1.6 geometry homework answers big ideas math

Big Ideas Math Geometry Answers Chapter 1 Basics of Geometry

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Big Ideas Math Book Geometry Answer Key Chapter 1 Basics of Geometry

Access the Big Ideas Math Geometry Ch 1 Answers for all the topics and prepare accordingly. Find complete assistance on Geometry Chapter 1 including questions from Lessons 1.1-1.6, Performance Tests, Review Tests, Cumulative Practice, Assessment Tests, etc. All you have to do is simply tap on the quick links available and clear your ambiguities in no time. To make it easy for you we have compiled the BIM Geometry Chapter 1 Basics of Geometry Textbook Solutions aligned as per the BIM Textbooks.

• Basics of Geometry Maintaining Mathematical Proficiency – Page 1
• Basics of Geometry Mathematical Practices – Page 2
• 1.1 Points, Lines, and Planes – Page(3-10)
• Lesson 1.1 Points, Lines, and Planes – Page(4-7)
• Exercise 1.1 Points, Lines, and Planes – Page(8-10)
• 1.2 Measuring and Constructing Segments – Page(11-18)
• Lesson 1.2 Measuring and Constructing Segments – Page(12-15)
• Exercise 1.2 Measuring and Constructing Segments – Page(16-18)
• 1.3 Using Midpoint and Distance Formulas – Page(19-26)
• Lesson 1.3 Using Midpoint and Distance Formulas – Page(20-23)
• Exercise 1.3 Using Midpoint and Distance Formulas -Page(24-26)
• Study Skills: Keeping Your Mind Focused – Page 27
• 1.1 – 1.3 Quiz – Page 28
• 1.4 Perimeter and Area in the Coordinate Plane – Page(28-36)
• Lesson 1.4 Perimeter and Area in the Coordinate Plane – Page(30-33)
• Exercise 1.4 Perimeter and Area in the Coordinate Plane – Pae(34-36)
• 1.5 Measuring and Constructing Angles – Page(37-46)
• Lesson 1.5 Measuring and Constructing Angles – Page(38-43)
• Exercise 1.5 Measuring and Constructing Angles – Page(43-46)
• 1.6 Describing Pairs of Angles – Page(47-54)
• Lesson 1.6 Describing Pairs of Angles – Page(48-51)
• Exercise 1.6 Describing Pairs of Angles – Page(52-54)
• 1.4 – 1.6 Performance Task: Comfortable Horse Stalls – Page 55
• Basics of Geometry Chapter Review – Page(56-58)
• Basics of Geometry Chapter Test – Page 59
• Basics of Geometry Cumulative Assessment – Page(60-61)

Basics of Geometry Maintaining Mathematical Proficiency

Simplify the expression.

Question 1. |8 – 12| Answer: The given absolute value expression is: | 8 – 12| We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 8 – 12 |= | -4| = 4 Hence, from the above, We can conclude that the value of the given absolute value expression is: 4

Question 2. |- 6 – 5| Answer: The given absolute value expression is: | -6 – 5| We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | -6 – 5| = | -11 | = 11 Hence, from the above, We can conclude that the value of the given absolute value expression is: 11

Question 3. |4 + (-9)| Answer: The given absolute value expression is: | 4 + (-9) | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 4 + (-9) | = | 4 – 9 | = | -5 | = 5 Hence, from the above, We can conclude that the value of the given absolute value expression is: 5

Question 4. |13 + (-4)| Answer: The given absolute value expression is: | 13 + (-4) | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 13 + (-4) | = | 13 – 4 | = | 9 | = 9 Hence, from the above, We can conclude that the value of the given absolute value expression is: 9

Question 5. |6 – (- 2)| Answer: The given absolute value expression is: | 6 – (-2) | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 6 – (-2) | = | 6 + 2 | = | 8 | = 8 Hence, from the above, We can conclude that the value of the given absolute value expression is: 8

Question 6. |5 – (- 1)| Answer: The given absolute value expression is: | 5 – (-1) | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 5 – (-1) | = | 5 + 1 | = | 6 | = 6 Hence, from the above, We can conclude that the value of the given absolute value expression is: 6

Question 7. |- 8 – (- 7)| Answer: The given absolute value expression is: | -8 – (-7) | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | -8 – (-7) | = | -8 + 7 | = | -1 | = 1 Hence, from the above, We can conclude that the value of the given absolute value expression is: 1

Question 8. |8 – 13| Answer: The given absolute value expression is: | 8 – 13 | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 8 – 13 | = | -5 | = 5 Hence, from the above, We can conclude that the value of the given absolute value expression is: 5

Question 9. |- 14 – 3| Answer: The given absolute value expression is: | -14 – 3 | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | -14 – 3 | = | -17 | = 17 Hence, from the above, We can conclude that the value of the given absolute value expression is: 17

Find the area of the triangle.

Question 13. ABSTRACT REASONING Describe the possible values for x and y when |x – y| > 0. What does it mean when |x – y| = 0 ? Can |x – y| < 0? Explain your reasoning. Answer: We know that, The value of the absolute expression must be  greater than or equal to 0 but not less than 0 So, The values for | x – y | do not exist Now, The possible values of | x – y | > 0 should be greater than 0 and maybe x > y and x < y The possible values of | x – y | = 0 should be only one value i.e., 0 as x and y must be equal to make the difference value 0

Basics of Geometry Mathematical Practices

Monitoring Progress

Question 3. The distance between the two cities is 120 miles. What is the distance in kilometers? Round your answer to the nearest whole number. Answer: It is given that the distance between the two cities is 120 miles. Now, We know that, 1 mile = 1.609 kilometer So, The distance between the two cities in kilometers = 120 × 1.609 = 193 kilometers Hence, from the above, We can conclude that the distance between the two cities in kilometers is: 193 kilometers

1.1 Points, Lines, and Planes

Exploration 1

Using Dynamic Geometry Software

Exploration 2

Intersections of Lines and Planes

Exploration 3

Exploring Dynamic Geometry Software

UNDERSTANDING MATHEMATICAL TERMS To be proficient in math, you need to understand definitions and previously established results. An appropriate tool, such as a software package, can sometimes help.

Work with a partner. Use dynamic geometry software to explore geometry. Use the software to find a term or concept that is unfamiliar to you. Then use the capabilities of the software to determine the meaning of the term or concept. Answer: The dynamic geometry software used is “GeoGebra” The URL for “GeoGebra – Geometry” is: https://www.geogebra.org/calculator So, Using the above URL, Check the term that you are unfamiliar  to you and find the meaning of that term using the above URL

Communicate Your Answer

Question 4. How can you use dynamic geometry software to visualize geometric concepts? Answer: Some of the uses dynamic geometry software to visualize geometric concepts are: a. Request dispatching for cheap energy prices in cloud data centers b. Springerlink training kit c. Luminosity measurements at Hadron colliders d. From word embeddings to document distances

Lesson 1.1 Points, Lines, and Planes

Exercise 1.1 Points, Lines, and Planes

Vocabulary and Core Concept Check

Monitoring Progress and Modeling with Mathematics

In Exercises 17 – 24, sketch the figure described.

In Exercises 35 – 38, name the geometric term modeled by the object.

Question 46. CRITICAL THINKING Is it possible for one point to be in two different planes? Explain your reasoning. Answer: Yes, it is possible for one point to be in two different planes. This can be explained in the following way: The Intersection of two planes is a line. Hence, yes, one point or an Infinite number of points can be common between two different planes.

b. Your friend claims he went north on Cherry Court. and you went east on Apple Avenue. Make an argument as to why you know this could not have happened. Answer: It is given that your friend claims he went north on Cherry Court. and you went east on Apple Avenue But, from the above, It is given that you and your friend have to travel in the opposite directions But, according to the given statement, You and your friend travels in the perpendicular directions Hence, We can conclude that the claim of your friend is not possible

MATHEMATICAL CONNECTIONS In Exercises 51 – 54. graph the inequality on a number line. Tell whether the graph is a segment a ray or rays. a point, or a line.

CRITICAL THINKING In Exercises 56 – 63. complete the statement with always, sometimes, or never. Explain your reasoning.

Question 56. A line ________________ has endpoints. Answer: The given statement is: A line ________________ has endpoints. Hence, The completed statement is: A line never has endpoints.

Question 58. A plane and a point ________________ intersect. Answer: The given statement is: A plane and a point ________________ intersect. Hence, The completed statement is: A plane and a point sometimes intersect.

Question 60. Two points ____________________ determine a line. Answer: The given statement is: Two points ____________________ determine a line. Hence, The completed statement is: Two points always determine a line.

Question 62. Any three points not on the same line ____________________ determine a plane. Answer: The given statement is: Any three points not on the same line ____________________ determine a plane. Hence, The completed statement is: Any three points not on the same line always determine a plane.

Question 66. |3 – 9| Answer: The given absolute value expression is: | 3 – 9 | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 3 – 9 | = | -6 | = 6 Hence, from the above, We can conclude that the value of the given absolute value expression is: 6

Question 68. |7 – 11| Answer: The given absolute value expression is: | 7 – 11 | We know that, | x | = x for x > 0 | -x | = x for x > 0 So, | 7 – 11 | = | -4 | = 4 Hence, from the above, We can conclude that the value of the given absolute value expression is: 4

Solve the equation

Question 70. 36 + x = 20 Answer: The given equation is: 36 + x = 20 So, x = 20 – 36 x = -16 Hence, from the above, We can conclude that the value of x is: -16

Question 72. x – 23 = 19 Answer: The given equation is: x – 23 = 19 So, x = 19 + 23 x = 42 Hence, from the above, We can conclude that the value of x is: 42

1.2 Measuring and Constructing Segments

Essential Question How can you measure and construct a line segment? Answer: The steps used to measure a line segment are: a. Pick up a scale to measure the length of a line segment. b. Identify the line segment you want to measure c. Place the tip of the ruler at the starting of the line segment

The steps used to construct a line segment are: a. Place the compass at one end of the line b. Adjust the compass to slightly longer than half of the line’s length c. Draw arcs above and below the line d. Keeping the same compass width, draw arcs from the other end of the line e. Place ruler where the arcs cross and draw the line segment

Measuring Line Segments Using Nonstandard Units

Work with a partner.

c. Measure the length and width of the index card in paper clips. Answer: We know that, The length of the index card in paper clips is: 4 cm (Approximately ) The width of the index card in the paper clips is: 1 cm

Measuring Heights Using Nonstandard Units

Consider a unit of length that is equal to the length of the diagonal you found in Exploration 2. Call this length “1 diag.” How tall are you in diags? Explain how you obtained your answer. Answer: From Exploration 2, The length of the diagonal we have found is approximately equal to 6 inches So, According to the given problem, 1 diag = 6 inches We know that, The height can be measured in the foot So, 1 inch = 0.08 feet So, 63.6 inch = 5.3 feet So, In terms of diags, 63.6inches = 60 + 3.6 inches = 10 (6 inches) + 6 (0.6 inches) = 10 diags + 0.6 diags = 10.6 diags Hence, from the above, We can conclude that you are 10.6 diags tall

Communicate Your Answer:

Question 4. How can you measure and construct a line segment? Answer: The steps used to measure a line segment are: a. Pick up a scale to measure the length of a line segment. b. Identify the line segment you want to measure c. Place the tip of the ruler at the starting of the line segment

The steps used to construct a line segment are: a. Place the compass at one end of line b. Adjust the compass to slightly longer than half of the line’s length c. Draw arcs above and below the line d. Keeping the same compass width, draw arcs from the other end of line e. Place ruler where the arcs cross and draw the line segment

Lesson 1.2 Measuring and Constructing Segments

Use a ruler to measure the length of the segment.

Exercise 1.2 Measuring and Constructing Segments

In Exercises 3 – 6, use a ruler to measure the length of the segment to the nearest tenth of a centimeter.

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in finding the length of \(\overline{A B}\).

b. The model airplane’s flight lasted nearly 38 hours. Estimate the airplane’s average speed in miles per hour. Answer: From part (a), We observed that, The total distance traveled by the model airplane is: 1881 miles It is given that The model airplane’s flight lasted nearly 38 hours We know that, Average speed = \(\frac{Total distance}{Total time}\) So, Average speed = \(\frac{1881}{38}\) = 49.5 miles per hour Hence, from the above, We can conclude that the average speed of the airplane is: 49.5 miles per hour

Question 29. MATHEMATICAL CONNECTIONS Point S is between points R and T on \(\overline{R T}\). Use the information to write an equation in term of x. Then Solve the equation and find RS, ST, and RT. a. RS = 2x + 10 ST = x – 4 RT = 21

b. RS = 3x – 16 ST = 4x – 8 RT = 60

c. RS = 4x – 9 ST=11 RTx+IO

Question 32. MAKING AN ARGUMENT Your friend and your Cousin discuss measuring with a ruler. Your friend says that you must always line up objects at the zero on a ruler. Your cousin says it does not matter. Decide who is correct and explain your reasoning. Answer: Your cousin is correct

Explanation: It is given that, Your friend said that you must always line up objects at the zero on a ruler and your cousin said that it does not matter Now, We just have to deduce the endpoint length from the start point length on the ruler Example: If you measure a segment that is 5 cm long, it is the same if you put zero on the start point or 5 on the start point, then 10 is on the endpoint Remember that the starting point will be any number on the ruler and the ending point will be n units away from that number So, 5 – 0 = 5 is same as 10 – 5 = 5

Maintaining Mathematical Proficiency

Question 38. \(\frac{-4+6}{2}\) Answer: The given expression is: \(\frac{-6 + 4}{2}\) So, \(\frac{-4+6}{2}\) = \(\frac{2}{2}\) = \(\frac{1}{1}\) = 1 Hence, from the above, We can conclude that the value of the given expression is: 1

Question 40. \(\sqrt{25+9}\) Answer: The given expression is: \(\sqrt{25 + 9}\) So, \(\sqrt{25+9}\) = \(\sqrt{34}\) Hence, from the above, We can conclude that the value of the given expression is: \(\sqrt{34}\)

Solve the equation.

Question 42. 5x + 7 = 9x – 17 Answer: The given equation is: 5x + 7 = 9x – 17 So, 5x – 9x = -17 – 7 -4x = -24 4x = 24 x = \(\frac{24}{4}[/larex] x = 6 Hence, from the above, We can conclude that the value of x for the given equation is: 6

Question 44. \(\frac{-5+x}{2}\) = – 9 Answer: The given equation is: \(\frac{-5+x}{2}\) = – 9 So, -5 + x = -9 (2) -5 + x = -18 x = -18 + 5 x = -13 Hence, from the above, We can conclude that the value of x for the given equation is: -13

1.3 Using Midpoint and Distance Formulas

EssentiaI Question How can you find the midpoint and length of a line segment in a coordinate plane? Answer: Let the line segment is formed by the points A (x1, y1), B (x2, y2) So, The coordinates of the midpoint of the line segment are given as: M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) The length of the line segment in a coordinate plane is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)

Finding the Midpoint of a Line Segment

c. What are the coordinates of the midpoint M? Answer: We know that, The coordinates of the midpoint of the line segment are given as: M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{3 – 5}{2}\),\(\frac{4 – 2}{2}\)) M = (\(\frac{-2}{2}\),\(\frac{2}{2}\)) M = (-1, 1) Hence, The coordinates of the midpoint M is: (-1, 1)

d. Compare the x-coordinates of A, B, and M. Compare the y-coordinates of A, B, and M. How arc the coordinates of the 4 midpoints M related to the coordinates of A and B? Answer: The coordinates of A, B, and M are: A (3, 4), B (-5, -2), and M (-1, 1) Compare the given coordinates with A (x1, y1), B (x2, y2), C (x3, y3) Hence, The x-coordinates and y-coordinates of M are related to the coordinates of A and B are: M= (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}[/latex}

b. Use the Pythagorean Theorem to find the length of AB. Answer: The given points are: A (3, 4), B (-5, -2), and C (3, 2) Compare the given points with A (x1, y1), B (x2, y2), and C (x3, y3) We know that, The length of a segment is: D = [latex]\sqrt{(x2 – x1)² + (y2 – y1)²}\) So, \(\overline{B C}\) = \(\sqrt{(-5 + 3)² + (-2 + 2)²}\) = \(\sqrt{(-2)² + (0)²}\) = \(\sqrt{4 + 0}\) =\(\sqrt{4}\) = 2 \(\overline{A C}\) = \(\sqrt{(-3 + 3)² + (-2 + 4)²}\) = \(\sqrt{0² + 2²}\) = \(\sqrt{2²}\) = 2 Now, According to Pythagoras theorem, Hypotenuse² = Side² + Side² So, From the graph, AB represents the hypotenuse AC represents the side BC represents the side So, AB² = AC² + BC² AB² = 2² + 2² AB² = 4 + 4 AB² = 8 AB = \(\sqrt{8}\) Hence, from the above, We can conclude that the length of AB is: \(\sqrt{8}\)

c. Use a centimeter ruler to verify the length you found in part (b). MAKING SENSE OF PROBLEMS To be proficient in math, you need to check your answers and continually ask yourself, “Does this make sense?’ Answer: From part (b), We found the length of \(\overline{A B}\) as: \(\sqrt{8}\) Now, To verify the length of \(\overline{A B}\), use the “Distance formula” We know that, The distance beween 2 points of a line segment is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) So, \(\overline{A B}\) = \(\sqrt{(-5 + 3)² + (-2 + 4)²}\) = \(\sqrt{(-2)² + (2)²}\) = \(\sqrt{4 + 4}\) =\(\sqrt{8}\)

d. Use the Pythagorean Theorem and point M from Exploration 1 to find the lengths of \(\overline{A M}\) and \(\overline{M B}\). What can you conclude? Answer:

Question 3. How can you find the midpoint and length ola line segment in a coordinate plane? Answer: Let the line segment is formed by the points A (x1, y1), B (x2, y2) So, The coordinates of the midpoint of the line segment are given as: M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) The length of the line segment in a coordinate plane is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)

Question 4. Find the coordinates of the midpoint M and the length of the line segment whose endpoints are given. a. D(- 10, – 4), E(14, 6) Answer: The given points are: D (-10, -4), E (14, 6) We know that, The coordinates of the midpoint of the line segment are given as: M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) M = (\(\frac{14 – 10}{2}\), \(\frac{6 – 4}{2}\)) M = (\(\frac{4}{2}\), \(\frac{2}{2}\)) M = (2, 1) Hence, from the above, We can conclude that the coordinates of the midpoint of the given points are: (2, 1)

b. F(- 4, 8), G(9, 0) Answer: The given points are: F (-4, 8), G (9, 0) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{9 – 4}{2}\), \(\frac{8 + 0}{2}\)) M = (\(\frac{5}{2}\), \(\frac{8}{2}\)) M = (\(\frac{5}{2}\), 4) Hence, from the above, We can conclude that the coordinates of the midpoint of the given points are: (\(\frac{5}{2}\), 4)

Lesson 1.3 Using Midpoint and Distance Formulas

Identify the segment, bisector of \(\overline{P Q}\). Then find PQ.

