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Contents Lesson 2-1Inductive Reasoning and Conjecture Lesson 2-2Logic Lesson 2-3Conditional Statements Lesson 2-4Deductive Reasoning Lesson 2-5Postulates and Paragraph Proofs Lesson 2-6Algebraic Proof Lesson 2-7Proving Segment Relationships Lesson 2-8Proving Angle Relationships
Lesson 1 Contents Example 1Patterns and Conjecture Example 2Geometric Conjecture Example 3Find a Counterexample
Example 1-1a Make a conjecture about the next number based on the pattern. 2, 4, 12, 48, 240 Answer: 1440 Find a pattern: ×2×2 The numbers are multiplied by 2, 3, 4, and 5. Conjecture: The next number will be multiplied by 6. So, it will be or ×3×3×4×4×5×5
Example 1-1b Make a conjecture about the next number based on the pattern. Answer: The next number will be
Example 1-2a Given: points L, M, and N; Examine the measures of the segments. Since the points can be collinear with point N between points L and M. Answer: Conjecture: L, M, and N are collinear. For points L, M, and N, and, make a conjecture and draw a figure to illustrate your conjecture.
Example 1-2b ACE is a right triangle with Make a conjecture and draw a figure to illustrate your conjecture. Answer: Conjecture: In ACE, C is a right angle and is the hypotenuse.
Example 1-3a UNEMPLOYMENT Based on the table showing unemployment rates for various cities in Kansas, find a counterexample for the following statement. The unemployment rate is highest in the cities with the most people. CountyCivilian Labor ForceRate Shawnee90,2543.1% Jefferson 9,9373.0% Jackson 8,9152.8% Douglas55,7303.2% Osage10,1824.0% Wabaunsee 3,5753.0% Pottawatomie11,0252.1% Source: Labor Market Information Services– Kansas Department of Human Resources
Example 1-3b Examine the data in the table. Find two cities such that the population of the first is greater than the population of the second while the unemployment rate of the first is less than the unemployment rate of the second. Shawnee has a greater population than Osage while Shawnee has a lower unemployment rate than Osage. Answer: Osage has only 10,182 people on its civilian labor force, and it has a higher rate of unemployment than Shawnee, which has 90,254 people on its civilian labor force.
Example 1-3c DRIVING The table on the next screen shows selected states, the 2000 population of each state, and the number of people per 1000 residents who are licensed drivers in each state. Based on the table, find a counterexample for the following statement. The greater the population of a state, the lower the number of drivers per 1000 residents.
Example 1-3c StatePopulationLicensed Drivers per 1000 Alabama 4,447, California33,871, Texas20,851, Vermont 608, West Virginia 1,808, Wisconsin 5,363, Source: The World Almanac and Book of Facts 2003 Answer: Alabama has a greater population than West Virginia, and it has more drivers per 1000 than West Virginia.
End of Lesson 1
Lesson 2 Contents Example 1Truth Values of Conjunctions Example 2Truth Values of Disjunctions Example 3Use Venn Diagrams Example 4Construct Truth Tables
Example 2-1a Use the following statements to write a compound statement for the conjunction p and q. Then find its truth value. p: One foot is 14 inches. q: September has 30 days. r: A plane is defined by three noncollinear points. Answer: One foot is 14 inches, and September has 30 days. p and q is false, because p is false and q is true.
Example 2-1b Use the following statements to write a compound statement for the conjunction. Then find its truth value. p: One foot is 14 inches. q: September has 30 days. r: A plane is defined by three noncollinear points. Answer: A plane is defined by three noncollinear points, and one foot is 14 inches. is false, because r is true and p is false.
Example 2-1c Use the following statements to write a compound statement for the conjunction. Then find its truth value. p: One foot is 14 inches. q: September has 30 days. r: A plane is defined by three noncollinear points. Answer: September does not have 30 days, and a plane is defined by three noncollinear points. is false because is false and r is true.
Example 2-1d Use the following statements to write a compound statement for the conjunction p r. Then find its truth value. p: One foot is 14 inches. q: September has 30 days. r: A plane is defined by three noncollinear points. Answer: A foot is not 14 inches, and a plane is defined by three noncollinear points. ~p r is true, because ~p is true and r is true.
