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Unit 3
Triangles
Chapter Objectives
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Classification of Triangles by Sides
Classification of Triangles by Angles
Exterior Angle Theorem
Triangle Sum Theorem
Adjacent Sides and Angles
Parts of Specific Triangles
5 Congruence Theorems for Triangles
Lesson 3.1
Classifying Triangles
Lesson 3.1 Objectives
• Classify triangles according to their side
lengths. (G1.2.1)
• Classify triangles according to their angle
measures. (G1.2.1)
• Find a missing angle using the Triangle Sum
Theorem. (G1.2.2)
• Find a missing angle using the Exterior Angle
Theorem. (G1.2.2)
Classification of Triangles by Sides
Name
Equilateral
Isosceles
Scalene
3 congruent
sides
At least 2
congruent
sides
No Congruent
Sides
Looks Like
Characteristics
Classification of Triangles by Angles
Name
Acute
Equiangular
Right
Obtuse
3 acute
angles
3 congruent
angles
1 right
angles
1 obtuse
angle
Looks Like
Characteristics
Example 3.1
Classify the following triangles by their
sides and their angles.
Scalene
Obtuse
Scalene
Right
Equilateral
Equiangular
Isosceles
Acute
Vertex
• The vertex of a triangle is any point at
which two sides are joined.
– It is a corner of a triangle.
• There are 3 in every triangle
Adjacent Sides and Adjacent Angles
• Adjacent sides are
those sides that
intersect at a common
vertex of a polygon.
– These are said to be
adjacent to an angle.
• Adjacent angles are
those angles that are
right next to each other
as you move inside a
polygon.
– These are said to be
adjacent to a specific
side.
More Parts of Triangles
• If you were to extend the sides you will
see that more angles would be formed.
• So we need to keep them separate
– The three angles are called interior angles
because they are inside the triangle.
– The three new angles are called exterior
angles because they lie outside the triangle.
Theorem 4.1: Triangle Sum Theorem
• The sum of the measures of the interior
angles of a triangle is 180o.
B
mA + mB + mC = 180o
C
A
Example 3.2
Solve for x and then classify the triangle
based on its angles.
Acute
75o
50o
3x + 2x + 55 = 180
Triangle Sum Theorem
5x + 55 = 180
Simplify
5x = 125
SPOE
x = 25
DPOE
Example 3.3
Solve for x and classify each triangle by angle measure.
1. mA  ( x  30) o
( x  30)  x  ( x  60)  180
mB  x o
mC  ( x  60) o
mA  60o
mB  30o
mC  90
2. mA  (6 x  11)o
o
3x  90  180
3x  90
x  30
Right
(6 x  11)  (3x  2)  (5 x  1)  180
mB  (3 x  2)o
mC  (5 x  1)o
mA  83o
mB  34o
mC  59o
14x 12  180
14x  168
x  12
Acute
Example 3.4
Draw a sketch of the triangle described.
1.
Mark the triangle with symbols to indicate the
necessary information.
Acute Isosceles
2.
Equilateral
3.
Right Scalene
Example 3.5
Draw a sketch of the triangle described.
1.
Mark the triangle with specific angle measures,
side lengths, or symbols to indicate the necessary
information.
Obtuse Scalene
2.
Right Isosceles
3.
Right Equilateral
(Not Possible)
Theorem 4.2: Exterior Angle Theorem
• The measure of an exterior angle of a
triangle is equal to the sum of the
measures of the two nonadjacent
interior angles.
B
A
C
m A +m B = m C
Example 3.6
Solve for x
6 x  7  2 x  (103  x)
6x  7  x  103
5x  7  103
5x  110
x  22
Exterior Angles Theorem
Combine Like Terms
Subtraction Property
Addition Property
Division Property
Corollary to the Triangle Sum Theorem
• A corollary to a theorem is a statement that
can be proved easily using the original
theorem itself.
– This is treated just like a theorem or a postulate in
proofs.
• The acute angles in a right triangle are
complementary.
A
mA + mB = 90o
B
C
Example 3.7
If you don’t like the Exterior
Angle Theorem, then find
m2 first using the Linear
Pair Postulate.
Find the unknown angle measures.
1.
90o  53o  m1  180o 3.
143o  m1  180o
102o  m2  180o
m1  37 o
m2  78o
2.
90  42  m1  180
o
o
VA
o
132o  m1  180o
m1  48o
90  33  m2  180
o
o
123o  m2  180o
m2  57o
VA
m2  m3  122o
m1  34
o
78o  68o  m1  180o
68o  34o  m2  180o
o
4.
68o  m1  102o
Then find m1 using the
Angle Sum Theorem.