Question 5. The endpoints of \(\overline{A B}\) are A (1, 2) and B(7, 8). Find the coordinates of the midpoint M. Answer: The given endpoints of \(\overline{A B}\) are: A (1, 2), B (7, 8) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{1 + 7}{2}\), \(\frac{8 + 2}{2}\)) M = (\(\frac{8}{2}\), \(\frac{10}{2}\)) M = (4, 5) Hence, from the above, We can conclude that the coordinates of the midpoint of \(\overline{A B}\) are: (4, 5)

Question 6. The endpoints of \(\overline{C D}\) are C( – 4, 3) and D(6, 5). Find the coordinates of the midpoint M. Answer: The given endpoints of \(\overline{C D}\) are: C (-4, 3), D (6, 5) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{6 – 4}{2}\), \(\frac{5 + 3}{2}\)) M = (\(\frac{2}{2}\), \(\frac{8}{2}\)) M = (1, 4) Hence, from the above, We can conclude that the coordinates of the midpoint of \(\overline{C D}\) are: (1, 4)

Question 7. The midpoint of \(\overline{T U}\) is M(2, 4). One endpoint is T(1, 1). Find the coordinates of endpoint U. Answer: The given points of \(\overline{T U}\) are: M (2, 4), T (1, 1) Let the one endpoint be T We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, (2, 4) = (\(\frac{x1 + 1}{2}\), \(\frac{y1 + 1}{2}\)) So, \(\frac{x1 + 1}{2}\) = 2              \(\frac{y1 + 1}{2}\) = 4 x1 + 1 = 2 (2)                                            y1 + 1 = 4 (2) x1 = 4 – 1                                                   y1 = 8 – 1 x1 = 3                                                         y1 = 7 Hence, from the above, We can conclude that the other endpoint of \(\overline{T U}\) is: (3, 7)

Question 8. The midpoint of \(\overline{V W}\) is M (- 1, – 2). One endpoint is W(4, 4). Find the coordinates of endpoint V. Answer: The given points of \(\overline{V W}\) are: M (-1, -2), W (4, 4) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, (-1, -2) = (\(\frac{x1 + 4}{2}\), \(\frac{y1 + 4}{2}\)) So, \(\frac{x1 + 4}{2}\) = -1              \(\frac{y1 + 4}{2}\) = -2 x1 + 4 = -1 (2)                                            y1 + 4 = -2 (2) x1 = -2 – 4                                                  y1 = -4 – 4 x1 = -6                                                         y1 = -8 Hence, from the above, We can conclude that the other endpoint V is: (-6, -8)

Question 9. In Example 4, a park is 3 miles east and 4 miles south of your apartment. Find the distance between the park and your school. Answer: Given, a park is 3 miles east and 4 miles south of your apartment. 3² + 4² = c² 9 + 16 = c² 25 = c² c = 5 Thus there is a 5-mile distance between the park and the apartment.

Exercise 1.3 Using Midpoint and Distance Formulas

Question 2. COMPLETE THE SENTENCE To find the length of \(\overline{A B}\), with endpoints A(- 7, 5) and B(4, – 6). you can use the _________ . Answer: We know that, To find the length with two endpoints, we use the “Distance formula” The distance between the 2 endpoints is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) Hence, from the above, We can conclude that To find the length of \(\overline{A B}\), with endpoints A(- 7, 5) and B(4, – 6). you can use the ” Distance formula ”

In Exercises 3 – 6. identify the segment bisector of \(\overline{R S}\). Then find RS.

In Exercises 7 and 8, identify the segment bisector of \(\overline{J K}\). Then find JM.

In Exercises 9 and 10. identify the segment bisector of \(\overline{X Y}\). Then find XY.

CONSTRUCTION In Exercises 11 – 14, copy the segment and construct a segment bisector by paper folding. Then label the midpoint M.

Question 16. C (4, 7) and D (0, – 3) Answer: The given endpoints of \(\overline{C D}\) are: C (4, 7), D (0, -3) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{4 + 0}{2}\), \(\frac{7 – 3}{2}\)) M = (\(\frac{4}{2}\), \(\frac{4}{2}\)) M = (2, 2) Hence, from the above, We can conclude that the coordinates of the midpoint of \(\overline{C D}\) are: (2, 2)

Question 18. C (- 8, – 6) and D (- 4, 10) Answer: The given endpoints of \(\overline{C D}\) are: C (-8, -6), D (-4, 10) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{-8 – 4}{2}\), \(\frac{10 – 6}{2}\)) M = (\(\frac{-12}{2}\), \(\frac{4}{2}\)) M = (-6, 2) Hence, from the above, We can conclude that the coordinates of the midpoint of \(\overline{C D}\) are: (-6, 2)

In Exercises 19 – 22. the midpoint M and one endpoint of \(\overline{G H}\) are given. Find the coordinates of the other endpoint.

Question 20. H (- 3, 7) and M (- 2, 5) Answer: The given points of \(\overline{G H}\) are: M (-2, 5), H (-3, 7) Let the point G be: (x1, y1) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, (-2, 5) = (\(\frac{x1 -3}{2}\), \(\frac{y1 + 7}{2}\)) So, \(\frac{x1 – 3}{2}\) = -2              \(\frac{y1 + 7}{2}\) = 5 x1 – 3 = -2 (2)                                            y1 + 7 = 5 (2) x1 = -4 – 3                                                  y1 = 10 – 7 x1 = -7                                                         y1 = 3 Hence, from the above, We can conclude that the other endpoint G is: (-7, 3)

Question 22. G (- 4, 1) and M (- \(\frac{13}{2}\), – 6) Answer: The given points of \(\overline{G H}\) are: M (- \(\frac{13}{2}\), – 6), G (-4, 1) Let the other endpoint be H (x2, y2) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, (- \(\frac{13}{2}\), – 6) = (\(\frac{x2 – 4}{2}\), \(\frac{y2 + 1}{2}\)) So, \(\frac{x2 – 4}{2}\) = – \(\frac{13}{2}\)           \(\frac{y2 + 1}{2}\) = -6 x2 – 4 = – \(\frac{13}{2}\) (2)                                         y2 + 1 = -6 (2) x1 = -13 + 4                                                                                 y2 = -12 – 1 x1 = -9                                                                                          y1 = -13 Hence, from the above, We can conclude that the other endpoint H is: (-9, -13)

In Exercises 23 – 30, find the distance between the two points.

Question 24. C (- 6, 5) and D (- 3, 1) Answer: The given points are: C (-6, 5), D (-3, 1) Compare the given points with A (x1, y1), B (x2, y2) So, The distance between points C and D is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) D = \(\sqrt{(6 – 3)² + (1 – 5)²}\) D = \(\sqrt{(3)² + (-4)²}\) D = \(\sqrt{9 + 16}\) D = \(\sqrt{25}\) D = 5 Hence, from the above, We can conclude that the distance between 2 given points is: 5

Question 26. G (- 5, 4) and H (2, 6) Answer: The given points are: G (-5, 4), H (2, 6) Compare the given points with A (x1, y1), B (x2, y2) So, The distance between points A and B is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) D = \(\sqrt{(2 + 5)² + (6 – 4)²}\) D = \(\sqrt{(7)² + (2)²}\) D = \(\sqrt{49 + 4}\) D = \(\sqrt{53}\) Hence, from the above, We can conclude that the distance between 2 given points is: \(\sqrt{53}\)

Question 28. L (7, – 1) and M (- 2, 4) Answer: The given points are: L (7, -1), M (-2, 4) Compare the given points with A (x1, y1), B (x2, y2) So, The distance between points A and B is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) D = \(\sqrt{(-2 – 7)² + (4 + 1)²}\) D = \(\sqrt{(-9)² + (5)²}\) D = \(\sqrt{81 + 25}\) D = \(\sqrt{106}\) Hence, from the above, We can conclude that the distance between 2 given points is: \(\sqrt{106}\)

Question 30. T (13, 1.6) and V (5.4, 3.7) Answer: The given points are: T (13, 1.6), V (5.4, 3.7) Compare the given points with A (x1, y1), B (x2, y2) So, The distance between points A and B is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) D = \(\sqrt{(13 – 5.4)² + (1.6 – 3.7)²}\) D = \(\sqrt{(7.6)² + (2.1)²}\) D = \(\sqrt{57.76 + 4.41}\) D = \(\sqrt{62.17}\) D = 7.884 Hence, from the above, We can conclude that the distance between 2 given points is: 7.884

ERROR ANALYSIS In Exercises 31 and 32, describe and correct the error in finding the distance between A(6, 2) and 8(1, – 4).

COMPARING SEGMENTS In Exercises 33 and 34, the endpoints of two segments are given. Find each segment length. Tell whether the segments are congruent. If they are not congruent, a state which segment length is greater.

Question 34. \(\overline{E F}\): E(1, 4),F(5, 1) and \(\overline{G H}\): G(-3, 1), H(1, 6) Answer: The given points of \(\overline{E F}\) are: E (1, 4), F (5, 1) Compare the given points with A (x1, y1), B (x2, y2) So, The distance between points E and F is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) D = \(\sqrt{(5 – 1)² + (1 – 4)²}\) D = \(\sqrt{(4)² + (-3)²}\) D = \(\sqrt{16 + 9}\) D = \(\sqrt{25}\) D = 5 Now, The given points are: G (-3, 1), H (1, 6) Compare the given points with A (x1, y1), B (x2, y2) So, The distance between points G and H is given as: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) D = \(\sqrt{(-1 – 3)² + (1 – 6)²}\) D = \(\sqrt{(-4)² + (-5)²}\) D = \(\sqrt{16 + 25}\) D = \(\sqrt{41}\) D = 6.4 Hence, from the above, We can conclude that The lengths of \(\overline{E F}\) and \(\overline{G H}\) are not congruent The comparison of the lengths is: \(\overline{E F}\) < \(\overline{G H}\)

Question 40. MAKING AN ARGUMENT Your friend claims there is an easier way to find the length of a segment than the Distance Formula when the x-coordinates of the endpoints are equal. He claims all you have to do is subtract the y-coordinates. Do you agree with his statement? Explain your reasoning. Answer: Yes, your friend is correct

Explanation: We know that, Distance between 2 points = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) It is given that the x-coordinates are equal So, x1 = x2 = 0 So, Distance between 2 points = y2 – y1 Hence, from the above, We can conclude that your friend is correct

Question 42. HOW DO YOU SEE IT? \(\overline{A B}\) contains midpoint M and points C and D. as shown. Compare the lengths. If you cannot draw a conclusion. write impossible to tell. Explain your reasoning. a. AM and MB b. AC and MB c. MC and MD d. MB and DB Answer: a. AM and MB can be measured by using the midpoint formula. Because M is the midpoint of AB Hence, this indicates that AM=MB Therefore, AM=MB b. M is the midpoint of AB. If M is the midpoint of AB, then AM = MB Point C is between A and M and point D is between M and B. So, for comparison of MC and MD, nothing can be stated because it is not given whether the point C and D are a midpoint or not. DB and MB can be measured by using the midpoint formula. M is the midpoint AB. Point D is between B and M So, MD < AB If MD < MB, then MB > DB

Find the perimeter and area of the figure.

Solve the inequality. Graph the solution.

Study Skills: Keeping Your Mind Focused

1.1 – 1.3 What did you learn

Mathematical Practices

Question 2. Explain how you arrived at your answer for Exercise 35 on page 18. Answer: we arrived at the answer for Exercise 35 on page 18 by using the distance formula between 2 points. We know that, The distance between the 2 points = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\)

Question 3. What assumptions did you make when solving Exercise 43 0n page 26? Answer: The assumptions we make when solving Exercise 43 on page 26 is: All the lengths between the 2 points in the segment are equal

1.1 – 1.3 Quiz

Use the diagram

Question 1. Name four points. Answer: From the above figure, The four points are: A, L, B, and K

Question 2. Name three collinear points. Answer: We know that, The points that lie in the same line are called  “Collinear points” Hence, from the above figure, The three collinear points are: A, L, and B

Question 3. Name two lines. Answer: We know that, A line has a starting point but no ending point From the above figure, The lines are GHF, ALB

Question 4. Name three coplanar points. Answer: We know that, The points that are present in the same plane are called “Co-planar points” Hence, from the above figure, The coplanar points are: G, H, and F

Question 5. Name the plane that is shaded green. Answer: From the above figure, The plane that is shaded green is named as Plane GHF

Question 6. Give two names for the plane that is shaded blue. Answer: From the figure, The two names for the plane that is shaded blue are: Plane AKL, Plane BCD

Question 7. Name three line segments. Answer: From the figure, The 3 line segments are: CD, HG, DE

Question 8. Name three rays. Answer: From the figure, The names of the 3 rays are: Ray GHF, ray CDE, Ray ALB

Sketch the figure described.

Plot the points in a coordinate plane. Then determine whether \(\overline{A B}\) and \(\overline{C D}\) are congruent.

Find AC. (Section 1.2)

Find the coordinates of the midpoint M and the distance between the two points.

Question 15. J(4, 3) and K(2, – 3) Answer: The given points are: J (4, 3), K (2, -3) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{4 + 2}{2}\), \(\frac{3 – 3}{2}\)) M = (\(\frac{6}{2}\), \(\frac{0}{2}\)) M = (3, 0) Hence, from the above, We can conclude that the coordinates of the midpoint are: (3, 0)

Question 16. L(- 4, 5) and N(5, – 3) Answer: The given points are: L (-4, 5), N (5, -3) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{-4 + 5}{2}\), \(\frac{5 – 3}{2}\)) M = (\(\frac{1}{2}\), \(\frac{2}{2}\)) M = (\(\frac{1}{2}\), 1) Hence, from the above, We can conclude that the coordinates of the midpoint are: (\(\frac{1}{2}\), 1)

Question 17. P(- 6, – 1) and Q(1, 2) Answer: The given points are: P (-6, -1), Q (1, 2) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, M = (\(\frac{-6 + 1}{2}\), \(\frac{2 – 1}{2}\)) M = (\(\frac{-5}{2}\), \(\frac{1}{2}\)) M = (-\(\frac{5}{2}\), \(\frac{1}{2}\)) Hence, from the above, We can conclude that the coordinates of the midpoint are: (-\(\frac{5}{2}\), \(\frac{1}{2}\))

Question 19. The midpoint of \(\overline{J K}\) is M(0, 1). One endpoint is J(- 6, 3). Find the coordinates of endpoint K. Answer: The given points of \(\overline{J K}\) are: M (0, 1), H (-6, 3) Let the point G be: (x1, y1) We know that, M = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, (0, 1) = (\(\frac{x1 -6}{2}\), \(\frac{y1 + 3}{2}\)) So, \(\frac{x1 – 6}{2}\) = 0              \(\frac{y1 + 3}{2}\) = 1 x1 – 12 = 0 (2)                                            y1 + 7 = 1 (2) x1 = 0 + 12                                                 y1 = 2 – 7 x1 = 12                                                        y1 = -5 Hence, from the above, We can conclude that the other endpoint G is: (12, -5)

Question 20. Your mom asks you to run some errands on your way home from school. She wants you to Stop at the post office and the grocery store, which is both on the same straight road between your school and your house. The distance from your school to the post office is 376 yards. the distance from the post office to your house is 929 yards. and the distance from the grocery store to your house is 513 yards. a. Where should you stop first? Answer: It is given that your mom asks you to run some errands on your way home from school. She wants you to stop at the post office and the grocery store, which is both on the same straight road between your school and your house. Hence, from the above, We can conclude that you should first stop at the post office

b. What is the distance from the post office to the grocery store? Answer: It is given that the distance from your school to the post office is 376 yards. the distance from the post office to your house is 929 yards. and the distance from the grocery store to your house is 513 yards. So, The distance from the post office to the grocery store = (The distance from the post office to your house) + (The distance from the grocery store to your house) = 929 + 513 = 1442 yards Hence, from the above, We can conclude that the distance from the post office to the grocery store is: 1442 yards

c. What is the distance from your school to your house? Answer: It is given that the distance from your school to the post office is 376 yards. the distance from the post office to your house is 929 yards. and the distance from the grocery store to your house is 513 yards. So, The distance from your school to your house = (The distance from your school to the post office) + (The distance from the post office to the grocery store) = 376 + 1442 = 1818 yards Hence, from the above, We can conclude that the distance from your school to your house is: 1818 yards

d. you walk at a speed of 75 yards per minute. How long does it take you to walk straight home from school? Explain your answer. Answer: We know that, Time = \(\frac{Distance}{Speed}\) From part (c), The total distance is: 1818 yards The given speed is: 75 yards per minute So, Time = \(\frac{1818}{75}\) Time = 24.24 minutes Hence, from the above, We can conclude that the time taken for you to walk straight home from school is: 24.24 minutes

1.4 Perimeter and Area in the Coordinate Plane

Essential Question

How can you find the perimeter and area of a polygon in a coordinate plane? Answer: When two-dimensional figures are shown on the coordinate plane, a mix of counting and the Pythagorean Theorem can be used to determine the lengths of each side. Then add up the lengths to determine the perimeter or use the basic area formulas for triangles and rectangles to determine the area of the figure.

