1. © 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 10 Geometry

2. © 2010 Pearson Prentice Hall. All rights reserved. 2 10.2 Triangles

3. © 2010 Pearson Prentice Hall. All rights reserved. Objectives 1.Solve problems involving angle relationships in triangles. 2.Solve problems involving similar triangles. 3.Solve problems using the Pythagorean Theorem. 3

4. © 2010 Pearson Prentice Hall. All rights reserved. Triangle A closed geometric figure that has three sides, all of which lie on a flat surface or plane. Closed geometric figures –If you start at any point and trace along the sides, you end up at the starting point. 4

5. © 2010 Pearson Prentice Hall. All rights reserved. Euclid’s Theorem Theorem: A conclusion that is proved to be true through deductive reasoning. Euclid’s assumption: Given a line and a point not on the line, one and only one line can be drawn through the given point parallel to the given line. 5

6. © 2010 Pearson Prentice Hall. All rights reserved. Euclid’s Theorem (cont.) Euclid’s Theorem: The sum of the measures of the three angles of any triangle is 180º. 6

7. © 2010 Pearson Prentice Hall. All rights reserved. Euclid’s Theorem (cont.) Proof: m  1 = m  2 and m  3 = m  4 (alternate interior angles) Angles 2, 5, and 4 form a straight angle (180º) Therefore, m  1 + m  5 + m  3 = 180º. 7

8. © 2010 Pearson Prentice Hall. All rights reserved. Two-Column Proof A Given: △ ABC Prove: ∠ A + ∠ B + ∠ ACB = 180° Statements 1.Through C, draw line m || to line AB. 2. ∠ 1 + ∠ 2 + ∠ 3 = 180° 3. ∠ 1 = ∠ A; ∠ 3 = ∠ B 4. ∠ A + ∠ ACB + ∠ B = 180° Reasons 1.Parallel postulate 2.Definition of straight angle 3.If 2 || lines, alt. int. ∠ ’s are equal 4.Substitution of equals axiom 1 A B C m 1 2 3 Theorem: The sum of ∠ ’s in a △ equals 180°.

9. © 2010 Pearson Prentice Hall. All rights reserved. Exterior ∠ Theorem Theorem: An exterior angle of a triangle equals the sum of the non-adjacent interior angles. ∠ CBD = ∠ A + ∠ C Since ∠ CBD + ∠ ABC = 180 -- def. of supplementary ∠ s ∠ A + ∠ C + ∠ ABC = 180 -- sum of △ ’s interior ∠ s ∠ A + ∠ C + ∠ ABC = ∠ CBD + ∠ ABC -- axiom ∠ A + ∠ C = ∠ CBD -- axiom 1 A C B D

10. © 2010 Pearson Prentice Hall. All rights reserved. Base ∠ s of Isosceles △ Theorem: Base ∠ s of an isosceles △ are equal Given: Isosceles △ ABC, which AC = BC Prove: ∠ A = ∠ B 1 A C B

11. © 2010 Pearson Prentice Hall. All rights reserved. Two-Column Proof B Given: Circle C with A, B, D on the circle Prove: ∠ ACB = 2 ∠ ADB Statements 1.Circle C, with central ∠ C, inscribed ∠ D 2.Draw line CD, intersecting circle at F 3. ∠ y = ∠ a + ∠ b = 2 ∠ a ∠ x = ∠ c + ∠ d = 2 ∠ c 4. ∠ x + ∠ y = 2( ∠ a + ∠ c) 5. ∠ ACB = 2 ∠ ADB Reasons 1.Given 2.2 points determine a line (postulate) 3.Exterior ∠ ; Isosceles △ angles 4.Equal to same quantity (axiom) 5.Substitution (axiom) 1 Theorem: An ∠ formed by 2 radii subtending a chord is 2 x an inscribed ∠ subtending the same chord. A B D C F c x b y a d

12. © 2010 Pearson Prentice Hall. All rights reserved. Example 1: Using Angle Relationships in Triangles Find the measure of angle A for the triangle ABC. Solution: m  A + m  B + m  C =180º m  A + 120º + 17º = 180º m  A + 137º = 180º m  A = 180º − 137º = 43º 12

13. © 2010 Pearson Prentice Hall. All rights reserved. Example 2: Using Angle Relationships in Triangles Find the measures of angles 1 through 5. Solution: m  1+ m  2 + 43º =180º 90º + m  2 + 43º = 180º m  2 + 133º = 180º m  2 = 47º 13

14. © 2010 Pearson Prentice Hall. All rights reserved. Example 2 continued m  3 + m  4+ 60º =180º 47º + m  4 + 60º = 180º m  4 = 180º − 107º = 73º m  4 + m  5 = 180º 73º + m  5 = 180° m  5 = 107º 14

15. © 2010 Pearson Prentice Hall. All rights reserved. Triangles and Their Characteristics 15

16. © 2010 Pearson Prentice Hall. All rights reserved. Similar Triangles 1 A B C Y Z X △ ABC ~ △ XYZ iff Corresponding angles are equal Corresponding sides are proportional ∠ A = ∠ X; ∠ B = ∠ Y; ∠ C = ∠ Z AB/XY = BC/YZ = AC/XZ If 2 corresponding ∠ s of 2 △ s are equal, then △ s are similar.

17. © 2010 Pearson Prentice Hall. All rights reserved. Example 1 Given: –AB || DE –AB = 25; CE = 8 Find AC Solution x 25 --- = ----- 8 12 8(25) 25 50 x = ------- = 2 ---- = ----- ≈ 16.66 12 3 3 1 A E D C B 25 8 12 x

18. © 2010 Pearson Prentice Hall. All rights reserved. Example 2 How can you estimate the height of a building, if you know your own height (on a sunny day)? 1

19. © 2010 Pearson Prentice Hall. All rights reserved. Example 2 1 x 400 6 10 6 10 --- = ------ x 400 6 400 = 10 x x = 240

20. © 2010 Pearson Prentice Hall. All rights reserved. Similar Triangles Similar figures have the same shape, but not necessarily the same size. In similar triangles, the angles are equal but the sides may or may not be the same length. Corresponding angles are angles that have the same measure in the two triangles. Corresponding sides are the sides opposite the corresponding angles. 20

21. © 2010 Pearson Prentice Hall. All rights reserved. Similar Triangles (continued) Triangles ABC and DEF are similar: Corresponding Angles Corresponding Sides Angles A and D Sides Angles C and F Sides Angles B and E Sides 21

22. © 2010 Pearson Prentice Hall. All rights reserved. Example 3: Using Similar Triangles Find the missing length x. Solution: Because the triangles are similar, their corresponding sides are proportional: 22

23. © 2010 Pearson Prentice Hall. All rights reserved. Example 3: Using Similar Triangles Find the missing length x. The missing length of x is 11.2 inches. 23

24. © 2010 Pearson Prentice Hall. All rights reserved. Pythagorean Theorem The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse. If the legs have lengths a and b and the hypotenuse has length c, then a² + b² = c² 24

25. © 2010 Pearson Prentice Hall. All rights reserved. Example 5: Using the Pythagorean Theorem Find the length of the hypotenuse c in this right triangle: Solution: Let a = 9 and b = 12 C 25