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5.5 Inequalities in One Triangle

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5.5 Inequalities in One Triangle

Geometry

Mrs. Spitz

Fall, 2004

- Use triangle measurements to decide which side is longest or which angle is largest.
- Use the Triangle Inequality

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.

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.

mA > mC

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.

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 < mH < m J

JH < JG < GH

100°

45°

35°

Write the measurements of the triangles from least to greatest.

QP < PR < QR

m R < mQ < m P

8

7

5

Given►AC > AB

Prove ►mABC > mC

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 mABC = m1+m3, it follows that mABC > m1. Substituting m2 for m1 produces mABC > m2. Because m2 = m3 + mC, m2 > mC. Finally because mABC > m2 and m2 > mC, you can conclude that mABC > mC.

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 m2 must be greater than the measure of either nonadjacent interior angle. This result is stated in Theorem 5.12

- The measure of an exterior angle of a triangle is greater than the measure of either of the two non adjacent interior angles.
- m1 > mA and m1 > mB

- 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, mB = mC. Because BC>AB, mA > mC by Theorem 5.10. By substitution, mA > mB. In addition, you can conclude that mA >60°, mB< 60°, and mC < 60°.

- Not every group of three segments can be used to form a triangle. The lengths of the segments must fit a certain relationship.

- 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

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 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.

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