58o  m2  180o
m2  122o
102o  m2  180o
146o  m1  180o
m1  34o
m2  78o
122o  22o  m1  180o 122o  20o  m4  180o
144o  m1  180o
m1  36o
142o  m4  180o
m4  38o
Homework 3.1
• Lesson 3.1 – All Sections
– p1-6
• Due Tomorrow
Lesson 3.2
Inequalities in One Triangle
Lesson 3.2 Objectives
• Order the angles in a triangle from
smallest to largest based on given side
lengths. (G1.2.2)
• Order the side lengths of a triangle from
smallest to largest based on given
angle measures. (G1.2.2)
• Utilize the Triangle Inequality Theorem.
Theorem 5.10: Side Lengths of a Triangle Theorem
• If one side of a triangle is longer than
another side, then the angle opposite
the longer side is larger than the angle
opposite the shorter side.
– Basically, the larger the side, the larger the
angle opposite that side.
2nd Largest
Angle
Smallest
Side
Largest Angle
2nd Longest Side
Smallest
Angle
Theorem 5.11: Angle Measures of a Triangle Theorem
• If one angle of a triangle is larger than
another angle, then the side opposite
the larger angle is longer than the side
opposite the smaller angle.
– Basically, the larger the angle, the larger
the side opposite that angle.
2nd Largest
Angle
Smallest
Side
Largest Angle
2nd Longest Side
Smallest
Angle
Example 3.8
Order the angles from largest to smallest.
2.
1.
B, A, C
Q, P, R
3.
A, C , B
Example 3.9
Order the sides from largest to smallest.
1.
ST , RS , RT
2.
DE , EF , DF
33o
Example 3.10
Order the angles from largest to smallest.
1.
In ABC
AB = 12
BC = 11
C , A, B
AC = 5.8
Order the sides from largest to smallest.
2.
In XYZ
mX = 25o
XY , XZ , YZ
mY = 33o
mZ = 122o
Theorem 5.13: Triangle Inequality
• The sum of the lengths of any two sides
of a triangle is greater than the length of
the third side.
4
3
1
3
6
6
4
2
6
Add each combination of two sides to make sure
that they are longer than the third remaining side.
Example 3.11
Determine whether the following could be
lengths of a triangle.
a) 6, 10, 15
a) 6 + 10 > 15
10 + 15 > 6
6 + 15 > 10
YES!
b) 11, 16, 32
b) 11 + 16 < 32
NO!
Hint: A shortcut is to make sure that the sum of the two smallest
sides is bigger than the third side.
The other sums will always work.
Homework 3.2
• Lesson 3.2 – Inequalities in One Triangle
– p7-8
• Due Tomorrow
• Quiz Friday, October 15th
Lesson 3.3
Isosceles,
Equilateral,
and
Right Triangles
Lesson 3.3 Objectives
• Utilize the Base Angles Theorem to
solve for angle measures. (G1.2.2)
• Utilize the Converse of the Base Angles
Theorem to solve for side lengths.
(G1.2.2)
• Identify properties of equilateral
triangles to solve for side lengths and
angle measures. (G1.2.2)
Special Parts of an Isosceles Triangle
• An isosceles triangle has only two
congruent sides
– Those two congruent sides are
called legs.
– The third side is called the base.
legs
base
Isosceles Triangle Theorems
•Theorem 4.6: Base
Angles Theorem
–If two sides of a
triangle are
congruent, then the
angles opposite them
are congruent.
•Theorem 4.7:
Converse of Base
Angles Theorem
–If two angles of a
triangle are
congruent, then the
sides opposite them
are congruent.
Example 3.12
Solve for x and y.
1.
4.
2(22)  11  44  11  55
3x 11  2x 11
x 11  11
x  22
55  55  2 y  180
110  2 y  180
2 y  70 y  35
55o
55o
x7
2.
5.
= 90o
45o= 45
+ = 45o
x  75o
3.
75o
x  75  75  180
x  150  180
x  30
= 90o
3x  45
x  15
y  7  45
y  38
Equilateral Triangles
•Corollary to Theorem 4.6
•Corollary to Theorem 4.7
–If a triangle is
equilateral, then it is
equiangular.
–If a triangle is
equiangular, then it is
equilateral.
Example 3.13
Solve for x and y.
1.
5xo
2.
5xo
5x  5x  5x  180
15x  180
x  12
Or…In order for a triangle to be
equiangular, all angles must
equal…
5x  60
x  12
It does not matter which two sides you
set equal to each other, just pick the
pair that looks the easiest!
2x  3  4x  5
3  2x  5
8  2x
x4
Special Parts in a Right Triangle
• Right triangles have special names that
go with its parts as well.
• For instance:
– The two sides that form the right angle are
called the legs of the right triangle.
– The side opposite the right angle is called
the hypotenuse.
• The hypotenuse is always the longest side of a
right triangle.
hypotenuse
legs
Homework 3.3
• Lesson 3.3 – Isosceles, Equilateral, and
Right Triangles
– p9-11
• Due Tomorrow
• Quiz Tomorrow
– Tuesday, October 19th
Lesson 3.4
Medians
And
Altitudes of Triangles
Lesson 3.4 Objectives
• Identify a median, an altitude, and a
perpendicular bisector of a triangle.