Example 2-1e Answer: June is the sixth month of the year, and a turtle is a bird; false. Answer: A square does not have five sides, and a turtle is not a bird; true. Use the following statements to write a compound statement for each conjunction. Then find its truth value. p: June is the sixth month of the year. q: A square has five sides. r: A turtle is a bird. a. p and r b.
Example 2-1f Answer: A square does not have five sides, and June is the sixth month of the year; true. Answer: A turtle is not a bird, and a square has five sides; false. c. d. Use the following statements to write a compound statement for each conjunction. Then find its truth value. p: June is the sixth month of the year. q: A square has five sides. r: A turtle is a bird.
Example 2-2a Use the following statements to write a compound statement for the disjunction p or q. Then find its truth value. p: is proper notation for “line AB.” q: Centimeters are metric units. r: 9 is a prime number. Answer: is proper notation for “line AB,” or centimeters are metric units. p or q is true because q is true. It does not matter that p is false.
Example 2-2b Answer: Centimeters are metric units, or 9 is a prime number. is true because q is true. It does not matter that r is false. Use the following statements to write a compound statement for the disjunction. Then find its truth value. p: is proper notation for “line AB.” q: Centimeters are metric units. r: 9 is a prime number.
Example 2-2c Answer: 6 is an even number, or a triangle as 3 sides; true. Answer: A cow does not have 12 legs, or a triangle does not have 3 sides; true. Use the following statements to write a compound statement for each disjunction. Then find its truth value. p: 6 is an even number. q: A cow has 12 legs r: A triangle has 3 sides. a. p or r b.
Example 2-3a DANCING The Venn diagram shows the number of students enrolled in Monique’s Dance School for tap, jazz, and ballet classes.
Example 2-3a How many students are enrolled in all three classes? The students that are enrolled in all three classes are represented by the intersection of all three sets. Answer: There are 9 students enrolled in all three classes
Example 2-3b How many students are enrolled in tap or ballet? The students that are enrolled in tap or ballet are represented by the union of these two sets. Answer: There are or 121 students enrolled in tap or ballet.
Example 2-3c How many students are enrolled in jazz and ballet and not tap? The students that are enrolled in jazz and ballet and not tap are represented by the intersection of jazz and ballet minus any students enrolled in tap. Answer: There are – 9 or 25 students enrolled in jazz and ballet and not tap.
Example 2-3d PETS The Venn diagram shows the number of students at Manhattan School that have dogs, cats, and birds as household pets.
Example 2-3e a. How many students in Manhattan School have one of three types of pets? b. How many students have dogs or cats? c. How many students have dogs, cats, and birds as pets? Answer: 311 Answer: 280 Answer: 10
Example 2-4a Step 1 Make columns with the headings p, q, ~p, and ~p Construct a truth table for. ~p ~p~p q p
Example 2-4b Step 2 List the possible combinations of truth values for p and q. Construct a truth table for. FF TF FT TT ~p ~p~p qp
Example 2-4c Step 3 Use the truth values of p to determine the truth values of ~p. Construct a truth table for. TFF TTF FFT FTT ~p ~p~p qp
Example 2-4d Step 4 Use the truth values for ~p and q to write the truth values for ~p q. Answer: Construct a truth table for. TTFF TTTF FFFT TFTT ~p ~p~p qp
Example 2-4f p (~q r)~q r~q r ~qrqp Step 1 Make columns with the headings p, q, r, ~q, ~q r, and p (~q r). Construct a truth table for.
Example 2-4g Step 2 List the possible combinations of truth values for p, q, and r. Construct a truth table for. FFF p (~q r)~q r~q r ~q F T T F F T T r TF TT FT TF FF FT TT qp
Example 2-4h Step 3 Use the truth values of q to determine the truth values of ~q. Construct a truth table for. TFFF p (~q r)~q r~q r F T F T F T F ~q F T T F F T T r TF TT FT TF FF FT TT qp
Example 2-4i Step 4 Use the truth values for ~q and r to write the truth values for ~q r. Construct a truth table for. FTFFF p (~q r) F T F F F T F ~q r~q r F T F T F T F ~q F T T F F T T r TF TT FT TF FF FT TT qp
Example 2-4j Step 5 Use the truth values for p and ~q r to write the truth values for p (~q r). Answer: Construct a truth table for. FFTFFF F T F T T T T p (~q r) F T F F F T F ~q r~q r F T F T F T F ~q F T T F F T T r TF TT FT TF FF FT TT qp
Example 2-4l Construct a truth table for (p q) ~r. (p q) ~rp qp q ~rrqp Step 1 Make columns with the headings p, q, r, ~r, p q, and (p q) ~r.