Finding the Perimeter and Area of a Quadrilateral

Work with a partner

b. Find the perimeter of quadrilateral ABCD. Answer: The given points are: A (1, 4), B (-3, 1), C (0, -3), and D (4, 0) We know that, The perimeter is defined as the sum of all the sides of a polygon So, In a coordinate plane, the perimeter of the polygon can be found by finding the lengths of the given sides We know that, The length of the side of a polygon in a coordinate plane is: D = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) So, AB = \(\sqrt{(1 + 3)² + (4 – 1)²}\) = \(\sqrt{(4)² + (3)²}\) = \(\sqrt{16 + 9}\) = \(\sqrt{25}\) = 5 cm BC = \(\sqrt{(0 + 3)² + (1 + 3)²}\) = \(\sqrt{(3)² + (4)²}\) =\(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 cm CD = \(\sqrt{(4 – 0)² + (0 + 3)²}\) = \(\sqrt{(4)² + (3)²}\) = \(\sqrt{(16 + 9}\) = \(\sqrt{25}\) = 5 cm DA = \(\sqrt{(4 – 1)² + (4 – 0)²}\) = \(\sqrt{(3)² + (4)²}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 cm Hence, The perimeter of the Quadrilateral ABCD = AB + BC + CD + DA = 5 + 5 + 5 + 5 = 20 cm

c. Are adjacent sides of quadrilateral ABCD perpendicular to each other? How can you tell? Answer: From part (a), The lengths of all the sides of a quadrilateral are: 5 cm We know that, For a square, the lengths of all the sides are equal and the angle between all the sides of the square is 90° Hence, from the above, We can conclude that the adjacent sides of quadrilateral ABCD are perpendicular to each other

d. What is the definition of a square? Is quadrilateral ABCD a square? Justify your answer. Find the area of quadrilateral ABCD. Answer: Definition of the square: A “Square” is a quadrilateral that has the length of all the sides equal and all the angles 90° From part (b), We found out the lengths of all the sides are equal and all the angles are 90° So, Quadrilateral ABCD is a square So, Area of a square = Side² = 5² = 25 cm²

Finding the Area of a Polygon Work with a partner.

a. Partition quadrilateral ABCD into four right triangles and one square, as shown. Find the coordinates of the vertices for the five smaller polygons. Answer: From the figure, The five smaller polygons are: ΔAQD, ΔBPA, ΔDRC, ΔCSB, and square PQRS Now, The coordinates for the given vertices polygons are: The coordinates of ΔAQD are: A (1, 4), Q (1, 0), and D (4, 0) The coordinates of ΔBPA are: B (-3, 1), P (1, 1), and A (1, 4) The coordinates of ΔDRC are: D (4, 0), R (0, 0), and C (0, -3) The coordinates of ΔCSB are: C (0, -3), S (0, 1), and B (-3, 1) The coordinates of square PQRS are: P (1, 1), Q (1, 0), R (0, 0), and S (0, 1)

The area of Δ BPA: The length of BP = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(1 + 3)² +  (1 – 1)²}\) = 4 The length of PA = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(1 – 1)² +  (4 – 1)²}\) = 3 The length of BA = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(4 – 1)² +  (1 + 3)²}\) = 5 So, The area of ΔBPA = \(\frac{1}{2}\) × Base × Height = \(\frac{1}{2}\) (4) (3) = 6

The area of Δ DRC: The length of DR = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(0 – 0)² +  (4 – 0)²}\) = 4 The length of RC = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(3 + 0)² +  (0 – 0)²}\) = 3 The length of CD = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(0 + 3)² +  (4 – 0)²}\) = 5 So, The area of ΔDRC = \(\frac{1}{2}\) × Base × Height = \(\frac{1}{2}\) (4) (3) = 6

The area of Δ CSB: The length of CS = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(0 – 0)² +  (1 + 3)²}\) = 4 The length of SB = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(1 – 1)² +  (0 + 3)²}\) = 3 The length of BC = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(1 + 3)² +  (0+ 3)²}\) = 5 So, The area of ΔCSB = \(\frac{1}{2}\) × Base × Height = \(\frac{1}{2}\) (4) (3) = 6

The area of square PQRS: The length of PQ = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(1 – 1)² +  (1 – 0)²}\) = 1 The length of QR = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(1 – 0)² +  (0 – 0)²}\) = 1 The length of RS = \(\sqrt{(x2 – x1)² +  (y2 – y1)²}\) = \(\sqrt{(1 – 0)² +  (0 – 0)²}\) = 1 The length of PS = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(1 – 0)² + (1 – 1)²}\) = 1 So, The area of square PQRS = Side² = 1

c. Is the sum of the areas of the five smaller polygons equal to the area of quadrilateral ABCD? Justify your answer. Answer: From the given figure, The coordinates of the quadrilateral ABCD are: A (1, 4), B(-3, 1), C (0, -3), and D (4, 0) Now, The length of AB = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(1 + 3)² + (4 – 1)²}\) = 5 The length of BC = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(0 + 3)² + (1 + 3)²}\) = 5 The length of CD = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 – 0)² + (0 + 3)²}\) = 5 The length of DA = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 – 0)² + (4 – 1)²}\) = 5 So, from the above We can observe that all the lengths of the sides are equal and the product of the slopes of the perpendicular sides is -1 Hence, The area of quadrilateral ABCD = Side² = 5² = 25 Now, The sum of the areas of the five smaller polygons = 6 + 6 + 6 + 6 + 1 = 25 Hence, from the above, We can conclude that the sum of the areas of the five smaller polygons equal to the area of quadrilateral ABCD

Question 3. How can you find the perimeter and area of a polygon in a coordinate plane? Answer: When two-dimensional figures are shown on the coordinate plane, a mix of counting and the Pythagorean Theorem can be used to determine the lengths of each side. Then add up the lengths to determine the perimeter or use the basic area formulas for triangles and rectangles to determine the area of the figure.

Question 4. Repeat Exploration 1 for quadrilateral EFGH. where the coordinates of the vertices are E(- 3, 6), F(- 7, 3), G(- 1, – 5), and H(3, – 2). Answer: The coordinates of the vertices of quadrilateral EFGH are: E (-3, 6), F (-7, 3), G (-1, -5), and H (3, -2) Now, The length of EF = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(7 – 3)² + (6 – 3)²}\) = 5 The length of FG = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(7 – 1)² + (3 + 5)²}\) = 10 The length of GH = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 + 1)² + (5 – 2)²}\) = 5 The length of HE = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 + 3)² + (6 + 2)²}\) = 10 Hence, The perimeter of quadrilateral EFGH = EF + FG + GH + HE = 5 + 5 + 10 +  10 = 30 From the lengths of the sides, We can observe that the opposite sides have the same lengths and the product of the slopes of the perpendicular sides is -1 So, The figure is: Rectangle We know that, The area of a rectangle = Length × Width So, The area of quadrilateral EFGH = 5 × 10 = 50 Hence, from the above, We can conclude that The perimeter of quadrilateral EFGH is: 30 The area of quadrilateral EFGH is: 50

Lesson 1.4 Perimeter and Area in the Coordinate Plane

Classify the polygon the number of sides. Tell whether it is convex or concave.

Find the perimeter of the polygon with the given vertices.

Question 3. D(- 3, 2), E(4, 2), F(4, – 3) Answer: The given vertices are: D (-3, 2), E (4, 2), F (4, -3) Now, The length of DE = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 + 3)² + (2 – 2)²}\) = 7 The length of EF = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 – 4)² + (3 + 2)²}\) = 5 The length of FD = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 + 3)² + (2 + 3)²}\) = 8.60 So, The perimeter of the given vertices = 7 + 5 + 8.60 = 20.60 Hence, from the above, We can conclude that the perimeter of the given vertices is: 20.60

Question 4. G(- 3, 2), H(2, 2), J(- 1, – 3) Answer: The given vertices are: G (-3, 2), H (2, 2), I (-1, -3) Now, The length of GH = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 3)² + (2 – 2)²}\) = 5 The length of HI = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 1)² + (3 + 2)²}\) = 5.8 The length of IG = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 – 1)² + (2 + 3)²}\) = 5.3 So, The perimeter of the given vertices = 5 + 5.8 + 5.3 = 16.1 Hence, from the above, We can conclude that the perimeter of the given vertices is: 16.1

Question 5. K( – 1, 1), L(4, 1), M(2, – 2), N(- 3, – 2) Answer: The given vertices are: K (-1, 1), L (4, 1), M (2, -2), N (-3, -2) Now, The length of KL = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 + 1)² + (1 – 1)²}\) = 5 The length of LM = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 – 2)² + (1 + 2)²}\) = 3.6 The length of MN = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 3)² + (2 – 2)²}\) = 5 The length of NK = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3- 1)² + (1 + 2)²}\) = 3.6 So, The perimeter of the given vertices = 5 + 5 + 3.6 + 3.6 = 17.2 Hence, from the above, We can conclude that the perimeter of the given vertices is: 17.2

Question 6. Q(- 4, – 1), R(1, 4), S(4, 1), T(- 1, – 4) Answer: The given vertices are; Q (-4, -1), R (1, 4), S (4, 1), T (-1, -4) Now, The length of QR = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 + 1)² + (4 + 1)²}\) = 7.07 The length of RS = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 – 1)² + (4 – 1)²}\) = 4.2 The length of ST = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(4 + 1)² + (1 + 4)²}\) = 7.07 The length of TQ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4- 1)² + (4 – 1)²}\) = 4.2 So, The perimeter of the given vertices = 7.07 + 7.07 + 4.2 + 4.2 = 22.54 Hence, from the above, We can conclude that the perimeter of the given vertices is: 22.54

Find the area of the polygon with the given vertices.

Question 7. G(2, 2), H(3, – 1), J(- 2, – 1) Answer: The given vertices are: G (2, 2), H (3, -1), I (-2, -1) Now, The length of GH = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 – 3)² + (2 + 1)²}\) = 3.16 The length of HI = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(1 – 1)² + (3 + 2)²}\) = 5 The length of IG = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 2)² + (2 + 1)²}\) = 5 So, The area of the given vertices = \(\frac{1}{2}\) × Base × Height = \(\frac{1}{2}\) × 5 × 5 = 12.5 Hence, from the above, We can conclude that the area of the given vertices is: 12.5

Question 8. N(- 1, 1), P(2, 1), Q(2, – 2), R(- 1, – 2) Answer: The given vertices are: N (-1, 1), P (2, 1), Q (2, -2), R (-1, -2) Now, The length of NP = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 1)² + (1 – 1)²}\) = 3 The length of PQ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 – 2)² + (1 + 2)²}\) = 3 The length of QR = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 1)² + (2 – 2)²}\) = 3 The length of RN = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(1- 1)² + (1 + 2)²}\) = 3 So, From the lengths of all of the sides, We can say that the vertices belong to a square So, The area of a square = Side² = 3² = 9 Hence, from the above, We can conclude that the area of the given vertices is: 9

Question 9. F(- 2, 3), G(1, 3), H(1, – 1), J(- 2, – 1) Answer: The given vertices are: F (-2, 3), G (1, 3), H (1, -1), J (-2, -1) Now, The length of FG = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 1)² + (3 – 3)²}\) = 3 The length of GH = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(1 – 1)² + (1 + 3)²}\) = 4 The length of HJ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 + 1)² + (1 – 1)²}\) = 3 The length of JF = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(2- 2)² + (1 + 3)²}\) = 4 So, From the lengths of all of the sides, We can say that the vertices belong to a rectangle since the lengths of the opposite sides are the same So, The area of a rectangle = Length × Width = 3 × 4 = 12 Hence, from the above, We can conclude that the area of the given vertices is: 12

Question 10. K(- 3, 3), L(3, 3), M(3, – 1), N(- 3, – 1) Answer: The given vertices are: K (-3, 3), L (3, 3), M (3, -1), N (-3, -1) Now, The length of KL = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 + 3)² + (3 – 3)²}\) = 6 The length of LM = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 – 3)² + (1 + 3)²}\) = 4 The length of MN = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 + 3)² + (1 – 1)²}\) = 6 The length of NK = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3- 3)² + (1 + 3)²}\) = 4 So, From the lengths of all of the sides, We can say that the vertices belong to a rectangle So, The area of a rectangle = Length × Width = 6 × 4 = 24 Hence, from the above, We can conclude that the area of the given vertices is: 24

Exercise 1.4 Perimeter and Area in the Coordinate Plane

Question 2. WRITING What formulas can you use to find the area of a triangle in a coordinate plane? Answer: We know that, The formula you can use to find the area of a triangle in a coordinate plane is: Area of a triangle (A) = \(\frac{1}{2}\) × Base × Height

In Exercises 3 – 6, classify the polygon by the number of sides. Tell whether it is convex or concave.

In Exercises 7 – 12. find the perimeter of the polygon with the given vertices.

Question 8. Q(- 3, 2), R(1, 2), s(1, – 2), T(- 3, – 2) Answer: The given vertices of a polygon are: Q (-3, 2), R (1, 2), S (1, -2), T (-3, -2) Now, The length of QR = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(1 + 3)² + (2 – 2)²}\) = 4 The length of RS = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(1 – 1)² + (2 + 2)²}\) = 4 The length of ST = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(2 – 2)² + (1 + 3)²}\) = 4 The length of TQ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3- 3)² + (2 + 2)²}\) = 4 So, From the lengths of all of the sides, We can say that the vertices belong to a square So, The perimeter of a square = 4 × Side = 4 × 4 = 16 Hence, from the above, We can conclude that the perimeter of the given polygon is: 16

Question 10. X(- 1, 3), Y(3, 0), Z(- 1, – 2) Answer: The given vertices of a polygon are: X (-1, 3), Y (3, 0), Z (-1, -2) Now, The length of XY = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 + 1)² + (3 – 0)²}\) = 5 The length of YZ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(3 + 1)² + (0 + 2)²}\) = 4.47 The length of ZX = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) =\(\sqrt{(1 – 1)² + (3 + 2)²}\) = 5 So, The perimeter of the given polygon = 5 + 5 + 4.7 = 14.47 Hence, from the above, WE can conclude that the perimeter of the given polygon is: 14.47

In Exercises 13 – 16. find the area of the polygon with the given vertices.

Question 14. J(- 3, 4), K(4, 4), L(3, – 3) Answer: The given vertices are: J (-3, 4), K (4, 4), L (3, -3) Now, The length of JK = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 + 3)² + (4 – 4)²}\) = 7 The length of KL = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 3)² + (4 + 3)²}\) = 7.07 The length of LJ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3 + 3)² + (3 + 4)²}\) = 9.21 So, The area of the given vertices = \(\frac{1}{2}]/latex] × JK ×KL = [latex/]frac{1}{2}\) × 7 × 7.07 = 24.74 Hence, from the above, We can conclude that the area of the given vertices is: 24.74

Question 16. N(- 2, 1), P(3, 1), Q(3, – 1), R(- 2, 1) Answer: The given vertices are: N (-2, 1), P (3, 1), Q (3, -1), R (-2, 1) Now, The length of NP = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3 + 2)² + (1 – 1)²}\) = 5 The length of PQ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3 – 3)² + (1 + 1)²}\) = 2 The length of QR = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3 + 2)² + (1 + 1)²}\) = 5.38 The length of RN = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(2 – 2)² + (1 – 1)²}\) = 0 So, The area of the given vertices = 5 × 2 × 5.38 = 28.97 Hence, from the above, We can conclude that the area of the given vertices is: 28.97

In Exercises 17 – 24, use the diagram.

Question 18. Find the perimeter of the rectangle BCEF. Answer: From the given figure, The coordinates of the rectangle BCEF are: B (0, 3) C (4, -1), E (2, -3), F (-2, 1) Now, The length of BC = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 0)² + (3 + 1)²}\) = 5.65 The length of CE = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 2)² + (3 – 1)²}\) = 2.82 The length of EF = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(1 + 3)² + (2 + 2)²}\) = 5.65 The length of FB = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(0 + 2)² + (3 – 1)²}\) = 2.82 So, The perimeter of the rectangle BCEF = BC + CE + EF + FB = 5.65 + 2.82 + 5.65 + 2.82 = 16.94 Hence, from the above, We can conclude that the perimeter of the rectangle BCEF is: 16.94

Question 20. Find the perimeter of quadrilateral ABCD. Answer: From the given figure, The given vertices of the quadrilateral ABCD are: A (-5, 4), B (0, 3), C (4, -1), D (4, -5) Now, The length of AB = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(0 + 5)² + (4 – 3)²}\) = 5.09 The length of BC = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 0)² + (1 + 3)²}\) = 5.65 The length of CD = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 4)² + (5 – 1)²}\) = 4 The length of DA = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 + 5)² + (4 + 5)²}\) = 12.72 So, The perimeter of quadrilateral ABCD = AB + BC + CD + DA = 5.09 + 5.65 + 4 + 12.72 = 31.37 Hence, from the above, We can conclude that the perimeter of quadrilateral ABCD is: 31.37

Question 22. Find the area of rectangle BCEF Answer: From the given figure, The coordinates of the rectangle BCEF are: B (0, 3) C (4, -1), E (2, -3), F (-2, 1) Now, The length of BC = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 0)² + (3 + 1)²}\) = 5.65 The length of CE = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 2)² + (3 – 1)²}\) = 2.82 The length of EF = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(1 + 3)² + (2 + 2)²}\) = 5.65 The length of FB = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(0 + 2)² + (3 – 1)²}\) = 2.82 So, The area of the rectangle BCEF = Length × Width = 2.82 × 5.65 = 15.93 Hence, from the above, We can conclude that the area of the rectangle BCEF is: 15.93

Question 24. Find the area of quadrilateral ABCD. Answer: The given vertices of the quadrilateral ABCD are: A (-5, 4), B (0, 3), C (4, -1), D (4, -5) Now, The length of AB = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(0 + 5)² + (4 – 3)²}\) = 5.09 The length of BC = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 0)² + (1 + 3)²}\) = 5.65 The length of CD = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 4)² + (5 – 1)²}\) = 4 The length of DA = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 + 5)² + (4 + 5)²}\) = 12.72 The length of AC = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 + 5)² + (4 + 1)²}\) = 10.29 The length of BD = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(4 – 0)² + (3 + 5)²}\) = 8.94 So, The area of the quadrilateral ABCD = \(\frac{1}{2}\) ×AC × BD = \(\frac{1}{2}\) × 10.29 × 8.94 = 45.99 Hence, from the above, We can conclude that the area f the quadrilateral ABCD is: 45.99

ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in finding the perimeter or area of the polygon.

In Exercises 27 and 28, use the diagram.

Question 28. Determine which points are the remaining vertices of a rectangle with a perimeter of 14 units. A. A(2, – 2) and B(2, – 1) B. C(- 2, – 2) and D(- 2, 2) C. E(- 2, – 2) and F(2, – 2) D. G(2, 0) and H(2, 0) Answer: From the given figure, The given vertices are: P (-2, 1), Q (2, 1) We know that, The perimeter is defined as the sum of all the sides of a given polygon Now, Let, A (x, y), and B(p, q) are the remaining 2 points of a rectangle Now, The length of PQ = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{2 + 2)² + (1 – 1)²}\) = 4 The length of QA = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(x – 2)² + (y – 1)²}\) The length of AB = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(x – p)² + (y – q)²}\) The length of BP = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(p + 2)² + (q – 1)²}\) So, The perimeter of the rectangle = PQ + QA + AB + BP = 4 + \(\sqrt{(x – 2)² + (y – 1)²}\) + \(\sqrt{(x – p)² + (y – q)²}\) + \(\sqrt{(p + 2)² + (q – 1)²}\) It is given that, The perimeter of the rectangle = 14 So, 4 + \(\sqrt{(x – 2)² + (y – 1)²}\) + \(\sqrt{(x – p)² + (y – q)²}\) + \(\sqrt{(p + 2)² + (q – 1)²}\) = 14 \(\sqrt{(x – 2)² + (y – 1)²}\) + \(\sqrt{(x – p)² + (y – q)²}\) + \(\sqrt{(p + 2)² + (q – 1)²}\) = 14 – 4 \(\sqrt{(x – 2)² + (y – 1)²}\) + \(\sqrt{(x – p)² + (y – q)²}\) + \(\sqrt{(p + 2)² + (q – 1)²}\) = 10 Now, from the given options, a. The given remaining vertices are: A (2, -2), B (2, -1) Compare the above vertices with A (x, y), and B (p, q) So, \(\sqrt{(2- 2)² + (-2 – 1)²}\) + \(\sqrt{(2 – 2)² + (-2 + 1)²}\) + \(\sqrt{(2 + 2)² + (-1 – 1)²}\) = 10 3 + 1 + 4.47 = 10 8.47 ≠ 10 b. The given remaining vertices are: C (-2, -2), and D (-2, 2) Compare the above vertices with A (x, y), and B (p, q) So, \(\sqrt{(-2 – 2)² + (-2 – 1)²}\) + \(\sqrt{(2 – 2)² + (-2 – 2)²}\) + \(\sqrt{(-2 + 2)² + (2 – 1)²}\) = 10 5 + 4 +1 = 10 10 = 10 Hence, from the above, We can conclude that the remaining two vertices which gives the perimeter of 14 units are: C (-2, -2), and D (-2, 2)

b. Zucchini plants require 9 square feet around each plant. How many zucchini plants can you plant? Answer: From part (a), The area of the entire garden is: 72 square feet It is given that zucchini plants require 9 square feet around each plant So, The number of zucchini plants you can plant = \(\frac{The area of the entire garden}{The area required for each zucchini plant}\) = \(\frac{72}{9}\) = 8 Hence, from the above, We can conclude that the number of zucchini plants you can plant is: 8 plants

c. You decide to use square TUVW to grow lettuce. you can plant lour heads of lettuce per square loot.  How many of each vegetable can you plant? Explain. Answer: From the figure, The coordinates of the vertices of the rectangle TUVW are: T (1, 1), U (1, 4), V (4, 4), W (4, 1) Now, The length of TU = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{1 – 1)² + (4 – 1)²}\) = 3 The length of UV = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{4 – 4)² + (4 – 1)²}\) = 3 The length of VW = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{4 – 4)² + (4 – 1)²}\) = 3 The length of WT = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{4 – 1)² + (1 – 1)²}\) = 3 So, The area of square TUVW = Side² = 3² = 9 square feet It is given that, 1 square foot = 4 heads of lettuce So, 9 square feet = 9 × 4 = 36 heads of lettuce Hence, from the above, We can conclude that you can plant 36 vegetables of each type

Question 31. MODELING WITH MATHEMATICS You are going for a hike in the woods. You hike to a waterfall that is 4 miles east of where you left your car. You then hike to a lookout point that is 2 miles north of your car. From the lookout point. you return to our ear.

