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# Properties of angles and triangles

Download Presentation ## Properties of angles and triangles

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1. Properties of angles and triangles Chapter 2

2. Getting started: Geometric act Open to page 68 of your textbook. • Follow the instructions on page 69, and answer questions C and D fully. (Include a chart like in C for part D). • Make sure your answers are recorded on paper that you can hand in to me. • This activity is a summative assessment, so make sure that your work is neat and complete. • Your mosaic will be displayed after you finish, so feel free to use any supplies you want. • Grading will be done as follows: • Part C: 15 % • Part D: 85% • Minimum Requirements: 15% • Creativity: 30% • Classifications: 30%

3. 2.1 – Exploring parallel lines Chapter 2: Properties of Angles and Triangles

4. TraNsversals A transversal is a line that intersects two or more other lines at distinct points. What happens when a transversal intersects two parallel lines? How are the angles related?

5. Transversals (continued) • What can we say about angles a through g? • 140° + a = 180° • a = 180° - 140° = 40° • 140° + c = 180° • c = 180° - 140° = 40° • c + b = 40° + b = 180° • b = 180° - 40° = 140° • We can see that: • b = e = f = 140° • a = c = d = g = 40°

6. Transversals (continued) • So, we found that: • b = e = f = 140° • a = c = d = g = 40° • b and e, a and d, and c and g,are considered corresponding angles. When a transversal intersects parallel lines, corresponding angles are always equal. • The converse is also true: if a transverse creates equal corresponding angles then the two lines are parallel.

7. Some definitions Interior angles are any angles formed by a transversal and two parallel lines that lie inside the parallel lines. Exterior angles are any angles formed by a transversal and two parallel lines that lie outside the parallel lines. Corresponding angles are the set of one interior angle and one exterior angle that are non-adjacent and on the same side of the transversal.

8. Pg. 72 # 1 – 6 Independent practice

9. 2.2 – Angles formed by parallel lines Chapter 2: Properties of Angles and Triangles

10. Example Question: Solution: Conjecture: When a transversal intersects a pair of parallel lines,the alternate interior angles are equal. Make a conjecture that involves the interior angles formed by parallel lines and a transversal. Prove your conjecture. Our conjecture is proven!

11. example Determine the measures of a, b, c, and d. a = 110° (corresponding) a = b = 110° (vertically opposite) c + a = 180° (supplementary) c + 110° = 180° c = 180° - 110° = 70° c = d = 70° (alternate interior) So, a = b = 110° c = d = 70° Supplementary angles are two angles that together form a straight line, or have a sum of 180°

12. example One side of a cellphone tower will be built as shown. Use the angle measures to prove that braces CG, BF, and AE are parallel. Solution: • ∠DCG = ∠BAE = 78° • Since these two angles are corresponding angles, and are equal, CG||AE. • ∠HGC = ∠GFB = 78° • Since these two angles are corresponding angles, and are equal, CG||BF. • CG||AE and CG||BF so CG||AE||BF. • The three braces are parallel! Can you see another possible way to prove that CG, BF, and AE are parallel?

13. Cheat sheet! pg 78.

14. Pg. 78 – 82, # 1, 2, 4, 6, 8, 10, 12, 14, 15, 16, 19 Independent practice

15. 2.3 – Angle properties in triangles Chapter 2: Properties of Angles and Triangles

16. What is the sum of the measures of the angles of the interior of any triangle? Can we prove that the sum of the measures of the angles of the the interior of any triangle must be 180°? Draw an acute triangle, ΔRED. Construct line PQ through vertex D, parallel to RE. What can we say about the sum of angles ∠PDR, ∠RDE, and ∠QDE?

17. The sum of angles in a triangle So, we know that ∠PDR + ∠RDE + ∠QDE = 180°, because they form a straight line. So: ∠PDR + ∠RDE + ∠QDE = 180° ∠QDE = ∠DER (opposite interior ∠PDR = ∠DRE angles) Substitution: ∠DRE + ∠RDE + ∠DER = 180° We have proven that the sum of the measures of the interior angles of any triangle must be 180°. • What’s the relationship between ∠QDE and ∠DER? (Hint: they are opposite interior angles). • What about ∠PDR and ∠DRE?

