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Splash Screen. Lesson Menu Five-Minute Check (over Lesson 2–4) CCSS Then/Now New Vocabulary Postulates: Points, Lines, and Planes Key Concept: Intersections.

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Presentation on theme: "Splash Screen. Lesson Menu Five-Minute Check (over Lesson 2–4) CCSS Then/Now New Vocabulary Postulates: Points, Lines, and Planes Key Concept: Intersections."— Presentation transcript:

1 Splash Screen

2 Lesson Menu Five-Minute Check (over Lesson 2–4) CCSS Then/Now New Vocabulary Postulates: Points, Lines, and Planes Key Concept: Intersections of Lines and Planes Example 1: Real-World Example: Identifying Postulates Example 2: Analyze Statements Using Postulates Key Concept: The Proof Process Example 3: Write a Paragraph Proof Theorem 2.1:Midpoint Theorem

3 Define the following terms: 1. Postulate 2. Proof 3. Theorem 4. Algebraic Proof 5. Two Column Proof Bellringer 9/24/12

4 Vocabulary postulate proof theorem algebraic proof two column proof

5 Concept

6

7 Example 1 Identifying Postulates ARCHITECTURE Explain how the picture illustrates that the statement is true. Then state the postulate that can be used to show the statement is true. A. Points F and G lie in plane Q and on line m. Line m lies entirely in plane Q. Answer: Points F and G lie on line m, and the line lies in plane Q. Postulate 2.5, which states that if two points lie in a plane, the entire line containing the points lies in that plane, shows that this is true.

8 Example 1 ARCHITECTURE Explain how the picture illustrates that the statement is true. Then state the postulate that can be used to show the statement is true. B. Points A and C determine a line. Answer: Points A and C lie along an edge, the line that they determine. Postulate 2.1, which says through any two points there is exactly one line, shows that this is true. Identifying Postulates

9 Example 1 A.Through any two points there is exactly one line. B.A line contains at least two points. C.A plane contains at least three noncollinear points. D.A plane contains at least two noncollinear points. ARCHITECTURE Refer to the picture. State the postulate that can be used to show the statement is true. A. Plane P contains points E, B, and G.

10 Example 1 A.Through any two points there is exactly one line. B.A line contains at least two points. C.If two lines intersect, then their intersection is exactly one point. D.If two planes intersect, then their intersection is a line. ARCHITECTURE Refer to the picture. State the postulate that can be used to show the statement is true. B. Line AB and line BC intersect at point B.

11 Example 2 Analyze Statements Using Postulates Answer:Always; Postulate 2.5 states that if two points lie in a plane, then the entire line containing those points lies in the plane. A. Determine whether the following statement is always, sometimes, or never true. Explain. If plane T contains contains point G, then plane T contains point G.

12 Example 2 Analyze Statements Using Postulates Answer: Never; noncollinear points do not lie on the same line by definition. B. Determine whether the following statement is always, sometimes, or never true. Explain. contains three noncollinear points.

13 Example 2 A. Determine whether the statement is always, sometimes, or never true. Plane A and plane B intersect in exactly one point. A.always B.sometimes C.never

14 Example 2 B. Determine whether the statement is always, sometimes, or never true. Point N lies in plane X and point R lies in plane Z. You can draw only one line that contains both points N and R. A.always B.sometimes C.never

15 Concept

16 Example 1 Justify Each Step When Solving an Equation Solve -5(x+4) = 70. Algebraic StepsProperties -5(x+4)=70Original equation -5x + (-5)4=70Distributive Property -5x - 20=70Substitution Property -5x - 20 + 20 =70 + 20Addition Property

17 Example 1 Justify Each Step When Solving an Equation Answer: a = –18 –5x=90Substitution Property Division Property x = -18 Substitution Property -5x = 90 -5

18 Example 1 Justify Each Step When Solving an Equation Solve 2(5 – 3a) – 4(a + 7) = 92. Algebraic StepsProperties 2(5 – 3a) – 4(a + 7)=92Original equation 10 – 6a – 4a – 28=92Distributive Property –18 – 10a=92Substitution Property –18 – 10a + 18 =92 + 18Addition Property

19 Example 1 Justify Each Step When Solving an Equation Answer: a = –11 –10a=110Substitution Property Division Property a=–11Substitution Property

20 Example 1 A.a = 12 B.a = –37 C.a = –7 D.a = 7 Solve –3(a + 3) + 5(3 – a) = –50.

21 Example 2 Write an Algebraic Proof Begin by stating what is given and what you are to prove.

22 Example 2 Write an Algebraic Proof 2. d – 5 = 20t2. Addition Property of Equality StatementsReasons Proof: 1. Given 1. d = 20t + 5 4.4. Symmetric Property of Equality 3.3. Division Property of Equality = t

23 Example 2 Which of the following statements would complete the proof of this conjecture? If the formula for the area of a trapezoid is, then the height h of the trapezoid is given by.

24 Example 2 StatementsReasons Proof: 3.3. Division Property of Equality 4.4. Symmetric Property of Equality 1. Given 1. 2._____________2. Multiplication Property of Equality ?

25 Example 2 A.2A = (b 1 + b 2 )h B. C. D.

26 Example 3 Write a Geometric Proof If  A  B, m  B = 2m  C, and m  C = 45, then m  A = 90. Write a two-column proof to verify this conjecture.

27 Example 3 5. m  A = 90 5. Substitution StatementsReasons Proof: 4. Substitution 4. m  A = 2(45) Write a Geometric Proof 2. m  A = m  B 2. Definition of angles 1. Given 1.  A  B; m  B = 2m  C; m  C = 45 3. Transitive Property of Equality 3. m  A = 2m  C

28 Example 3

29 StatementsReasons Proof: 1. Given 1. 2.2. _______________ ? 3. AB = RS3. Definition of congruent segments 4. AB = 124. Given 5. RS = 125. Substitution

30 Example 3 A. Reflexive Property of Equality B. Symmetric Property of Equality C.Transitive Property of Equality D. Substitution Property of Equality

31 End of the Lesson


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