This is an instructional material for math teachers.

1. Geometry
Lesson 5 – 6
Inequalities in Two Triangles
Objective:
Apply the Hinge Theorem or its converse to make comparisons in
two triangles.
Prove triangle relationships using the Hinge Theorem or its converse.

2. Inequalities in Two Triangles
Hinge Theorem
 If two sides of a triangle are congruent to two
sides of another triangle, and the included angle
of the first is larger than the included angle of the
second triangle, then the third side of the first
triangle is longer than the third side of the
second triangle.

3. Converse of Hinge Theorem
If two sides of a triangle are congruent to two
sides of another triangle, and the third side in
the first triangle is longer than the third side in
the second triangle, then the include angle
measure of the first triangle is greater than
the included angle measure in the second.

4. Compare the given measures
WX and XY
BFC
m
and
FCD
m 

WX < XY
BFC
m
FCD
m 



5. Compare the given measures
JK and MQ
VRT
m
and
SRT
m 

JK > MQ
VRT
m
SRT
m 



6. AD and BD
Compare the given measures
BDC
m
and
ABD
m 

AD > BD
BDC
m
ABD
m 



7. Real World
Two groups of snowmobilers leave from the same
base camp. Group A goes 7.5 miles due west and
then turns 35 degrees north of west and goes 5
miles. Group B goes 7.5 miles due east then turns
40 degrees north of east and goes 5 miles. At this
point, which group is farther from the base camp?
Explain.
Draw a picture:
Group A is farther from camp
since the included angle is
larger than Group B.

8. Find the range of possible values for x.
6x + 15 > 65 6x + 15 > 0 6x + 15 < 180
Angle has to be greater than 0, but less than 180.
6x > 50
3
1
8

x
6x < 165
2
1
27

x
2
1
27
3
1
8
: 
 x
Range
Don’t have to solve since
we already said has to be
greater than 65.
Double check each time!

9. Find the range of possible values for x.
9a + 15 < 141 9a + 15 > 0 9a + 15 < 180
9a < 126
a < 14
9a > -15
3
2
1


a
Don’t have to solve since
we already said has to be
less than 141.
Double check each time!
14
3
2
1
: 

 a
Range

10. Find the range of possible values for x.
5x + 2 < 47 5x + 2 > 0 The length of a side
must be positive.
Do not need < 180
since 180 is for an angle
not a side, and side has
no limit on length.
5x < 45 5x > -2
5
2


x
x < 9
9
5
2
: 

 x
Range

12. Homework
Pg. 371 1 – 8 all, 10 – 22 E, 38,
44 – 58 E