18. example ∠MTA + 155° = 180° ∠MTA = 180° - 155 ° = 25° ∠MAT + ∠AMT + ∠MTA = 180° (sum of angles in a triangle) ∠MAT + 40° + 25°= 180° ∠MAT = 180° - 40°- 25° = 115° ∠MTA = 25°; ∠MAT = 115°; ∠AMT = 40°

19. example Determine the relationship between an exterior angle of a triangle and its non-adjacent interior angles? • ∠d + ∠c = 180° (straight line) • ∠c + (∠a + ∠b) = 180° (sum of angles in a triangle) •  ∠d + ∠c = ∠c + (∠a + ∠b) • ∠d = ∠a + ∠b • So, an exterior angle is equal to the sum of the measures of its non-adjacent interior angles.

20. Try it! Hint: ∠LMN and ∠MNP are opposite interior angles!

21. Pg. 90-93 # 2, 3, 5, 6, 7, 9, 10, 11, 13, 15, 16, 18. Independent practice

22. 2.4 – Angle properties in polygons Chapter 2: Properties of Angles and Triangles

23. Sum of angles of a quadrilateral How can we use what we know about triangles to help us figure out the angles of this quadrilateral? 180° 180° By breaking a quadrilateral into two triangles, we know that its interior angles must have a sum of 360°. Complete the handout using this method. Answer all of the questions as fully as possible, as this is a summative assessment.

24. example Outdoor furniture and structures like gazebos sometimes use a regular hexagon in their building plan. Determine the measure of each interior angle of a regular hexagon. Use the formula we learned to find the sum of the interior angles of a hexagon. S(n) = 180°(n – 2) S(6) = 180°(6-2) = 180° x 4 = 720° We know that the interior angles must have a sum of 720°, and “regular hexagon” means that all of the interior angles are equal. There are six equal angles in this hexagon. How can we figure out what each angle measures? The measure of each interior angle of a regular hexagon is 120°. 720° / 6 = 120°

25. Try it! Determine the measure of each interior angle of this 15 sided polygon.

26. example A floor tiler designs custom floors using tiles in the shape of regular polygons. Can the tiler use congruent regular octagons and congruent squares to tile a floor, if they have the same side length? Two octagons fit together, forming an angle of 2(135°) = 270°. This leaves a gap of 360° - 270° = 90°  A square can fit perfectly, as long as they have the same side lengths! Octagon: S(n) = 180°(n– 2 ) S(8) = 180°(8-2) = 1080° 1080°/8 = 135° Each angle in a regular octagon has an angle of 135°. Each interior angle in a square has an angle of 90°

27. Pg. 99 – 103, #2, 3, 4, 7, 8, 10, 11, 13, 15, 16, 18, 20 Independent practice

28. 2.5 – Exploring congruent triangles Chapter 2: Properties of Angles and Triangles

29. Congruent triangles activity Complete the handout. Make sure that as you do the activity you draw your completed triangles in the last two pages. It may be helpful to separate the last two pages. Make sure that you answer all the questions fully, as this is a summative assessment.

30. Congruent triangles There are minimum sets of angle and side measurements that, if known, allow you to conclude that two triangles are congruent. • SSS: Known as side-side-side congruence, if all three pairs of corresponding sides are equal, then the triangles are congruent. • SAS: Known as side-angle-side congruence, if two pairs of corresponding sides and the contained angles are equal, then the triangles are congruent. • ASA: Known as angle-side-angle congruence, if two pairs of corresponding angles and the contained sides are equal, then the triangles are congruent.

31. Congruent triangles SSS SAS ASA

32. Pg. 106, # 1 – 4 Independent practice

33. 2.6 – Proving Congruent triangles Chapter 2: Proving Congruent Triangles

34. Use sss, asa, and sasto prove congruence Given: TP ⊥ AC AP = CP Prove: ΔTAC is isosceles ∠TPA = ∠TPC = 90° (since TP and AC are perpendicular) AP = CP (given) TP = TP ∠TPA = ∠TPC AP = CP That’s SAS so therefore ΔTAP ≅ΔTCP, so TA = TC. Therefore, ΔTAC is isosceles. (Isosceles triangles have two equal sides.)

35. example Given: AE and BD bisect each other at C. AB = ED Prove: ∠A = ∠E AC = CE (since AE is bisected by C) BC = CD (since BD is bisected by C) AB = ED (given) That’s SSS, so ΔABC ≅ΔCDE The triangles are congruent, so all corresponding sides and angles are equal. ∠A = ∠E (since they are corresponding angles)

36. Pg. 112-115, #2, 4, 5, 7, 8, 9, 11, 13, 14, 17, 19 Independent practice