Explanation: It is given that your friend claims that a rectangle with the same perimeter as △QRS will have the same area as the triangle. But, We know that, The size of the rectangle is greater than the size of the triangle So, The area of the rectangle will be greater than the area of the triangle even though the perimeter of the rectangle and the triangle are the same Hence, from the above, We can conclude that your friend is not correct

Question 38. 3x – 7 = 2 Answer: The given equatio is: 3x – 7 = 2 So, 3x = 2 + 7 3x = 9 x = \(\frac{9}{3}\) x = 3 Hence, from the above, We can conclude that the value of x to the given equation is: 3

Question 40. x + 4 = x – 12 Answer: The given equation is: x + 4 = x – 12 x – x + 4 = -12 4 = -12 Hence, from the above, We can conclude that the given equation has no solution

Question 42. 11 – 2x = 5x – 3 Answer: The given equation is: 11 – 2x = 5x – 3 11 + 3 = 5x + 2x 7x = 14 x = \(\frac{14}{7}\) x = 2 Hence, from the above, We can conclude that the value of x to the given equation is: 2

1.5 Measuring and Constructing Angles

How can you measure and classify an angle? Answer: An angle is formed when two rays have the same endpoint or vertex. An “Angle” will be measured by using a protractor Classification of angles: An angle can be classified according to the size of the opening between its rays. An acute angle measures greater than 0° and less than 90°. A right-angle forms a square corner and measures 90°. An obtuse angle measures greater than 90° and less than 180°

Measuring and Classifying Angles

b. ∠AOC From the above figure, ∠AOC = 65° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠AOC is an acute angle

c. ∠BOC From the above figure, ∠BOC = |65° – 35°| = 30° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠BOC is an acute angle

d. ∠BOE From the above figure, ∠BOE = |145° – 35°| = 110° We know that, The angle that is greater than 90° and less than 180° is an “Obtuse angle” Hence, ∠BOE is an acute angle

e. ∠COE From the above figure, ∠COE = |145° – 65°| = 80° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠COE is an acute angle

f. ∠COD From the above figure, ∠COD =|70° –  65°| = 5° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠COD is an acute angle

g. ∠BOD From the above figure, ∠BOD = |70° –  35°| = 35° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠BOD is an acute angle

h. ∠AOE From the above figure, ∠AOE = 145° We know that, The angle that is greater than 90° and less than 180° is an “Obtuse angle” Hence, ∠AOE is an obtuse angle

Drawing a Regular Polygon

C. The sum of the angle measures of a polygon with n sides is equal to 180(n – 2)°. Do the angle measures at your hexagons agree with this rule? Explain. ATTENDING TO PRECISION To be proficient in math, you need to calculate and measure accurately and efficiently. Answer: It is given that the sum of the angle measures of a polygon with n sides is equal to 180 (n – 2)° This rule will be applicable for all the angle measures of all polygons Hence, We can conclude that the above rule will be applicable for the angle measures of the hexagon

Question 3. How can you measure and classify an angle? Answer: An angle is formed when two rays have the same endpoint or vertex. An “Angle” will be measured by using a protractor Classification of angles: An angle can be classified according to the size of the opening between its rays. An acute angle measures greater than 0° and less than 90°. A right-angle forms a square corner and measures 90°. An obtuse angle measures greater than 90° and less than 180°

Lesson 1.5 Measuring and Constructing Angles

Write three names for the angle.

Use the diagram in Example 2 to find the angle measure. Then classify the angle.

Question 4. ∠JHM Answer: From the above figure, ∠JHM =|145° –  0°| = 145° We know that, The angle that is greater than 90° and less than 180° is an “Obtuse angle” Hence, ∠JHM is an obtuse angle

Question 5. ∠MHK Answer: From the above figure, ∠MHK =|145° –  55°| = 110° We know that, The angle that is greater than 90° and less than 180° is an “Obtuse angle” Hence, ∠MHK is an obtuse angle

Question 6. ∠MHL Answer: From the above figure, ∠MHL =|145° –  90°| = 55° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠MHL is an acute angle

Question 7. Without measuring, is ∠DAB ≅ ∠FEH in Example 3? Explain your reasoning. Use a protractor to verify your answer. Answer: From the given figure, We can observe that, ∠DAB ≅ 90° ∠FEH > 90° Hence, from the above, We can conclude that ∠DAB is not congruent to ∠FEH

Find the indicated angle measures.

Exercise 1.5 Measuring and Constructing Angles

In Exercises 3 – 6. write three names for the angle.

In Exercises 7 and 8, name three different angles in the diagram.

In Exercises 9 – 12 find the angle measure. Then classify the angle.

Question 10. m∠BOD Answer: From the above figure, ∠BOD = |0° –  65°| = 65° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠BOD is an acute angle

Question 12. m∠EOD Answer: From the above figure, ∠EOD = |40° –  65°| = 25° We know that, The angle that is greater than 0° and less than 90° is an “Acute angle” Hence, ∠EOD is an acute angle

ERROR ANALYSIS In Exercises 13 and 14. describe and correct the error in finding the angle measure. Use the diagram from Exercises 9 – 12.

CONSTRUCTION In Exercises 15 and 16. use a compass and straightedge to copy the angle.

In Exercises 17 – 20, in m∠AED = 34° and m∠EAD = 112°.

Question 18. Identify the angles congruent to ∠EAD. Answer: From the given figure, The angles congruent to ∠EAD are: ∠DBC, ∠ADB, and ∠DAE

Question 20. Find m∠ADB. Answer: From the given figure, ∠EAD = 112° Hence, By the property of Congruence, ∠EAD ≡ ∠ADB Hence, from the above, We can conclude that ∠ADB = 112°

In Exercises 21 – 24, find the indicated angle measure.

In Exercises 25 – 30. find the indicated angle measures.

Question 34. m∠RQS = 71°. Find m∠PQS and m∠PQR. Answer: We know that, An “Angle Bisector” makes the two angles congruent i.e., equal So, From the above figure, We can say that, ∠PQS = ∠RQS Now, By using the Angle Addition Postulate, ∠PQR = ∠PQS + ∠RQS It is given that, ∠RQS = 71° So, ∠PQS = 71° So, ∠PQR = 71 + 71 = 142° Hence, from the above, We can conclude that ∠PQS = 71° and ∠PQR = 142°

Question 36. m∠PQR = 119°. Find m∠PQS and m∠RQS. Answer: We know that, An “Angle Bisector” makes the two angles congruent i.e., equal So, From the above figure, We can say that, ∠PQS = ∠RQS Now, By using the Angle Addition Postulate, ∠PQR = ∠PQS + ∠RQS It is given that, ∠PQR = 119° So, 119° = 2 (∠PQS) ∠PQS = 59.5° Hence, from the above, We can conclude that ∠PQS = 59.5° and ∠RQS = 59.5°

In Exercises 37 – 40, \(\overrightarrow{B D}\) bisects ∠ABC. Find m∠ABD, m∠CBD, and m∠ABC.

Question 44. In the roof truss, \(\vec{B}\)G bisects ∠ABC and ∠DEF. m∠ABC = 112°. and ∠ABC ≅ ∠DEF Find the measure of each angle. a. m∠DEF Answer: It is given that \(\vec{B}\)G bisects ∠ABC and ∠DEF It is also given that ∠ABC = ∠DEF ∠ABC = 112° Hence, ∠DEF = 112°

b. m∠ABG Answer: It is given that \(\vec{B}\)G bisects ∠ABC and ∠DEF So, For ∠ABC, By congruence property, ∠ABG = ∠CBG Hence, ∠ABG = ∠ABC / 2 = 112 / 2 = 56°

c. m∠CBG Answer: It is given that \(\vec{B}\)G bisects ∠ABC and ∠DEF So, For ∠ABC, By congruence property, ∠ABG = ∠CBG Hence, ∠CBG = ∠ABC / 2 = 112 / 2 = 56°

d. m∠DEG Answer: It is given that \(\vec{B}\)G bisects ∠ABC and ∠DEF So, For ∠DEF, By congruence property, ∠DEG = ∠EGF Hence, ∠DEG = ∠DEF / 2 = 112 / 2 = 56°

Question 46. Name an example of each of the four types of angles according to their measures in the diagram. Answer: There are 4 types of angles. They are: a. Acute angle: The angle that is greater than 0° and less than 90° is an “Acute angle” b. Obtuse angle: The angle that is greater than 90° and less than 180° is an “Obtuse angle” c. Straight angle: The angle 180° is called a “Straight angle” d. Right angle: The angle 90° is called a “Right angle”

Question 48. THOUGHT-PROVOKING The angle between the minute hand and the hour hand of a clock is 90° What time is it? Justify your answer. Answer: From the clock, We can observe that the angle between the minute hand and the hour hand of a clock is 90° when the time will be 3:00 and 9:00

Question 50. ABSTRACT REASONING Classify the angles that result from drawing a ray in the interior of each type of angle. Include all possibilities and explain your reasoning. a. acute angle b. right angle c. obtuse angle d. straight angle Answer: The classification of angles that result from drawing a ray in the interior of each type of angle is: a. Acute angle: The angle that is greater than 0° and less than 90° is an “Acute angle” b. Obtuse angle: The angle that is greater than 90° and less than 180° is an “Obtuse angle” c. Straight angle: The angle 180° is called a “Straight angle” d. Right angle: The angle 90° is called a “Right angle”

Question 52. MAKING AN ARGUMENT Your friend claims it is possible for a straight angle to Consist of two obtuse angles. Is your friend correct? Explain your reasoning. Answer: No, your friend is not correct We know that, A straight angle consists of 2 Right angles or a Right angle and an obtuse angle So, The claim of your friend which is a straight angle consists of two obtuse angles is not correct Hence, from the above, We can conclude that the claim of your friend is not correct

b. What can you change in the diagram so that ∠XYZ is a straight angle? Answer: From the given figure, ∠XYZ is already a straight angle So, There is no change to make ∠XYZ a straight line

Question 56. ANALYZING RELATIONSHIPS \(\vec{S}\)Q bisects ∠RST, \(\vec{S}\)P bisects ∠RSQ. and \(\vec{S}\)V bisects ∠RSP. The measure of ∠VSP is 17°. Find in m∠TSQ. Explain. Answer:

Question 58. x + 67 = 180 Answer: The given equation is: x + 67 = 180 So, x = 180 – 67 x = 113 Hence, from the above, We can conclude that the value of x is: 113

Question 60. 16 + x = 90 Answer: The given equation is: 16 + x = 90 So, x = 90 – 16 x = 74 Hence, from the above, We can conclude that the value of x is: 74

Question 61. 109 + x = 180 Answer: The given equation is: 109 + x = 180 So, x = 180 – 109 x = 71 Hence, from the above, We can conclude that the value of x is: 71

Question 62. (6x + 7) + (13x + 21) = 180 Answer: The given equation is: (6x + 7) + (13x + 21) = 180 So, 19x + 28 = 180 19x = 180 – 28 19x = 152 x = \(\frac{152}{9}\) x = 8 Hence, from the above, We can conclude that the value of x is: 8

Question 64. (11x – 25) + (24x + 10) = 90 Answer: The given equation is: (11x – 25) + (24x + 10) = 90 So, 35x – 15 = 90 35x = 90 + 15 35x = 105 x = \(\frac{105}{35}\) x = 3 Hence, from the above, We can conclude that the value of x is: 3

1.6 Describing Pairs of Angles

How can you describe angle pair relationships and use these descriptions to find angle measures? Answer: Two adjacent angles are a linear pair when their non-common sides are opposite rays. The angles in a linear pair are supplementary angles. common side L1+22=180°. Two angles are vertical angles when their sides form two pairs of opposite rays

a. What do you notice about the following angle pairs? x° and y° Answer: From the given figure, x and y share a common line We know that, The angles that share a common side are called “Adjacent angles” Hence, x° and y° are called “Adjacent angles”

y° and z° Answer: From the given figure, y and z share a common line We know that, The angles that share a common side are called “Adjacent angles” Hence, y° and z° are called “Adjacent angles”

x° and z° Answer: From the given figure, x and z share a common vertex We know that, The angles that share a common vertex are called “Vertical angles” Hence, x° and z° are called “Vertical angles”

Finding Angle Measures

a. What do you notice about the following angle pairs? a° and b° Answer: From the above figure, We can observe that a° and b° are adjacent angles So, We can say that a° + b° = 90°

c° and d° Answer: From the above figure, We can observe that c° and d° are adjacent angles So, We can say that c° + d° = 90°

c° and e° Answer: From the above figure, We can observe that c and e are the opposite angles So, We can say that, c° = e°

Question 3. How can you describe angle pair relationships and use these descriptions to find angle measures? Answer: Two adjacent angles are a linear pair when their non-common sides are opposite rays. The angles in a linear pair are supplementary angles. common side L1+22=180°. Two angles are vertical angles when their sides form two pairs of opposite rays

Question 4. What do you notice about the angle measures of complementary angles, supplementary angles, and vertical angles? ATTENDING TO PRECISION To be proficient in math, you need to communicate precisely with others. Answer: The “Complementary angles” are the angles that the measure of angles is 90° The “Supplementary angles” are the angles that the measure of angles is 180° The opposite angles are the angles that have a common side and opposite to each other

Lesson 1.6 Describing Pairs of Angles

In Exercises 1 and 2. use the figure.

Question 1. Name a pair of complementary angles, a pair of supplementary angles, and a pair of adjacent angles. Answer: We know that, The pair of angles that the sum of the angle measures is 90° is called “Complementary angles” The pair of angles that the sum of the angle measures is 180° is called “Supplementary angles” The angle that is the same in more than 1 angle-pair is called “Adjacent Angle” So, From the given figure, The pair of complementary angles is: ∠FGK and ∠LKG The pair of supplementary angles is: ∠HGK and ∠GKL The pair of adjacent angles is: ∠FGK and ∠HGK

Question 2. Are ∠KGH and ∠LKG adjacent angles? Are ∠FGK and ∠FGH adjacent angles? Explain. Answer: We know that, The angle that is the same in more than 1 angle-pair is called “Adjacent Angle” But, In ∠KGH and ∠LKG, the angle is not the same So, ∠KGH and ∠LKG are not the adjacent angles In ∠FGK and ∠FGH, the angle is the same So, ∠FGK and ∠FGH are adjacent angles

Question 3. ∠1 is a complement of ∠2. arid m∠2 = 5°. Find m∠1. Answer: We know that, The pair of angles that the sum of the angle measures is 90° is called “Complementary angles” It is given that ∠1 is a complement of ∠2 So, ∠1 + ∠2 = 90° It is also given that ∠2 = 5° So, ∠1 = 90° – ∠2 ∠1 = 90° – 5° ∠1 = 85° Hence, from the above, We can conclude that ∠1 = 85°

Question 4. ∠3 is a supplement of ∠4. and m∠3 = 148°. Find m∠4. Answer: We know that, The pair of angles that the sum of the angle measures is 180° is called “Supplementary angles” It is given that ∠3 is a supplement of ∠4 So, ∠3 + ∠4 = 180° It is also given that ∠3 = 148° So, ∠4 = 180° – ∠3 ∠4 = 180° – 148° ∠4 = 32° Hence, from the above, We can conclude that ∠4 = 32°

Question 5. ∠LMN and ∠PQR are complementary angles. Find the measures of the angles when in m∠LMN = (4x – 2)° and m∠PQR = (9x + 1)°. Answer: We know that, The pair of angles that the sum of the angle measures is 90° is called “Complementary angles” It is given that ∠LMN and ∠PQR are complementary angles So, ∠LMN + ∠PQR = 90° It is also given that ∠LMN = (4x – 2)° and ∠PQR = (9x + 1)° So, (4x – 2)° + (9x + 1)° = 90° 13x – 1 = 90° 13x = 90 + 1 x = \(\frac{91}{13}\) x = 7° So, ∠LMN = (4x – 2)° = 4 (7) – 2 = 28 – 2 = 26° ∠PQR = (9x + 1)° = 9 (7) + 1 = 63 + 1 = 64° Hence, from the above, We can conclude that ∠LMN = 26° and ∠PQR = 64°

Exercise 1.6 Describing Pairs of Angles

In Exercises 3-6, use the figure.

Question 4. Name a pair of adjacent supplementary angles. Answer: We know that, The angles that have the sum of the angle measures 180° are called “Supplementary angles Hence, from the figure, A pair of adjacent supplementary angles are: ∠LJN + ∠LJK

Question 6. Name a pair of nonadjacent supplementary angles. Answer: We know that, The angles that have the sum of the angle measures 180° are called “Supplementary angles Hence, from the figure, A pair of adjacent supplementary angles are: ∠LJN + ∠LJK and ∠NGP + ∠HGF

In Exercises 7 – 10. find the angle measure.

Question 8. ∠3 is a complement of ∠4. and m∠3 = 46°. Find, m∠4. Answer: We know that, The pair of angles that the sum of the angle measures is 90° is called “Complementary angles” It is given that ∠3 is a complement of ∠4 So, ∠3 + ∠4 = 90° It is also given that ∠3 = 46° So, ∠4 = 90° – ∠3 ∠4 = 90° – 46° ∠4 = 44° Hence, from the above, We can conclude that ∠4 = 44°

Question 10. ∠7 is a supplement of ∠8. and m∠7 = 109°. Find m∠8. Answer: We know that, The pair of angles that the sum of the angle measures is 180° is called “Supplementary angles” It is given that ∠7 is a supplement of ∠8 So, ∠7 + ∠8 = 180° It is also given that ∠7 = 109° So, ∠8 = 180° – ∠7 ∠8 = 180° – 109° ∠8 = 71° Hence, from the above, We can conclude that ∠8 = 71°

In Exercises 11 – 14. find the measure of each angle.

