
given. Find the length of PH. A B
Triangles and Construction 113
41. The base and a leg of an isosceles triangle 48. In the figure, A
E
measure 14 cm and 25 cm respectively. Find the ED = AC,
height of the altitude drawn to the base. BD = DC and
? 63°
m(C) = 63°. Find
B D C
m(EDC).
42. A rectangular opening in a wall has dimensions
21 cm by 20 cm. Can an empty circular service
tray with a diameter of 28 cm pass through the
49. In the figure, A
opening?
AB = AC = 18 cm,
PH = 5 cm and
43. Two ships are in the same port. One starts to PD = 4 cm.
H
travel due west at 40 km/h at 3 p.m. Two hours Find m(HPD). D
5 ?
later the second ship leaves port, traveling due 4
south at 60 km/h. How far apart will the ships be B P C
at 7 p.m.?
50. In the figure, A
44. The median drawn from the vertex at the right ABC is an equilateral
angle of a right triangle divides the right angle in triangle, AH = ED, ? E
the ratio 13 : 5. Find the smallest angle in this AE = EC and
triangle. CD = 4 cm. 4
B H C D
Find the length of
AB.
45. In a triangle DEF, m(DEF) = 120°, DE = 2ñ5
and DF = 8. Find the length of the side EF.
51. In the figure, C
AB = AC, ?
46. In the figure, A
AH = HB, P
BD = DC and
m(A) = 120° and 8
BC = 12. ? 120°
PB = 8 cm. Find
Find the length of AD. A H B
the length of CP.
B D C
47. In the figure, C 52. In the figure, A
AD = EB, AT = DB = DC and
? 36°
CD = DB and D m(A) = 36°. T
m(DEB) = 80°. Find Find m(DBT). E
80°
the measure of ACB. ?
A E B
B D C
114 Geometriy 7
53. A line m divides a segment KN with the ratio 2 : 3 59. In the figure, M
at the point M. Find the distances from the points NL = LK and K
K and N to m if KN = 40 and the angle between NK is the bisector L
m and the segment KN is 30°. of N. If 120°
?
m(NMP) = 90° N S P
and m(MSP) = 120º, what is m(P)?
54. In a triangle KMN, T is the intersection point of
the three angle bisectors and the distance from T
to the side MN is 4. Find the distance from T to 60. In the figure, D
the vertex K if m(K) = 60°. BE = EC, BD AC, A
30°
m(A) = 45°,
45° E
m(D) = 30° and x
6
55. Prove that in a right triangle ABC with m(A) = 90° DC = 6. Find the
and acute angles 15° and 75°, the altitude to the length x.
1 B C
hypotenuse measures of the length of the
4
hypotenuse.
D. The Triangle Angle Bisector Theorem
56. In the figure, C 15° 61. In a triangle ABC, D AB and CD is the interior
AC BD, angle bisector of C. Given AC = 9 cm, BD = 8 cm,
CE = 2AB and D AD = m and BC = n, find the value of m n.
E
m(C) = 15°.
?
Find m(CAF). F A B
62. In a triangle ABC, D lies on the side BC and AD is
the interior angle bisector of A. If AC = BD,
57. Prove that if AH is an A AB = 9 cm and DC = 4 cm, find the length of BD.
altitude and AD is a
median of a triangle
ABC then 63. In a triangle ABC, D lies on the side AC and BD is
|b2 – c2| = 2a HD. B H D C the interior angle bisector of B. If AD = 5 cm,
DC = 6 cm and the sum of the lengths of the sides
a and c is 22 cm, find the value of a.
58. In the figure, A
2
ABC is an equilateral D
?
triangle, m(BCE) = 15°, 64. In a triangle ABC, points C, A and D are collinear
EF = FC, DF EC and E and C, B and E are also collinear. BD is the angle
AD = 2 cm are given. F bisector of EBA. If AC = AD and AB = 6 cm,
Find AE. B C
find the length of BC.
15°
Triangles and Construction 115
65. In a triangle EFM, points F, N and M are collinear 71. In a triangle ABC, D AB, E BC and AE and CD
and EN is the interior angle bisector of E. are the angle bisectors of A and C respectively,
If EN = 6, NM = 3 and FN = 4, find the length
intersecting at the point K. If AD = 4, DB = 6 and
of EF. AK
AC = 8, find the value of .
KE
66. In a triangle ABC, D lies on the side BC and AD is
the interior angle bisector of A. If BD = 3 cm, 72. In a triangle ABC, points D, B and C are collinear
DC = 4 cm and AB + AC = 14 cm, find the length and AD is the angle bisector of A. If AB = 2m,
of AD. AC = 2m + 6, BC = 2m + 2 and DB = 28, find
the value of m.
67. In a triangle ABC, D lies on the side BC and AD is
the interior angle bisector of A. If AB = AD = 12 cm 73. In the figure, AN and BE A
and AC = 16 cm, find the length of BD. are the angle bisectors of
A and B respectively. 6
4
If AB = 4, E
?
AC = 6 and BC = 5,
68. In a triangle DEF, E, F and K are collinear and find the length of BE. B N C
DK is the exterior angle bisector of the angle D. If 5
DE = 4 cm, DF = 3 cm and EF = 2 cm, find DK.
74. In the triangle
A
69. In the figure, AD and A
ABC at the right,
BE are the angle AD is the angle
bisectors of A and bisector of A 4 N
12
B, respectively, E and BN is the 2
AE 3 angle bisector of
= and D m B 3 C
ED 2 B
B. DB = m,
D ? C
AC = 12 cm. BC = 3, AB = 4 and NC = 2 are given. Find the
Find the length of segment DC. value of m.
70. In a triangle ABC, points B, C and D are collinear 75. In a triangle ABC, points D, B and C are collinear
and AD is the angle bisector of A. If CD = 3 BC and AD is the angle bisector of A. If AB = 12
and AB = 12 cm, find the length of AC. and BD = 4 BC, find the length of AC.
116 Geometriy 7
Objectives
After studying this section you will able to
1. Describe and use relations between angles
2. Describe and use relations between angles
In this section we will look at some basic rules related to the basic and auxiliary elements of
a triangle.
Activity Angles in a Triangle
Do the following activities and then try to
A. RELATIONS BETWEEN ANGLES
K
M
find a common conjecture.
1. Cut out three identical triangles and N
K
M K
label their vertices K, M and N. Draw a M
® K
straight line and place the triangles N M
K
along the line as in the diagram at the N MN K N
far right. What can you say about the
sum of K, M and N? N M
2
2. Cut out an acute triangle, a right triangle
®
and an obtuse triangle. Number the
1 3 1 2 3
angles of each triangle 1, 2 and 3 and
tear them off. Finally, put the three 1
angles of each triangle adjacent to each ®
other to form one angle as in the 1
2 2
figures at the far right. What can you
1
‘It is indeed wonderful say about the sum of the angles in each
®
that so simple a figure triangle?
as the triangle is so 2 3 1 2 3
inexhaustible in 3. Cut out a triangle ABC and label its
properties. longest side BC. Fold the triangle so that A lies on the fold line and C lies on BC. Label
How many as yet
unknown properties of the intersection T of BC and the fold line. Unfold. Now fold the paper so that A, B and
other figures may there C coincide with T. How does this activity support the results of activities 1 and 2 above?
not be?’
August Crelle
(1780-1856),
civil engineer and
mathematician
Triangles and Construction 117
Theorem Triangle Angle-S
Sum Theorem
The sum of the measures of the interior angles of a triangle is 180°.
D A E
Proof Given: ABC
4 2 5
Prove: m(1) + m(2) + m(3) = 180°
We begin by drawing an auxiliary line DE
1 3
through A, parallel to BC. Then we continue B C
with a two-column proof.
Statements Reasons
1. 1 4; 3 5 1. Alternate Interior Angles Theorem
An auxiliary line is a
line which we add to a 2. m(DAE)=m(4)+m(2)+m(5)=180° 2. Angle Addition Postulate, by the definition of straight angle
figure to help in a proof.
3. m(1) + m(2) + m(3) = 180° 3. Substitute 1 into 2
EXAMPLE 49 In the figure, points A, B, F and E, B, C are A
respectively collinear. Given that C and F
a
are right angles, m(E) = 40° and m(A) = ,
find the value of . E
40° B C
Solution In EFB by the Triangle Angle-Sum Theorem, F
m(E) + m(F) + m(B) = 180°
A
40° + 90° + m(B) = 180°
a
m(B) = 50°.
By the Vertical Angles Theorem, E
40° C
50° B
m(EBF) = m(ABC)
50° = m(ABC). F
In ABC by the Triangle Angle-Sum Theorem,
m(A) + m(B) + m(C) = 180°
+ 50° + 90° = 180°
= 40°.
118 Geometriy 7
EXAMPLE 50 In a triangle ABC, m(A) = 3x – 10°, m(B) = 2x + 20° and m(C) = 5x.
Find the value of x.
Solution m(A) + m(B) + m(C) = 180° (Triangle Angle-Sum Theorem)
3x – 10° + 2x + 20° + 5x = 180°
10x + 10° = 180°
x = 17°
Activity
Complete the table for the figure at the right, using the
Triangle Angle-Sum Theorem and the Linear Pair Postulate. 3
m(4) m(3) m(2) m(1) m(3) + m(4)
75° 55°
1 2 4
63° 135°
77° 46°
39° 85°
What do you notice about the values in the last two columns of the table?
Theorem Triangle Exterior Angle Theorem
The measure of an exterior angle in a triangle is equal to the sum of the measures of its two
nonadjacent interior angles.
Proof Given: ABC 3
Prove: m(1) = m(3) + m(4)
1 2 4
The two interior angles
which are not adjacent
Statements Reasons
to an exterior angle in a
triangle are sometimes 1. m(1) + m(2) = 180°
called remote angles. 1. Linear Pair Postulate
m(2) = 180° – m(1)
exterior C
angle 2. m(2) + m(3) + m(4) = 180° 2. Triangle Angle-Sum Theorem
3. 180° – m(1) + m(3) + m(4) = 180° 3. Substitute 1 into 2.
A B
remote
angles 4. m(1) = m(3) + m(4) 4. Simplify.
Triangles and Construction 119
EXAMPLE 51 In a triangle ABC, AB AC and m(B) = 136°. Find m(C).
A
Solution m(B) = m(A) + m(C)
136° = 90° + m(C)
46° = m(C) 136°
B C
EXAMPLE 52 In the figure, AB = BD, AD = DC A
and m(DAC) = 35°. Find m(B). 35°
B D C
Solution m(DCA) = m(DAC) (Base angles of an isosceles triangle)
= 35°
m(BDA) = m(DAC) + m(DCA) (Triangle Exterior Angle Theorem)
= 35° + 35°
= 70°
m(BAD) = m(BDA) (Base angles of an isosceles triangle)
= 70°
m(B) + m(BAD) + m(BDA) = 180° (Triangle Angle-Sum Theorem)
m(B) + 70° + 70° = 180°
m(B) = 40°
EXAMPLE 53 In a triangle KMN, D lies on side KM. Decide whether each statement about the figure is
possible or impossible. If it is possible, sketch an example.
a. Triangles KDN and MDN are both acute triangles.
b. Triangles KDN and MDN are both right triangles.
c. Triangles KDN and MDN are both obtuse triangles.
d. Triangle KDN is obtuse and triangle KNM is acute.
Solution a. impossible b. possible c. possible d. possible
N N
N
10°
100° 55°
50° 30°
120° 80°
20° 50° 45°
K D M K D M K D M
120 Geometriy 7
Theorem Triangle Exterior Angle-S
Sum theorem
The sum of the measures of the exterior angles of a triangle is equal to 360°.
Proof Given: ABC
A A¢
Prove: m(A) + m(B) + m(C) = 360°
We will give a flow-chart proof.
B¢
C
B C¢
m(ÐA¢) = m(ÐB) + m(ÐC)
Triangle Exterior Angle Theorem
m(ÐA¢)+m(ÐB¢)+m(ÐC¢) = m(ÐA¢)+m(ÐB¢)+m(ÐC¢) =
m(ÐB¢) = m(ÐA) + m(ÐC)
= 2(m(ÐA)+m(ÐB)+m(ÐC)) =2 × 180° = 360°
Triangle Exterior Angle Theorem
Addition Property of Equality Triangle Angle-Sum Theorem
m(ÐC¢) = m(ÐA) + m(ÐB)
Triangle Exterior Angle Theorem
EXAMPLE 54 In the figure, m(TCA) = 120°,
K
m(KAB) = 5x and m(PBC) = 7x.
5x A
a. Find the value of x.
b. Find m(BAC).
120°
B
P C T
7x
Solution a. m(A) + m(B) + m(C) = 360° (Triangle Exterior Angle-Sum Theorem)
5x + 7x + 120° = 360°
12x = 240°
x = 20°
b. m(KAB) + m(BAC) = 180° (Linear Pair Postulate)
(5 20°) + m(BAC) = 180°
m(BAC) = 80°
Triangles and Construction 121
EXAMPLE 55 In the figure, A
m(A) = m, m
m(B) = n and m(C) = k.
Find the value of m if m + n + k = 280°. n
C
P B k T
Solution m(A) + m(A) = 180° (Linear Pair Postulate)
m(A) = 180° – m
180° – m + n + k = 360° (Triangle Exterior Angle-Sum Theorem)
n + k = 180° + m (1) A
180° m
m
Also, m + n + k = 280° (Given)
m + 180° + m = 280° (Substitute (1)) n
C
2m = 100° P B k T
m = 50°.
Check Yourself 14
1. The two acute angles in a right triangle measure 0.2x + 6.3° and 3.8x – 2.7. Find x.
2. The measures of the interior angles of a triangle are in the ratio 4 : 6 : 8. Find the degree
measures of these angles.
3. The vertex angle of an isosceles triangle measures 42°. An altitude is drawn from a base
angle to one of the legs. Find the angle between this altitude and the base of the triangle.
4. In an isosceles triangle, the angle between the altitude drawn to the base of the triangle
and one leg of the triangle measures 16° less than one of the base angles of the triangle.
Find the measure of the vertex angle of this triangle.
5. Two points E and F are drawn on the extension of the side MN of a triangle MNP such
that point M is between the points E and N and point N is between points M and F.
State which angle is the smallest angle in EPF if EM = MP, NF = NP, m(PMN) = 30°
and m(PNM) = 40°.
122 Geometriy 7
6. In a triangle DEF, point M lies on the side DF such that MDE and DEM are acute
angles. Decide whether each statement about the figure is possible or impossible. If it is
possible, sketch an example.
a. FME is an acute triangle.
b. FME is a right isosceles triangle.
c. FME and DME are both acute triangles.
d. DME is equilateral and EMF is isosceles.
e. DME is isosceles and DEF is isosceles.
7. In the figure, KLN is an isosceles triangle in a plane, N
m(KLN) = 120°, and L is the midpoint of the segment KM.
A point P is taken in the same plane such that MP = KL. Find
the measure of LPM when
120°
a. the distance between N and P is at its maximum. K L M
b. the distance between N and P is at its minimum.
8. One of the exterior angles of an isosceles triangle measures 85°. Find the measure of the
vertex angle of this triangle.
9. State whether each triangle is a possible or impossible figure. If it is possible, sketch an
example. If it is impossible, give a reason why.
a. A triangle with two obtuse exterior angles.
b. A triangle with one acute exterior angle.
c. A triangle with two right exterior angles.
d. A triangle with two acute exterior angles.
10.Find the value of x in each figure, using the information given.
a. A b. c. M
d. C
80° D
A
70° 72°
105° x x
3x 40° B 60°
B x B x
115° C N P A
E
Triangles and Construction 123
Answers
1. 21.6° 2. 40°, 60° and 80° 3. 21° 4. 74° 5. PEF
6. a. possible b. possible c. not possible d. possible e. possible
E E E
E
65° 60° 30°
40° 45°
20° 30°
40° 60° 55° 50° 45° 60° 60° 30° 30° 30°
D M F D M F D M F D M F
7. a. 30° b. 60° 8. 95°
9. a. possible b. possible
100°
130° 50° 150°
30° 15°
c. impossible because the third exterior angle would be 180°
d. impossible because the third exterior angle would have to be more than 180°
10. a. 45° b. 25° c. 27° d. 80°
So far we have looked at some basic properties of angles. Now we will study some other
useful and important properties.
Properties 3
1. For any triangle, the following statements are true:
a. The measure of the angle formed by the bisectors of two interior angles of the triangle
is 90° more than half of the third angle, i.e. in the figure,
m( BAC )
m(BKC) = 90° + . A
S
2
a
b. The measure of the angle formed by a
a 2
the bisectors of two exterior angles of a 90° +
2
triangle is 90° minus half of the third K
angle, i.e. in the figure,
B
m( BAC ) C
m(BTC) = 90° – .
2
c. The measure of the angle which is
formed by the bisector of one interior 90° a
2
angle and the bisector of a second T
exterior angle is the half the measure
of the third interior angle.
We can refer to properties 1a, 1b and 1c as the Angle Bisector Relations Theorem.
124 Geometriy 7
2. In any triangle, the measure of the angle
A
formed by the altitude and the angle
bisector which both extend from the
x
same vertex is equal to the half the
absolute value of the difference of the
other two angles of the triangle. B H N C
|m(ÐB) m(ÐC)|
x=
2
3. In any triangle KLM, if N is any point in
K
the interior of KLM then
a. m(LNM) = m(LKM) + m(KLN) N
+ m(KMN).