Example 2-4m Step 2 List the possible combinations of truth values for p, q, and r. FFF (p q) ~rp qp q ~r F T T F F T T r TF TT FT TF FF FT TT qp Construct a truth table for (p q) ~r.
Example 2-4n Step 3 Use the truth values of r to determine the truth values of ~r. Construct a truth table for (p q) ~r. TFFF (p q) ~rp qp q T F F T T F F ~r F T T F F T T r TF TT FT TF FF FT TT qp
Example 2-4o Step 4 Use the truth values for p and q to write the truth values for p q. Construct a truth table for (p q) ~r. FTFFF (p q) ~r T F T T T T T p qp q T F F T T F F ~r F T T F F T T r TF TT FT TF FF FT TT qp
Example 2-4p Step 5 Use the truth values for p q and ~r to write the truth values for (p q) ~r. Answer: Construct a truth table for (p q) ~r. FFTFFF F F T F T F T (p q) ~r T F T T T T T p qp q T F F T T F F ~r F T T F F T T r TF TT FT TF FF FT TT qp
Example 2-4r Construct a truth table for the following compound statement. a. FFFFFF F F T F T F T F F T F F F T F F F F T F T TFF F T F F T T r TT FT TF TF FT TT qp Answer:
Example 2-4s b. Answer: FFFFFF T F T F T T T T T T F T T T T F T T T T T TFF F T F F T T r TT FT TF TF FT TT qp Construct a truth table for the following compound statement.
Example 2-4t c. Answer: FFFFFF T F T T T T T F F T F F F T T F T T T T T TFF F T F F T T r TT FT TF TF FT TT qp Construct a truth table for the following compound statement.
End of Lesson 2
Lesson 3 Contents Example 1Identify Hypothesis and Conclusion Example 2Write a Conditional in If-Then Form Example 3Truth Values of Conditionals Example 4Related Conditionals
Example 3-1a Identify the hypothesis and conclusion of the following statement. If a polygon has 6 sides, then it is a hexagon. Answer: Hypothesis: a polygon has 6 sides Conclusion: it is a hexagon hypothesis conclusion If a polygon has 6 sides, then it is a hexagon.
Example 3-1b Tamika will advance to the next level of play if she completes the maze in her computer game. Answer: Hypothesis: Tamika completes the maze in her computer game Conclusion: she will advance to the next level of play Identify the hypothesis and conclusion of the following statement.
Example 3-1c Identify the hypothesis and conclusion of each statement. a. If you are a baby, then you will cry. b. To find the distance between two points, you can use the Distance Formula. Answer: Hypothesis: you are a baby Conclusion: you will cry Answer: Hypothesis: you want to find the distance between two points Conclusion: you can use the Distance Formula
Example 3-2a Identify the hypothesis and conclusion of the following statement. Then write the statement in the if-then form. Answer: Hypothesis: a distance is determined Conclusion: it is positive If a distance is determined, then it is positive. Sometimes you must add information to a statement. Here you know that distance is measured or determined. Distance is positive.
Example 3-2b A five-sided polygon is a pentagon. Answer: Hypothesis: a polygon has five sides Conclusion: it is a pentagon If a polygon has five sides, then it is a pentagon. Identify the hypothesis and conclusion of the following statement. Then write the statement in the if-then form.
Example 3-2c Identify the hypothesis and conclusion of each statement. Then write each statement in if-then form. a. A polygon with 8 sides is an octagon. b. An angle that measures 45º is an acute angle. Answer: Hypothesis: a polygon has 8 sides Conclusion: it is an octagon If a polygon has 8 sides, then it is an octagon. Answer: Hypothesis: an angle measures 45º Conclusion: it is an acute angle If an angle measures 45º, then it is an acute angle.
