5.5 Inequalities
in One Triangle
Geometry
Mrs. Spitz
Fall, 2004
Objectives:
• Use triangle measurements to
decide which side is longest or which
angle is largest.
• Use the Triangle Inequality
Assignment
pp. 298-300 #1-25, 34
Objective 1: Comparing
Measurements of a Triangle
• In activity 5.5, you
may have
discovered a
relationship
between the
positions of the
longest and
shortest sides of a
triangle and the
position of its
angles.
The diagrams illustrate Thms. 5.10
and 5.11.
largest angle
longest side
shortest side
smallest angle
Theorem 5.10
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.
B
3
5
A
C
mA
> mC
Theorem 5.11
D
If one ANGLE of a
E
60°
40°
triangle is larger
than another
ANGLE, then the
SIDE opposite the
larger angle is
F
longer than the
EF > DF
side opposite the
smaller angle.
You can write the measurements
of a triangle in order from least to
greatest.
Ex. 1: Writing Measurements in Order
from Least to Greatest
Write the
measurements
of the triangles
from least to
greatest.
a. m G < mH <
m J
JH < JG < GH
J
100°
45°
H
35°
G
Ex. 1: Writing Measurements in Order
from Least to Greatest
Write the
measurements
of the triangles
from least to
greatest.
8
Q
7
5
b. QP < PR < QR
m R < mQ < m
P
R
P
Paragraph Proof – Theorem 5.10
A
Given►AC > AB
Prove ►mABC > mC
2
1
B
D
3
C
Use the Ruler Postulate to locate a point D on AC
such that DA = BA. Then draw the segment BD.
In the isosceles triangle ∆ABD, 1 ≅ 2.
Because mABC = m1+m3, it follows that
mABC > m1. Substituting m2 for m1
produces mABC > m2. Because m2 = m3
+ mC, m2 > mC. Finally because mABC >
m2 and m2 > mC, you can conclude that
mABC > mC.
NOTE:
The proof of 5.10 in the slide previous
uses the fact that 2 is an exterior
angle for ∆BDC, so its measure is
the sum of the measures of the two
nonadjacent interior angles. Then
m2 must be greater than the
measure of either nonadjacent
interior angle. This result is stated in
Theorem 5.12
Theorem 5.12-Exterior
Angle Inequality
• The measure of an exterior angle of
a triangle is greater than the
measure of either of the two non
adjacent interior angles.
• m1 > mA and m1 > mB
A
1
C
B
Ex. 2: Using Theorem 5.10
• DIRECTOR’S CHAIR. In the
director’s chair shown, AB ≅ AC and
BC > AB. What can you conclude
about the angles in ∆ABC?
A
B
C
Ex. 2: Using Theorem 5.10
Solution
• Because AB ≅ AC,
∆ABC is isosceles, so
B ≅ C. Therefore,
mB = mC.
Because BC>AB,
mA > mC by
Theorem 5.10. By
substitution, mA >
mB. In addition, you
can conclude that
mA >60°, mB< 60°,
and mC < 60°.
A
B
C
Objective 2: Using the
Triangle Inequality
• Not every group of three segments
can be used to form a triangle. The
lengths of the segments must fit a
certain relationship.
Ex. 3: Constructing a Triangle
a. 2 cm, 2 cm, 5 cm
b. 3 cm, 2 cm, 5 cm
c. 4 cm, 2 cm, 5 cm
Solution: Try drawing triangles with
the given side lengths. Only group
(c) is possible. The sum of the first
and second lengths must be
greater than the third length.
Ex. 3: Constructing a Triangle
a. 2 cm, 2 cm, 5 cm
b. 3 cm, 2 cm, 5 cm
c. 4 cm, 2 cm, 5 cm
2
2
5
C
D
D
3
4
2
A
5
2
B
A
5
B
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.
AB + BC > AC
AC + BC > AB
AB + AC > BC
A
C
B
Ex. 4: Finding Possible
Side Lengths
• A triangle has one
side of 10 cm and
another of 14 cm.
Describe the possible
lengths of the third
side
• SOLUTION: Let x
represent the length
of the third side.
Using the Triangle
Inequality, you can
write and solve
inequalities.
x + 10 > 14
x>4
10 + 14 > x
24 > x
►So, the length of the
third side must be
greater than 4 cm and
less than 24 cm.
#24 - homework
• Solve the
inequality:
AB + AC > BC.
A
x+ 2
B
x+ 3
3x - 2
C
(x + 2) +(x + 3) > 3x – 2
2x + 5 > 3x – 2
5>x–2
7>x
5. Geography
AB + BC > AC
MC + CG > MG
99 + 165 > x
264 > x
x + 99 < 165
x < 66
66 < x < 264
Masbate
99 miles
Guiuan
165 miles
Cadiz