Question 14. ∠EFG and ∠LMN are supplementary angles. m∠EFG = (3x + 17)°, and m∠LMN = (\(\frac{1}{2} x\) – 5)° Answer: We know that, The pair of angles that the sum of the angle measures is 180° is called “Supplementary angles” It is given that ∠EFG and ∠LMN are supplementary angles So, ∠LMN + ∠EFG = 180° It is also given that ∠LMN = (\(\frac{1}{2} x\) – 5)° and ∠EFG = (3x + 17)° So, (\(\frac{1}{2} x\) – 5)° + (3x + 17)° = 180° \(\frac{7}{2} x\) + 12 = 180° \(\frac{7}{2} x\) = 180 – 12 \(\frac{7}{2} x\) = 168° 7x = 168 (2) 7x = 336 x = \(\frac{336}{7}\) x = 48° So, ∠LMN = (\(\frac{1}{2} x\) – 5)° = \(\frac{1}{2} \) × 48 – 5 = 24 – 5 = 19° ∠EFG = (3x + 17)° = 3 (48) + 17 = 144 + 17 = 161° Hence, from the above, We can conclude that ∠LMN = 19° and ∠EFG = 161°

In Exercises 15 – 18. use the figure.

Question 16. Identify the linear pair(s) that include ∠7. Answer: We know that, The angle that is the same in more than 1 angle-pair is called “Adjacent Angle” Hence, from the figure, We can conclude that The linear pairs that include ∠7 are: ∠6 and ∠7 and ∠7 and ∠8

Question 18. Are ∠2 and ∠5 vertical angles? Explain your reasoning. Answer: We know that, The angles that have the same vertex are called “Vertical angles” Hence, from the figure, We can conclude that ∠2 and ∠5 are vertical angles

In Exercises 19 – 22, find the measure of each angle.

Question 20. Two angles form a linear pair. The measure of one angle is \(\frac{1}{3}\) the measure of the other angle. Answer: We know that, A linear pair of angles are formed when two lines intersect. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines So, The given angles are the supplementary angles It is given that, The measure of one angle is \(\frac{1}{3}\) times the measure of the other angle Let the angle be x° So, x + \(\frac{1}{3}\)x = 180° \(\frac{4}{3}\)x = 180° 4x = 180° (3) 4x = 540° x = \(\frac{540}{4}\) x = 135° Hence, from the above, We can conclude that The measure of one angle is 135° and the measure of the other angle is 270°

Question 22. The measure of an angle is \(\frac{1}{4}\) the measure of its complement. Answer: We know that, A linear pair of angles are formed when two lines intersect. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines So, The given angles are the complementary angles It is given that, The measure of one angle is \(\frac{1}{4}\) times the measure of the other angle Let the angle be x° So, x + \(\frac{1}{4}\)x = 90° \(\frac{5}{4}\)x = 90° 5x = 90° (4) 5x = 360° x = \(\frac{360}{5}\) x = 72° Hence, from the above, We can conclude that The measure of one angle is 72° and the measure of the other angle is 144°

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in identifying pairs of angles in the figure.

Question 26. Find the measure of the supplement of ∠2. Answer: Let the other angle with ∠2 be x° We know that, The sum of the angle measures of supplementary angles is: 180° So, ∠2 + x° = 180° Fro the above, It is given that ∠2 = 24° So, 24° + x° = 180° x° = 180° – 24° x° = 156° Hence, from the above, We can conclude that the supplement of ∠2 is: 156°

Question 28. REASONING The foul lines of a baseball field intersect at home plate to form a right angle. A batter hits a fair ball such that the path of the baseball forms an angle of 27° with the third base foul line. What is the measure of the angle between the first base foul line and the path of the baseball? Answer: It is given that the foul lines of a baseball field intersect at home plate to form a right angle. A batter hits a fair ball such that the path of the baseball forms an angle of 27° with the third base foul line. So, We can say that The angle between the path of the baseball and the third base foul line is complementary i.e., 90° with the angle between the path of the baseball and the base of the first foul line So, The angle between the first base foul line and the path of the baseball = 90° – (The angle between the path of the baseball and the third base foul line) =90° – 27° = 63° Hence, from the above, We can conclude that the angle between the first base foul line and the path of the baseball is: 63°

MATHEMATICAL CONNECTIONS In Exercises 32 – 35, write and solve an algebraic equation to find the measure of each angle based on the given description.

Question 32. The measure of an angle is 6° less than the measure of its complement. Answer: It is given that the measure of an angle is 6° less than the measure of its complement Let the angle be x° S0, The complement of the angle is: (90 – x)° So, (90 – x)° = x – 6° 90 + 6 = x + x 2x = 96° x = \(\frac{96}{2}\) x = 48° So, The measures of the two angles are: 48°, 90° – 48° = 42° Hence, from the above, We can conclude that the measures of the angles are: 48°, 42°

Question 34. The measure of one angle is 3° more than \(\frac{2}{3}\) the measure of its supplement. Answer: It is given that the measure of an angle is 3° more than \(\frac{2}{3}\) the measure of its supplement Now, Let the measure of an angle be x° So, The measure of the other angle is (180 – x)° So, x° = 3° + \(\frac{2}{3}\) (180 – x)° 3x° = 9° + 360° – 2x° 5x° = 369° x = \(\frac{369}{5}\) x° = 73.4° So, The measures of the angles are: 73.4° and 180° – 73.4° = 106.6° Hence, from the above, We can conclude that the measures of the angles are: 73.4v and 106.6°

CRITICAL THINKING In Exercises 36 – 41. tell whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 36. Complementary angles are adjacent. Answer: The given statement is “Complementary angles are adjacent” We know that, The complementary angles sometimes will be vertical angles or sometimes adjacent angles or sometimes non-adjacent angles Hence, from the above, We can conclude that the given statement is sometimes true

Question 38. Vertical angles are adjacent. Answer: The given statement is “Vertical angles are adjacent” We know that, The vertical angles are the angles formed by the intersection of two lines whereas adjacent angles share a common side and a common vertex Hence, from the above, We can conclude that vertical angles are never adjacent

Question 40. If an angle is acute. then its complement is greater than its supplement. Answer: The given statement is “If an angle is acute, then its complement is greater than its supplement” we know that, The sum of complementary angles is 90° i.e., the measures of the angles are always acute The sum of the supplementary angles is 180° i.e., the measure of 1 angle is acute and the measure of the other angle is obtuse Hence, from the above, We can conclude that if an angle is acute, then its complement will never be greater than its supplement

Question 42. WRITING Explain why the supplement of an acute angle must be obtuse. Answer: We know that, The sum of the measures of the supplementary angles is: 180° So, If one angle is acute i.e., the angle less than 90°, then the other angle must be obtuse to make the sum 180° Hence, The supplement of an acute angle must be obtuse

b. Points C, A, and F are collinear Answer: We know that, The points that are present on the same line are called “Collinear points” From the figure, We can observe that C, A, and F lie on the same line Hence, from the above, We can conclude that C, A, and F are collinear points

c. ∠CAD ≅ ∠EAF. Answer: From the figure, We can observe that point A is the angle bisector of C and D Hence, from the above, We can conclude that ∠CAD ≅ ∠EAF

d. \(\overline{B A}\) ≅ \(\overline{A E}\) Answer: We know that, C is the Bisector” A “Bisector” divides anything into 2 congruent parts Hence, from the above, We can conclude that \(\overline{B A}\) ≅ \(\overline{A E}\)

f. ∠BAC and ∠CAD are complementary angles. Answer: From the figure, We can observe that ∠A has a complementary angle Hence, from the above, We can conclude that ∠BAC and ∠CAD are complementary angles

g. ∠DAE is a right angle. Answer: From the figure, We can observe that, D is perpendicular to E and A has a complementary angle Hence, from the above, We can conclude that ∠DAE is a right angle

Explanation: It is given that the light from a flashlight strikes a mirror and is reflected so that the angle of reflection is congruent to the angle of incidence. Your classmate claims that ∠QPR is congruent to ∠TPU regardless of the measure of ∠RPS. We know that, The vertical angle is the intersection of two lines and \(\overline{S P}\) is the angle bisector So, ∠QPR is congruent to ∠TPU regardless of the measure of ∠RPS. Hence, from the above, We can conclude that the claim of your friend is correct

Question 50. MATHEMATICAL CONNECTIONS Let m∠1 = x°, m∠2 = y 1 °, and m∠3 = y 2 ° ∠2 is the complement of ∠1, and ∠3 is the supplement of ∠1.

a. Write equations for y 1 as a function of x and for y 2 as a function of x. What is the domain of each function? Explain. Answer: It is given that m∠1 = x°, m∠2 = y 1 °, and m∠3 = y 2 ° ∠2 is the complement of ∠1, and ∠3 is the supplement of ∠1 So, ∠1 + ∠2 = 90° and ∠3 + ∠1 = 180° x° + y1° = 90° and y2° + x° = 180° So, y1° = 90° – x° and y2° = 180° – x° Hence, from the above, The domain of y1 can be given as: 0° ≤ x ≤ 90° The domain of y2 can be given as: 0° ≤ x ≤ 180°

Determine whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 52. An integer is a whole number. Answer: The given statement is “An integer is a whole number” We know that, All whole numbers must be integers but all integers are not whole numbers Example: –6 is an integer but not a whole number 3 is a whole number and also an integer Hence, from the above, We can conclude that an integer is sometimes a whole number

Question 54. An irrational number is a real number. Answer: The given statement is “An irrational number is a real number” We know that, “Real numbers” are the numbers that can be written in the form of \(\frac{p}{q}\) whereas “Irrational numbers” can’t be written in the form of \(\frac{p}{q}\) Hence, from the above, We can conclude that an irrational number is never a real number

Question 56. A rational number is an integer. Answer: The given statement is “A rational number is an integer” We know that, Rational numbers are the numbers that can be written in the form of \(\frac{p}{q}\) Hence, from the above, We can conclude that a rational number is sometimes an integer

Question 58. A whole number is a rational number. Answer: The given statement is “A whole number is a rational number” We know that, A whole number can be written in the form of \(\frac{p}{q}\) Hence, from the above, We can conclude that a whole number is always a rational number

1.4 – 1.6 Performance Task: Comfortable Horse Stalls

Question 1. How could you explain your answers to Exercise 33 on page 36 to a friend who is unable to hear’? Answer: In Exercise 33 on page 36, The given vertices of the triangle are in the form of the standard linear equation So, Compare the given vertices with the standard linear equation and find the slopes and x and y-intercepts for the coordinates of the vertices of the triangle

Question 2. What tool(s) could you use to verify your answers to Exercises 25 – 30 on page 44? Answer: To verify the answers to Exercise 25 – 30 on page 44, We can use “Angle Addition Postulate”

Question 3. Your friend says that the angles in Exercise 28 on page 53 are supplementary angles. Explain why you agree or disagree. Answer: In Exercise 28 on page 53, It is given that a right-angled triangle is formed We know that, The sum of the complementary angles is: 90° Hence, from the above, We can conclude that you disagree with your friend about supplementary angle

Basics of Geometry Chapter Review

Use the diagram.

Question 1. Give another name for plane M. Answer: From the given figure, We can conclude that Another name for plane M is: Plane XNZ or plane g

Question 2. Name a line in the plane. Answer: From the given figure, We can conclude that The names of lines in the pane are: Line PY or Line XZ

Question 3. Name a line intersecting the plane. Answer: From the above figure, We can conclude that The line intersecting the plane is: Line PY

Question 4. Name two rays. Answer: From the above figure, We can conclude that The two rays are: Ray g and Ray h

Question 5. Name a pair of opposite rays. Answer: From the above figure, We can conclude that The opposite rays are: Rays g and h

Question 6. Name a point, not in plane M. Answer: From the above figure, We can conclude that The point that is not in plane M is: point p

Find the coordinates of the midpoint M. Then find the distance between points S and T.

Question 10. S(- 2, 4) and T(3, 9) Answer: The given points are: S (-2, 4) and T (3, 9) We know that, The coordinates of the Midpoint = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, The coordinates of the Midpoint = (\(\frac{3 – 2}{2}\), \(\frac{9 + 4}{2}\)) The coordinates of the Midpoint = (\(\frac{1}{2}\), \(\frac{13}{2}\) ) Hence, from the above, We can conclude that the coordinates of the Midpoint are: (\(\frac{1}{2}\), \(\frac{13}{2}\))

Question 11. S(6, – 3) and T(7, – 2) Answer: The given points are: S (6, -3) and T (7, -2) We know that, The coordinates of the Midpoint = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, The coordinates of the Midpoint = (\(\frac{7 + 6}{2}\), \(\frac{-2 – 3}{2}\)) The coordinates of the Midpoint = (\(\frac{13}{2}\), \(\frac{-5}{2}\)) Hence, from the above, We can conclude that the coordinates of the Midpoint are: (\(\frac{13}{2}\), \(\frac{-5}{2}\))

Question 12. The midpoint of \(\overline{J K}\) is M(6, 3). One endpoint is J(14, 9). Find the coordinates of endpoint K. Answer: The given points of \(\overline{J K}\) are: M (6, 3) and J (14, 9) Let H be (x. y) We know that, The coordinates of the Midpoint = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, (6, 3) = (\(\frac{x + 14}{2}\), \(\frac{y + 9}{2}\)) \(\frac{x + 14}{2}\) = 6        \(\frac{y + 9}{2}\) = 3 x + 14 = 6 (2)                                       y + 9 = 3 (2) x = 12 – 14                                           y = 6 – 9 x = -2                                                    y = -3 Hence, from the above, We can conclude that the coordinates of H are: (-2, -3)

Question 13. Point M is the midpoint of \(\overline{A B}\) here AM = 3x + 8 and MB = 6x – 4. Find AB. Answer: It is given that Point M is the midpoint of \(\overline{A B}\) here AM = 3x + 8 and MB = 6x – 4. Now, By using the Segment Addition Postulate, AB = AM + MB AB = (3x + 8) + (6x – 4) AB = 9x – 4

Find the perimeter and area of the polygon with the given vertices.

Question 14. W(5, – 1), X(5, 6), Y(2, 1) Z(2, 6) Answer: The given points are: W (5, -1), X (5, 6), Y (2, 1), Z (2, 6) Now, \(\overline{W X}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(5 – 5)² + (1 + 6)²}\) = 7 \(\overline{X Y}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(5 – 2)² + (6 – 1)²}\) = 5.8 \(\overline{Y Z}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(2 – 2)² + (6 – 1)²}\) = 5 \(\overline{Z W}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(5 – 2)² + (6 + 1)²}\) = 7.6 \(\overline{W Y}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(5 – 2)² + (1 + 1)²}\) = 3.6 \(\overline{X Z}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(5 – 2)² + (6 – 6)²}\) = 3 Hence, The perimeter of the given polygon = WX + XY + YZ + ZW = 7 + 5.8 + 5 + 7.6 = 25.4 Since all the lengths of the sides are different, The area of the given polygon = \(\frac{1}{2}\) (d1) (d2) Where, d1 and d2 are the diagonals So, The area of the polygon = \(\frac{1}{2}\) (WY) (XZ) = \(\frac{1}{2}\) (3) (3.6) = 5.4 Hence, from the above, We can conclude that The perimeter of the given polygon is: 25.4 The area of the given polygon is: 5.4

Question 15. E(6, – 2) , F(6, 5), G(- 1, 5) Answer: The given points are: E (6, -2), F (6, 5), G (-1, 5) Now, \(\overline{E F}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(6 – 6)² + (2 + 5)²}\) = 7 \(\overline{F G}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(6 + 1)² + (5 – 5)²}\) = 7 \(\overline{G E}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(6 + 1)² + (5 + 2)²}\) = 9.8 Hence, The perimeter of the given polygon = EF + FG + GE = 7 + 7 + 9.8 = 23.8 Now, The area of the given polygon = \(\frac{1}{2}\) (Base) (Height) So, The area of the polygon = \(\frac{1}{2}\) (EF) (FG) = \(\frac{1}{2}\) (7) (7) = 24.5 Hence, from the above, We can conclude that The perimeter of the given polygon is: 23.8 The area of the given polygon is: 24.5

Find m∠ABD and m∠CBD.

∠1 and ∠2 are complementary angles. Given m∠1, find m∠2.

Question 19. m∠1 = 12° Answer: It is given that ∠1 and ∠2 are the complementary angles ∠1 = 12° So, ∠1 + ∠2 = 90° So, ∠2 = 90° – 12° ∠2 = 78° Hence, from the above, We can conclude that ∠2 = 78°

Question 20. m∠1 = 83° Answer: It is given that ∠1 and ∠2 are the complementary angles ∠1 = 83° So, ∠1 + ∠2 = 90° So, ∠2 = 90° – 83° ∠2 = 7° Hence, from the above, We can conclude that ∠2 = 7°

∠3 and ∠4 are supplementary angles. Given m∠3, find m∠4.

Question 21. m∠3 = 116° Answer: It is given that ∠3 and ∠4 are the supplementary angles ∠3 = 116° So, ∠3 + ∠4 = 180° So, ∠4 = 180° – 116° ∠4 = 64° Hence, from the above, We can conclude that ∠4 = 64°

Question 22. m∠3 = 56° Answer: It is given that ∠3 and ∠4 are the supplementary angles ∠3 = 56° So, ∠3 + ∠4 = 180° So, ∠4 = 180° – 56° ∠4 = 124° Hence, from the above, We can conclude that ∠4 = 124°

Basics of Geometry Chapter Test

Find the length of \(\overline{Q S}\). Explain how you found your answer.

Find the coordinates of tile midpoint M. Then find the distance between the two points.

Question 3. A(- 4, – 8) and B(- 1, 4) Answer: The given points are: A (-4, -8), and B (-1, 4) We know that, The coordinates of the midpoint = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, The coordinates of the midpoint = (\(\frac{-1 – 4}{2}\), \(\frac{-8 + 4}{2}\)) The coordinates of the midpoint = (\(\frac{-5}{2}\), \(\frac{-4}{2}\)) The coordinates of the midpoint = (\(\frac{-5}{2}\), \(\frac{-2}\)) Hence, from the above, We can conclude that the coordinates of the midpoint are: (\(\frac{-5}{2}\), -2)

Question 4. C(- 1, 7) and D(- 8, – 3) Answer: The given points are: C (-1, 7), and D (-8, -3) We know that, The coordinates of the midpoint = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, The coordinates of the midpoint = (\(\frac{-1 – 8}{2}\), \(\frac{-3 + 7}{2}\)) The coordinates of the midpoint = (\(\frac{-9}{2}\), \(\frac{4}{2}\)) The coordinates of the midpoint = (\(\frac{-9}{2}\), 2) Hence, from the above, We can conclude that the coordinates of the midpoint are: (\(\frac{-9}{2}\), 2)

Question 5. The midpoint of \(\overline{E F}\) is M(1, – 1). One endpoint is E(- 3, 2). Find the coordinates of endpoint F. Answer: The given points of \(\overline{E F}\) are: M (1, -1), and E (-3, 2) Let the other endpoint be F (x, y) We know that, The coordinates of the midpoint = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) So, (1, -1) = (\(\frac{x – 3}{2}\), \(\frac{y + 2}{2}\)) \(\frac{x – 3}{2}\) = 1                   \(\frac{y + 2}{2}\) = -1 x – 3 = 2                                           y + 2 = -2 x = 5                                                y = -4 Hence, from the above, We can conclude that the coordinates of the other endpoint are: (5, -4)

Use the diagram to decide whether the statement is true or false.