Example 3-3a Determine the truth value of the following statement for each set of conditions. If Yukon rests for 10 days, his ankle will heal. Answer: Since the result is not what was expected, the conditional statement is false. The hypothesis is true, but the conclusion is false. Yukon rests for 10 days, and he still has a hurt ankle.
Example 3-3b Determine the truth value of the following statement for each set of conditions. If Yukon rests for 10 days, his ankle will heal. Answer: In this case, we cannot say that the statement is false. Thus, the statement is true. The hypothesis is false, and the conclusion is false. The statement does not say what happens if Yukon only rests for 3 days. His ankle could possibly still heal. Yukon rests for 3 days, and he still has a hurt ankle.
Example 3-3c Determine the truth value of the following statement for each set of conditions. If Yukon rests for 10 days, his ankle will heal. Answer: Since what was stated is true, the conditional statement is true. The hypothesis is true since Yukon rested for 10 days, and the conclusion is true because he does not have a hurt ankle. Yukon rests for 10 days, and he does not have a hurt ankle anymore.
Example 3-3d Determine the truth value of the following statement for each set of conditions. If Yukon rests for 10 days, his ankle will heal. Answer: In this case, we cannot say that the statement is false. Thus, the statement is true. The hypothesis is false, and the conclusion is true. The statement does not say what happens if Yukon only rests for 7 days. Yukon rests for 7 days, and he does not have a hurt ankle anymore.
Example 3-3e Determine the truth value of the following statements for each set of conditions. If it rains today, then Michael will not go skiing. a. It does not rain today; Michael does not go skiing. b. It rains today; Michael does not go skiing. c. It snows today; Michael does not go skiing. d. It rains today; Michael goes skiing. Answer: true Answer: false
Example 3-4a Write the converse, inverse, and contrapositive of the statement All squares are rectangles. Determine whether each statement is true or false. If a statement is false, give a counterexample. Conditional: If a shape is a square, then it is a rectangle. The conditional statement is true. First, write the conditional in if-then form. Write the converse by switching the hypothesis and conclusion of the conditional. Converse: If a shape is a rectangle, then it is a square. The converse is false. A rectangle with = 2 and w = 4 is not a square.
Example 3-4b Inverse: If a shape is not a square, then it is not a rectangle. The inverse is false. A 4-sided polygon with side lengths 2, 2, 4, and 4 is not a square, but it is a rectangle. The contrapositive is the negation of the hypothesis and conclusion of the converse. Contrapositive: If a shape is not a rectangle, then it is not a square. The contrapositive is true.
Example 3-4b Write the converse, inverse, and contrapositive of the statement The sum of the measures of two complementary angles is 90. Determine whether each statement is true or false. If a statement is false, give a counterexample. Answer: Conditional: If two angles are complementary, then the sum of their measures is 90; true. Converse: If the sum of the measures of two angles is 90, then they are complementary; true. Inverse: If two angles are not complementary, then the sum of their measures is not 90; true. Contrapositive: If the sum of the measures of two angles is not 90, then they are not complementary; true.
End of Lesson 3
Lesson 4 Contents Example 1Determine Valid Conclusions Example 2Determine Valid Conclusions From Two Conditionals Example 3Analyze Conclusions
Example 4-1a The following is a true conditional. Determine whether the conclusion is valid based on the given information. Explain your reasoning. If two segments are congruent and the second segment is congruent to a third segment, then the first segment is also congruent to the third segment. Conclusion: The hypothesis states that Answer: Since the conditional is true and the hypothesis is true, the conclusion is valid. Given:
Example 4-1b The following is a true conditional. Determine whether the conclusion is valid based on the given information. Explain your reasoning. If two segments are congruent and the second segment is congruent to a third segment, then the first segment is also congruent to the third segment. Answer: According to the hypothesis for the conditional, you must have two pairs of congruent segments. The given only has one pair of congruent segments. Therefore, the conclusion is not valid. The hypothesis states that is a segment and Conclusion: Given:
Example 4-1c The following is a true conditional. Determine whether each conclusion is valid based on the given information. Explain your reasoning. If a polygon is a convex quadrilateral, then the sum of the interior angles is 360. a. Given: Conclusion: If you connect X, N, and O with segments, the figure will be a convex quadrilateral. b. Given: ABCD is a convex quadrilateral. Conclusion: The sum of the interior angles of ABCD is 360. Answer: not valid Answer: valid
Example 4-2a PROM Use the Law of Syllogism to determine whether a valid conclusion can be reached from the following set of statements. (2) Mark is a 17-year-old student. (1) If Salline attends the prom, she will go with Mark. Answer: There is no valid conclusion. While both statements may be true, the conclusion of each statement is not used as the hypothesis of the other.
