# 5.5 Inequalities in One Triangle - PowerPoint PPT Presentation

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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.

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## 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

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.

### Objective 1: Comparing Measurements of a Triangle

The diagrams illustrate Thms. 5.10

and 5.11.

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.

### Theorem 5.10

mA > mC

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.

### Theorem 5.11

60°

40°

EF > DF

You can write the measurements

of a triangle in order from least to

greatest.

Write the measurements of the triangles from least to greatest.

m G < mH < m J

JH < JG < GH

### Ex. 1: Writing Measurements in Order from Least to Greatest

100°

45°

35°

Write the measurements of the triangles from least to greatest.

QP < PR < QR

m R < mQ < m P

8

7

5

### Paragraph Proof – Theorem 5.10

Given►AC > AB

Prove ►mABC > m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

### 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?

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°.

### 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

• 2 cm, 2 cm, 5 cm

• 3 cm, 2 cm, 5 cm

• 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.

2 cm, 2 cm, 5 cm

3 cm, 2 cm, 5 cm

4 cm, 2 cm, 5 cm

### Ex. 3: Constructing a Triangle

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

### Theorem 5.13: Triangle Inequality

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.

### Ex. 4: Finding Possible Side Lengths

Solve the inequality:

AB + AC > BC.

(x + 2) +(x + 3) > 3x – 2

2x + 5 > 3x – 2

5 > x – 2

7 > x

AB + BC > AC

MC + CG > MG

99 + 165 > x

264 > x

x + 99 < 165

x < 66

66 < x < 264