Question 6. Points A, R, and B are collinear. Answer: We know that, The points that lie in the same line are called “Collinear points” From the figure, We can observe that A, R, and B are in the same plane but not in the same line Hence, from the above, We can conclude that A, R, and B are not collinear

Question 9. Plane D could also be named plane ART. Answer: We know that, The points that are coplanar in the given plane can also be named for the name of the plane Hence, from the above, We can conclude that plane D can also be named plane ART

Find the perimeter and area of the polygon with the given vertices. Explain how you found your answer.

Question 10. P(- 3, 4), Q(1, 4), R(- 3, – 2), S(3, – 2) Answer: The given points are: P(- 3, 4), Q(1, 4), R(- 3, – 2), S(3, – 2) Now, \(\overline{P Q}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(1 + 3)² + (4 – 4)²}\) = 4 \(\overline{Q R}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(1 + 3)² + (4 + 2)²}\) = 7.2 \(\overline{R S}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(2 – 2)² + (3 + 3)²}\) = 6 \(\overline{S P}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3 + 3)² + (4 + 2)²}\) = 7.2 Hence, The perimeter of the given polygon = PQ + QR + RS + SP = 4 + 6 + 7.2 + 7.2 = 24.4 The area of the given polygon = \(\frac{1}{2}\) (a + b) (h) Where, a and b are the sides that are not equal So, The area of the polygon = \(\frac{1}{2}\) (4 + 6) (7.2) = \(\frac{1}{2}\) (10) (7.2) = 36 Hence, from the above, We can conclude that The perimeter of the given polygon is: 24.4 The area of the given polygon is: 36

Question 11. J(- 1, 3), K(5, 3), L(2, – 2) Answer: The given points are: J(- 1, 3), K(5, 3), L(2, – 2) Now, \(\overline{J K}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3 – 3)² + (1 + 5)²}\) = 6 \(\overline{K L}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(3 + 2)² + (5 – 2)²}\) = 5.8 \(\overline{L J}\) = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(2 + 1)² + (3 + 2)²}\) = 5.8 Hence, The perimeter of the given polygon = JK + KL + LJ = 6 + 5.8 + 5.8 = 17.8 Now, The area of the given polygon = \(\frac{1}{2}\) (Base) (Height) So, The area of the polygon = \(\frac{1}{2}\) (JK) (KL) = \(\frac{1}{2}\) (6) (5.8) = 17.4 Hence, from the above, We can conclude that The perimeter of the given polygon is: 17.8 The area of the given polygon is: 17.4

b. What is the measure of the obtuse angle created when the clock is at 5:00? Answer: From the figure, We can observe that there are 4 sectors between 12 and  at 5:00 So, The acute angle created when the clock is at 5:00 = 4 (30) = 120° So, The obtuse angle created when the clock is at 5:00 = 360 – 120 = 240° Hence, from the above, We can conclude that the obtuse angle created when the clock is at 5:00 is: 240°

c. Find a time where the hour and minute hands create a straight angle. Answer: we know that, A straight angle is an angle that is 180° Now, From the figure, We can observe that the hour and minute hands create a straight angle is: 6:00, 2:45 Hence, from the above, We can conclude that the time where the hour and minute hands create a straight angle is: 6:00 and 2:45

Basics of Geometry Cumulative Assessment

Question 2. Order the terms so that each consecutive term builds off the previous term. plane           segment      line     point    ray Answer: The given terms are: a. Plane b. Segment c. Line d. Point e. Ray Hence, The order of the given terms are: Point, Line, Ray, Segment, and Plane

Question 3. The endpoints of a line segment are (- 6, 13) and (11, 5). Which choice shows the correct midpoint and distance between these two points? (A) \(\left(\frac{5}{2}, 4\right)\); 18.8 units (B) \(\left(\frac{5}{2}, 9\right)\); 18.8 units (C) \(\left(\frac{5}{2}, 4\right)\); 9.4 units (D) \(\left(\frac{5}{2}, 9\right)\); 9.4 units Answer: The given endpoints of a line segment are: (-6, 13) and (11, 5) We know that, The coordinates of the midpoint = (\(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\)) = (\(\frac{11 – 6}{2}\), \(\frac{13 + 5}{2}\)) = (\(\frac{5}{2}\), \(\frac{18}{2}\)) = (\(\frac{5}{2}\), 9) We know that, The distance between the coordinates = \(\sqrt{(x2 – x1)² + (y2 – y1)²}\) = \(\sqrt{(11 + 6)² + (13 – 5)²}\) = 18.78 ≈ 18.8 Hence, from the above, We can conclude that option (B) shows the correct midpoint and distance

b. Identify all linear pairs. Answer: From the given figure, The linear pairs are: ∠LJK, ∠MJP, ∠MJN

C. you are traveling east on Buffalo Road and decide to turn left onto Carter Hill. Name the angle of the turn you made. ∠KJL     ∠KJM     ∠KJN      ∠KJP       ∠LJM ∠LJN     ∠LJP      ∠MJN     ∠MJP     ∠NJP Answer: From the above figure, The angle made by you when you are traveling east n BuffaloRoad and decide to turn left onto Carter hill is: ∠KJL

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Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles

Need instant homework help for solving all chapter 6 Relationships Within Triangles Questions? Then, don’t worry we have come up with a great study guide and one-stop destination for looking at what you require ie., Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles . This BIM Geometry Solution key covered all Chapter 6 Relationships Within Triangles Exercises Questions, Practices, Chapter Review, Chapter Test, Assessments, etc. All you can attain for free of cost and make use of this Ch 6 Relationships Within Triangles Big Ideas Math Geometry Answers for better practice and learning.

Big Ideas Math Book Geometry Answer Key Chapter 6 Relationships Within Triangles

Improve your subject knowledge & clear all your exams with flying colors by taking the help of the BIM Geometry solution key of Ch 6 Relationships Within Triangles . All the questions covered in this study material is very beneficial for students to understand the concept thoroughly. Students who need to learn how to Answer Ch 6 Relationships Within Triangles Questions should definitely go ahead with this page and score maximum marks in the exams. So, here are the links to access Topic-wise Big Ideas Math Geometry Answers Chapter 6 Relationships Within Triangles & ace up your preparation.

• Relationships Within Triangles Maintaining Mathematical Proficiency – Page 299
• Relationships Within Triangles Mathematical Practices – Page 300
• 6.1 Perpendicular And Angle Bisectors – Page 301
• Lesson 6.1 Perpendicular And Angle Bisectors – Page(302-308)
• Exercise 6.1 Perpendicular And Angle Bisectors – Page(306-308)
• 6.2 Bisectors of Triangles – Page 309
• Lesson 6.2 Bisectors of Triangles – Page(310-318)
• Exercise 6.2 Bisectors of Triangles – Page(315-318)
• 6.3 Medians and Altitudes of Triangles – Page 319
• Lesson 6.3 Medians and Altitudes of Triangles – Page(320-326)
• Exercise 6.3 Medians and Altitudes of Triangles – Page(324-326)
• 6.1 and 6.3 Quiz – Page 328
• 6.4 The Triangle Midsegment Theorem – Page 329
• Lesson 6.4 The Triangle Midsegment Theorem – Page (330-334)
• Exercise 6.4 The Triangle Midsegment Theorem – Page(333-334)
• 6.5 Indirect Proof and Inequalities in One Triangle – Page 335
• Lesson 6.5 Indirect Proof and Inequalities in One Triangle – Page(336-342)
• Exercise 6.5 Indirect Proof and Inequalities in One Triangle – Page(340-342)
• 6.6 Inequalities in Two Triangles – Page 343
• Lesson 6.6 Inequalities in Two Triangles – Page(344-348)
• Exercise 6.6 Inequalities in Two Triangles – Page(347-348)
• Relationships Within Triangles Chapter Review – Page(350-352)
• Relationships Within Triangles Chapter Test – Page 353
• Relationships Within Triangles Cumulative Assessment – Page(354-355)

Relationships Within Triangles Maintaining Mathematical Proficiency

Write an equation of the line passing through point P that is perpendicular to the given line.

Question 1. P(3, 1), y = \(\frac{1}{3}\)x – 5

Answer: y = -3x + 10

Explanation: The slope of the given line is \(\frac { 1 }{ 3 } \). Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line. The slope of the perpendicular line = -3 Substitute the values in y = mx + c 1 = -3(3) + c 1 = -9 + c c = 1 + 9 c = 10 use the slope intercept form of a linear equation again substitute m, c y = -3x + 10

Question 2. P(4, – 3), y = – x – 5

Answer: y = \(\frac { 1 }{ 3 } \)x – \(\frac { 5 }{ 3 } \)

Explanation: The slope of the given line is -3. Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line. The slope of the perpendicular line = \(\frac { 1 }{ 3 } \) Substitute the values in y = mx + c -3 = –\(\frac { 1 }{ 3 } \)(4) + c c = -3 + \(\frac { 4 }{ 3 } \) = \(\frac { -9 + 4 }{ 3 } \) = \(\frac { -5 }{ 3 } \) use the slope intercept form of a linear equation again substitute m, c y = \(\frac { 1 }{ 3 } \)x + \(\frac { -5 }{ 3 } \) y = \(\frac { 1 }{ 3 } \)x – \(\frac { 5 }{ 3 } \)

Question 3. P(- 1, – 2), y = – 4x + 13

Answer: y = \(\frac { 1 }{ 4 } \)x – \(\frac { 7 }{ 4 } \)

Explanation: The slope of the given line is -4. Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line. The slope of the perpendicular line = \(\frac { 1 }{ 4 } \) Substitute the values in y = mx + c -2 = \(\frac { 1 }{ 4 } \)(-1) + c c = -2 + \(\frac { 1 }{ 4 } \) = \(\frac { -8 + 1 }{ 4 } \) = \(\frac { -7 }{ 4 } \) use the slope intercept form of a linear equation again substitute m, c y = \(\frac { 1 }{ 4 } \)x + \(\frac { -7 }{ 4 } \) y = \(\frac { 1 }{ 4 } \)x – \(\frac { 7 }{ 4 } \)

Write the sentence as an inequality.

Question 4. A number w is at least – 3 and no more than 8.

Answer: -3 ≤ w < 8

Explanation: At least means ≥ and no more than means < w ≥ -3 and w < 8 -3 ≤ w < 8

Question 5. A number m is more than 0 and less than 11.

Answer: 0 < m < 11

Explanation: more than means > and less than means < m > 0 and m < 11 0 < m < 11

Question 6. A number s is less than or equal to 5 or greater than 2.

Answer: 2 < s ≤ 5

Explanation: less than or equal to means ≤ and greater than means > s ≤ 5 or s > 2 2 < s ≤ 5

Question 7. A number d is fewer than 12 or no less than – 7.

Answer: -7 < d < 12

Explanation: fewer than means < no less than means > d < 12 or d > -7 -7 < d < 12

Question 8. ABSTRACT REASONING Is it possible for the solution of a compound inequality to be all real numbers? Explain your reasoning.

Answer: Yes

Relationships Within Triangles Mathematical Practices

Monitoring Progress

Refer to the figures at the top of the page to describe each type of line, ray, or segment in a triangle.

Question 1. perpendicular bisector

Answer: The perpendicular bisector of a side of a triangle is a line perpendicular to the side and passes through its midpoint. Three perpendicular bisectors of sides meet at a point in a triangle.

Question 2. angle bisector

Answer: The angle bisector of an angle in a triangle is a straight line that divides the angle into two congruent angles.

Question 3. median

Answer: It is a line segment from a vertex to the midpoint of the opposite side.

Question 4. altitude

Answer: It is a segment from a vertex of the triangle, perpendicular to the side opposite that vertex of the triangle.

Question 5. midsegment

Answer: It is a segment that connects the mid points of two sides in a triangle.

6.1 Perpendicular and Angle Bisectors

Exploration 1

Points on a Perpendicular Bisector

Work with a partner. Use dynamic geometry software.

a. Draw any segment and label it \(\overline{A B}\). ConStruct the perpendicular bisector of \(\overline{A B}\). Answer:

b. Label a point C that is on the perpendicular bisector of \(\overline{A B}\) but is not on \(\overline{A B}\). Answer:

c. Draw \(\overline{C A}\) and \(\overline{C B}\) and find their lengths. Then move point C to other locations on the perpendicular bisector and note the lengths of \(\overline{C A}\) and \(\overline{C B}\). Answer:

d. Repeat parts (a) – (c) with other segments. Describe any relationships(s) you notice. Answer:

Exploration 2

Points on an Angle Bisector

a. Draw two rays \(\vec{A}\)B and \(\vec{A}\)C to form ∠BAC. Construct the bisector of ∠BAC. Answer:

b. Label a point D on the bisector of ∠BAC. Answer:

c. Construct and find the lengths of the perpendicular segments from D to the sides of ∠BAC. Move point D along the angle bisector and note how the lengths change. Answer:

d. Repeat parts (a)-(c) with other angles. Describe an relationship(s) you notice. USING TOOLS STRATEGICALLY To be proficient in math, you need to visualize the results of varying assumptions, explore consequences, and compare predictions with data. Answer:

Communicate Your Answer

Question 3. What conjectures can you make about a point on the perpendicular bisector of a segment and a point on the bisector of an angle? Answer: It is given a point on the perpendicular bisector of a segment and a point on the bisector of an angle. It is required to give conjectures about a point on the perpendicular bisector of a segment and a point on the bisector of an angle. To do these use the perpendicular bisector and angle bisector theorem.

• By the perpendicular bisector theorem, a perpendicular bisector divides the line into two equal halves and makes a right angle with it. So, it can be said that every point on the perpendicular bisector will be equidistant from both the ends of the line it is bisecting.
• By the angle bisector theorem, a point is equidistant from the sides of the angle if a point is on the bisector of an angle.

Question 4. In Exploration 2. what is the distance from point D to \(\vec{A}\)B when the distance from D to \(\vec{A}\)C is 5 units? Justify your answer. Answer:

Lesson 6.1 Perpendicular and Angle Bisectors

Use the diagram and tile given information to find the indicated measure.

Answer: WZ = 13.75

Explanation: Using the perpendicular bisector theorem YZ = WZ 13.75 = WZ

Answer: YZ = 27

Explanation: Using the perpendicular bisector theorem YZ = WZ n + 17 = 4n – 13 4n – n = 17 + 13 3n = 30 n = 10 YZ = 10 + 17 = 27

Question 3. Find WX when WZ = 20.5. WY = 14.8. and YZ = 20.5.

Answer: WX = 7.4

Explanation: Because WZ = ZY and ZX is the perpendicular bisector of WY by the converse of perpendicular bisector theorem. By the definition of segment sector, WY = 2WX 14.8 = 2WX WX = 7.4

Use the diagram and the given information to find the indicated measure.

Question 4. \(\vec{B}\)D bisects ∠ABC, and DC = 6.9, Find DA.

Answer: DA = 6.9

Explanation: By using the angle bisector theorem DC = DA DA = 6.9

Question 5. \(\vec{B}\)D bisects ∠ABC, AD = 3z + 7, and CD = 2z + 11. Find CD.

Answer: z = 4

Explanation: By using the angle bisector theorem DC = DA 2z + 11 = 3z + 7 3z – 2z = 11 – 7 z = 4

Question 6. Find m∠ABC when AD = 3.2, CD = 3.2, and m∠DBC = 39°.

Answer: m∠DBC = 78°

Explanation: m∠DBC = 39 = m∠DBA m∠ABC = m∠DBC + m∠DBA = 39 + 39 = 78°

Question 8. Write an equation of the perpendicular bisector of the segment with endpoints (- 1, – 5) and (3, – 1).

Answer: y + 5 = 0

Explanation: Midpoint = (\(\frac { -1 + 3 }{ 2 } \), \(\frac { -5 – 1 }{ 2 } \)) = (\(\frac { 2 }{ 2 } \), \(\frac { -6 }{ 2 } \)) = (1, -3) Slope of the line = \(\frac { -1 + 1 }{ 3 + 5 } \) = 0 Slope of the perpendicular line = 0 Equation of the perpendicular bisector is (y + 5) = 0(x + 1) y + 5 = 0

Exercise 6.1 Perpendicular and Angle Bisectors

Vocabulary and Core Concept Check

Question 2. DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.

Is point B the same distance from both X and Z?

Answer: Yes.

Is point B equidistant from X and Z?

Is point B collinear with X and Z?

Answer: No.

Is point B on the perpendicular bisector of \(\overline{X Z}\)?

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, find the indicated measure. Explain your reasoning.

Answer: QR = RS So, QR = 1.3

Answer: UW = 55

Explanation: UV = WU 9x + 1 = 7x + 13 9x – 7x = 13 – 1 2x = 12 x = 6 UW = 7(6) + 13 = 42 + 13 = 55

In Exercises 7-10. tell whether the information in the diagram allows you to conclude that point P lies on the perpendicular bisector of \(\overline{L M}\). Explain your reasoning.

Answer: PN ⊥ LM. So, LN – MN Therefore, point P lies on the LM.

In Exercises 11-14. find the indicated measure. Explain your reasoning.

Answer: PS = RS PS = 12

Answer: FG = 16

Explanation: FG = GH x + 11 = 3x + 1 3x – x = 11 – 1 2x = 10 x = 5 FG = 5 + 11 = 16

In Exercises 15 and 16, tell whether the information in the diagram allows you to conclude that \(\vec{E}\)H bisects ∠FEG. Explain your reasoning.

Answer: Yes, because H is equidistant from EF and EG, EH bisects ∠FEG by the Angle Bisector Theorem.

In Exercises 17 and 18, tell whether the information in the diagram allows you to conclude that DB = DC. Explain your reasoning.

Answer: Yes, because AD is perpendicular to BD and DC. using perpendicular bisector theorem DB = DC.

In Exercises 19-22, write an equation of the perpendicular bisector of the segment with the given endpoints.

Question 20. Q(- 2, 0), R(6, 12)

Answer: An equation of the perpendicular bisector is y = –\(\frac { 3 }{ 7 } \)x + \(\frac { 36 }{ 7 } \)

Explanation: Slope of QR = \(\frac { 12 + 2 }{ 6 – 0 } \) = \(\frac { 7 }{ 3 } \) The slope of the perpendicular line m = –\(\frac { 3 }{ 7 } \) Midpoint of QR = (\(\frac { -2 + 6 }{ 2 } \), \(\frac { 0 + 12 }{ 2 } \) = (-2, 6) y = mx + b 6 = –\(\frac { 3 }{ 7 } \)(-2) + b 6 = \(\frac { 6 }{ 7 } \) + b b = \(\frac { 36 }{ 7 } \) An equation of the perpendicular bisector is y = –\(\frac { 3 }{ 7 } \)x + \(\frac { 36 }{ 7 } \)

Question 22. Y( 10, – 7), Z(- 4, 1)

Answer: An equation of the perpendicular bisector is y = \(\frac { 7 }{ 4 } \)x – \(\frac { 33 }{ 4 } \)

Explanation: Slope of YZ = \(\frac { 1 + 7 }{ -4 – 10 } \) = \(\frac { -4 }{ 7 } \) The slope of the perpendicular line m = \(\frac { 7 }{ 4 } \) Midpoint of YZ = (\(\frac { 10 – 4 }{ 2 } \), \(\frac { -7 + 1 }{ 2 } \)) = (3, -3) y = mx + b -3 = \(\frac { 7 }{ 4 } \)(3) + b b = -3 – \(\frac { 21 }{ 4 } \) = \(\frac { -12 – 21 }{ 4 } \) = \(\frac { -33 }{ 4 } \) An equation of the perpendicular bisector is y = \(\frac { 7 }{ 4 } \)x – \(\frac { 33 }{ 4 } \)

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in the student’s reasoning.