Example 4-2b PROM Use the Law of Syllogism to determine whether a valid conclusion can be reached from the following set of statements. (2) The Peddler Steakhouse stays open until 10 P.M. (1) If Mel and his date eat at the Peddler Steakhouse before going to the prom, they will miss the senior march. Answer: There is no valid conclusion. While both statements may be true, the conclusion of each statement is not used as the hypothesis of the other.
Example 4-2c Use the Law of Syllogism to determine whether a valid conclusion can be reached from each set of statements. a. (1) If you ride a bus, then you attend school. (2) If you ride a bus, then you go to work. b. (1) If your alarm clock goes off in the morning, then you will get out of bed. (2) You will eat breakfast, if you get out of bed. Answer: invalid Answer: valid
Example 4-3a Determine whether statement (3) follows from statements (1) and (2) by the Law of Detachment or the Law of Syllogism. If it does, state which law was used. If it does not, write invalid. (1) If the sum of the squares of two sides of a triangle is equal to the square of the third side, then the triangle is a right triangle. (3) XYZ is a right triangle. (2) For XYZ, (XY) 2 + (YZ) 2 = (ZX) 2.
Example 4-3b p: the sum of the squares of the two sides of a triangle is equal to the square of the third side q: the triangle is a right triangle By the Law of Detachment, if is true and p is true, then q is also true. Answer: Statement (3) is a valid conclusion by the Law of Detachment
Example 4-3c (1) If Ling wants to participate in the wrestling competition, he will have to meet an extra three times a week to practice. (3) If Ling wants to participate in the wrestling competition, he cannot take karate lessons. (2) If Ling adds anything extra to his weekly schedule, he cannot take karate lessons. Determine whether statement (3) follows from statements (1) and (2) by the Law of Detachment or the Law of Syllogism. If it does, state which law was used. If it does not, write invalid.
Example 4-3d p: Ling wants to participate in the wrestling competition q: he will have to meet an extra three times a week to practice r: he cannot take karate lessons By the Law of Syllogism, if and are true. Then is also true. Answer: Statement (3) is a valid conclusion by the Law of Syllogism.
Example 4-3e Determine whether statement (3) follows from statements (1) and (2) by the Law of Detachment of the Law of Syllogism. If it does, state which law was used. If it does not, write invalid. a. (1) If a children’s movie is playing on Saturday, Janine will take her little sister Jill to the movie. (3) If a children’s movie is playing on Saturday, Jill will get popcorn. (2) Janine always buys Jill popcorn at the movies. Answer: Law of Syllogism
Example 4-3f b. (1) If a polygon is a triangle, then the sum of the interior angles is 180. (3) The sum of the interior angles of polygon GHI is 180. (2) Polygon GHI is a triangle. Answer: Law of Detachment
End of Lesson 4
Lesson 5 Contents Example 1Points and Lines Example 2Use Postulates Example 3Write a Paragraph Proof
Example 5-1a SNOW CRYSTALS Some snow crystals are shaped like regular hexagons. How many lines must be drawn to interconnect all vertices of a hexagonal snow crystal? Explore The snow crystal has six vertices since a regular hexagon has six vertices. Plan Draw a diagram of a hexagon to illustrate the solution.
Example 5-1b Solve Label the vertices of the hexagon A, B, C, D, E, and F. Connect each point with every other point. Then, count the number of segments. Between every two points there is exactly one segment. Be sure to include the sides of the hexagon. For the six points, fifteen segments can be drawn.
Example 5-1b Answer: 15 Examine In the figure, are all segments that connect the vertices of the snow crystal.
