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5.5 Inequalities in One Triangle

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

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  1. 5.5 Inequalities in One Triangle Geometry Mrs. Spitz Fall, 2004

  2. Objectives: • Use triangle measurements to decide which side is longest or which angle is largest. • Use the Triangle Inequality

  3. Assignment pp. 298-300 #1-25, 34

  4. 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.

  5. 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

  6. 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.

  7. 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°

  8. Write the measurements of the triangles from least to greatest. QP < PR < QR m R < mQ < m P Ex. 1: Writing Measurements in Order from Least to Greatest 8 7 5

  9. 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.

  10. 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

  11. 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

  12. 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?

  13. 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°. Ex. 2: Using Theorem 5.10Solution

  14. 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.

  15. 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.

  16. 2 cm, 2 cm, 5 cm 3 cm, 2 cm, 5 cm 4 cm, 2 cm, 5 cm Ex. 3: Constructing a Triangle

  17. 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

  18. 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

  19. Solve the inequality: AB + AC > BC. (x + 2) +(x + 3) > 3x – 2 2x + 5 > 3x – 2 5 > x – 2 7 > x #24 - homework

  20. AB + BC > AC MC + CG > MG 99 + 165 > x 264 > x x + 99 < 165 x < 66 66 < x < 264 5. Geography

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