(G1.2.5)
• Identify a centroid of a triangle.
• Utilize medians and altitudes to solve
for missing parts of a triangle. (G1.2.5)
• Identify the orthocenter of a triangle.
Perpendicular Bisector
• A segment, ray, line, or plane that is perpendicular to
a segment at its midpoint is called the
perpendicular bisector.
Triangle Medians
• A median of a triangle is a segment that does the
following:
– Contains one endpoint at a vertex of the triangle,
and
– Contains its other endpoint at the midpoint of the
opposite side of the triangle.
A
B
D
C
Centroid
Remember: All medians intersect the
midpoint of the opposite side.
• When all three medians are drawn in, they
intersect to form the centroid of a triangle.
– This forms a point of concurrency which is defined as a
point formed by the intersection of two or more lines.
• The centroid happens to find the balance point
for any triangle.
• In Physics, this is how we locate the center of mass.
Obtuse
Acute
Right
Theorem 5.7: Concurrency of Medians of a Triangle
• The medians of a triangle intersect at a point that is
two-thirds of the distance from each vertex to the
midpoint of the opposite side.
– The centroid is 2/3 the distance from any vertex to the
opposite side.
• Or said another way, the centroid is twice as far away from the
opposite angle as it is to the nearest side.
AP = 2/3AE
BP = 2/3BF
CP = 2/3CD
Example 3.14
S is the centroid of RTW, RS = 4, VW = 6, and TV = 9. Find the following:
a)
RV
a)
b)
6
RU
b)
6
•
Half of 4 is 2, and …
•
c)
4+2=6
•
Works every time!!
SU
c)
d)
2
RW
d)
e)
12
TS
e)
6
•
f)
6 is 2/3 of 9
SV
f)
3
•
Half of 6, which is the other part of the median.
Altitudes
• An altitude of a triangle is the perpendicular
segment from a vertex to the opposite side.
– It does not bisect the angle.
– It does not bisect the side.
• The altitude is often thought of as the height.
– While true, there are 3 altitudes in every triangle but only 1
height!
Orthocenter
• The three altitudes of a triangle intersect at a point
that we call the orthocenter of the triangle.
• The orthocenter can be located:
– inside the triangle
– outside the triangle, or
– on one side of the triangle
Obtuse
Right
Acute
The orthocenter of a right triangle
will always be located at the vertex
that forms the right angle.
Example 3.15
Is segment BD a median, altitude, or
perpendicular bisector of ABC?
Hint: It could be more than one!
1.
3.
Perpendicular
Bisector
Median
Altitude
Median
4.
2.
None
None
Homework 3.4
• Lesson 3.4 – Altitudes and Medians
– p12-13
• Due Tomorrow
Lesson 3.5
Area
and
Perimeter of Triangles
Lesson 3.5 Objectives
• Find the perimeter and area of
triangles. (G1.2.2)
Reviewing Altitudes
Determine the size of the altitudes of the
following triangles.
I.
6
II.
III.
16
If it is a right triangle, then
you can use Pythagorean
Theorem to solve for the
missing side length.
c
?
a
b
a 2  b2  c2
a 2  62  102
a 2  36  100
a 2  64
a  64  8
Area
• The area of a figure is defined as “the amount of space inside
the boundary of a flat (2-dimensional) object”
– http://www.mathsisfun.com/definitions/area.html
• Because of the 2-dimensional nature, the units to measure
area will always be “squared.”
– For example:
•
•
•
•
•
in2 or square inches
ft2 or square feet
m2 or square meters
mi2 or square miles
The area of a rectangle has up until now been found by taking:
• length x width (l x w)
•
We will now change the wording slightly to fit a more general pattern
for all shapes, and that is:
• base x height (b x h)
•
bl
That general pattern will exist as long as
the base and height form a right angle.
– Or said another way, the base and
height both touch the right angle.
w
h
Area of a Triangle
• The area of a triangle is found by taking one-half
the base times the height of the triangle
• Again the base and height form a right angle.
– Notice that the base is an actual side of the triangle, and…
– The height is nothing more than the altitude of the triangle drawn
from the base to the opposite vertex.
1
A( )   b  h
2
b
h
h
b
Perimeter of a Triangle
• The perimeter of a triangle is found by taking the sum of all
three sides of the triangle.
– So basically you need to add all three sides together.
• The perimeter is a 1-dimensional measurement, so the units should
not have an exponent on them.
– Example:
» in
» ft
» m
» mi
P( )  a  b  c
b
a
h
c
Example 3.16
Find the area and perimeter of the following triangles.
1.
1
A( )   b  h
2
1
A( )  (24)(10)
2
A( )  (12)(10)
A( )  120 sq. units
Homework 3.5
• Lesson 3.5 – Area and Perimeter of
Triangles
– p14-15
• Due Tomorrow