L M
b. m(KNM) = m(KLM) + m(LKN)
+ m(LMN).
c. m(KNL) = m(KML) + m(MLN) + m(MKN).
56
A
EXAMPLE The triangle ABC at the right has incenter O.
Find m(AOC).
80°
B C
Solution Since the incenter is the intersection of the A
angle bisectors, both AO and CO are bisectors.
By Property 3.1a,
m( B) 80°
m(AOC)= 90° + = 90° + =130 . O
2 2
80°
B C
EXAMPLE 57 In the figure, K is the intersection point of
the bisectors of the exterior angles at A
K
vertices A and B with m(A) = 120°
120°
and m(B) = 40°. Find m(BKA).
40°
B C
Triangles and Construction 125
Solution m(A) + m(B) + m(C) = 180° (Triangle Angle-Sum Theorem in ABC)
120° + 40° + m(C) = 180°
m(C) = 20° (1)
m( C )
m(BKA)= 90° – (Property 3.1b)
2
20°
= 90° – (Substitute (1))
2
= 80°
EXAMPLE 58 Find the value of x in the figure. A
3x
B
C
x
K E
Solution m(C) + m(C) = 180° (Linear Pair Postulate)
m(C) = 180° – 3x (1)
m( C )
m(AEB)= (Property 3.1c)
2
180° – 3 x
x= (Substitute (1))
2
5x = 180°
x = 36°
EXAMPLE 59 In the figure at the right, A
AN is an angle bisector,
m(ANC) = 100° and
m(B) = 2m(C).
100°
Find the value of x. x
B N C
126 Geometriy 7
Solution m(C) = x is given, so m(B) = 2x.
A
Let us draw the altitude AH BC.
Since ANC is an exterior angle of AHN, 10°
m(HAN) + m(AHN) = m(ANC)
2x 100° x
m(HAN) + 90° = 100°
B H N C
m(HAN) = 100° – 90° = 10°
| m( B) – m( C )|
m(HAN) = (Property 3.2)
2
| 2x x|
10° =
2
x
10° =
2
x = m(C) = 20°.
EXAMPLE 60 One of the acute angles in a right triangle measures 20°. Find the angle between the altitude
and the angle bisector which are drawn from the vertex of the right angle of the triangle.
Solution Let us draw an appropriate figure. In the A
figure at the right, A is the right angle, AN 20°
45°
is the angle bisector and 25°
m(NAB) = m(NAC) = 45°.
70° 20°
Let m(C) = 20°, then m(B) = 70° and
B H N C
m(HAB) = 20°.
Therefore, m(HAN) = m(NAB) – m(HAB) = 45° – 20° = 25°. This is the required angle
measure.
Note that we can also solve this example by using Property 3.2. This is left as an exercise for
you.
EXAMPLE 61 In a triangle KLM, prove that if N is a point in the interior of KLM then
a. m(LNM) = m(LKM) + m(KLN) + m(KMN),
b. m(KNM) = m(KLM) + m(LKN) + m(LMN) and
c. m(KNL) = m(KML) + m(MLN) + m(MKN).
Triangles and Construction 127
K
Solution Let us draw an appropriate figure.
3
Given: N is an interior point of KLM T
Prove: m(1) = m(2) + m(3) + m() 3+4 N
2 1 4
Proof:
Let us extend segment MN through N and L M
label the intersection point T of ray MN and segment KL.
m(LTN) = m(3) + m(4) (Triangle Exterior Angle Theorem)
m(1) = m(2) + m(3) + m(4) (Triangle Exterior Angle Theorem)
This means m(LNM) = m(LKM) + m(KLN) + m(KMN).
The proofs of b. and c. are similar. They are left as an exercise for you.
EXAMPLE 62 ABC is an equilateral triangle and a point D int ABC such that AD DC and
m(DCA) = 42°. Find m(BAD).
Solution Let us draw an appropriate figure. A
m(B) = 60° ( ABC is equilateral)
m(BCD) = 60° – 42°
D
= 18° (m(BCA) = 60°) 42°
60°
B C
m(ADC) = m(B) + m(BAD) + m(BCD) (Property 3.3)
90° = 60° + m(BAD) + 18°
m(BAD) = 12°
128 Geometriy 7
Check Yourself 15
1. Each figure shows a triangle with two or more angle bisectors. Find the indicated angle
measures in each case.
a. b. c.
P M
P M
x
50°
S
x
S
40° x 70°
Q R T K Q R
x=? x=? x=?
d. e. f.
S
E S
p M
y
P 2x
70°
x
Q R
M
x
y T K
T K
3x+70°
M
x=? x y = ? (in terms of p) y=?
2. In the triangle ABC at the right, AN is an angle bisector A
and AH is an altitude. Given m(C) – m(B) = 36°, find
m(HAN).
B N H C
3. A student draws the altitude and the angle bisector at the vertex of the right angle of a right
triangle. The angle between them is 18°. Find the measure of the larger acute angle in the
right triangle.
4. Find the value of x in the figure.
4x
x
2x 105°
Answers
1. a. 110° b. 80° c. 35° d. 40° e. p f. 80° 2. 18° 3. 63° 4. 15°
Triangles and Construction 129
B. RELATIONS BETWEEN ANGLES AND SIDES
Theorem longer side opposite larger angle
If one side of a triangle is longer than another side of the triangle then the measure of the angle
opposite the longer side is greater than the measure of the angle opposite the shorter side. In
other words, if two sides of a triangle have unequal lengths then the measures of the angles
opposite them are also unequal and the larger angle is opposite the longer side.
Proof Given: ABC with AB > AC A
Prove: m(C) > m(B)
K 3
2
1
B C
We begin by locating K on AB such that AK = AC. We then draw CK and continue with a two-
column proof.
Statements Reasons
1. AB > AC 1. Given
2. AKC is isosceles 2. Definition of isosceles triangle (AK = AC)
3. 3 2 3. Base angles in an isosceles triangle are congruent.
4. m(ACB) = m(2) + m(1) 4. Angle Addition Postulate
5. m(ACB) > m(2) 5. Definition of inequality
6. m(ACB) > m(3) 6. Substitution property
7. m(3) > m(B) 7. Triangle Exterior Angle Theorem
8. m(ACB) > m(B) 8. Transitive property of inequality
EXAMPLE 63 Write the angles in each triangle in order of a. b. D
their measures.
A
5 5
5
3
B 7 C E 4 F
Solution a. Since 7 > 5 > 3, m(A) > m(B) > m(C).
b. Since 5 = 5 > 4, m(E) = m(F) > m(D).
130 Geometriy 7
Theorem larger angle opposite longer side
If two angles in a triangle have unequal measures then the sides opposite them have unequal
lengths and the longer side is opposite the larger angle.
Proof Given: ABC with m(B) > m(C) A
Prove: AC > AB
We will give a proof by contradiction in
Trichotomy property paragraph form.
For any two real numbers
a and b, exactly one of
According to the trichotomy property, exactly
the following is true: one of three cases holds: AC < AB, AC = AB B C
a < b, a = b, a > b.
or AC > AB.
Let us assume that either AC = AB or AC < AB and look for a contradiction.
If AC < AB then m(B) < m(C) by the previous theorem. Also, if AC = AB then
m(B) = m(C) by the definition of an isosceles triangle.
In both cases we have a contradiction of the fact that m(B) > m(C). That means that
our assumption AC AB must be false. By the trichotomy property, it follows that AC > AB.
EXAMPLE 64 Order the sides of triangle in the figure A
according to their length.
c (2x + 40°) b
20° (3x 10°)
Solution m(A) + m(B) + m(C) = 180°
B a C
2x + 40° + 20° + 3x – 10° = 180°
5x = 130°
x = 26°
So m(A) = (2 26°) + 40° = 92° and m(C) = (3 26°) – 10° = 68°.
Since m(A) > m(C) > m(B), by the previous theorem we have a > c > b.
EXAMPLE 65 In the figure at the right, KN = KM.
K
Prove that KT > KM.
4
1 2 3
T N M
Triangles and Construction 131
Solution Given: KN = KM
Prove: KT > KM
Proof:
Statements Reasons
1. KN = KM 1. Given
2. 2 3 2. Base angles of isosceles triangle KNM
3. m(2) = m(1) + m(4) 3. Triangle Exterior Angle Theorem
4. m(2) > m(1) 4. By 3
5. m(3) > m(1) 5. Substitute 2 into 4.
6. KT > KM 6. By the previous theorem
EXAMPLE 66 Prove that in any triangle ABC, a + b + c > ha + hb + hc, where ha, hb and hc are the altitudes
to the sides a, b and c, respectively.
A
Solution Given: ABC with altitudes ha, hb and hc
Prove: (a + b + c) > (ha + hb + hc) D
E
Proof:
B H C
Look at the figure. By the previous theorem,
in right triangle BCD, BC > BD, i.e. a > hb; (1)
in right triangle AEC, AC > CE, i.e. b > hc; (2)
in right triangle ABH, AB > AH, i.e. c > ha. (3)
Adding inequalities (1), (2) and (3) gives (a + b + c) > (ha + hb + hc).
EXAMPLE 67 Find the longest line segment in the figure
A D
using the given angle measures. 59°
60°
62°
61°
63° 55°
B C
Solution In ABC, m(B) > m(A) > m(C) so AC > BC > AB. (1) (By the previous theorem)
In ADC, m(C) > m(A) > m(D) so AD > CD > AC. (2) (By the previous theorem)
Combining (1) and (2) gives us AD > DC > AC > BC > AB. So AD is the longest segment in
the figure.
132 Geometriy 7
Check Yourself 16
1. Write the measures of the angles in each triangle in increasing order.
a. A b. c.
P K
4
8 10 5 6
C
B 7 M N
9 M 8 N
2. Write the lengths of the sides of each triangle in increasing order.
a. A
b. N c.
50° S K
K 40°
80° 80° 70°
60° 40° 50°
B C
T
M
3. Find the longest line segment in each figure using the given angle measures.
a. A b. P c.
E D
60° 62° K A 57° 66°
60° D 60° 67° 59° 61°
B 65°
61° 60°
60° C
C M N B
Answers
1. a. m(B) < m(A) < m(C) b. m(M) < m(P) < m(N) c. m(N) < m(K) < m(M)
2. a. c < b < a b. n = m < k c. k < s = t 3. a. CD b. PK c. BC
Activity Triangle Inequality
For this activity you will need a piece of string and a ruler.
Cut the string into eight pieces of different lengths. Measure the lengths and label or
mark each piece with its length.
Take any three pieces of string and try to form a triangle with them.
Make a table to note the lengths of the pieces of string and whether or not they formed
a triangle.
Repeat the activity until you have two successes and two failures at making a triangle.
Look at your table. Which lengths of string together made a triangle? Which lengths
didn’t make a triangle? What conjecture can you make about the sides of a triangle?
Triangles and Construction 133
Properties 4 Triangle Inequality Theorem
In any triangle ABC with sides a, b and c, the following inequalities are true:
|b – c| < a < (b + c),
|a – c| < b < (a + c),
|a – b| < c < (a + b).
The converse is also true. This property is also called the Triangle Inequality Theorem.
EXAMPLE 68 Is it possible for a triangle to have sides with the lengths indicated?
a. 7, 8, 9 b. 0.8, 0.3, 1 c. 1 , 1 , 1
2 3
Solution We can check each case by using the Triangle Inequality Theorem.
a. |9 – 8| < 7 < (8 + 9) b. |0.8 – 0.3| < 1 < (0.8 + 0.3) c. This is impossible,
|8 – 9| < 8 < (7 + 9) |1 – 0.3| < 0.8 < (1 + 0.3) since
1 1
|7 – 8| < 9 < (7 + 8). |1 – 0.8| < 0.3 < (1 + 0.8). 1< + .
2 3
This is true, so by the This is true, so by the
Triangle Inequality Triangle Inequality Theorem
Theorem this is a this is a possible triangle.
possible triangle.
EXAMPLE 69 Find all the possible integer values of x in the A
figure.
10
5
B C
x
4
7
D
Solution In ABC, |10 – 5| < x < (10 + 5) (Triangle Inequality Theorem)
5 < x < 15. (1)
In DBC, |7 – 4| < x < (7 + 4) (Triangle Inequality Theorem)
3 < x < 11. (2)
The possible values of x are the elements of the common solution of inequalities (1) and (2),
i.e. 5 < x < 11.
So x {6, 7, 8, 9, 10}.
134 Geometriy 7
EXAMPLE 70 Find the greatest possible integer value of m A
in the figure, then find the smallest possible
integer value of n for this case. 9
n
m
B
6
D 8 C
Solution In ABD, |9 – 6| < m < (9 + 6) (Triangle Inequality Theorem)
3 < m < 15.
So the greatest possible integer value of m is 14.
In ADC, |8 – m| < n < (m + 8) (Triangle Inequality Theorem)
|8 – 14| < n < (14 + 8) (m = 14)
6 < n < 22.
So when m = 14, the smallest possible integer value of n is 7.
Properties 5
1. In any triangle ABC,
a. if m(A) = 90° then b2 + c2 = a < ( b + c).
b. if m(A) < 90° then |b – c| < a < b2 + c2 .
c. if m(A) > 90° then b2 + c2 < a < (b + c).
2. In any triangle ABC, if P int ABC A
then (BP + PC) < (BA + AC).
B C
3. In any triangle ABC, if m(B) > m(C) or m(B) < m(C) then ha < nA < Va.
A A
ha nA Va nA ha
Va
B H N D C B D N H C
Triangles and Construction 135
EXAMPLE 71 In a triangle ABC, m(A) > 90°, c = 6 and b = 8. Find all the possible integer lengths of a.
Solution Since m(A) > 90°, b2 + c2 < a < (b + c) by Property 5.1.
Substituting the values in the question gives 82 +6 2 < a < (8 + 6), i.e.
10 < a < 14. So a {11, 12, 13}.
EXAMPLE 72 In the triangle ABC shown opposite, A
P int ABC, AB = 10, AC = 8 and BC = 9.
Find the sum of all the possible integer val- 10 8
P
ues of PB + PC.
Solution In PBC, BC < (BP + PC) by the Triangle B 9 C
Inequality Theorem.
So 9 < BP + PC. (1)
In ABC, (PB + PC) < (AB + AC) by Property 5.2.
So PB + PC < 10 + 8. (2)
Combining (1) and (2) gives 9 < (PB + PC) < 18.
So the possible integer values for PB + PC are 10, 11, 12, 13, 14, 15, 16 and 17.
The required sum is therefore 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 = 108.
EXAMPLE 73 Prove that the sum of the lengths of the medians of a triangle is greater than half of the
perimeter and less than the perimeter.
Solution Let us draw an appropriate figure. A
Given: ABC with centroid G
a+ b+ c
Prove: < ( Va + Vb + Vc ) < ( a + b + c). E
G
F
Remember! 2
The centroid of a Proof:
triangle is the point of
intersection of its We need to prove two inequalities. B D C
medians.
136 Geometriy 7
a+ b+ c
Proof that < Va + Vb + Vc : We will use the Triangle Inequality Theorem three times.
2
c
In CEB, (CE + EB) > BC, i.e. (Vc + ) > a. (Triangle Inequality Theorem)
2
a
In ADC, (AD + DC) > AC, i.e. (Va + ) > b. (Triangle Inequality Theorem)
2
b
In ABF, (BF + FA) > AB, i.e. (Vb + ) > c. (Triangle Inequality Theorem)
2
c a b
So (Vc + Va+ Vb + + + ) > (a + b + c). (Addition Property of Inequality)
2 2 2
a+ b+ c
So ( Va + Vb + Vc ) > . (1) (Subtraction Property of Inequality)
2
Proof that Va + Vb + Vc < a + b + c: A
For the second part of the inequality, let us
draw another figure as shown at the right b
c Va
and extend the median AD through D to a
point K such that
B a D a C
AD = DK. (2) 2 2
Then join K and B. Now, A
BD = DC (AD is a median)
b
m(BDK) = m(ADC) (Vertical angles) c
Va
AD = DK. (By (2))
B D C
So by the SAS Congruence Postulate, a a
2 2
DBK DCA and so
b Va
|BK| = |CA| = b.
Then, in ABK,
K
2Va < b + c. (3) (Triangle Inequality
Theorem)
By considering the other medians in a similar way
we get 2Vb < (a + c) and 2Vc < (a + b). (4)
Adding the inequalities from (3) and (4) side by side gives us
2(Va + Vb + Vc) < 2(a + b + c). So (Va + Vb + Vc) < (a + b + c). (5)
a+ b+ c
Finally, by (1) and (5), < ( Va + Vb + Vc ) < ( a + b + c) as required.
2
Triangles and Construction 137
Remark
The result we have just proved does not mean that for a given triangle, the sum of the
medians can be anything between the half perimeter and full perimeter of the triangle. This
is because the lengths of the medians are directly related to the lengths of the sides. As we
will see in the next chapter, once we know the lengths of the three sides of a triangle then
we can calculate the lengths of its medians. Their sum is a fixed number.