Answer: CP is not perpendicular to BP. So, x = 5 is wrong.

Answer: PG should be the perpendicular bisector of ∠APB.

b. How does m∠APB change as the puck gets closer to the goal? Does this change make it easier or more difficult for the goalie to defend the goal? Explain your reasoning. Answer: ∠APB = 2 ∠APG ∠APG = 10° ∠APG = 20°

Question 28. WRITING Explain how the Converse of the Perpendicular Bisector Theorem (Theorem 6.2) is related to the construction of a perpendicular bisector.

Answer: The converse of the perpendicular bisector theorem states if a point is equidistant from both the endpoints of the line segments in the same plane, then that point is on the perpendicular bisector of the line segment.

Question 30. REASONING Which point lies on the perpendicular bisector of the segment with endpoints M(7, 5) and m(- 1, 5)? (A) (2, 0) (B) (3, 9) (C) (4, 1) (D) (1, 3)

Answer: (B) (3, 9)

Explanation: The distance from M to (3, 9) is = √(7 – 3)² + (5 – 9)² = √16 + 16 = √32 The distance from m to (3, 9) = √(-1 – 3)² + (5 – 9)² = √16 + 16 = √32

b. Is the museum approximately equidistant from Wilson School and Roosevelt School? Explain your reasoning. Answer: The museum is closer to the Roosevelt school, while for Wilson school it is two blocks north and then two blocks east.

Question 36. THOUGHT PROVOKING The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, is it possible for two lines to be perpendicular but not bisect each other? Explain your reasoning. Answer:

• A line is a great circle that divides the sphere into two equal halves in spherical geometry.
• Any two lines intersecting in two points represent two poles of the spherical lines.
• So, it is not possible for two spherical lines to be perpendicular without bisecting each other.

Answer: a. By using the perpendicular bisector theorem \(\overline{X W} \cong \overline{Z W}\) b. By using the perpendicular bisector theorem \(\overline{X V} \cong \overline{Z V}\)

Maintaining Mathematical Proficiency

Classify the triangle by its sides.

Answer: The trinagle is scalene as all the sides have different lengths.

Classify the triangles by its angles.

Answer: The triangle is an obtuse triangle.

6.2 Bisectors of Triangles

Properties of the Perpendicular Bisectors of a Triangle

Work with a partner: Use dynamic geometry software. Draw any ∆ABC. a. Construct the perpendicular bisectors of all three sides of ∆ABC. Then drag the vertices to change ∆ABC. ‘What do you notice about the perpendicular bisectors? Answer:

b. Label a point D at the intersection of the perpendicular bisectors. Answer:

c. Draw the circle with center D through vertex A of ∆ABC. Then drag the vertices to change ∆ABC. What do you notice? Answer:

Properties of the Angle Bisectors of a Triangle

Work with a partner. Use dynamic geometry software. Draw any ∆ABC.

a. Construct the angle bisectors of all three angles of ∆ABC, Then drag the vertices to change ∆ABC. What do you notice about the angle bisectors? Answer:

b. Label a point D at the intersection of the angle bisectors. Answer:

c. Find the distance between D and \(\overline{A B}\). Draw the circle with center D and this distance as a radius. Then drag the vertices to change ∆ABC. What do you notice? LOOKING FOR STRUCTURE To be proficient in math, you need to see complicated things as single objects or as being composed of several objects. Answer:

Question 3. What conjectures can you make about the perpendicular bisectors and the angle bisectors of a triangle? Answer:

Lesson 6.2 Bisectors of Triangles

Find the coordinates of the circumcenter of the triangle with the given vertices.

Question 2. R(- 2, 5), S(- 6, 5), T(- 2, – 1)

Answer: The circumcenter is (4, 7.33)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let R(- 2, 5), S(- 6, 5), T(- 2, – 1) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PR = PS = PT PR² = PS² = PT² PR² = PS² (x + 2)² + (y – 5)² = (x + 6)² + (y – 5)² x² + 4x + 4 + y² – 10x + 25 = x² + 12x + 36 + y² -10y + 25 4x + 4 = 12x + 36 12x – 4x = 36 – 4 8x = 32 x = 4 PS² = PT² (x + 6)² + (y – 5)² = (x + 2)² + (y + 1)² x² + 12x + 36 + y² -10y + 25 = x² + 4x + 4 + y² + 2y + 1 12x – 4x – 10y – 2y + 61 = 5 8x – 12y = -56 8(4) – 12y = -56 32 – 12y = -56 -12y = -56 – 32 -12y = -88 y = 7.33 The circumcenter is (4, 7.33)

Question 3. W(- 1, 4), X(1, 4), Y( 1, – 6)

Answer: The circumcenter is (0.25, -1)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let W(- 1, 4), X(1, 4), Y( 1, – 6) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PW = PX = PY PW² = PY² = PX² PX² = PY² (x – 1)² + (y – 4)² = (x – 1)² + (y + 6)² x² -2x + 1 + y² – 8y + 16 = x² -2x + 1 + y² + 12y + 36 -8y – 12y = 36 – 16 -20y = 20 y = -1 PW² = PX² (x + 1)² + (y – 4)² = (x – 1)² + (y – 4)² x² + 2x + 4 + y² – 8y + 16 = x² – 2x + 1 + y² – 8y + 16 4x = 1 x = 0.25 The circumcenter is (0.25, -1)

Answer: QP = \(\frac { 23 }{ 2 } \)

Explanation: QM = QN 3x + 8 = 7x + 2 7x – 3x = 8 – 2 4x = 6 x = \(\frac { 3 }{ 2 } \) QP = QN = 7(\(\frac { 3 }{ 2 } \)) + 2 = \(\frac { 23 }{ 2 } \)

Question 5. Draw a sketch to show the location L of the lamppost in Example 4.

Exercise 6.2 Bisectors of Triangles

Answer: The third triangle does not belong with the other three. Because the point P in the remaining triangles is the circumcenter. But P is not circumcenter in the third triangle.

In Exercises 3 and 4, the perpendicular bisectors of ∆ABC intersect at point G and are shown in blue. Find the indicated measure.

Answer: Because G is the circumcenter of ∆ABC, AG = BG = CG AG = BG = 11 So, AG = 11

In Exercises 5 and 6, the angle bisectors of ∆XYZ intersect at point P and are shown in red. Find the indicated measure.

Answer: Because P is the incenter of ∆XYZ, PH = PF = PK Therefore, PK = 15 HP = 15

In Exercises 7-10. find the coordinates of the circumcenter of the triangle with the given vertice

Question 8. D(- 7, – 1), E(- 1, – 1), F(- 7, – 9)

Answer: The circumcenter is (-4, -5)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let D(- 7, – 1), E(- 1, – 1), F(- 7, – 9) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PD = PE = PF PD² = PE² = PF² PD² = PE² (x + 7)² + (y + 1)² = (x + 1)² + (y + 1)² x² + 14x + 49 + y² + 2y +1 = x² + 2x + 1 + y² + 2y + 1 14x – 2x = 1 – 49 12x = -48 x = -4 PD² = PF² (x + 7)² + (y + 1)² = (x + 7)² + (y + 9)² x² + 14x + 49 + y² + 2y + 1 = x² + 14x + 49 + y² + 18y + 81 18y – 2y = 1 – 81 16y = -80 y = -5 The circumcenter is (-4, -5)

Question 10. L(3, – 6), M(5, – 3) , N (8, – 6)

Answer: The circumcenter is (5.5, -5.5)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let L(3, – 6), M(5, – 3) , N (8, – 6) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PL = PM = PN PL² = PM² = PN² PL² = PN² (x – 3)² + (y + 6)² = (x – 8)² + (y + 6)² x² – 6x + 9 + y² + 12y + 36 = x² -16x + 64 + y² + 12y + 36 -16x + 6x = 9 – 64 -10x = -55 x = 5.5 PL² = PM² (x – 3)² + (y + 6)² = (x – 5)² + (y + 3)² x² – 6x + 9 + y² + 12y + 36 = x² – 10x + 25 + y² + 6y + 9 -6x + 10x + 45 = 6y – 12y + 34 4x = -6y -11 4(5.5) = -6y – 11 22 + 11 = -6y 33 = -6y y = -5.5 The circumcenter is (5.5, -5.5)

In Exercises 11-14, N is the incenter of ∆ABC. Use the given information to find the indicated measure.

Answer: NJ = 9

Explanation: NG = NH = NJ x + 3 = 2x – 3 2x – x = 3 + 3 x = 6 By Incenter theorem, NG = NH = NJ NJ = 6 + 3 = 9

Answer: NS = 4

Explanation: NQ = NR 2x = 3x – 2 3x – 2x = 2 x = 2 NQ = 2 (2) = 4 By Incenter theorem NS = NR = NQ So, NS = 4

Answer: PY = 31

Explanation: PX = PZ 4x + 3 = 6x – 11 6x – 4x = 3 + 11 2x = 14 x = 7 PZ = 6(7) – 11 = 42 – 11 = 31 By the incenter theorem, PX = PZ = PY PY = 31

CONSTRUCTION In Exercises 17-20. draw a triangle of the given type. Find the circumcenter. Then construct the circumscribed circle.

Question 18. obtuse

Question 20. equilateral

CONSTRUCTION In Exercises 21-24, copy the triangle with the given angle measures. Find the incenter. Then construct the inscribed circle.

ERROR ANALYSIS In Exercises 25 and 26. describe and correct the error in identifying equal distances inside the triangle.

Answer: Because point T is the intersection of the angle bisectors, it is the incenter. But, because TV and TF are not necessarily perpendicular to a side of the trinagle, there is not sufficient evidence to conclude that TV and TZ are congruent. Point T is equidistance from the sides of the triangle.

CRITICAL THINKING In Exercises 29-32, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 30. If the perpendicular bisector of one side of a triangle intersects the opposite vertex. then the triangle is ______________ isosceles.

Answer: If the perpendicular bisector of one side of a triangle intersects the opposite vertex. then the triangle is always isosceles.

Explanation: The Vertex is equidistant to other vertices then.

Question 32. The angle bisectors of a triangle intersect at a point that is ______________ equidistant from the sides of the triangle.

Answer: The angle bisectors of a triangle intersect at a point that is circumcenter equidistant from the sides of the triangle.

CRITICAL THINKING In Exercises 33 and 34, find the coordinates of the circumcenter of the triangle with the given vertices.

Question 34. D(- 9, – 5), E(- 5, – 9), F(- 2, – 2)

Answer: The circumcenter is (4.45, 4.45)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let D(- 9, – 5), E(- 5, – 9), F(- 2, – 2) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PD = PE = PF PD² = PE² = PF² PD² = PE² (x + 9)² + (y + 5)² = (x + 5)² + (y + 9)² x² + 18x + 81 + y² + 10y + 25 = x² + 10x + 25 + y² + 18y + 81 18x – 10x = 18y – 10y 8x = 8y x = y PE² = PF² (x + 5)² + (y + 9)² = (x + 2)² + (y + 2)² x² + 10x + 25 + y² + 18y + 81 = x² + 4x + 4 + y² + 2y + 4 10x – 4x + 106 = 8 + 2y – 18y 6x + 106 = 8 – 16y 6x + 98 = -16x 22x = -98 x = 4.45 y = 4.45 The circumcenter is (4.45, 4.45)

MATHEMATICAL CONNECTIONS In Exercises 35 and 36. find the a1ue of x that makes N the incenter of the triangle.

Answer: N the incenter is \(\frac { 12 }{ 2 } \).

Explanation: 25² = 24² + (14x)² 625 = 576 + 196x² 49 = 196x² x = \(\frac { 7 }{ 14 } \) = \(\frac { 1 }{ 2 } \) The value of x will make N the incenter is \(\frac { 12 }{ 2 } \).

Answer: Given ∆ABC, \(\overline{A D}\) bisects∠CAB, \(\overline{B D}\) bisects ∠CBA, \(\overline{D E}\) ⊥ \(\overline{A B}\), \(\overline{D F}\) ⊥ \(\overline{B C}\), and \(\overline{D G}\) ⊥ \(\overline{C A}\) To Prove: The angle bisectors intersect at D, which is equidistant from \(\overline{A B}\), \(\overline{B C}\), and \(\overline{C A}\) It is also given that, AD bisects ∠CAD. So according to the definition of an angle bisector, we can write that∠CAD≅∠BAD. ∠ADP≅∠AFP [Corresponding right angles] And, AP is the common side Thus according to the AAS congruence theorem, we have that, ΔAED≅ΔAGD And, DE≅DG [corresponding parts of congruent triangles] Also, we can write that DE≅DF From the above result, we obtain that the point D is equidistant from the sides AB, BC and CA

Question 40. REASONING Is the incenter of a triangle ever located outside the triangle? Explain your reasoning.

Explanation: Because the incenter is the center of an inscribed circle, it must always be inside the triangle.

b. Copy the diagram and find the approximate coordinates of the point at which the archaeologists should look for the fire pit.

COMPARING METHODS In Exercises 49 and 50. state whether you would use perpendicular bisectors or angle bisectors. Then solve the problem.

Question 50. On a map of a camp. You need to create a circular walking path that connects the pool at (10, 20), the nature center at (16, 2). and the tennis court at (2, 4). Find the coordinates of the center of the circle and the radius of the circle.

Answer: The center of the circular path is located at (10, 10) and the radius of the circular path is 10 units.

Let the centre of the circle be at O (x, y) Slope of AB = \(\frac { 20 – 4 }{ 10 – 2 } \) = 2 The slope of XO must be \(\frac { -1}{ 2 } \) the negative reciprocal of the slope of AB as the 2 lines are perpendicular Slope of XO = \(\frac { y – 12 }{ x – 6 } \) = \(\frac { -1 }{ 2 } \) y – 12 = -0.5x + 3 0.5x + y = 12 + 3 = 15 x + 2y = 30 The slope of BC = \(\frac { 2 – 20 }{ 16 – 10 } \) = -3 The slope of XO must be \(\frac { 1 }{ 3 } \) = \(\frac { 11 – y }{ 13 – x } \) 33 – 3y = 13 – x x – 3y = -33 + 13 = -20 Subtrcat two equations x + 2y – x + 3y = 30 + 20 y = 10 x – 30 = -20 x = 10 r = √(10 – 2)² + (10 – 4)² r = 10

The endpoints of \(\overline{A B}\) are given. Find the coordinates of the midpoint M. Then find AB. Question 52. A(- 3, 5), B(3, 5)

Answer: AB = 6

Explanation: Midpoint of AB = (\(\frac { -3 + 3 }{ 2 } \), \(\frac { 5 + 5 }{ 2 } \)) = (0, 5) AB = √(3 + 3)² + (5 – 5)² = 6

Question 54. A(- 5, 1), B(4, – 5)

Answer: AB = 10.81

Explanation: Midpoint of AB = (\(\frac { -5 + 4 }{ 2 } \), \(\frac { 1 – 5 }{ 2 } \)) = (\(\frac { -1 }{ 2 } \), -2) AB = √(4 + 5)² + (-5 – 1)² = √81 + 36 =  10.81

Write an equation of the line passing through point P that is perpendicular to the given line. Graph the equations of the lines to check that they are perpendicular. Question 56. P(2, 8), y = 2x + 1

Answer: y = \(\frac { -1 }{ 2 } \)x + 9

Question 58. P(- 8, – 6), 2x + 3y = 18

Answer: The perpendicular equation is y = \(\frac { 3 }{ 2 } \)x + 6

6.3 Medians and Altitudes of Triangles

Finding Properties of the Medians of a Triangle

a. Plot the midpoint of \(\overline{B C}\) and label it D, Draw \(\overline{A D}\), which is a median of ABC. Construct the medians to the other two sides of ∆ABC. Answer:

b. What do you notice about the medians? Drag the vertices to change ∆ABC. Use your observations to write a conjecture about the medians of a triangle. Answer:

c. In the figure above, point G divides each median into a shorter segment and a longer segment. Find the ratio of the length of each longer segment to the length of the whole median. Is this ratio always the same? Justify your answer. Answer:

Finding Properties of the Altitudes of a Triangle

a. Construct the perpendicular segment from vertex A to \(\overline{B C}\). Label the endpoint D. \(\overline{A D}\) is an altitude of ∆ABC. Answer:

b. Construct the altitudes to the other two sides of ∆ABC. What do you notice? Answer:

c. Write a conjecture about the altitudes of a triangle. Test your conjecture by dragging the vertices to change ∆ABC. LOOKING FOR STRUCTURE To be proficient in math, you need to look closely to discern a pattern or structure. Answer:

Lesson 6.3 Medians and Altitudes of Triangles

There are three paths through a triangular park. Each path goes from the midpoint of one edge to the opposite corner. The paths meet at point P.

Question 1. Find PS and PC when SC = 2100 feet.

Answer: PS = 700, PC = 2100

Explanation: PC = \(\frac { 2 }{ 3 } \)SC SC = PS + PC SC = PS + \(\frac { 2 }{ 3 } \)SC PS = \(\frac { 1 }{ 3 } \)SC So, PS = \(\frac { 1 }{ 3 } \)(2100) = 700 PC = \(\frac { 2 }{ 3 } \)(2100) = 1400

Question 2. Find TC and BC when BT = 1000 feet.

Answer: BC = 2000 ft, TC = 1000 ft

Explanation: BT = TC So, TC = 1000 ft BC = BT + TC BC = 1000 + 1000 = 2000 ft

Question 3. Find PA and TA when PT = 800 feet.

Answer: PA = 2400 ft, TA = 1600 ft

Explanation: PT = \(\frac { 1 }{ 3 } \)PA PA = 3PT = 3 • 800 = 2400 TA = \(\frac { 2 }{ 3 } \)PA TA = \(\frac { 2 }{ 3 } \)(2400) = 1600

Find the coordinates of the centroid of the triangle with the given vertices.

Question 4. F(2, 5), G(4, 9), H(6, 1)

Answer: The centroid is (4, 5).

Explanation: The centroid of the triangle FGH is O = (\(\frac { 2 + 4 + 6 }{ 3 } \), \(\frac { 5 + 9 + 1 }{ 3 } \)) = (\(\frac { 12 }{ 3 } \), \(\frac { 15 }{ 3 } \)) = (4, 5)

Question 5. X(- 3, 3), Y(1, 5), Z(- 1, – 2)

Answer: The centriod of given triangle is (-1, 2)

Explanation: The centroid of the triangle XYZ is O = (\(\frac { -3 + 1 – 1 }{ 3 } \), \(\frac { 3 + 5 – 2 }{ 3 } \)) = (\(\frac { -3 }{ 3 } \), \(\frac { 6 }{ 3 } \)) = (-1, 2)

Tell whether the orthocenter of the triangle with the given vertices is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 6. A(0, 3), B(0, – 2), C(6, -3)

Answer: The orthocenter lies outside the triangle.