Example 5-1c ART Jodi is making a string art design. She has positioned ten nails, similar to the vertices of a decagon, onto a board. How many strings will she need to interconnect all vertices of the design? Answer: 45
Example 5-2a 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. 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.
Example 5-2b Answer: Sometimes; planes Q and R can be parallel, and can intersect both planes. For, if X lies in plane Q and Y lies in plane R, then plane Q intersects plane R. Determine whether the following statement is always, sometimes, or never true. Explain.
Example 5-2c Answer: Never; noncollinear points do not lie on the same line by definition. contains three noncollinear points. Determine whether the following statement is always, sometimes, or never true. Explain.
Example 5-2d Determine whether each statement is always, sometimes, or never true. Explain. a. Plane A and plane B intersect in one point. b. 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. Answer: Never; Postulate 2.7 states that if two planes intersect, then their intersection is a line. Answer: Always; Postulate 2.1 states that through any two points, there is exactly one line.
Example 5-2e c. Two planes will always intersect a line. Answer: Sometimes; Postulate 2.7 states that if the two planes intersect, then their intersection is a line. It does not say what to expect if the planes do not intersect. Determine whether each statement is always, sometimes, or never true. Explain.
Example 5-3a Prove: ACD is a plane. Given intersecting, write a paragraph proof to show that A, C, and D determine a plane. Given: intersects Proof: must intersect at C because if two lines intersect, then their intersection is exactly one point. Point A is on and point D is on Therefore, points A and D are not collinear. Therefore, ACD is a plane as it contains three points not on the same line.
Example 5-3b Given is the midpoint of and X is the midpoint of write a paragraph proof to show that
Example 5-3c Proof: We are given that S is the midpoint of and X is the midpoint of By the definition of midpoint, Using the definition of congruent segments, Also using the given statement and the definition of congruent segments, If then Since S and X are midpoints, By substitution, and by definition of congruence,
End of Lesson 5
Lesson 6 Contents Example 1Verify Algebraic Relationships Example 2Write a Two-Column Proof Example 3Justify Geometric Relationships Example 4Geometric Proof
Example 6-1a Original equation Algebraic StepsProperties Solve Distributive Property Substitution Property Addition Property
Example 6-1b Substitution Property Division Property Substitution Property Answer:
Example 6-1c Original equation Algebraic StepsProperties Distributive Property Substitution Property Subtraction Property Solve
Example 6-1d Substitution Property Division Property Substitution Property Answer:
Example 6-2a If Write a two-column proof. then StatementsReasons Proof: 1. Given Multiplication Property 3.3. Substitution 4.4. Subtraction Property 5.5. Substitution 6.6. Division Property 7.7. Substitution
Example 6-2c Write a two-column proof. If then 1. Given Multiplication Property Distributive Property Subtraction Property Substitution Addition Property 6. Proof: Statements Reasons
Proof: Statements Reasons Example 6-2e 8. Division Property8. 9. Substitution9. 7. Substitution 7. Write a two-column proof. If then
Example 6-2f Write a two-column proof for the following. a.
Example 6-2g 1. Given 2. Multiplication Property 3. Substitution 4. Subtraction Property 5. Substitution 6. Division Property 7. Substitution Proof: Statements Reasons
Example 6-2h Prove: b. Given: Write a two-column proof for the following.
Example 6-2i Proof: Statements Reasons 1. Given 2. Multiplication Property 3. Distributive Property 4. Subtraction Property 5. Substitution 6. Subtraction Property 7. Substitution
Example 6-3a Read the Test Item Determine whether the statements are true based on the given information. A I only B I and II C I and III D I, II, and III MULTIPLE- CHOICE TEST ITEM then which of the following is a valid conclusion? I II III Ifand
Example 6-3b Solve the Test Item Statement II: Since the order you name the endpoints of a segment is not important, and TS = PR. Thus, Statement II is true. Statement I: Examine the given information, GH JK ST and. From the definition of congruence of segments, if, then ST RP. You can substitute RP for ST in GH JK ST to get GH JK RP. Thus, Statement I is true.
Example 6-3c Because Statements I and II only are true, choice B is correct. Answer: B Statement III If GH JK ST, then. Statement III is not true.