EXAMPLE 74 Two towns K and N are on the same side of the river Nile. The residents of the two towns
want to construct a water pumping station at a point A on the river. To minimize the cost of
constructing pipelines from A to K and N, they wish to locate A along the Nile so that the
distance AN + AK is as small as possible. Find the corresponding location for A and show that
any other location requires a path which is longer than the path through A.
l l
K l K A C K A
C
X
B B
N N N
figure 1 figure 2 figure 3
Solution Let M and N be as in figure 1, and let be a line representing the river. Then we can use the
following method to locate A:
1. Draw a ray from N perpendicular to , intersecting at point B.
2. Locate point C on the extension of NB such that NB = BC.
3. Draw KC.
4. Locate A at the intersection of KC and , as shown in figure 2.
Now we need to show that A is really the location which makes AN + AK as small as
possible. Figure 3 shows an alternative location X on l. Notice that in KXC, (KX + XC) > KC
by the Triangle Inequality Theorem. So (KX + XC) > (KA + AC) (1) by the Segment Addition
Postulate. Since AB NC and NB = BC, NXC is isosceles with XC = NX (2). By the same
reasoning, NAC is isosceles with NA = AC (3). Substituting (2) and (3) into (1) gives us
(KX + XN) > (KA + AN). So A is the best location for the station.
EXAMPLE 75 Prove that for any triangle ABC, A
if P int ABC and x, y and z are as shown
in the figure then z
c b
(x + y + z) < (a + b + c) < 2(x + y + z), i.e. y
x P
a+ b+ c
< ( x + y + z) < ( a + b + c).
2 B a C
138 Geometriy 7
Solution In ABP, c < (x + z). (Triangle Inequality Theorem)
In APC, b < (y + z). (Triangle Inequality Theorem)
In BPC, a < (x + z). (Triangle Inequality Theorem)
So (a + b + c) < 2(x + y + z). (1) (Addition property of inequality)
Also, (x + y) < (c + b), (Property 5.2)
(y + z) < (a + c) and (Property 5.2)
(x + z) < (b + a). (Property 5.2)
So (x + y + z) < (a + b + c) (2) (Addition property of inequality)
As a result, (x + y + z) < (a + b + c) < 2(x + y + z), (By (1) and (2))
a+ b+ c
or equivalently, < ( x + y + z) < ( a + b + c).
2
Remark
The example that we have just seen shows an application of triangle inequality. But the result
we obtained does not mean that the value of x + y + z can be any number less than
a + b + c. In other words, the maximum value of x + y + z may be a lot less than a + b + c.
In fact, the maximum value of x + y + z is always less than the sum of the lengths of the
two longer sides of the triangle, because as the interior point moves towards one of the
vertices, two distances increase but the third distance decreases. When this interior point
reaches the vertex point, the distance to this point becomes zero and the sum of the distances
becomes the sum of the two sides which include this vertex. So the maximum value of
x + y + z will always be less than the sum of the length of the two longer sides.
Check Yourself 17
1. Two sides of a triangle measure 24 cm and 11 cm respectively. Find the perimeter of the
triangle if its third side is equal to one of other two sides.
2. Determine whether each ratio could be the ratio of the lengths of the sides of a triangle.
a. 3 : 4 : 5 b. 4 : 3 : 1 c. 10 : 11 : 15 d. 0.2 : 0.3 : 0.6
3. The lengths of the sides DE and EF of a triangle DEF are 4.5 and 7.8. What is the greatest
possible integer length of DF?
4. The base of an isosceles triangle measures 10 cm and the perimeter of the triangle is an
integer length. What is the smallest possible length of the leg of this triangle?
5. In an isosceles triangle KLM, KL = LM = 7 and m(K) < 60°. If the perimeter of the
triangle is an integer, how many possible triangle(s) KLM exist?
6. In a triangle ABC, AB = 9 and BC = 12. If m(B) < 90°, find all the possible integer
lengths of AC.
Answers
1. 59 cm 2. a. yes b. no c. yes d. no 3. 12 4. 5.5 cm 5. six triangles
6. AC {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
Triangles and Construction 139
EXERCISES 3 .3
A. Relations Between Angles 3. The two acute angles in a right triangle measure
2x x
1. Each figure below shows triangles with two or ( +5) and ( – 25) respectively. Find x and
3 4
more congruent sides. Find the value of x in each the measures of these angles.
figure, using the information given.
a. A b. 4. In a triangle ABC, the angle bisector of the interior
A
angle C makes an angle of 40° with the side AB.
48° Find the angle between the bisector of the exterior
x angle C and the extension of the side AB.
B D C B D E C
m(ÐBAC) = x
T
c. d. M
5. In a triangle KMN, the altitudes to sides KM and
x S
MN intersect each other at a point P. Find
x
m(KPN) if m(KNM) = 72° and m(NKM) = 64°.
9°
2x
6. In the triangle MNP M
M TM = TN N N S Q
opposite, 94°
S
?
e. S f. MS = MP,
M
26°
ST = TP, N T P
30°
15° m(M) = 94° and
m(N) = 26°. Find m(MST).
x x
T M J N R P
7. In the triangle ABC at A
M
g. Q h. the right,
52° BD = BE = BC and D
T
x
K
P
segment EB bisects
S F E
x B.
P M R If m(ACD) = 18°, 18°
J N K
find m(ABC). B C
8. In the triangle ABC at A
2. An angle in a triangle measures 20° less than the the right, 114° D
measure of the biggest angle in the triangle. The AB = AD = BE,
?
measure of the third angle is half the measure of m(A) = 114° and 60°
B E C
the biggest angle. Find the measures of all three m(B) = 60°. Find
angles. m(EDC).
140 Geometriy 7
9. In the figure, P 14. x, y and z are the exterior angles of a triangle.
PQ = PS = PR and Determine whether each ratio is a possible ratio
m(SPR) = 24°. Find 24° of x : y : z.
m(SQR). a. 2 : 3 : 5
b. 1 : 2 : 3
Q ? E R
c. 6 : 11 : 19
S d. 12 : 15 : 21
10. In the figure opposite, M
m(NMT) = 16°, 15. Find the value of x in
16° 44°
m(TMP) = 44°, the figure. 140°
S
m(P) = 38° and
145°
m(SNT) = 22°. 38°
x 130°
N T P
Find m(TSN). 22°
16. In an isosceles triangle KMN, the bisectors of the
11. The bisectors of the interior angles D and F in
base angles K and M intersect each other at a
a triangle DEF intersect at the point T. Find the
point T. Prove that m(KTM) = m(K).
measure of DEF if its measure is one-third of
m(DTF).
B. Relations Between Angles and Sides
17. Each figure shows the lengths of two sides of a
12. a, b and b are the measures of the interior angles
triangle. Write an interval for the possible length
of an isosceles triangle such that a and b are
of the third side.
integers and 24° < b < 38°. Find the smallest
possible value of a. a. b. c. K
A P
3
8 12
6 8 m
13. In a triangle DEF, point M is on the side DF and M
MDE and DEM are acute angles. 4
B a C R p S L
Draw an appropriate figure for each of the
following, if it is possible.
18. For each figure, state the interval of possible
a. FME is obtuse
values for the length x.
b. FME is equilateral
a. b. c.
A A
c. DME is equilateral and DEF is isosceles K
6 10 4 12
8 6
d. DME is isosceles and EMF is isosceles x M P x
B D x
5 7 7 8 B C
e. DME is isosceles and DEF is equilateral C N
5
D
4
Triangles and Construction 141
19. A triangle ABC has sides a, b and c with integer 24. In the figure, A
lengths. How many triangles can be formed such AB = 8 cm,
that b = c and a b = 18? AC = 10 cm, 8
10
BD = 3 cm,
CD = 7 cm and 2x + 1
20. In the figure, a, b 5 A B C
D
and c are integers. 9 BC = 2x + 1 cm. 3 7
c b D
Calculate the smallest 6 F
Find the sum of
possible value of 6 all the possible integer values of x.
B
a + b + c, using the a C
5 7
information given. E
25. In the figure, C
AB = 8 cm,
6
BC = 12 cm, 12
21. In the figure, A D
CD = 6 cm and
AC = 9 cm,
DA = 4 cm. 4 B
BC = a, c 9 8
Find the number A
AB = c and
of possible integer lengths of AC.
m(BAC) > 90°.
B a C
Find the smallest
26. In each case determine whether it is possible for
possible value of a + c if a, c .
a triangle to have sides with the lengths given.
a. 13, 9, 5 b. 5, 5, 14
22. In the figure, A
m(A) > 90° and c. 8, 8, 16.1 d. 17, 11, 6
6 10
m(C) > 90°. e. 0.5, 0.6, 1 f. 18, 18, 0.09
If AB = 6 cm, B D
AD = 10 cm, 12 5 27. A triangle has side lengths 2x + y, 2y + 3x and 2x.
BC = 12 cm and Which one is bigger: x or y?
C
CD = 5 cm, find
the sum of the all the possible integer lengths of 28. For each figure, order the numbered angles
the side BD. according to their size.
a. b. n1
2 1
23. In the triangle ABC A
20 n 2
at the right, 18
3 n+1
AD = 9 cm, x 1 3
y
20
BD = 6 cm, 9
DC = 8 cm,
c. d.
AC = x cm and B
8
C 1
6
D 4.20 ò17 15
AB = y cm. 16
Find the sum of the smallest and largest possible 3
2 3 1 2
integer values of x + y. 82
20
142 Geometriy 7
29. Determine whether each statement is true or 31. A student has five sticks, each with an integer
false. length. He finds that he cannot form a triangle
a. In a triangle ABC, if the measure of A is 57° using any three of these sticks. What is the shortest
and the measure of B is 64° then the shortest possible length of the longest stick, if
side of ABC is AB. a. the lengths of the sticks can be the same?
b. In a triangle KMN, if the measure of K is 43 b. all the sticks have different lengths?
and the measure of M is 47 then the shortest
(Hint: Use the Triangle Inequality Theorem.)
side of KMN is KM.
c. In a triangle ABC, if B is an obtuse angle and
AH BC then HA < AB.
d. If an isosceles triangle KTA with base KA has
32. How many distinct isosceles triangles have
TA < KA then the measure of T is always less
integer side lengths and perimeter 200 cm?
than 90°.
e. An angle bisector in an equilateral triangle is
shorter than any of the sides.
f. All obtuse triangles are isosceles.
33. How many triangles can be drawn if the length of
g. Some right triangles are equilateral. the longest side must be 11 units and all side
lengths must be integer values?
34. In the figure, D
AD = 5 cm, 5
8
AB = 12 cm, A
30. State the longest line segment in each figure. BC = 9 cm and
C
DC = 8 cm. ?
a. C b.
A If m(A) > 90° and
12
m(C) < 90°, find 9
(x 8)°
B all possible integer
10° values of BD.
(x + 10)° 110° 20°
B
A B C D
c. A d. A
20° 65° 70° 70°
B C
80° D
35. In a triangle ABC, AB = 8 cm, BC = x and AC = y.
60° If m(A) > 90° and x, y , find the
B C
D 10° smallest possible value of x + y.
Triangles and Construction 143
36. How many distinct triangles have integer side 38.
lengths and perimeter 11? x K
y 130°
M N
37. Prove each theorem. a. 130°, x b. y, 90°
a. Hinge Theorem:
c. y, x d. KM, MN
If two sides of one triangle are congruent to
two sides of another triangle, and if the included
angle of the first triangle is larger than the 39. A
included angle of the second, then the third
y
side of the first triangle is longer than the third
side of the second. M
30° x
b. Converse of the Hinge Theorem:
If two sides a and b of one triangle are B C
congruent to two sides d and e of another a. (x – 10)°, (y + 20)°
triangle, and if the third side of the first b. MB + MC, AB + AC
triangle is longer than the third side of the
second, then the angle between a and b is
larger than the angle between d and e. 40. In a triangle DTF, m(D) = 90° and m(F) < 45°.
a. 2 m(T), m(D)
b. FT, 2DF
c. FD, DT
Instructions for questions 38 to 42 d. m(F), 2m(T)
Each question gives two quantities to be compared,
separated by a comma. In each case, use the figure or
41. In a triangle DEF, m(D) > m(E) = m(F).
extra information to compare the quantities. Write
A if the first quantity is greater than the second, a. m(D), 60°
B if the first quantity is smaller than the second, b. m(E), 60°
C if the quantities are equal, or
D if the extra information is not enough for you to
42. In a triangle DEF, m(E) = 120° and EF > DE.
be able to compare the quantities.
All variables represent real numbers. Figures are a. 120°, 3 m(D)
generally not drawn to scale. b. 2 m(E), 3 m(D)
144 Geometriy 7
Objectives
After studying this section you will be able to:
1. Find the distance from a point to a line.
2. Find the distance between two parallel lines.
A. DISTANCE FROM A POINT TO A LINE
Theorem distance from a point to a line
Let A(x1, y1) be a point and d: ax + by + c = 0 be a line, then the distance from A to the
line d is
| ax1 + by1 + c |
l= .
a2 + b 2
Proof Let the distance of A(x1, y1) to the line y A(x1, y1)
d: ax + by + c = 0 be l = AH. d
a
Take C(x2, y2) = AD d. x2 = x1 and y2 = CD H
C is a point on the line ax + by + c = 0, so C(x1, |CD|)
ax1 + b CD + c = 0
a
b CD = –a x1 – c x
O B D
a c
CD x1 .
b b
a c
So we have the coordinates of C, C( x1 , – x1 – ).
b b
Now, is the inclination of d and = m(CBD) = m(CAH) (angles with perpendicular sides).
AH
In the right triangle ACH, cos = and AH = AC cos ...(1)
AC
Now, let’s find the equivalent expressions for AC and cos.
a c a c
AC = AD – CD = y1 ( x1 ), so AC = y1 x1 ...(2)
b b b b
We know sec2 = 1 + tan2
1 a
so cos = , and tan = m
2
1+ tan b
1 1
so cos = = ...(3)
a a
1+( )2 1 ( )2
b b
Triangles and Construction 145
Substituting (2) and (3) in (1),
l = AH = AC cos , and since l is the distance,
a c
y1 + x1 +
a c
l = ( y1 x1 )
1
= b b = ax1 + by1 + c .
b b a2 1 2 a2 + b 2
1+ a + b2
b2 b
EXAMPLE 76 Find the distance from the point O(0, 0) to the line x – y + 4 = 0.
Solution O(0, 0) = O(x1, y1). Using the formula,
| ax1 + by1 + c | |0 0+ 4| 4 4 2
l= = = = = 2 2.
a +b2 2 2
1 +( 1) 2
2 2
EXAMPLE 77 Find the distance from A(5, 2) to the line 3x – 4y + 5 = 0.
Solution A A(x1, y1). Using the formula,
| ax1 + by1 + c | | 3 5 4 2+5| 12
l= = = = 2.4.
a +b2 2
3 +42 2 5
EXAMPLE 78 The distance from A(12, 5) to the line 5x – 12y + 5k = 0 is ten units. Find the possible values
of k.
| ax1 + by1 + c | |5 12 – 12 5+5 k|
Solution l= = =10
a +b2 2
25+144
|5k |
l= =10, so |5 k|=130, i.e. 5 k= 130, and so k= 26.
13
Check Yourself 18
1. Find the distance from the point P(–2, 3) to the line 3x + 4y + 9 = 0.
2. Find the distance from the point A(1, 4) to the line y = 3x – 4.
3. The distance between the point P(k, 3) and the line 4x – 3y + 5 is 4 units. Find k.
Answers
10
1. 3 2. 3. k {–4, 6}
2
146 Geometriy 7
B. DISTANCE BETWEEN TWO PARALLEL LINES
Let d1: a1x + b1y + c1 = 0
d2: a2x + b2y + c2 = 0 be two parallel lines.
a b
Since d1 d2, we can write 1 1 k , so a1 = k a2 and b1 = k b2.
a2 b2
Now, let’s substitute these values into d1:
k a2x + k b2y + c1 = 0
c
k(a2x + b2y + 1 ) = 0.
k
c1
k 0, so we get d1: a2x + b2y + = 0.
k
When we compare d1 with d2, we see that their difference is a constant number.
In general, we can write two parallel lines d1 and d2 as:
d1: ax + by + c1 = 0
d2: ax + by + c2 = 0.
Proof The distance of any point A(x, y) on line d1 to the line d2 is
A(x, y)
| ax + by + c2 |
. d1
a2 + b 2 l
In the equation of d1: ax + by + c = 0
| c2 c1 | d2
ax + by = – c1, and so l = .
2 2
a +b
Remark
It is important to notice that to find the distance between two parallel lines, first of all we
need to equalize the coefficients of x and y.
EXAMPLE 79 Find the distance between the parallel lines x – 2y + 5 = 0 and 3x – 6y + 9 = 0.
Solution d1: x 2y +5 = 0 3x 6 y +15 = 0
d1 was multiplied by 3.
d2 : 3x 6 y +9 = 0 3x 6 y +9 = 0
Now, we have c1 =15 | c1 c2 | 15 – 9 6 6 2 5
l= = = = = .
c2 = 9 a b
2 2
9+ 36 45 3 5 5
Theorem distance between two parallel lines
Let d1: ax+ by + c1 = 0 and d2: ax + by + c2 = 0 be two parallel lines. Then the distance
|c c |
between d1 and d2 is l= 2 1 .
a2 + b 2
Triangles and Construction 147
EXAMPLE 80 Find the distance between the parallel lines 3x – 2y + 5 = 0 and –3x + 2y + 8 = 0.