Example 6-3d If and then which of the following is a valid conclusion? I. II. III. MULTIPLE- CHOICE TEST ITEM A I only B I and II C I and III D II and III Answer: C
Example 6-4a SEA LIFE A starfish has five legs. If the length of leg 1 is 22 centimeters, and leg 1 is congruent to leg 2, and leg 2 is congruent to leg 3, prove that leg 3 has length 22 centimeters. Given: m leg 1 22 cm Prove: m leg 3 22 cm
Example 6-4a 1. Given1. 2. Transitive Property2. Proof: Statements Reasons 3. Definition of congruencem leg 1 m leg Givenm leg 1 22 cm4. 5. Transitive Propertym leg 3 22 cm5.
Example 6-4b DRIVING A stop sign as shown below is a regular octagon. If the measure of angle A is 135 and angle A is congruent to angle G, prove that the measure of angle G is 135.
Example 6-4c Proof: StatementsReasons 1. Given 2. Given 3. Definition of congruent angles 4. Transitive Property
End of Lesson 6
Lesson 7 Contents Example 1Proof With Segment Addition Example 2Proof With Segment Congruence
Example 7-1a Given: PR = QS Prove the following. Prove: PQ = RS 1. Given PR = QS Subtraction Property PR – QR = QS – QR Segment Addition Postulate PR – QR = PQ; QS – QR = RS Substitution PQ = RS 4. Proof: Statements Reasons
Example 7-1b Prove the following. Prove: Given:
Example 7-1c Proof: Statements Reasons 1. Given 2. Transitive Property 3. Given 4. Addition Property AC = AB, AB = BX AC = BX CY = XD AC + CY = BX + XD 5. Segment Addition Property AC + CY = AY; BX + XD = BD AY = BD 6. Substitution
Example 7-2a Prove the following. Prove: Given:
Example 7-2b 1. Given Definition of congruent segments Given Transitive Property Transitive Property 5. Proof: Statements Reasons
Example 7-2c Prove the following. Prove: Given:
Example 7-2d Proof: Statements Reasons 1. Given 2. Transitive Property 3. Given 4. Transitive Property 5. Symmetric Property
End of Lesson 7
Lesson 8 Contents Example 1Angle Addition Example 2Supplementary Angles Example 3Use Supplementary Angles Example 4Vertical Angles
Example 8-1a If the second hand stops where the angle is bisected, then the angle between the minute and second hands is one-half the measure of the angle formed by the hour and minute hands, or. TIME At 4 o’clock, the angle between the hour and minute hands of a clock is 120º. If the second hand stops where it bisects the angle between the hour and minute hands, what are the measures of the angles between the minute and second hands and between the second and hour hands?
Example 8-1b By the Angle Addition Postulate, the sum of the two angles is 120, so the angle between the second and hour hands is also 60º. Answer: They are both 60º by the definition of angle bisector and the Angle Addition Postulate.
Example 8-1b Answer: 50 QUILTING The diagram below shows one square for a particular quilt pattern. If andis a right angle, find
Example 8-2a Supplement Theorem Subtraction Property Answer: 14 form a linear pair andfindIfand
Example 8-2b Answer: 28 are complementary angles and. andIf find
Example 8-3a Given: form a linear pair. Prove: In the figure, form a linear pair, and Prove that are congruent. and
Example 8-3b 1. Given Definition of supplementary angles Subtraction Property Substitution Definition of congruent angles 6. Proof: Statements Reasons 2. Linear pairs are supplementary. 2.
Example 8-3c In the figure, NYR and RYA form a linear pair, AXY and AXZ form a linear pair, and RYA and AXZ are congruent. Prove that RYN and AXY are congruent.
Example 8-3d Proof: Statements Reasons 1. Given 2. If two s form a linear pair, then they are suppl. s. 3. Given linear pairs.
Example 8-4a Substitution Add 2d to each side. Add 32 to each side. Divide each side by 3. If 1 and 2 are vertical angles and m 1 and m 2 find m 1 and m 2. Vertical Angles Theorem 11 22 Definition of congruent angles m1m1m2m2
Example 8-4b Answer: m 1 = 37 and m 2 = 37
Example 8-4b Answer: m A = 52; m Z = 52 find and If and are vertical angles andand
End of Lesson 8
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