Solution d1: 3x 2y +5 = 0 d1: 3x 2 y +5 = 0
d2 : 3x + 2 y +8 = 0 d2: 3 x 2 y 8 = 0
So c1 = 5 | c1 c2 | |5+8| 13
l= = = = 13.
c2 = –8 a b
2 2 2
3 +(–2)2
13
There is also another way to solve the problem:
The distance between d1 and d2 is the same as the distance of any point on d1 or d2 to the
other line.
For example, A(0, – 4) is one point on d2, and the distance of A to d1 is
| 3 x1 2 y1 +5| | 3 0 2(– 4)+5| 13
l= = = 13. The solution is the same.
2
3 +(–2) 2
13 13
Check Yourself 19
1. Find the distance between the lines 4x – 3y – 5 = 0 and –12x + 9y + 4 = 0.
2. The lines x + 2y + 1 = 0 and 3x + 6y + k = 0 are parallel and the distance between
them is ñ5. Find k.
3. Find the area of the square whose two sides are on the parallel lines 2x + y – 2 = 0 and
4x + 2y + 6 = 0.
Answers
11
1. 2. k {–12, 18} 3. 5
15
148 Geometriy 7
EXERCISES 3.4
A. Distance from a Point to a Line 8. The distance between the parallel lines
1. Find the distance from the point A(–2, 3) to the 12x + 9y – 2 = 0 and ax + 3y + c = 0 is three
a
line 8x + 6y – 15 = 0. units. Find the ratio , if c 0.
c
2. The distance between B(2, 3) and the line 9. Write the equations of the lines which are four
units away from the line 3x + 4y + 10 = 0.
5
12y – 5x = k is . Find k.
13
10 . The distance between the parallel lines
3x + 4y – 6 = 0 and 4x – ky + 4 = 0 is p. Find
3. The distance from a line with equation k + p.
y – 4 = m(x + 2) to the origin is 2. Find m.
4. The distance between P(1, –2) and the line
7x – y + k = 0 is 4ñ2 units. Find k.
5. The points A(1, 3), B(–2, 1) and C(3, –1) are the
vertices of the triangle ABC. Find the length of
the altitude of BC.
1
6. The distance from P( , k) to the line
2
12x + 9y – 1 = 0 is 2 units. Find k.
B. Distance Between Two Parallel
Lines
7. Find the distance between each pair of parallel
lines.
a. –2x + 3y – 4 = 0 and –2x + 3y – 17 = 0
b. x – y – 4 = 0 and –2x + 2y – 7 = 0
c. y = 2x + 1 and 2y = 4x – 3
Triangles and Construction 149
CHAPTER 3 REVIEW TEST A
1. In the triangle ABC in A 5. Two sides of a triangle have lengths 8 and 12.
the figure, 4x What is the sum of the minimum and maximum
m(A) = 4x, x possible integer values of the length of the third
30°
m(B) = x and B C side?
m(C) = 30°. Find the value of x.
A) 24 B) 22 C) 19 D) 18 E) 16
A) 10° B) 15° C) 20° D) 25° E) 30°
6. Which is the longest D
C
2. In a triangle MNP, the interior angle bisectors of side in the figure, 62°
61°
M and P intersect at the point S. Given that according to the given
N measures 40°, find m(PSM). angle measures? 60° 60°
E
A) 95° B) 100° C) 105° D) 110° E) 120° A B
A) BC B) AB C) BD D) CD E) BE
3. In the triangle STK S
opposite, N TK and 7. In a triangle DEF, DE = EF and DF > EF.
SN is the interior angle Which statement is true?
bisector of S.
If m(T) – m(K) = 40°, ?
A) DE < (DF – EF) B) m(E) > m(F)
find m(SNK). T N K C) m(E) < m(D) D) m(E) = 60°
A) 110° B) 105° C) 100° D) 95° E) 90° E) m(E) = m(D)
4. In the figure, P 8. What is the sum of A
m(P) = 45°, the smallest and 8 10
45°
m(N) = 36° and largest possible integer
x
B C
m(R) = 25°. Find x values of x in the
the value of x. figure? 10
25° S 36° 14
N
R D
A) 260° B) 256° C) 254° D) 248° E) 244° A) 26 B) 24 C) 22 D) 20 E) 17
Chapter 3 Review Test A Angles
9. In a triangle ABC, D BC and AD bisects A. If 13. a, b and c are the lengths of the sides of a triangle
AB = 6 cm, BD = 3 cm and DC = 2 cm, find the ABC. Given that a, b and c are integers and
length of AD. a2 – b2 =17, what is the sum of the minimum and
maximum possible values of c?
A) 5ñ2 cm B) 4ñ3 cm C) 3ñ2 cm
A) 7 B) 13 C) 17 D) 18 E) 23
D) 2ñ3 cm E) ñ3 cm
14. In the figure, C
m(BAC) = 90°, 60°
10. In the figure, m(C) = 60° and D
6x 1
A
BD = DC, BD = DC. 2x + 3
E
AD = AE and Find BC if
? 20° AD = 2x + 3 and A B
m(C) = 20°. Find B D C
m(EDC). AC = 6x – 1.
A) 70° B) 65° C) 60° D) 45° E) 30° A) 6 B) 8 C) 10 D) 12 E) 14
15. In the figure, A
12°
m(BAC) = 90°,
11. If MNP STK, which of the following 4
m(BAD) = 12°,
statements is false? ?
BC = 8 cm and D B 8 C
A) MN ST B) MP TK C) NP TK AD = 4 cm. What is m(ABC)?
D) PNM STK E) KT PN A) 52° B) 54° C) 58° D) 60° E) 64°
12. In the figure, A 16. In the figure, M
BD bisects B, 80° ND = DP and
BD = BE and D HD 3
= .
DE = EC. 20° MH 3
If m(A) = 80° and, N H D P
B E C MH
m(ACD) = 20°, what What is ?
NP
is m(BDC)?
3 3 3 3 3
A) B) C) D) E)
A) 100° B) 110° C) 120° D) 140° E) 150° 2 3 4 5 6
Chapter Review Test 1A 151
CHAPTER 3 REVIEW TEST B
1. In the figure, 15° 5. In a triangle ABC, points D and E are the
A
AB = AD, midpoints of the sides AB and AC respectively.
C
AC = BC and ? DE = (x + 5)/4 and BC = 8x – 5 are given. What is
D
m(DAC) = 15°. Find the value of x?
m(C).
A) 1 B) 2 C) 3 D) 4 E) 5
B
A) 40° B) 45° C) 50° D) 60° E) 65°
6. ABC is a right triangle with m(A) = 90°, and AH
is the altitude to the hypotenuse. If m(C) = 30°
and BH = 2 cm, find HC.
2. In the triangle MNP M
in the figure, A) 4 cm B) 5 cm C) 6 cm
MS = NS and K
117° D) 7 cm E) 8 cm
KN = KP. R
If m(MRP) = 117°, ?
N S P
what is m(MNP)?
A) 39° B) 41° C) 43° D) 45° E) 47°
7. In the figure, A
AB = AD.
DE E
What is ?
EC
B D C
3. In a triangle ABC, D is a point on the side AB and CD
is the interior angle bisector of C. If AB = 15 cm 1 2 3 4
A) B) 1 C) D) E)
and 3 AC = 2 BC, find the length of DB. 2 3 2 3
A) 2 cm B) 3 cm C) 4 cm D) 6 cm E) 9 cm
8. In the figure, M
MS = SN and 40°
4. In the figure, P MP = PN. S
12°
MN = MP and If m(P) = 20°, ? K
K L 30° 20°
ML = MK. m(KMP) = 40° and
N P
If m(PLK) = 12°, m(KNP) = 30°,
?
what is m(LMN)? M N what is m(SKN)?
A) 18° B) 20° C) 24° D) 30° E) 36° A) 50° B) 45° C) 40° D) 35° E) 30°
Chapter 3 Review Test B Angles
9. In the figure, A 13. In the triangle MNP M
HK = KN, shown opposite, point
40°
m(DAC) = 40° and O is the center of the
m(HKB) = 20°. B D inscribed circle of O
C K S
Find m(BKD). MNP. ?
H N 6 8
If KS NP,
?
KN = 6 and N P
20°
K SP = 8, what is the length of KS?
A) 20° B) 30° C) 40° D) 60° E) 70°
A) 10 B) 12 C) 14 D) 16 E) 18
14. In the figure, C
10. In the figure, BD A
CH = HB,
bisects angle B. Given AD = 3 and H
12 ?
m(ADB) = 90°, E
DB = 8. What is the
D ?
DE BC, sum of the all
AB = 12 and B 16 C
possible integer values
A 3 D 8 B
BC = 16, find the length of DE. of the length AC?
3 5
A) 1 B) C) 2 D) E) 3 A) 30 B) 34 C) 40 D) 42 E) 51
2 2
15. In the figure,
A E
ACDE is a square,
11. In the figure, M
m m(ABC) = 60° and
N BD = 2 cm. Find the
MK = KL, 60°
MN = m, ? 3m length of one side of B C D
2m
KN = 2m and the square. 2
NL = 3m. Find K L
A) (3 – ñ3) cm B) (ñ3 – 1) cm C) (ñ3 + 1) cm
m(KNL).
D) (4 – 2ñ3) cm E) (2ñ5 – 3) cm
A) 45° B) 50° C) 60° D) 70° E) 75°
16. In the figure, E
AE = BD = DC and A
12. The lengths of the sides of a triangle ABC are AB = AC. F
integers a, b and c such that b = c and
What is m(FDC)?
(a + b + c) (a + b – c ) = 15. ?
Find the value of a. B D C
A) 1 B) 2 C) 3 D) 5 E) 7 A) 45° B) 50° C) 60° D) 62.5° E) 67.5°
Chapter Review Test 1A 153
CHAPTER 3 REVIEW TEST C
1. In the figure, A 5. In the figure, ABC, A
m(KBC) = m(KCA). CDE, and FEG are F
and m(LKB) = 80°. equilateral triangles.
L
What is the measure of K If BG = 16, what is
80°
ACB? the sum of the B C E G
B C perimeters of the
D
three triangles?
A) 40° B) 60° C) 70° D) 75° E) 80°
A) 32 B) 36 C) 42 D) 46 E) 48
2. Which is the largest S
6. In the triangle ABC in A
N
the figure, CD is the
angle in the figure, 8
8 D
bisector of C, AE is F
according to the given P 2ò22
12 the median to BC and 50°
lengths?
6 B E C
9 DE AC.
M
If m(B) = 50°, what is m(BAC)?
K
A) 30° B) 35° C) 40° D) 45° E) 50°
A) M B) N C) S D) SPK E) K
7. Which of the A
10 13
following is a possible E
3. In an isosceles triangle XYZ, m(Y) = m(Z) and sum of the lengths of 7 C
B
m(X) < m(Y). What is the largest possible AB and BC in the 6 12
D
integer measure of the angle Y? figure?
A) 59° B) 60° C) 89° D) 90° E) 110° A) 11 B) 12 C) 13 D) 14 E) 37
8. In the figure, M
4. In a triangle ABC, points B, C and D are collinear MP = PS = SN = PT S ?
and AD is the angle bisector of the exterior angle and ST = TN.
A. If AC = BC, DB = 12 and AB = 4, find the What is m(NMP)?
length of BC. N T P
A) 2 B) 3 C) 4 D) 5 E) 8 A) 36° B) 60° C) 72° D) 84° E) 108°
Chapter 3 Review Test C Angles
9. In the figure, ABC is A 13. In the figure, A
an equilateral triangle. E
AF = FB and
F E
If BD = AE, what is AE = EC.
the measure of EFC? D F ?
If EH + FH = 12,
what is AB + AC? B H C
B C A) 16 B) 18 C) 22 D) 24 E) 36
A) 45° B) 60° C) 75° D) 90° E) 120°
14. In the triangle ABC in A
10. In the figure, A
18° the figure, BH is the
AC = BC and E
exterior angle bisector ?
AB = AD.
of B and B
If m(CAD) = 18° C C
? AE = EC.
and m(EBD) = 12°, E
If m(BHC) = 90°,
what is m(AEB)? 12° H
B D BC = 8 and
EH = 7, what is the length of AB?
A) 82° B) 80° C) 78° D) 72° E) 42°
A) 6 B) 7 C) 8 D) 10 E) 12
11. In the figure, A
O is the incenter of
ABC, AB OT and 15. In the figure, M
AC OV. O MK = NK = PK. x
If BT = 6 cm, What is x + y + z?
K
TV = 7 cm and z
y
VC = 5 cm, what is B 6 T 7 V 5 C
N P
the perimeter of the triangle OTV?
A) 270° B) 180° C) 90° D) 60° E) 45°
A) 12 cm B) 15 cm C) 16 cm
D) 18 cm E) 20 cm
16. P
12. In the figure, ABC is N
T
an equilateral triangle. A
10
If PB = 16 and 150°
?
P ?
PN = 10, what is the M S N
16 H
length of AH?
In the figure, m(M) = 90°, m(MST) = 150°
K B C and PM = MS = ST. What is m(N)?
A) 2ñ3 B) 3ñ3 C) 4ñ3 D) 5ñ3 E) 6ñ3 A) 5° B) 10° C) 15° D) 22.5° E) 30°
Chapter Review Test 1A 155
CHAPTER 3 REVIEW TEST D
1. In the triangle ABC in A 5. In the triangle MNP M
the figure, BN is the opposite, MK = TK,
bisector of ABC and N
NS = TS and
110° T K
H is the intersection H m(KTS) = 50°. 50°
point of the altitudes 20° What is m(MPN)?
?
of ABC. B C
N S P
If m(AHC) = 110° and m(HBN) = 20°,
A) 70° B) 65° C) 60° D) 55° E) 50°
what is m(BAC)?
A) 50° B) 55° C) 65° D) 75° E) 80° 6. According to the D
m
figure, what is A
2. In the figure, K n
the value of 8
MN = MP, KP = KT, M
k (2 m + n ) ?
m(NMP) = m, E B 12 A
m m
m(PKT) = k, and
? 1 1 1
points N, P and T are A) 5 B) 2 C) 2 D) 1 E)
2 2 2
collinear. N P T
If m + k = 130°, what is m(MPK)?
7. The measure of one angle in a triangle is equal to
A) 50° B) 55° C) 60° D) 65° E) 70°
the sum of the measures of the other two angles.
Which statement about this triangle is always
3. In the figure, A
true?
AB = AC = b,
BC = a, and a < b. A) The triangle is equilateral.
What is the largest b b
B) The triangle is acute.
possible integer value
C) The triangle is a right triangle.
of m(A)?
B a C D) The triangle is obtuse.
A) 59° B) 60° C) 44° D) 30° E) 29° E) The triangle is isosceles.
4. In the figure, E 8. In the triangle XYZ in X
AD = BD, 110°
A the figure,
P
m(DAC) = x and m(YZX) = 90°,
x
m(BCE) = 2x. XZ = PK and ?
If m(EAB) = 110°, XP = PY. Y K Z
2x
what is the value of x? What is m(PKZ)?
B D C
A) 30° B) 35° C) 40° D) 45° E) 50° A) 120° B) 135° C) 140° D) 150° E) 160°
Chapter 3 Review Test D Angles
9. In the figure, PM is the K 13. In the figure, A
M
angle bisector of NPK, m(DAC) = m(B) 130°
MN = MP, NS = SP T and
?
and m(MKP) = 90°. ? m(EAB) = m(C). B D E C
What is m(STP)? N S P
If m(AEC) = 130°, what is m(ADE)?
A) 90° B) 85° C) 80° D) 75° E) 60° A) 50° B) 55° C) 60° D) 70° E) 80°
10. In the figure, A
14. In the figure, M
KS = KN,
ABC is an equilateral 70°
m(M) = 70°, K 2x
triangle and BD = CE.
6ñ3 m(P) = x and
If AD = 6ñ3, what is
m(MKS) = 2x. x
the length of DE? C
B D What is the value of x? N S P
?
E A) 55° B) 60° C) 65° D) 70° E) 75°
A) 6 B) 8 C) 4ñ3 D) 13 E) 6ñ3
15. In the figure, C
A
AD and CB bisect
11. In the figure, A 75°
angles A and C,
AB = BC,
respectively. 30° E ?
DE = BE
D If m(AEC) = 75° and
m(ABD) = 36° and B D
48° m(B) = 30°,
m(EDC) = 48°. 36°
? what is m(ADC)?
What is m(ACB)? B E C
A) 5° B) 10° C) 15° D) 20° E) 25°
A) 76° B) 72° C) 68° D) 58° E) 52°
16. In the figure, M
12. In the equilateral A
KL = LM and
triangle ABC in the LH = MH. N
figure, AF = FC and F If NH = 5 and 5 H
AH = BD. What is the E
m(K) = 30°,
measure of EDC? ? what is KM? 30°
D B H C K L
A) 5° B) 10° C) 15° D) 20° E) 30° A) 15 B) 20 C) 25 D) 30 E) 40
Chapter Review Test 1A 157
CHAPTER 3 REVIEW TEST E
1. In the figure, A 5. In the figure, points K, A
DE = DC and 45° E S, T, M, N and P are
F
DB = BF. the midpoints of the
K M T
If m(A) = 45°, ? sides on which they lie.
D B C
what is m(ABC)? If AB = 12, N P
AC = 8 and
A) 30° B) 45° C) 50° D) 60° E) 75° B S C
BC = 16, what is
P(MNP)?
2. In the triangle ABC at A A) 6 B) 8 C) 9 D) 10 E) 12
x+13°
the right,
AB = AC,
6. In the figure, KL RS P
m(A) = x + 13° and
and KM bisects RKL. ?
m(B) = y – 38°. y38°
If KL = 6, K 6 L
What is the sum of the B C
KR = 4 and
minimum integer value of y and the maximum 4
MS = 8, what is the
integer value of x? R M S
length of PK? 8
A) 248° B) 243° C) 240° D) 233° E) 204°
A) 4 B) 5 C) 6 D) 7 E) 8
3. According to the
7. In the figure, A
figure, what is the 130° m(BAC) = 90°,
value of x? 105°
m(C) = 15° and ?
x 140°
BC = 24. What is 15°
B H C
the length of AH?
A) 10° B) 15° C) 20° D) 25° E) 30° 24
A) 4 B) 5 C) 6 D) 8 E) 12
4. In the figure, A
ABC is an equilateral 8. According to the 4 A
8
D
D figure, what is the F
triangle and
c b
AD = EC = CF. E smallest possible value 5 5
If BC = 12, of a + b + c if a, b C
B a
what is the length of B 12 C ? F and c are integers? 4
6
CF? E
A) 2 B) 3 C) 4 D) 5 E) 6 A) 7 B) 8 C) 9 D) 10 E) 11
Chapter 3 Review Test E Angles
9. In the figure, A 13. In the figure, C
AH = BH = HC. BD = DC,
If AC = 1, D 1 CE = 3AE and
D
what is HD? ? 2AB = AC. ?
B H C If m(A) = 90°, what E
is m(DEC)?
1 1 1 3
B) 1
A B
A) C) D) E)
2 3 4 5 5 A) 30° B) 40° C) 45° D) 50° E) 60°
10. In the triangle ABC in A 14. In the figure, A
the figure, m(A) = 90°,
? 6+3ñ3 ?
CD AB and D E m(B) = 15° and
F
BE AC. AB = 6 + 3ñ3. 15°
140° B C
If m(BFC) = 140°, What is the length
what is m(A)? B C of AC?
A) 20° B) 30° C) 40° D) 45° E) 50° A) 1 B) 2 C) 3 D) 4 E) 5
15. In the figure, M
11. In the figure, ABC is A
MK = KP, K
an equilateral triangle m(M) = 90°,
and DEFH is a square. ? H ?
D NS = 9 cm and
K
Find the measure of SP = 3 cm. Find the N 9 S 3 P
AKD. length of KS.
B E F C
A) 2 cm B) 3 cm C) 4 cm D) 5 cm E) 6 cm
A) 65° B) 67.5° C) 70° D) 75° E) 80°
16.In the figure, CD is the A
15°
12. In the figure, D C bisector of C.
60°
ABCD is a square and F
If m(BAE) = 15°, D
? E
ABF and BEC are m(EAC) = 60° and
?
equilateral triangles. m(B) = 45°, what is 45°
What is m(FEC)? A B m(DEA)? B E C
A) 5° B) 10° C) 15° D) 20° E) 22.5° A) 10° B) 15° C) 20° D) 22.5° E) 30°
Chapter Review Test 1A 159
160 Angles
Objectives
After studying this section you will be able to:
1. Define the concept of a circle and its basic elements.
2. Describe and use the properties of chords.
3. Describe and use the properties of tangents.
4. Describe the possible relative positions of two circles in the same plane.
A. BASIC CONCEPTS
1. Definition
You can see many circular or ring-shaped geometric figures all around you. For example,
wheels, gears, compact discs, clocks, and windmills are all basic examples of circles in the
world around us.
wheels compact disc gears
It is easy to recognize a circle, but how can we define it as a shape? Let us look at a
geometric definition.
Note
The word ‘circle’ is derived from the latin word circus, which means ‘ring’ or ‘racecourse’.
Definition circle
A circle is the set of all the points in a plane that are at the same distance from a fixed point
in the plane. The distance is called the radius of the circle (plural radii), and the fixed point
is called the center of the circle.
All radii of a circle are congruent. A circle is named by its
center. For example, the circle on the left is named center
circle O. O
We write a circle with center O and with radius r as radius
circle
or C(O, r).
In this book, the point O in a circle is always the center
of the circle.
162 Geometriy 7
To construct a circle, fix a pin on a piece of paper, connect a string of any length
to the pin, tie the other end of the string to your pencil, and turn your pencil
on the paper around the pin for one
complete revolution, keeping the string
taut. You will get a circle.
You can also use a compass to draw a circle.
Mark a point O as the center and set your compass to the r
O
length of the radius. Turn your compass around the
center for one complete revolution. You will get a circle.
2. Regions Separated by a Circle in a Plane
A circle divides a plane into three separate regions. The set of points whose dis-
tance from the center of a circle is less than the radius of the circle is called the
interior of the circle.
For example, if R is a point in the plane and P
Q
interior region circle
|OR| < r, then R is in the interior of the interior
circle. O P
The set of points whose distance from the R
center is greater than the radius of the exterior circle
exterior region circle is called the exterior of the circle.
For example, if Q is a point in the plane and |OQ| > r, then the point Q is in the exterior of
the circle. The set of points whose distance from the center is equal to the radius is called
the circle itself, and the points are on the circle. For example, if P is a point in the plane and
|OP| = r, the point P is on the circle.
Note
The union of a circle and its interior is called a circular closed region or a disc.
EXAMPLE 1 Name the points in the figure which are
E
a. in the interior of the circle. B
G O
b. on the circle. C
A
c. in the exterior of the circle. D
F
Solution a. Since |OA| < r and |OB| < r, points A and B are in
the interior of the circle.
b. Since |OC| = |OD| = r, points C and D are on the circle.
c. Since |OE| > r, |OF| > r and |OG| > r, points E, F, and G are in the exterior of the circle.
Circles 163
3. Auxiliary Elements of a Circle
Definition chord
A line segment which joins two different points on a B
circle is called a chord. A chord
C O
diameter D
For example, [AB] and [CD] in the figure are chords.
Definition diameter
A chord which passes through the center of a circle is called a diameter of the circle.
In the figure, chord [CD] passes through the center of the circle, so [CD] is a diameter.
We can see that the length of every diameter in a given circle is the same. For this reason,
we usually talk about ‘the diameter’ of a circle to mean the length of any diameter in the
circle.
The length of the diameter of a circle is twice the radius. For example, if r is the radius of a
d
circle and d is the diameter, then d = 2 r, or r = .
2
The diameter of a circle is the longest chord in the circle.
EXAMPLE 2 1. Find the length of the diameter for each 2. The length of the diameter of a circle is
given radius. 20 cm and the radius is 2x – 4. Find x.
a. 3 1 cm b. 3x cm
2
c. 2x + 5 cm d. 7x – 12 cm
Solution 1
1. a. d = 2 r d = 2 3 = 7 cm
2
b. d = 2 (3x) = 6x cm
c. d = 2 (2x + 5) = 4x + 10 cm
d. d = 2 (7x – 12) = 14x – 24 cm
2. d=2r
20 = 2 (2x – 4)
2x – 4 = 10
2x = 14
x = 7 cm
164 Geometriy 7
4. Relative Position of a Line and a Circle in the Same
Plane
A line and a circle in the same plane can have one of
H l
three different positions relative to each other.
If the distance from the center of the circle to the line O
is greater than the radius of the circle, then the line r
does not intersect the circle.
In the figure, [OH] l and |OH| > r,
and l C(O, r) = . tangent point of
tangency
If the distance from the center of the circle to the line H
l
is equal to the radius, then we say that the line is
tangent to the circle. In the figure, |OH| l, O
|OH| = r, and l C(O, r) = {H}. H is the only point of
intersection of the line and the circle.
Definition tangent
A line which intersects a circle at exactly one point is called a tangent of the circle. The inter-
section point is called the point of tangency.
If the distance from the center of the circle to the line
is less than the radius, then the line intersects the A H B
l
circle at two points. O
In the figure, [OH] and |OH| < r,
and C(O, r) = {A, B}.
Definition secant
A line which intersects a circle at two different points is called a secant of the circle.
For example, line is a secant in the figure on the left.
EXAMPLE 3 Name all the radii, diameters, chords, secants, and
E
tangents of the circle in the figure.
D
Solution [OF], [OC], and [OB] are radii. O
F C l
[FC] is a diameter. l is a secant line.
G
[EF], [ED], and [FC] are chords. GH is a
A B
tangent, and A is a point of tangency. H
Circles 165
Check Yourself 1
1. Define the terms center, radius, chord, diameter, tangent, and secant. Show them in a
figure.
2. How many regions does a circle divide the plane into?
3. Sketch all the possible relative positions of a circle and a line in the same plane.
4. Look at the figure on the right.
E
a. Name the tangents.
B
b. Name the secants.
A
c. Name the chords. O
D C
d. Name the radii. F
e. Name the diameters.
Answers
1. center: a point inside the circle that is equidistant from all the tangent
points on the circle.
radius: a distance from the center to a point on the circle. chord
secant diameter
chord: a line segment joining two different points of a circle.
diameter: a chord passing through the center of a circle
ra
di
us
tangent: a line intersecting a circle at exactly one point.
secant: a line intersecting a circle at two different points.
2. three parts: the interior of the circle, the circle, and the exterior of the circle.
n
3.
m B
C
l
A
4. a. EF, EB b. BC, DB c. [AB], [DB], [BC] d. [OD], [OA], [OB], [OC] e. [BD]
166 Geometriy 7
B. CHORDS
Remember that a chord is a line segment which joins two different points on a circle. In this
section we will look at the properties of chords.
Property
A radius that is perpendicular to a chord bisects the chord.
A H B
For example, in the figure, if [OH] [AB] then
|AH| = |HB|. O
EXAMPLE 4 A chord of length 10 cm is 12 cm away from the center
5 H
of a circle. Find the length of the radius. A B
12
r
Solution Look at the figure. O
| AB|
2
10
5 cm
2
In AHO, r2 = 52 + 122
r2 = 25 + 144
r2 = 169
r = 13 cm.
Property
In the same circle or in congruent circles, two chords which are equidistant from the center
are congruent.
C
For example, in the figure, if |OM| = |ON|, B N
then |AB| = |CD|.
O D
The converse of this property is also true: M
if |AB| = |CD|, then |OM| = |ON|.
A
Circles 167
EXAMPLE 5 In the figure, |AB| = 8 cm,
B C
|CN| = 4 cm, and 4
x
N
|OM| = 3 cm. Find |OC| = x. M
3 O
A D
Solution |CD| = 8 cm, since |CN| = 4 cm. So |AB| = |CD|,
and by the property, |OM| = |ON| = 3 cm.
Let us use the Pythagorean Theorem to find the length of [OC]:
|OC|2 = |ON|2 + |NC|2
x2 = 32 + 42
x = 5 cm.
Property
In the same circle or in congruent circles, if two chords have different lengths, then the
longer chord is nearer to the center of the circle.
C
B
For example, in the figure,
if |CD| > |AB|, then |OF| < |OE|. The converse F
E
of this property is also true: if |OF| < |OE|, O
then |CD| > |AB|. A D
EXAMPLE 6 In the circle in the figure,
D M
|OM| < |ON| and r = 9 cm. C
B
|AB| = 3x + 2 cm and
O
|CD| = 5x – 2 cm are given.
N
Find the possible integer
A
values of x.
Solution If |OM| < |ON|, then |CD| > |AB|.
5x – 2 > 3x + 2
2x > 4
x > 2 (1)
Since the longest chord is the diameter, the greatest possible value of |CD| is the diameter.
d = 2r, d = 2 9 = 18 cm
|CD| 18
5x – 2 18
5x 20
x 4 (2)
From (1) and (2), the possible integer values of x are 3 and 4.
168 Geometriy 7
Check Yourself 2
1. In the figure, the radius of the circle is 5 cm and
A F
|AB| = |CD| = 8 cm. Find|OE|. B
C
O
E
D
2. In the figure, |AB| = |CD|, [OM] [AB], [ON] [CD], and N
C
|ON| = |OM| = 4 cm. Given |AB| = 5x + 1 cm and D
B
|CD| = 4x + 2 cm, find the radius of the circle. O
M
3. In the figure, |AP| = 12 cm, |PB| = 4 cm, and D
|OP| = 11 cm. Find the radius of the circle.
A
12 O
P 4
C B
4. In the figure, |AB| = 12 cm, |DC| = 2 cm, D
A B
[OD] [AB]. C
Find the radius of the circle. O
Answers
1. 3 cm 2. 5 cm 3. 13 cm 4. 10 cm
C. TANGENTS T H
Remember that a tangent is a line in the plane which
intersects a circle at exactly one point. The point is
called the point of tangency. In this section we will look
at the properties of tangents.
Property
If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point
of tangency.
H
l
For example, in the figure, if l is tangent to the circle C
O
at point H, then [OH] l.
Circles 169
Property
If two segments from the same exterior point are tangent to a circle, then they are congruent.
A
For example, in the figure, if [PA and [PB are tangent to
the circle at points A and B respectively, then
O P
|PA| = |PB|.
B
Property
Two tangent line segments from the same external point determine an angle that is bisected by
the ray from the external point through the center of the circle.
A
For example, in the figure, if [PA and [PB are tangent to
the circle then [PO is the angle bisector of APB, i.e.
P
mAPO = mBPO. O
EXAMPLE 7 The circle in the figure is inscribed in the triangle ABC. A
|AK| = x + 5 cm,
L
|BM| = 2x + 3 cm,
K
|CL| = 2x + 5 cm, and the perimeter of triangle ABC O
is 46 cm.
B M C
Find |MC|.
Solution |AK| = |AL|, |BK| = |BM|, and |CM| = |CL|.
P(ABC) = |AB| +|BC| +|AC|
= |AK| + |KB| + |BM| + |MC| + |CL| + |LA|
= 2 |AK| + 2 |BM| + 2 |CL|
= 2 (x + 5) + 2 (2x + 3) + 2 (2x + 5)
= 2x + 10 + 4x + 6 + 4x + 10
= 10x + 26
P(ABC) = 10x + 26 = 46 x = 2 cm
So |MC| = 2x + 5 = 9 cm.
170 Geometriy 7
D. RELATIVE POSITION OF TWO CIRCLES IN THE SAME
PLANE
Definition nonintersecting circles
Two circles which have no common point are called nonintersecting circles.
If two or more circles
share the same center,
then they are called a r2
r1 r2
concentric circles.
O1 O2 O
O1 O2
a r1
C1 Ç C2= Æ C1 Ç C2= Æ C1 Ç C2= Æ
r1 + r2< a r1 r2> a a=0
concentric circles nonintersecting circles
Definition tangent circles
Two circles which have only one common point are called tangent circles.
Tangent circles can be externally tangent or internally tangent, as shown in the figure.
l l
r1
r1 r2
A O2 A
O1 O1
a a O2 r2
C Ç C2 = {A} and r1 + r2 = a C1 Ç C2 = {A} and r1 r2 = a
externally tangent circles: internally tangent circles:
l is the common tangent l is the common tangent
Circles 171
Definition intersecting circles
Two circles which have two common points are called intersecting circles.
A A
r1
O1 O2 O1 H O2
r2
a
B B
C1 Ç C2 = {A, B} and r1 + r2 > a [AB] is the common chord
[O1O2] ^ [AB] and |AH| = |HB|
intersecting circles
EXAMPLE 8 The circles in the figure with centers A, B, and C are
C
externally tangent to each other.
|AB| = 16 cm, |BC| = 12 cm, and |AC| = 14 cm are
B
given. Find the radii of the circles.
Solution Let the radii of circles A, B, and C be r1, r2 and r3 A
respectively. Then we can write,
|AB| = r1 + r2 = 16
|BC| = r2 + r3 = 12
+ |CA| = r1 + r3 = 14
2 (r1 + r2 + r3) = 42
r1 + r2 + r3 = 21 r1 + r2 + r3 = 21 r1 + r2 = 16 r1 + r2 + r3 = 21
r1 + 12 = 21 9 + r2 = 16 16 + r3 = 21
r1 = 9 cm. r2 = 7 cm. r3 =5 cm.
172 Geometriy 7
EXERCISES 4.1
1. Describe each line and E 6. In the figure, A
C
D
line segment in the [OA] [BC], 2 4
O K
figure as an element of F C |AK| = 2 cm, and B
the circle. G B |KC| = 4 cm. O
Find |OK|.
H A
2. The points in the figure A E
F 7. In the figure, C
are in the same plane as C
|AC| = 6 cm and
the circle. State the D
O |AB| = 3 cm.
position of each point A B r O
G Find |OB| = r.
with respect to the
circle. B
C
8. In the figure,
3. In the figure, the radius D
|BC| = 12 cm and
E
D 12
of circle O is 15 cm, B
|AD| = 8 cm. 8
I
|CD| = 24 cm, and Find the radius of the A B
H O
|OH| = 12 cm. O
circle.
C
a. Find |OI|.
A
b. Find |AB|. 9. In the figure, A
|AP| = 6ñ3 cm and
4. Complete each state- A mAPB = 60°. C
O
P
ment about the figure F Find the radius of the
B
with a suitable symbol. C B circle.
O
a. If |OE| = |OF|, then
E
|AB|...|CD| 10. In the figure,
r2
b. If |OE| > |OF|, then D |O1O2| = 3 cm and r1
A B
|AB|...|CD|. r1 + r2 = 11 cm. O1 O2
Find r1 and r2.
5. In the figure, X, Y, and Z are C
points of tangency. 11. In the figure, C
|AX| = 6 cm, |AB| = 3x + 4, N
X A D
|CZ| = 4 cm, and |CD| = 2x + 9, and O
|BY| = 2 cm. O Z
|OM| > |ON|. M
Find the perimeter of Find the greatest possible
ABC. A Y B B
integer value of x.
Circles 173
A. ARCS AND CENTRAL ANGLES
Definition arc of a circle
An arc of a circle consists of two points on the circle and the unbroken part of the circle
between these two points.
We use the sign over two or more points to denote the A
Objectives ï
After studying this arc which includes the points. For example, in the fig-
C minor arc
section you will be O
ure, we write AïB to denote the arc between A and B, and
able to:
AùCB to denote the arc ACB.
1. Describe the B
concepts of arc and
central angle. Notice that any two points of a circle divide the circle into
2. Name inscribed two arcs. If the arcs are unequal, the smaller arc is called
angles and calcu- A
the minor arc and the larger arc is called the major arc.
late their measure.
3. Use the properties O
In the figure on the right, AïB is the minor arc and AùCB
of arcs, central
angles, and is the major arc.
inscribed angles to B
solve problems.
Definition central angle of a circle
An angle whose vertex is at the center of a circle is called a central angle of the circle.
We use the sign over two or more points to denote
ï C
the arc which includes the points. For example, in the
figure, we write AïB to denote the arc between A and B,
and AùCB to denote the arc ACB.
A
Notice that any two points of a circle divide the circle O B
into two arcs. If the arcs are unequal, the smaller arc is
called the minor arc and the larger arc is called the
major arc.
In the figure on the right, AïB is the minor arc and AùCB is the major arc.
174 Geometriy 7
EXAMPLE 9 Find the measure of the a. b. c.
indicated central angle of 50° 120° 180°
each circle. A B
C D
Solution Remember that the measure O O A O B
of a minor arc is equal to the
measure of its central angle.
a. mAOB = mAïB = 50°
b. mCOD = mCïD = 120°
c. mAOB = mAïB = 180°
Property
In the same circle or in congruent circles, if two chords D
are congruent, then their corresponding arcs and cen-
r
tral angles are also congruent. O
r r
A
For example, in the figure, if [AB] [CD] then AïB CïD r C
and mAOB mCOD.
B
Property
If a line through the center of a circle is perpendicular K
to a chord, it bisects the arcs defined by the endpoints
of that chord.
O
For example, in the figure, if [PK] [AB] then r r
[AH] [HB]
A H B
AïP PïB P
AïK KïB.
B. INSCRIBED ANGLES
Definition inscribed angle of a circle
An angle whose vertex is on a circle and whose sides
contain chords of the circle is called an inscribed angle. C
For example, angle ABC in the figure is an inscribed intercepted
B arc
angle. [AB] and [BC] are both chords of the circle. O
The arc AïC in the figure is called the intercepted arc of
A
the inscribed angle ABC.
Circles 175
Property
The measure of an inscribed angle is half of the measure
of the central angle which intercepts the same arc. C
a°
a° O 2a° E
B
Proof b° 2b°
b°
In the figure, let mBCO = a° and mBAO = b°.
A
Since the triangles BOC and AOB are isosceles
triangles, we can write
mOBC = mOCB = a° mCOE = mCBO + mOCB = 2a°
mOAB = mOBA = b° mAOE = mOAB + mOBA = 2b°
mABC = mOBA + mOBC mAOC = mAOE + mEOC
= a° + b° = 2 (a° + b°)
mAOC
So mABC = .
2
Now remember that the measure of a minor arc is the same as the measure of its central
angle. So we can write the property in a slightly different way:
Property
The measure of an inscribed angle is equal to the half
C
the measure of its intercepted arc.
B a O 2a 2a
mACï
For example, in the figure, mABC = .
2
A
EXAMPLE 10 Find the measure of x in each figure.
a. b. c.
C C C
B 50° B B x
x x O
O
O
120°
A A
A
176 Geometriy 7
mACï mACï mAOC
a. mABC = b. mABC = c. mABC =
Solution 2 2 2
120 90
mACï mx = =
50 = 2 2
2
mAïC = 100° mx = 60° mx = 45°
mx = 100°
Property
If two inscribed angles intercept the same arc of a
circle, then the angles are congruent. C
For example, in the figure, ABC ADC, because
B
they both intercept AïC. A
EXAMPLE 11 Find the value of x and y in each figure.
a. b. c.
D
A D
x° x°
40° y°
A O 20° O y°
A
x°
y° C
C C
50°
B B B
mCDï mBOC mBCï
Solution a. mCAD = mCBD = b. mBAC = =
2 2 2
40 y ° 50
x° = y° = x° = =
2 2 2
x = y = 20 x = 25 and y = 50
Circles 177
mBOC mBOC
c. mBAC = mBDC =
2 2
y° 40°
20 = x° =
2 2
y = 40 x = 20
Property
An angle inscribed in a semicircle is a right angle.
N
M
For example, in the figure,
if mAùLB = mAùMB = mAùNB = 180°, then B
A
O
mALB = mAMB = mANB = 90° or
mL = mM = mN = 90°.
L
EXAMPLE 12 Find the value of x in each figure.
a. b. c.
B 60° D
D C
x° x°
x° y°
20° O
A C A B A B
O O
10°
C
Solution a. Since AC is the diameter, the arc AùBC is a semicircle.
So ABC is inscribed in a semicircle, and therefore mABC = x° = 90°, x = 90.
mDCï y = 2 mBAC mAïD + mDïC + mCïB = 180
b. mDAC =
2
= 2 20 x° + 60° + 40° = 180°
60°
= = 40 x = 80
2
= 30°
178 Geometriy 7
c. Let us draw the chord [BD]. D
mADB = 90° x°
mCAB = mCDB = 10° O
A B
mADC + mCDB = 90°
10°
x + 10 = 90 C
x = 80
Property
The measure of the angle formed by a tangent and a chord is equal to the half of the meas-
ure of its intercepted arc.
1
For example, in the figure, mCAB = mAOB.
2
Proof C
A
Let us draw the diameter [AD] and the chord [BD].
[AC] [AD] (a radius is perpendicular to a tangent at
the point of tangency) O
B
[AB] [BD] (definition of a semicircle)
mDAB + mBAC = 90° D
mDAB + mADB = 90° (in triangle ABD)
C
mADB = mBAC A
mABï
mADB = (inscribed angle rule)
2 B
O
mABï
mBAC = (inscribed angle rule)
2
Rule
Let [AB] and [CD] be two chords of a circle.
A B
If [AB] [CD], then
mABC = mBCD (alternate interior angles).
So mAïC = mBïD. C D
Circles 179
EXAMPLE 13 Find the value of x and y in each figure.
a. b. c.
A C B x°
D E A F
x°
y°
O B C
30°
O O
B x°
15° E 30°
A C
D
[AB] || [CD]
Solution a. Let us draw the radius [OA].
A C
Then [AC] [AO]. AOB is isosceles triangle.
x° y°
r 30°
mOAB = 30° and
30°
mOAB + mBAC = 90° O r B
30° + x° = 90°
x = 60
y = 2 60 = 120
mBDï
b. [AB] [CD] and mBAD =
2
mBïD = 2 15° = 30° B 30°
D
mAïC + mCïD + mDïB = 180°
30 + mCïD + 30 = 180° O
mCïD = 120° x°
15° E
mC ïD A 30° C
mDCE = x° =
2
120°
= = 60
2
mACï E
x°
A F
c. mADC =
2
mAïC = 2 30° = 60° B C
mACï O
mABC = 30°
2
60°
mABC = = 30° D
2
[AD] [BC] and [AF] [AD]. So [AF] [BC]. Therefore, mABC = mBAE and
mEAB = x° = 30°, x = 30.
180 Geometriy 7
Rule
The measure of an angle formed by two secants, a
secant and a tangent, or two tangents drawn from a B
point in the exterior of a circle is equal to half of the
difference of the measures of the intercepted arcs. A a b b O
mÐBAC + mÐBOC = 180°
mÐBAC + mBïC = 180°
a + b = 180°
C A
C
A A
P O P P O C
O
B
D B B
angle formed by two secants: angle formed by a secant angle formed by two
and a tangent: tangents:
mAùCB mAïB
mCïD mAïB mCïB mAïB mÐP =
mÐP = mÐP = 2
2 2
mÐP + mAïB = 180°
Rule
The measure of an angle formed by two chords that
a° B
intersect in the interior of a circle is equal to half the A
x°
sum of the measures of the intercepted arcs. a
E
y b
C
D b
ï + mAD
mBC ï
For example, in the figure, mAED = mBEC = and
2
mA ïB + mC ïD
mAEB = mCED =
2
x+ y a+ b
= and = .
2 2
Circles 181
EXAMPLE 14 Find the value of x in each figure.
a. b. c.
C Q
A
B
30° B 15°
E 70° A x° O 100° P 60° S x° O T
C x° D
D E
R
mA ïC + mBD
ï
Solution a. mBED =
2
mA ïC +60
70° = ï = 2 mBAD )
( mBD
2
mAïC = 140 – 60 = 80
mADC = mA ïC = 80 = 40°, x = 40
2 2
mC ïE – mB D
ï 100 – 30
b. mCAE = = (mB ïD = 2× mBCD)
2 2
70°
mCAE = x° = = 35°, x = 35
2
c. mQPR + mQïR = 180°
60 + mQïR = 180
mQïR = 120°
mQïR + mQùTR = 360°
mQùTR = 240°
ù
mQTR 240°
mQSR = x° = = = 120°, x = 120
2 2
182 Geometriy 7
EXERCISES 4.2
1. In the figure, A B 7. In the figure, C
mAOC = 120°. x° mBCï D
C mAïD = and
Find mABC. 120° 2 75°
O mDPC = 75°. P
Find mBAC. x°
B
A
2. In the figure, C 8. In the figure,
mBAC = 30° and [AE is tangent to the D
A x° C
70° circle at the point B, O
mBKC = 70°. O B
K and mEBC = 75°. 75°
Find mOCA. 30° B
Find mA. E
A C
C 9. In the figure,
3. In the figure, mAOC = mABC = 3x°. B 3x° 3x° O
mCBD = 120°. Find the value of x.
x°
Find AOC. O
120°
A A
B D
B
10. In the figure,
4. In the figure, C
mAOC = 100° and A 70° x° C
mBOC = 100° and 20° 100°
A O 100° mOAB = 70°.
mACO = 20°. O
x° Find mOCB.
Find mAOC.
B
C
11. In the figure, A
5. In the figure, mAPD = 30°, a° B
[AB] is a diameter and x° 40° mDKA = 60°,
A 60° K 30° P
mOCB = 40°. O mBAC = a°, and
B b°
Find mOAC. mDCA = b°. C
D
Find a and b.
D
6. In the figure, 12. In the figure, A
C
mCDB = 10° 10° K
mDPA = 50°.
x° P O
and mABD = 50°. O Find mBCA. D
50°
50° B P
Find mP. B x°
A C
Circles 183
Objectives
After studying this section you will be able to:
1. Describe the concepts of circumference and arc length.
2. Find the area of a circle, an annulus, a sector, and a segment.
A. CIRCUMFERENCE AND ARC LENGTH
1. Circumference of a Circle
Remember from chapter 4 that the distance around a polygon is called the perimeter of the
polygon.
Note
d
Pi (, pronounced like the English word ‘pie’) is a Greek
letter. It is the first letter of a Greek word that means r r
‘measure around.’
C
Definition circumference
The distance around a circle is called the circumference of the circle.
If you measure the circumference and diameter of a circle and divide the circumference by
the diameter, you always get the same constant. This constant is approximately equal to
3.14, and denoted by .
Property
For all circles, the ratio of the circumference to the diameter is always the same number. The
number is called (pronounced like‘pie’).
C
So if the circumference of a circle with a diameter d is C, then we can write = or C = d or
d
C = 2 r .
This is the formula for the measure of the circumference of a circle.
184 Geometriy 7
1. Find five different circular objects. Use a piece of string to measure their
circumference (C), and use a ruler to measure their diameter (d). Write
the values in a table.
2. For each circular object calculate the ratio C and then calculate the
d
average of all the ratios.
3. How do the number and the formula C = d relate to this activity?
EXAMPLE 15 a. Find the diameter of a circle with circumference 24 cm.
b. Find the circumference of a circle with radius 5 cm.
c. Find the circumference of a circle with diameter 9 cm.
Solution a. Let the diameter of the circle be d, then the circumference of the circle is C = d:
24 = d
d = 24 cm.
b. C = 2 5 = 2 5 = 10cm
c. C = 2r = 2r = d = 9 cm
2. Arc Length
Remember that an arc is a part of a circle. The measure of an arc is equal to the measure of
its central angle.
Rule
In a circle, the ratio of the length of a given arc AïB to A
the circumference is equal to the ratio of the measure
of the arc to 360°.
O a
r
arc length of A ïB mA ïB
= , B
circumference of the circle 360°
arc length of A ïB
so = .
2ð r 360°
We can rewrite this as arc length of AïB = 2 r .
360°
In the above formula the measure of the arc is given in degrees. The length of the arc is given
in a linear unit such as centimeters.
Circles 185
EXAMPLE 16 Find the length of each arc.
a. b. c. d.
C 210°
T
G 18 cm
6 cm O
A B D 12 cm
O O 10 cm O
60°
E
F
H
Solution a. The length of a semicircle is half b. The length of a 90° arc is a quarter of
of the circumference. the circumference.
arc length of AïB arc length of CïD
180° 90°
= 2r = 2 6 = 2r = 2 10
360 360° 360 360°
1 1
= 2 6 = 6cm = 2 10 = 5cm
2 4
c. arc length of EïF d. arc length of GùTH
60° 210°
= 2r = 2 12 = 2r = 2 18
360 360° 360 360°
1 21
= 2 12 = 4cm = 2 18 = 21cm
6 36
Check Yourself 3
1. Find the circumference of the circle with the given radius.
a. r = 3 cm b. r = 5 cm c. r = 7 cm d. r = 10 cm
2. Find the radius of the circle with the given circumference.
a. 12 cm b. 24 cm c. 36 cm d. cm
3. Find the length of the minor arc in each figure.
a. b. c. E d. G
C
F 9 cm
8 cm 60°
O 200°
3 cm O O
120° O K
5 cm L
A B
D H
Answers
1. a. 6 cm b. 10 cm c. 14 cm d. 20 cm
1
2. a. 6 cm b. 12 cm c. 18 cm d. cm
2
3 10 8
3. a. cm b. cm c. cm d. 8 cm
2 3 3
186 Geometriy 7
B. AREA OF A CIRCLE, A SECTOR, AND A SEGMENT
1. Area of a Circle
Property
The area of a circle is times the square of the radius.
r2
A =
r
A
O
To understand why this property is true, let us divide a
circle into 16 equal parts, and rearrange them as
follows:
r r
O
C = pr
2
As the number of equal parts increases, the area of the circle gets closer and closer to the
area of a parallelogram.
C 2r
The area of a parallelogram is A = r = r = r 2 .
2 2
So the area of a circle with radius r is A = r2.
EXAMPLE 17 a. Find the area of a circle with radius r = 6 cm.
b. Find the radius of a circle with area 16 cm2.
c. Find the area of a circle with circumference 10 cm.
Solution a. Let the area of the b. Let the radius of c. The formula for the
circle be A, then the circle be r, then circumference of a
A = r2 A = r2 circle is C = 2 r:
A = 62 16 = r2, 10 = 2 r
A = 36cm2. r2 = 16 r = 5 cm.
r = 4 cm. So the area of the
circle is
A = r2
= 52
= 25cm2.
Circles 187
2. Area of an Annulus
Definition annulus
An annulus is a region bounded by two concentric circles.
How can we find the area of an annulus? Look at the
R
diagram.
O
r
an annulus
area of the big circle area of the small circle = area of the annulus
R
R r r
O O = O
A(annulus) = R2 – r2
= (R2 – r2)
EXAMPLE 18 Find the area of the annulus bounded by concentric
circles with radii 5 cm and 3 cm long.
3 cm
Solution The radius of the big circle is R = 5 cm. 5 cm O
The radius of the small circle is r = 3 cm.
A = R2 – r2
A = (R2 – r2)
A = (52 – 32)
A = (25 – 9)
A = 16 cm2
188 Geometriy 7
3. Area of a Sector
Definition sector of a circle
A sector of a circle is the region bounded by two radii of the circle and their intercepted arc.
For example, in the figure, the smaller region AOB is a sec-
A
tor of the circle. If the degree measure of arc AB is r
a
a O
mAïB = a° then the area of sector AOB = p r2 .
360 B
We can also calculate the area of a sector in a different
way:
Rule
The area of a sector of a circle is half the product of the length of the arc and the length of
its radius.
In the figure,
a
A ïB r A
360
r
| A ïB | a a l
= r O
2 360
B
| A ïB | r a
= r2
2 360
| A ïB| r lr
A= or A= (|AïB| = l ).
2 2
In the above formula the measure of the arc is given in degrees. The length of the arc is given
in a linear unit such as centimeters.
EXAMPLE 19 Find the area of each shaded sector.
a. b. P c.
A B
5 cm 8 cm 6p
72° 15°
O A C
O 6 cm O
B S
Circles 189
Solution a. r = 5 cm and ma = 72°.
a 72 1
r2 =
A (sector AOB ) = 5 2 = 25 = 5 cm 2
360 360 5
b. r = 8 cm and l = 6 cm.
l r 6 8
A (sector POS ) = = = 24 cm 2
2 2
c. mBOC = 2 mBAC, so
mBOC = 30° and r = 6 cm.
mBOC 30 1
A (sector BOC ) = ð r2 = ð 62 = ð 36 = 3ð cm 2 .
360 360 12
4. Area of a Segment
Definition segment of a circle
A segment of a circle is a region bounded by a chord and its intercepted arc.
A a°
b a° A a°
A A
h
b
O r
B h =
O B O B B
area of area of
area of segment = –
sector triangle
A = A(sector AOB) – A(AOB)
a bh
A= ð r2 –
360 2
EXAMPLE 20 Find the area of each shaded segment.
A C
a. A b. c.
120°
45°
O O A B
12 cm
12
6
cm
cm
B
B
190 Geometriy 7
Solution a. Since mAOB = 90°,
90 1
A (sector AOB ) = r 2 = 36 = 9 cm 2
360 4
r 2 36
A ( AOB ) = = =18 cm 2
2 2
A(segment) = 9 – 18 cm2.
b. |OH| = 6 cm and
|AB| = 12ñ3 cm. A
120
A(sector AOB) = ð 12 2
360 6 H
60° 6ñ3
1 O 30°
= 144 12
3 B
= 48 cm2.
| AB|| OH | 12 3 6
A(AOB) = 36 3 cm 2
2 2 C
2
A(segment) = 48 – 36ñ3 cm . 6
2 45°
ð r ð 36 B
c. A(sector AOC) = = = 9ð cm 2 A
6 O 6
4 4
6 6 36
A(AOC) = 18 cm 2
2 2
A(segment) = 9 – 18 cm2.
Check Yourself 4
1. Find the area of a circle with the given radius.
a. r = 3 cm b. r = 5 cm c. r = 12 cm d. r = 16 cm
2. Find the area of a circle with the given circumference.
a. 4 cm b. 12 cm c. 20 cm d. cm
3. Find the circumference of a circle with area 36 cm2.
4. The ratio of the radii of two circles is 5 : 3. What is the ratio of their areas?
R
2
5. The area of the shaded region in the figure is 32 cm and r
O
R = 9 cm. Find r.
Circles 191
6. Find the area of the shaded region in each circle.
A A
A
r r r
60° 120° O
O O O r
B A B
B
B
r = 5 cm r = 8 cm r = 12 cm r = 5 cm
|AB| = 5ñ2 cm
R r
r r r 120°
O
A P O Q B O r B
R A O B
r = 5 cm
r = 8 cm
R = 10 cm R = 10 cm
r = 8 cm r = 7 cm
Answers
1. a. 9 cm2 b. 25 cm2 c. 144 cm2 d. 256 cm2
2. a. 4 cm2 b. 36 cm2 c. 100 cm2 d. cm2
4
3. 12 cm
25
4.
9
5. 7 cm
25 64 25 50
6. a. cm2 b. cm2 c. 36 cm2 d. cm2 e. 36 cm2 f. 25 cm2 g. 17 cm2 f. 16 cm2
6 3 4
192 Geometriy 7
EXERCISES 4.3
1. In the figure, 7. In the figure,
A
mAOB = 30° and r mBAC = 30° and C
A
r = 6 cm.
30° the radius of the 30°
B
O circle is 6 cm. Find the area
Find the area of the
of the shaded region.
shaded region. B
2. In the figure, A
mAOB = 45° and r 8. In the figure, ABCD is a D C
45° rectangle and
r = 10 cm. B
O r = 10 A and B are the
Find the area of the centers of two circles.
shaded region. Given |AD| = 6 cm, A E B
X find the area of the
3. In the figure, A
shaded region.
mAOB = 120° and
120° 9. In the figure, A, B, and C
r = 6 cm. B
Find the length of arc O 6 cm are the centers of three A B
AùXB. congruent tangent circles.
If the sum of the
4. In the figure, B is the D C circumferences of the C
circles is 24 cm, find
center of a circle and
the area of the shaded
ABCD is a square with region.
|AD| = 5 cm.
Find the area of the
10.
shaded region.
A B B C D
A P K
5. In the figure, mAOB = mCOD = mEOF = 20°
and r = 6 cm. Find the E 20°
sum of the areas of the F D
O B, C, P, and K are the centers of four circles in the
shaded regions. C
20° figure. Given |AB| = |BC| = |CD| = 4 cm,
20°
find the area of shaded region.
A B
11. In the figure, ABCD is a D C
6. In the figure, C
square with perimeter
|OB| = 5 cm, 64 cm. Find the area of
mDOB = 60°, and D the shaded region.
|BA| = 3 cm.
Find the area of the 60°
shaded region. O B A
A B
Circles 193
12. In the figure, B and D D F C 18. In the figure, B A
are the centers of two mOAB = 50° and 50°
circles. P mBCO = 35°. O
E x°
If ABCD is a square Find mAOC. 35°
and the shaded area
is 16 cm2, find C
|DE|. A B
19. In the figure, D
13. In the figure, mBCD = 130° and C 130°
A C B mOAC = 40°.
A(AOB) = 48 cm2, 8 cm
r Find mCBO.
|OC| = 8 cm, and O x°
O 40°
[OC] [AB]. Find r.
A B
14. In the figure,
A 8 cm B 20. In the figure, B
|AB| = |CD| = 8 cm and A
r mOAB = 45° and 45°
|OH| = 3 cm. O x° 60°
mOCB = 60°.
Find the radius of O
C
Find mAOC.
the circle. C H D
15. In the figure, C E
A
|CE| = 3x – 2, D 21. In the figure,
|FB| = x + 4, and mAOC = 160° and
O
O
|OE| = |OF|. mABC = x°. 160°
A x°
B
Find x. F Find mABC.
B
C
16. In the figure, A
B
D
mAOB = 60° and C
r C 22. In the figure,
|AB| = 5 cm.
mOAD = 40° and x°
Find the radius of the 40° 50°
O mBOC = 50°. A B
O
circle.
Find mCOD.
17. In the figure, B
C 23. In the figure, C
|AB| = 9 cm,
B is the point of x°
|BC| = 8 cm, and
tangency and 30°
|CA| = 5 cm. A
A B mOAB = 30°. O
Find the radius of
Find mABC.
circle A.
194 Geometriy 7
24. A and C are points of tan- C
30. In the figure, the radius X
gency on the circle in the of the circle is 6 cm and A B
D
figure. A
70° the length of arc AùXB is r=
6c
Given mABC = 60° and m
x°
60° 4 cm. Find the area of O
mBCD = 70°, E
the shaded region.
B
find mBAE.
31 . In the figure, A
D
25. In the figure, |OB| = r = 4 cm.
C
mBAD = 60° and x° Find the area of the
60°
|AD| = |DC|. A B shaded region.
O B C
Find mBCD. O
32 . In the figure, A
|AB| = 8 cm and
6c
E m
26. In the figure, |AC| = 6 cm. 8c
m
B C
mAOE = 60° and D Find the area of the O
60° shaded region.
|OA| = |DC|. A
x°
O B C
Find mACE.
33 . ABCD is a square with sides 10 cm long. Find the
area of each shaded region.
a. D C b. D C
27. In the figure,
mBAD = 30°. O
A B
30°
Find mACD.
x°
C
D
A B A B
E
28. In the figure,
D C D C
mBAC = 20° and 30° c. d.
F D
mDFE = 30°. x°
O
Find mCOD. 20°
A C
B
A B A B
29. In the circle in the E
D
figure, e. D C d. D C
F
40° C
|OA| = 6 cm, 30°
mAOB = 50°, O
50°
mCOD = 30°,
and mEOF = 40°. A B
Find the sum of the areas of the shaded regions. A B A B
Circles 195
CHAPTER 4 REVIEW TEST
1. In the figure, l 6. In the figure, l
A C
|OA| = 4 cm and line l is tangent to the
C x
|OC| = 7 cm. circle at point C and
B
What is |BC|? O |OA| = |AB| = 5 cm. B 5 A 5 O
Find |BC| = x.
A) 2 cm B) 3 cm C) 4 cm D) 5 cm A) 4 cm B) 5 cm C) 5ñ3 cm D) 6 cm
2. Find |AB|in the figure if D 7. Find mABC in the figure. C
|CH| = 4 cm. H
O C 98°
A O
A x°
G B
B
A) 8 cm B) 7 cm C) 6 cm D) 5 cm A) 49° B) 50° C) 51° D) 52°
3. In the figure, the radius A
8. In the figure,
of the circle is 10 cm D
H mABD = 60° and A
and |OH| = 6 cm. O
mCED = 80°. 80° x°
Find |AB|. O
B Find mCDE. 60° E
C
B
A) 8 cm B) 12 cm C) 16 cm D) 20 cm
A) 10° B) 20° C) 25° D) 40°
4. In the figure, A
C 1 cm 9. In the figure, B
|OC| = 3ñ2 cm,
3ñ2 cm mBDC = 70°.
|AC| = 1 cm, and
m
O C
Find mACB.
7c
|BC| = 7 cm. O A
What is the length B 70°
of the radius? D
A) 20° B) 25° C) 30° D) 40°
A) 3 cm B) 3ñ3 cm C) 4ñ2 cm D) 5 cm
5. In the figure, D 10. In the figure, line l is l
|OK| = |OH| = 5 cm, K tangent to the circle at A
|AB| = 2a + 2 cm, and point A and 80° D
O C B
|CD| = a + 13 cm. |AB| = |AC|. O
A
What is the length of H Find mCAD.
B
the radius? C
A) 13 cm B) 12 cm C) 11 cm D) 10 cm A) 65° B) 55° C) 50° D) 45°
196 Angles
11. In the figure, E 16. In the figure, A
A
[PE and [PD are mABC = 35°,
tangent to the mACB = 55°, and 35° 55°
P 50° x° C B C
circle at the points O |BC| = 4 cm. 4 cm
A and B, respectively. What is the area of the
Find mACB if B circle?
mAPB = 50°. D
A) 2 cm2 B) 3 cm2 C) 4 cm2 D) 8 cm2
A) 60° B) 65° C) 70° D) 75° 17. Find the length of the A
arc AïB in the figure if
12. In the figure, B
A
the radius is 3 cm and 3 cm
mAPC = 35° and O B
mACB = 60°. 60°
mBïD = 100°. P 35° O
x° C
Find mADC. C
D 3 5
A) cm B) cm C) 2 cm D) cm
A) 15° B) 20° C) 30° D) 40° 2 2
18. In the figure, ABCD is a D H C
13. In the figure, A
square with sides 6 cm
mDCE = 30° and long. Find the area of
x° B E G
mAïB = 80°. E the shaded region.
O
Find the value of x. O
A F B
30° C
2 9
D A) 9 – 2 cm B) 16 – cm2
4
A) 65 B) 70 C) 75 D) 80 C) 36 – 9 cm2 D) 49 – 12 cm2
14. In the figure, 19. In the figure, the circle C
A
has radius 6 cm and
|AB|= 2 cm and
2ñ3 cm
mABC = 75°. 6 cm
|AC| = 2ñ3 cm. 2 cm A
O
B C Find the area of the
What is the length of O 75°
shaded region.
the circumference? B
2 2 2
A) 6 cm B) 9 cm C) 12 cm D) 15 cm2
A) 3 cm B) 4 cm C) 6 cm D) 8 cm
20 . In the figure, ABCD is a D 6 cm F C
15. In the figure, square. |BE| = 4 cm,
the perimeter of the |DF| = 6 cm, and B and 6 cm G
circle is 10 cm and O D are the centers of two 4 cm
3 cm H
|OH| = 3 cm. A circles. Find the area of
A E 4 cm B
Find |AB|. H the shaded region.
B
A) 40 – 10 cm2 B) 50 – 13 cm2
A) 5 cm B) 6 cm C) 7 cm D) 8 cm C) 36 – 12 cm2 D) 64 – 20 cm2
Chapter Review Test 1A 197
EXERCISES 1. 1
1. Because there are no simpler concepts for us to buid on. Therefore, we need to understand these concepts
without a precise definition.
3. A ray has closed enpoint but a half line has an open endpoint.
4. 2 5. 3 7. a. size, length, width, thickness b. line c. plane d. skew lines
8. a. true b. true c. true d. false e. true
9. a. 10 b. 21 c. 210 d. 5050
10. lines: HL, HG rays: [LC, [LH, [HL, [HG, [GH half lines: ]LC, ]LH, ]HL, ]HG, ]GH
11. a. line segment CD b. half open line segment PQ c. open line segment AB d. ray KL e. half line MN
f. line EF
12. a. l (E) = l b. d (F) = {C} c. n (G) = 13. ‘M, N and P’, ‘R and S’, and ‘L and K’, are coplanar
points.
14. l Q
15. d (D) (E) = m
l (D) (F) = l
F
P (E) (F) = d
E m d l = {O}
D
O
m
16. a. 5 b. (P) (Q) = EB, (P) (S) = EA, (P) (T) = AB, (Q) (T) = BC, (Q) (R) = EC, (T) (R) = DC,
(S) (R) = ED, (T) (S) = AD c. 3 lines pass through point A, B, C, and D, 4 lines pass through point E.
EXERCISES 2. 1
1. a. b. c. {K, O, M} d. e. {N} f. g. h. {P} i. j. k.
2. a. {A} [CD b. ]AC[ ]AD[ c. ]CD[ d. ]CE ]DF e. ]AB] [BC[ ]AH ]DG
5. a. acute angle b. right angle c. obtuse angle d. straight angle e. complete angle
6. a. 20° b. 12° c. 20° 7. a. 32° b. 20° c. 10°
8. a. 115° b. 65° c. 115° d. 65° e. 115° f. 65° g. 65°
9. 130° 10. 40° 11. 25° 12. 50° 13. 100° 14. 70° 15. x = y + z 16. 160° 17. 35° 18. 140° 19. 80°
20. 90° 21. 35°
196 Geometriy 7
EXERCISES 3. 1
1. ADE, DEK, DKF, BDF, CKF, CKE, DEC, ADC, DFC, BDC, CEF, ABC
2. F eight triangles: GDT, DTE, ETF, FTG, GDE, GFE, GDF, DEF 3. 51 cm 4. 10 cm
G
T 5. 28.2 cm 6. a. B, E, F, C b. F c. segment AC and point E d. segment FC without
D E endpoints
7. A B 8. 12 9. 7 cm
A B
A B A
E C F C
C
B
D C D E D
10. a. A b. A c. A 11. a. Hint: Construct medians for each
r1 r1
ha Va na side. b. Hint: Construct angle bisectors for
r2
r2
B
r2 r2
C B C B C each angle. c. Hint: Construct altitudes
r r
for each vertex.
56 168
d. Hint: Construct perpendicular bisectors for each side. 14. hb = ; hc = 15. a. BFC b. CEF, BEF,
5 13
121
ABC c. BFC d. ABF e. ABF 17. a. yes b. no c. yes d. yes e. no 18. a. x {4, 9, 14} b. none 19. cm 2
2
20. a. in the interior b. in the interior c. in the interior d. in the interior e. on the triangle f. in the interior
g. in the interior h. on the triangle i. in the exterior j. in the interior k. in the interior l. in the exterior
21. a. sometimes b. always c. never d. sometimes e. never f. never g. always h. always
22. a. b. c. d. e. f.
equilateral scalene right isosceles obtuse triangle equilateral isosceles
23. a. b.
Answers to Exercises 197
EXERCISES 3. 2
2. a. 70° b. 1 d. 6 e. 3 3. A K; D L; E N; AD KL;
c. 60° DE LN; AE KN
11
4. a. 6 b. 20° c. 22° d. 5. a. BC = 3, MN = 8 9. 10 cm 11. 2 12. m(OKM) = 10°, m(OML) = 60°,
8
5 1
m(OLK) = 20° 13. 8 14. 15 15. 84° 16. 17. 84° 18. 38° 19. 9.6 cm 20. 20° 21. 22. 3ñ3 23. 8 cm
2 4
24. 3ñ3 25. 4 cm 26. 14 27. 15 28. 8ñ3 29. 3 cm 30. 16 31. 2 cm 32. 6 cm 33. 2 cm 34. 2 cm
35. 4ñ3 cm 36. 70° 37. 12 38. 99° 39. 9 cm 40. 3ñ3 – 3 41. 24 cm 42. yes 43. 200 km 44. 25° 45. 7 – ñ5
46. 6 47. 70° 48. 57° 49. 150° 50. 8 cm 51. 16 cm 52. 18° 53. 8 and 12 54. 8 55. 45° 58. 4 cm 59. 10°
60. 3ñ2 60. 72 62. 6 cm 63. 12 cm 64. 12 cm 65. 8 66. 6 cm 67. 6 cm 68. 6 cm 69. 8 cm 70. 9 cm
3
71. 72. 6 73. 2 74. 18 75. 15
2
EXERCISES 3. 3
1. a. 36° b. 114° c. 54° d. 20° e. 100° f. 50° g. 90° h. 64° 2. 40°, 60°, 80° 3. x = 120°, acute angles: 85°, 5°
4. 50° 5. 136° 6. 60° 7. 72° 8. 117° 9. 12° 10. 8° 11. 36° 12. 106°
13. a. E b. E c. not possible d. E e. not possible
20° 40°
60° 80°
80°
70° 150° 40° 60° 60° 20° 80° 40°
D M F D M F D M F
14. a. no b. no c. no d. yes 15. 55° 17. a. 2 < a < 14 b. 4 < p < 20 c. 1 < m < 7 18. a. 4 < x < 12
b. 4 < x < 11 c. 3 < x < 10 19. three 20. 9 21. 12 22. 29 23. 36 24. 9 25. five 26. a. yes b. no c. no
d. no e. yes f. yes 27. x 28. a. 1 = 2 > 3 b. 1 > 2 > 3 c. 3 > 2 > 1 d. 1 > 3 > 2 29. a. false b. false
c. true d. false e. true f. false g. false 30. a. AC b. AC c. DC d. BC 31. a. 5 b. 8 32. 49 33. 25
34. there are no values 35. 11 36. four triangles with side lengths (1, 5, 5), (2, 4, 5), (3, 4, 4), (3, 3, 5) 38. a. A
b. D c. D d. D 39. a. A b. B 40. a. A b. B c. A d. A 41. a. A b. B 42. a. D b. A
EXERCISES 3. 4
13 3 16 35 25 5 12
1. 2. {21, 31} 3. – 4. {–49, 31} 5. 6. {– , } 7. a. ò13 b. 15 2 c. 8.
10 4 29 9 9 4 2 43
53
9. 3x + 4y – 10 = 0 ; 3x + 4y + 30 = 0 10. –
15
198 Geometriy 7
EXERCISES 4. 1
1. radii: [OF], [OC], [OA], [OB] diameter: [FC] chords: [ED], [FC], [GB] tangent: AH secant: GB center: O
2. Points C, O, and D are in the interior region of the circle. Point E is on the circle. Points A, B, G and F are in the
9
exterior region of the circle 3. a. 9 cm b. 18 cm 4. a. = b. > 5. 24 cm 6. 3 cm 7. cm 8. 4ñ6 cm
2
9. 6 cm 10. r1 = 4 cm, r2 = 7 cm 11. 4 cm
EXERCISES 4. 2
1. 120° 2. 10° 3. 120° 4. 30° 5. 50° 6. 40° 7. 70° 8. 60° 9. 40 10. 60° 11. a = 15° b = 45°
12. 20°
EXERCISES 4. 3
25 13
1. 33 cm2 2. cm2 3. 4 cm 4. (25 – 25 ) cm2 5. 6 cm2 6. cm2 7. (6 – 9ñ3) cm2
2 4 2
8. (72 – 18) cm2 9. (16ñ3 – 8) cm2 10. 12 cm2 11. (256 – 64) cm2 12. (8ñ2 – 8) cm
5 3
13. 10 cm 14. 5 cm 15. 3 16. cm 17. 3 cm 18. 170° 19. 10° 20. 150° 21. 100° 21. 30° 22. 60° 23. 50°
3
25 - 48
24. 120° 25. 20° 26. 120° 27. 80° 28. 12 cm2 29. 12 cm2 30. 4 cm2 31. cm2 32. a. 50 cm2
2
25 25
b. (50 – 100) cm2 c. (100 – 25) cm2 d. (50 – 100) cm2 e. ( + 25) cm2 f. (75 – ) cm2
4 2
Answers to Exercises 199
TEST 1 TEST 2
1. D 9. D 1. C 9. C
2. A 10. D 2. A 10. C
3. B 11. C 3. D 11. D
4. C 12. D 4. C
5. C 5. B
6. D 6. B
7. C 7. B
8. B 8. B
TEST 2A TEST 2B TEST 2C
1. E 9. C 1. C 9. B 1. E 9. B
2. D 10. C 2. A 10. C 2. D 10. A
3. A 11. B 3. E 11. C 3. C 11. D
4. C 12. C 4. C 12. A 4. B 12. C
5. A 13. D 5. A 13. C 5. E 13. D
6. E 14. C 6. C 14. C 6. E 14. A
7. B 15. E 7. B 15. A 7. D 15. C
8. C 16. C 8. A 16. E 8. C 16. C
TEST 2D TEST 2E
1. D 9. A 1. D 9. A
2. D 10. E 2. E 10. C
3. A 11. E 3. B 11. D
4. C 12. C 4. C 12. C
5. E 13. A 5. C 13. C
6. B 14. A 6. A 14. C
7. C 15. C 7. C 15. B
8. D 16. D 8. C 16. B
200 Geometriy 7
Symbol Meaning Symbol Meaning
= is equal to is congruent to
is not equal to is not congruent to
is greater than is parallel to
is greater than or equal to is not parallel to
is less than is perpendicular to
is less than or equal to is similar to
is approximately equal to ABC triangle with vertices A, B and C
|x| absolute value of x ha length of the altitude to side a
pi
is an element of
ñ square root is not an element of
A angle A union
A exterior angle of A in a triangle intersection
mA measure of angle A in degrees is contained by
degrees A B A is contained by B
minutes A B A is not contained by B
seconds A.S.A angle-side-angle
right angle S.A.S side-angle-side
mABC measure of angle ABC in degrees S.S.S side-side-side
AïB minor arc with endpoints A and B A.A angle-angle
mAïB measure of minor arc AB in degrees (E) plane E
AùCB major arc with endpoints A and B (int ABC) interior of the triangle ABC
(ext ABC) exterior of the triangle ABC
mAùCB measure of major arc ACB in degrees
A(ABC) area of the triangle ABC
AB line AB, passing through the points A
P(ABC) perimeter of the triangle ABC
and B
ABCD quadrilateral ABCD
[AB] line segment AB or segment AB, with
endpoints A and B ABCD paralelogram ABCD
|AB| O circle with center O
length of segment AB
C circumference
[AB ray AB with initial point A, passing
through B sin sine
]AB half line AB cos cosine
]AB] half-open line segment AB, excluding tan tangent
point A and including point B cot cotangent
]AB[ open line segment sec secant
[AB] closed line segment cosec cosecant
angle bisector of a triangle: An angle A
bisector of a triangle is a segment that
acute angle: An acute angle is bisects one of the angles of the
an angle with measure greater triangle. Its endpoints are points on the
than 0° and less than 90°. 20°
85° triangle. B N C
horizon
acute triangle: An acute triangle has angle of depression: The angle formed (horizontal)
60°
three acute angles. by the horizontal and the line of sight to angle of
50° 70° depression
an object below the horizontal.
adjacent angles: Two angles
are adjacent if they share a com- a b
mon vertex and side, but have c angle of elevation : The angle formed
no common interior points. Ðb and Ðc are adjacent angles by the horizontal and the line of sight to
Ða and Ðc are not adjacent angles
an object above the horizontal. angle of
adjacent sides: In a triangle or elevation
other polygon, two sides that share a common vertex are horizontal
adjacent sides.
area: The number of square units that cover a given surface.
alternate exterior angles: Two angles t
a b
are alternate exterior angles if they lie c d
l
outside l and m on opposite sides of t,
e f m
such as b and g. g h
base: The lower face or side of a geometric shape.
alternate interior angles : Two angles are alternate
interior angles if they lie between l and m on opposite sides
of t, such as d and e. (See figure for alternate exterior center of a circle : The center of a circle is
r
angles.) the point inside the circle that is O
altitude of a triangle: An altitude of a equidistant from all the points on the
altitude O is the center of
triangle is a segment from a vertex that circle. the circle
is perpendicular to the opposite side or
to the line containing the opposite side. central angle of a circle: A central angle of a circle is an
An altitude may lie inside or outside the triangle. angle whose vertex is the center of the circle.
C
angle: An angle consists of two different circle: A circle is the set of all points in a plane that are
rays that have the same initial point. The equidistant from a given point, called the center of the
rays are the sides of the angle and the ini- A B circle.
tial point is the vertex of the angle. vertex: point A circumference of a circle: The circumference of a circle is
sides: [AC and [AB
the distance around the circle.
collinear : Points, segments,
A B C D
C or rays that are on the same
A, B, C, and D are collinear
angle bisector: An angle bisector is a B
line are collinear.
ray that divides the angle into two con- O
complementary angles: Two angles are complementary if
gruent angles. A the sum of their measures is 90°. Each angle is a
[OB is an angle bisector complement of the other.
[AN] is an angle bisector
concave polygon: See non-convex polygon.
202 Geometriy 7
concurrent : Two or more lines or segments are
concurrent if they intersect at a single point.
congruent angles: Two angles are congruent if they have decagon: A decagon is a polygon that has
the same measure. ten sides.
congruent arcs: On the same circle or on congruent circles, degree: A unit of angle and arc measure.
two arcs are congruent if they have the same measure. diameter of a circle: A diameter of a circle is a chord that
congruent polygons : Two polygons are congruent if there is passes through the center. The diameter, d, is twice the
a correspondence between their angles and sides such that radius: d = 2r.
corresponding angles are congruent and corresponding sides diagonal : A line segment joining two al
gon
are congruent. Congruent polygons have the same size and dia
non-adjacent vertices of a polygon.
the same shape.
congruent segments: Two segments are congruent if they
have the same length. t
equiangular triangle: An equiangular triangle has three
consecutive interior angles : Two d a
c b congruent angles, each with a measure of 60°.
angles are consecutive interior angles if l
m equilateral triangle: An equilateral triangle has three con-
they lie between l and m on the same side h e
g f gruent sides.
of t, such as b and e.
exterior angles of a triangle :
convex polygon: A polygon is convex if exterior exterior
When the sides of a triangle are angle angle
no line that contains a side of the
extended, the angles that are
polygon contains a point in the interior of interior
adjacent to the interior angles of the angle
the polygon. convex polygons
triangle are the exterior angles. Each
coplanar: Points, lines, segments or rays that lie in the same vertex has a pair of exterior angles.
plane.
t exterior of an angle: A point D is in the C
corresponding angles: Two angles are d a exterior of A if it is not on the angle or in D
c b
corresponding angles if they occupy cor- l
the interior of the angle.
h e m
responding positions, such as a and e g f A B
in the figure.
concentric circles : Circles that have
different radii but share the same center half line: A ray without an endpoint A B
are called concentric circles. (initial point). ]AB = half line AB
d
half planes: Two halves of a plane that are P
cone: A solid figure that has a circular separated by a line P1 P2
base and a point at the top.
P = P 1 È P2 È P
cube: A square prism that has six equal hexagon: A hexagon is a polygon with six
sides.
square sides.
hy
hypotenuse: In a right triangle, the side po
cylinder: A solid with circular ends and ten
us
opposite the right angle is the hypotenuse e
straight sides.
of the triangle.
Answers to Exercises 203
major arc: On circle P, if mAPB < 180°, C
A
inscribed angle of a circle: An angle is then the points A and B together with the P
O points of the circle that lie in the exterior
an inscribed angle of a circle if its vertex is
on the circle and its sides are chords of the B of mAPB form a major arc of the circle. A B
ÐAOB is an Major arcs are denoted by three letters, as AùCB is an major
circle. inscribed angle arc of the circle
in AùCB.
C
interior of an angle: A point D is in the midpoint of a segment: The midpoint |RS| = |ST|
D R S T
interior of A if it is between points that lie of a segment is the point that divides the
on each side of the angle. segment into two congruent segments. midpoint
A B
intersecting lines : Coplanar lines which have only one minor arc: On circle P, if mAPB < 180°, C
point in common. then the points A and B, together with the P
intersecting planes: Planes which have one common line. points of the circle that lie in the interior of
isosceles triangle : An isosceles triangle has at least two mAPB form a minor arc of the circle. A B
congruent sides. Minor arcs are denoted by two letters, such AïB is a minor
arc of the circle
as AïB.
isosceles trapezoid : A quadrilateral with one pair of
parallel sides and at least two sides the same length.
noncollinear : Points, segments, or rays that are not
collinear.
kite: A convex quadrilateral with two pairs of
non-cconvex polygon : A polygon is
equal adjacent sides. non-convex (concave) if at least one
line that contains a side of the polygon
contains a point in the interior of the
polygon.
legs of a right triangle: Either of the two
leg non-ccoplanar : Not coplanar.
sides that form a right angle of a right tri-
angle. leg
legs of an isosceles triangle: One of the oblique lines: Lines are oblique if they
two congruent sides in an isosceles leg leg intersect and do not form right angles.
triangle.
obtuse angle: An obtuse angle is an angle with measure
line: A line is an undefined term in geometry. In Euclidean greater than 90° and less than 180°.
geometry a line is understood to be straight, to contain an obtuse triangle: An obtuse triangle has exactly one obtuse
infinite number of points, to extend infinitely in two direc- angle.
tions, and to have no thickness. octagon: An octagon is a polygon with
line segment : See segment. eight sides.
204 Geometriy 7
postulate : A postulate is a statement that is accepted as true
without proof.
parallel lines: Two lines are parallel if proof: A proof is an organized series of statements that show
that the statement to be proved follows logically from known
they are coplanar and do not intersect. facts (given statements, postulates, and previously proven
parallel planes: Two planes are parallel theorems).
80° 90° 100°
protractor: A device used to determine the
70° 110°
60° 120°
if they do not intersect. E
50° 130°
90° 80° 70°
40° 110°100°
120° 60° 140°
130° 50°
30° 150°
140° 40°
20° 150° 30°
A and B are measures of angles.
160°
160° 20°
10° 170°
170° 10°
parallel planes
0° 180° 0° 180°
0 1 2 3 4 5 6 7 8 9 10
pythagorean triple: A set of three positive c b
parallelogram : A quadrilateral with integers a, b, and c that satisfy the equation
a2 + b2 = c2 is a pythagorean triple. a
opposite sides parallel, and hence equal in
a2 + b2 = c2
length.
prism: A solid figure that has two
bases that are parallel, congruent
pentagon: A pentagon is a polygon with
polygons and with all other faces
five sides. that are parallelograms.
pyramid: A solid figure with a
perimeter of a polygon: The perimeter of a polygon is the polygon base and whose other
faces are triangles that share a
sum of the length of its sides. common vertex.
k
perpendicular lines : Two lines are
perpendicular if they intersect to form a l
right angle. quadrilateral: A polygon with four sides. The sum of the
k^l angles is 360°.
45°
perpendicular line and plane: A line l
100° 120° 135°
135°
80° 60° 45°
is perpendicular to a plane if it is
perpendicular to each line in the plane. P
l^P
radius of circle: A radius of a circle is a segment that has
the center as one endpoint and a point on the circle as the
plane: A plane is an undefined term in geometry. In other endpoint.
Euclidean geometry it can be thought of as a flat surface that ray: The ray AB, or [AB, consists of the
[AB [BA
extends infinitely in all directions. initial point A and all points on line that
A B A B
lie on the same side of A as B lies.
point: A point is an undefined term in geometry. It can be
rectangle : A rectangle is a parallelogram that has four right
thought of as a dot that represents a location in a plane or in
angles.
space.
rectangular prism : A solid figure
polygon: A polygon is a plane figure formed by three or more that with two bases that are rectangles
segments called sides, such that the following are true: and with all other faces that are
1. each side intersects exactly two other sides, once at each parallelograms.
endpoint, and
2. no two sides with a common endpoint are collinear.
Answers to Exercises 205
regular polygon : A polygon whose square: A square is a parallelogram that
sides are equal and whose angles are is both a rhombus and a rectangle; that
equal. is, it has four congruent sides and four
right angles.
right prism: A prism that has two straight angle: A straight angle is an
angle that measures 180°. straight angle
special characteristics: all lateral
edges are perpendicular to the bases supplementary angles: Two angles are supplementary if
and all lateral faces are rectangular. the sum of their measures is 180°. Each angle is a
supplement of the other.
rhombus: A rhombus is a parallelogram that has four
congruent sides.
right triangle: A triangle with exactly one right angle.
tangent: The ratio of the length of the C
|AC|
side opposite an angle to the length of the tan a =
|AB|
side adjacent to the angle in a right trian-
scale factor : In two similar polygons or two similar solids, gle.
a
the scale factor is the ratio of corresponding linear A B
measures.
scalene triangle: A scalene triangle is a triangle that has no tangent to a circle: A line is tangent to A
congruent sides. a circle if it intersects the circle at exact-
C
segment: A segment AB, or [AB], AB or BA
ly one point.
B
consists of the endpoints A and B and A B
[AB] or [BA]
all points on the line AB that lie theorem: A theorem is a statement that must be proved to
A B
between A and B. be true.
l
similar polygons : Two polygons are similar if their transversal: A transversal is a line that d
corresponding angles are congruent and the lengths of intersects two or more coplanar lines at
their corresponding sides are proportional. different points.
C k
sine: The ratio of the length of the side sin a =
|AC|
|BC| l is the tranversal
opposite an angle to the length of the
hypotenuse in a right triangle. trapezoid : A quadrilateral with exactly
a
one pair of opposite parallel sides. The
A B
sum of the angles is 360°.
surface area: The sum of all the areas of the surfaces of a
solid figure. d
skew lines: Two lines are skew if they
do not lie in the same plane.
vertex of a polygon: A vertex of a polygon is a common
k endpoint of two of its sides.
d and k vertical angles : Two
are skew lines
angles are vertical if their a
b Ða and Ðc are vertical angles
space: The set of all points. sides form two pairs of d
c Ðb and Ðd are vertical angles
sphere: A sphere is the set of all points opposite rays.
in space that are a given distance r from
O
a point called the center. The distance r volume: The number of cubic units needed to occupy a
is the radius of the sphere. given space.