Explanation: The slope of the line BC = \(\frac { -3 + 2 }{ 6 – 0 } \) = \(\frac { -1 }{ 6 } \) The slope of the perpendicular line = 6 The perpendicular line is (y – 3) = 6(x – 0) y – 3 = 6x y = 6x + 3 The slope of AC = \(\frac { -3 – 3 }{ 6 – 0 } \) = \(\frac { -6 }{ 6 } \) = -1 The slope of the perpendicular line = 1 The equation of perpendicular line (y + 2) = 1(x – 0) y + 2 = x Substitute y = 6x + 3 in above equation 6x + 3 + 2 = x 5x = -5 x = -1 Put x = -1 in y + 2 = x y + 2 = -1 y = -3 The orthocenter is (-1, -3) The orthocenter lies outside the triangle.

Question 7. J(- 3, – 4), K(- 3, 4), L(5, 4)

Answer: The orthocenter lies on the vertex of the triangle.

Explanation: The slope of JL = \(\frac { 4 + 4 }{ 5 + 3 } \) = 2 The slope of perpendicular line is \(\frac { -1 }{ 2 } \) The equation of perpendicular line is (y – 4) = \(\frac { -1 }{ 2 } \)(x + 3) 2y – 8 = -x – 3 x + 2y – 5 = 0 The slope of JK = \(\frac { 4 + 4 }{ -3 + 3 } \) = 0 The slope of the perpendicular line is 0 the equation of perpendicular line is (y – 4) = 0 y = 4 Substitute y = 4 in x + 2y – 5 = 0 x + 2(4) – 5 = 0 x + 3 = 0 x = -3 So, the ortho center is (-3, 4) It lies on the vertex of the triangle.

Question 8. WHAT IF? In Example 4, you want to show that median \(\overline{B D}\) is also an angle bisector. How would your proof be different? Answer: Given ABC is an isosceles triangle and BD is the median. AB = BC AD = CD BD is the common sides ΔABD ≅ ΔBDC ∠ABD = ∠CBD So BD bisects ∠ABC BD is the angle bisector.

Exercise 6.3 Medians and Altitudes of Triangles

Question 2. COMPLETE THE SENTENCE The length of a segment from a vertex to the centroid is ______________ the length of the median from that vertex.

Answer: The length of a segment from a vertex to the centroid is one-third of the length of the median from that vertex.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, point P is the centroid of ∆LMN. Find PN and QP.

Answer: PN = 14 units, QP = 7 units

Explanation: PN = \(\frac { 2 }{ 3 } \)QN PN = \(\frac { 2 }{ 3 } \)(21) PN = 14 QP = \(\frac { 1 }{ 3 } \)QN = \(\frac { 1 }{ 3 } \)(21) = 7

Answer: PN = 28 units, QP = 14 units

Explanation: PN = \(\frac { 2 }{ 3 } \)QN PN = \(\frac { 2 }{ 3 } \)(42) PN = 28 QP = \(\frac { 1 }{ 3 } \)QN = \(\frac { 1 }{ 3 } \)(42) = 14

In Exercises 7-10. point D is the centroid of ∆ ABC. Find CD and CE.

Answer: CD = 22 units, CE = 33 units

Explanation: DE = \(\frac { 1 }{ 3 } \)CE 11 = \(\frac { 1 }{ 3 } \) CE CE = 33 CD = \(\frac { 2 }{ 3 } \) CE CD = \(\frac { 2 }{ 3 } \)(33) CD = 22

Answer: CE = 45 units, CD = 30

Explanation: DE = \(\frac { 1 }{ 3 } \)CE 15 = \(\frac { 1 }{ 3 } \) CE CE = 45 CD = \(\frac { 2 }{ 3 } \) CE CD = \(\frac { 2 }{ 3 } \)(45) CD = 30

In Exercises 11-14. point G is the centroid of ∆ABC. BG = 6, AF = 12, and AE = 15. Find the length of the segment.

Question 12. \(\overline{B F}\)

Answer: \(\overline{B F}\) = 15

Explanation: All the medians have same length. so, AE = 15 = BF

Question 14. \(\overline{G E}\)

Answer: \(\overline{G E}\) = 5 units

Explanation: AE = AG + GE 15 = 10 + GE GE = 5

In Exercises 15-18. find the coordinates of the centroid of the triangle with the given vertices.

Question 16. F(1, 5), G( – 2, 7), H(- 6, 3)

Answer: The centroid is (\(\frac { -7 }{ 3 } \), 5)

Explanation: The centroid of the trinagle = (\(\frac { 1 – 2 – 6 }{ 3 } \), \(\frac { 5 + 7 + 3 }{ 3 } \)) = (\(\frac { -7 }{ 3 } \), 5)

Question 18. X(1, 4), Y(7, 2), Z(2, 3)

Answer: The centroid of the triangle is (\(\frac { 10 }{ 2 } \), 3)

Explanation: The centroid of the trinagle = (\(\frac { 1 + 7 + 2 }{ 3 } \), \(\frac { 4 + 2 + 3 }{ 3 } \)) = (\(\frac { 10 }{ 2 } \), 3)

In Exercises 19-22. tell whether the orthocenter is inside, on, or outside the triangle. Then find the coordinates of the orthocenter.

Question 20. X(- 3, 2), Y(5, 2), Z(- 3, 6)

Explanation: The slope of YZ = \(\frac { 6 – 2 }{ -3 – 5 } \) = \(\frac { -1 }{ 2 } \) The slope of the perpendicular line is 2 The equation of perpendicular line is (y – 2) = 2(x + 3) y – 2 = 2x + 6 2x – y + 8 = 0 The slope of XZ = \(\frac { 6 – 2 }{ -3 + 3 } \) = 0 The equation of perpendicular line is (y – 2) = 0 y = 2 Substitute y = 2 in 2x – y + 8 = 0 2x – 2 + 8 = 0 2x + 6 = 0 x = -3 the orthocenter is (-3, 2) The orthocenter lies on the vertex of the triangle.

Question 22. T(-2, 1), U( 2, 1), V(0, 4)

Answer: The orthocenter lies inside the triangle ABC.

Explanation: The slope of UV = \(\frac { 4 – 1 }{ 0 – 2 } \) = \(\frac { -3 }{ 2 } \) The slope of the perpendicular line is \(\frac { 2 }{ 3 } \) The equation of the perpendicular line is (y – 1) = \(\frac { 2 }{ 3 } \)(x + 2) 3(y – 1) = 2(x + 2) 3y – 3 = 2x + 2 2x – 3y + 5 = 0 — (i) The slope of TV = \(\frac { 4 – 1 }{ 0 + 2 } \) = \(\frac { 3 }{ 2 } \) The slope of the perpendicular line is \(\frac { -2 }{ 3 } \) The equation of the perpendicular line is (y – 1) = \(\frac { -2 }{ 3 } \)(x – 2) 3(y – 1) = -2(x – 2) 3y – 3 = -2x + 4 2x + 3y – 7 = 0 —(ii) Add two equations 2x – 3y + 5 + 2x + 3y – 7 = 0 4x – 2 = 0 x = 0.5 2x – 1.5 + 5 = 0 x = -1.75 So, the orthocenter is (0, 2.33) The orthocenter lies inside the triangle ABC.

Question 24. obtuse scalene triangle

Question 26. acute isosceles triangle

ERROR ANALYSIS In Exercises 27 and 28, describe and correct the error in finding DE. Point D is the centroid of ∆ABC.

Answer: The length of DE should be 1/3 of length of AE because it is the shorter segment from the centroid DE = \(\frac { 1 }{ 3 } \)AE = \(\frac { 1 }{ 3 } \)(24) = 8 DE = 8

PROOF In Exercises 29 and 30, write a proof of the statement.

Question 30. The altitude from the vertex angle to the base of an isosceles triangle is also a perpendicular bisector. Answer:

CRITICAL THINKING In Exercises 31-36, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 32. The orthocenter is _____________ outside the triangle.

Answer: The orthocenter is always outside the triangle opposite the longest leg, on the same side as the largest angle. The only time all three of these centers fall in the same spot is in the case of an equilateral triangle. In fact, in this case, the incenter falls in the same place as well.

Question 34. An altitude is ______________ the same line segment as an angle bisector.

Answer: An altitude is sometimes the same line segment as an angle bisector.

Question 36. The centroid is ______________ formed by the intersection oÍ the three medians.

Answer: The centroid is a point formed by the intersection oÍ the three medians.

Question 38. WRITING Compare a median. an altitude, and an angle bisector of a triangle.

Answer: Median is a line segment joining a vertex of a triangle with the midoint of the opposite side. Angle bisector is a line segment joining a vertex of a trianglr with the opposite side such that the angle at the vertex split into two equal parts. While altitude is also a line segment joining a vertex of a triangle with opposite side such that the segment is perpendicular to the opposite side. In general, medians, angle bisectors and altitudes drawn from the same vertex of a triangle are different line segments. In an equilateral triagle, altitude, median, and angle bisector drawn from the same vertex overlap. In an isosceles trianglle, the altitude drawn to the base is the median and the angle bisector, the median drawn to the base is the altitude and the angle bisector, the bisector of the angle opposite to the base is the altitude and the median.

Question 40. ANALYZING RELATIONSHIPS Copy and complete the statement for ∆DEF with centroid K and medians \(\overline{D H}\), \(\overline{E J}\), and \(\overline{F G}\).

a. EJ = ____ KJ

Answer: EJ = 2KJ

b. DK = ____ KH

Answer: DK = 2 KH Since KH is the shorter distance between the two parts of the median, its length is 1/3 of the DH.

c. FG = ___ KF

Answer: FG = \(\frac { 2 }{ 3 } \)KF Since KF is the longest distance between two parts, its length is \(\frac { 2 }{ 3 } \) of FG

d. KG = ___ FG

Answer: KG = 2FG Since KG is the shorter distance between the two parts of the median, its length is 1/3 of the FG.

MATHEMATCAL CONNETIONS In Exercises 41-44, point D is the centroid of ∆ABC. Use the given information to find the value of x.

Question 42. GD = 2x – 8 and GC = 3x + 3

Answer: x = 9

Explanation: GD = \(\frac { 1 }{ 3 } \)GC 2x – 8 = \(\frac { 1 }{ 3 } \)(3x + 3) 2x – 8 = x + 1 x = 9

Question 44. DF = 4x – 1 and BD = 6x + 4

Answer: x = 3

Explanation: BD = 2DF 6x + 4 = 2(4x – 1) 6x + 4 = 8x – 2 2x = 6 x = 3

Question 46. CRITICAL THINKING In what types of triangles can a vertex be one of the points of concurrency of the triangle? Explain your reasoning.

Answer: KM is the median drawn from K to JL.

b. What type of segments is \(\overline{K N}\)? Which point of concurrency lies on \(\overline{K N}\)?

Answer: KN is the altitude drawn from K to JL.

c. Compare the areas of ∆JKM and ∆KLM. Do you think the areas of the triangles formed by the median of any triangle will always compare this way? Explain your reasoning.

Question 50. DRAWING CONCLUSIONS The center of gravity of a triangle, the point where a triangle can balance on the tip of a pencil, is one of the four points of concurrency. Draw and cut out a large scalene triangle on a piece of cardboard. Which of the four points of concurrency is the center of gravity? Explain.

Answer: The median is the balancing line of the triangle The centroid is the point of concurrency of three medians The centroid is the center of gravity of the triangle.

Question 52. THOUGHT PROVOKING Construct an acute scalene triangle. Find the orthocenter, centroid, and circumcenter. What can you conclude about the three points of concurrency?

Answer: The centroid, orthocenter and circumcenter are collinear.

Question 54. PROOF Prove the statements in parts (a)-(c). Given \(\overline{L P}\) and \(\overline{M Q}\) are medians of scalenc ∆LMN. Point R is on \(\vec{L}\)P such that \(\overline{L P} \cong \overline{P R}\). Point S is on \(\vec{M}\)Q such that \(\overline{M Q} \cong \overline{Q S}\). Prove a. \(\overline{N S} \cong \overline{N R}\) b. \(\overline{N S}\) and \(\overline{N R}\) are both parallel to \(\overline{L M}\). c. R, N, and S are collinear. Answer:

Determine whether \(\overline{A B}\) is parallel to \(\overline{C D}\).

Question 56. A(- 3, 6), B(5, 4), C(- 14, – 10), D(- 2, – 7)

Answer: AB is not parallel to CD.

Explanation: Slope of AB = \(\frac { 4 – 6 }{ 5 + 3 } \) = \(\frac { -1 }{ 4 } \) Slopeof CD = \(\frac { -7 + 10 }{ -2 + 14 } \) = \(\frac { 1 }{ 4 } \) The slopes of AB and CD are not equal. So AB and CD are not parallel.

Question 58. A(- 5, 6), B(- 7, 2), C(7, 1), D(4, – 5)

Answer: AB is parallel to CD.

Explanation:, Slope of AB = \(\frac { 2 – 6 }{ -7 + 5 } \) = 2 Slopeof CD = \(\frac { -5 – 1 }{ 4 – 7 } \) = 2 The slopes of AB and CD are equal. So, AB is parallel to CD.

6.1 and 6.3 Quiz

Find the indicated measure. Explain your reasoning.

Answer: UV = 15

Explanation: SV = VU 2x + 11 = 8x – 1 8x – 2x = 11 + 1 6x = 12 x = 2 UV = 8(2) – 1 = 15

Answer: QP = 18

Explanation: QP = QR 6x = 3x + 9 3x = 9 x = 3 QP = 6(3) = 18

Answer: m∠GJK = 59°

Explanation: 5x – 4 = 4x + 3 x = 7 ∠JGK = 4(7) + 3 = 31 m∠GJK = 180 – (31 + 90) = 180 – 121 = 59

Question 4. A(- 4, 2), B(- 4, – 4), C(0, – 4)

Answer: The circumcenter is (-2, -1)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(- 4, 2), B(- 4, – 4), C(0, – 4) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PA = PB = PC PA² = PB² = PC² PA² = PB² (x + 4)² + (y – 2)² = (x + 4)² + (y + 4)² x² + 8x + 16 + y² – 4y + 4 = x² + 8x + 16 + y² + 8y + 16 12y = -12 y = -1 PB² = PC² (x + 4)² + (y + 4)² = (x – 0)² + (y + 4)² x² + 8x + 16 + y² + 8y + 16 = x² + y² + 8y + 16 8x = -16 x = -2 The circumcenter is (-2, -1)

Question 5. D(3, 5), E(7, 9), F(11, 5)

Answer: The circumcenter is (7, 5)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let D(3, 5), E(7, 9), F(11, 5) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PD = PE = PF PD² = PE² = PF² PD² = PE² (x – 3)² + (y – 5)² = (x – 7)² + (y – 9)² x² – 6x + 9 + y² – 10y + 25 = x² – 14x + 49 + y² – 18y + 81 -6x + 14x – 10y + 18y = 130 – 34 8x + 8y = 96 x + y = 12 — (i) PE² = PF² (x – 7)² + (y – 9)² = (x – 11)² + (y – 5)² x² – 14x + 49 + y² – 18y + 81 = x² – 22x + 121 + y² – 10y + 25 -14x + 22x – 18y + 10y = 146 – 130 8x – 8y = 16 x – y = 2 — (ii) Add (i) & (ii) x + y + x – y = 12 + 2 2x = 14 x = 7 Put x = 7 in (i) 7 + y = 12 y = 5 The circumcenter is (7, 5)

The incenter of ∆ABC is point N. Use the given information to find the indicated measure.

Answer: NS = 11

Explanation: NQ = NR = NS 2x + 1 = 4x – 9 4x – 2x = 10 2x = 10 x = 5 NQ = 10 + 1 = 11 NS = 11

Answer: NT = 15

Explanation: NU = NV = NT -3x + 6 = -5x -3x + 5x = -6 2x = -6 x = -3 NT = -5(-3) = 15

Answer: NW = 26

Explanation: NZ = NY = NW 4x – 10 = 3x – 1 x = 9 NZ = 4(9) – 10 = 36 – 10 = 26 NW = 26

Find the coordinates of the centroid of the triangle wilt the given vertices. Question 9. J(- 1, 2), K(5, 6), L(5, – 2)

Answer: Centroid of the triangle is (3, 2)

Explanation: Centroid of the triangle = (\(\frac { -1 + 5 + 5 }{ 3 } \), \(\frac { 2 + 6 – 2 }{ 3 } \)) = (3, 2)

Question 10. M(- 8, – 6), N(- 4, – 2), P(0, – 4)

Answer: Centroid of the triangle is (-4, -4)

Explanation: Centroid of the triangle = (\(\frac { -8 – 4 + 0 }{ 3 } \), \(\frac { -6 – 2 – 4 }{ 3 } \)) = (-4, -4)

Tell whether the orthocenter is inside, on, or outside the triangle. Then find its coordinates.

Question 11. T(- 2, 5), U(0, 1), V(2, 5)

Answer: The orthocenter lies inside the triangle TUV

Explanation: The slope of TU = \(\frac { 1 – 5 }{ 0 + 2 } \) = -2 The slope of the perpendicular line is \(\frac { 1 }{ 2 } \) The perpendicular line is y – 5 = \(\frac { 1 }{ 2 } \)(x – 2) 2y – 10 = x – 2 x – 2y + 8 = 0 The slope of UV = \(\frac { 5 – 1 }{ 2 – 0 } \) = 2 The slope of the perpendicular line is \(\frac { -1 }{ 2 } \) The perpendicular line is y – 5 = \(\frac { -1 }{ 2 } \)(x + 2) 2y – 10 = -x – 2 x + 2y – 8 = 0 equate both equations x – 2y + 8 = x + 2y – 8 -4y = -16 y = 4 x – 2(4) + 8 = 0 x = 0 So, the orthocenter is (0, 4) The orthocenter lies inside the triangle TUV

Question 12. X(- 1, – 4), Y(7, – 4), Z(7, 4)

Answer: The orthocenter lies on the vertex.

Explanation: The slope of XY = \(\frac { -4 + 4 }{ 7 + 1 } \) = 0 The slope of the perpendicular line is 0 The perpendicular line is y – 4 = 0 y = 4 The slope of XZ =\(\frac { 4 + 4 }{ 7 + 1} \) = 1 The slope of the perpendicular line is -1 The perpendicular line is y + 4 = -1(x – 7) y + 4 = -x + 7 x + y – 3 = 0 x + 4 – 3 = 0 x = -1 So, the orthocenter is (-1, 4) The orthocenter lies on the vertex.

a. Which point of concurrency is the center of the circle? What type of segments are \(\overline{B G}\), \(\overline{C G}\), and \(\overline{A G}\)? Answer:

• The incircle of a triangle touches all sides of the triangle and has the incentre of the triangle as its center. Therefore, it is the largest possible circle that can be fitted inside a triangle.
• So, the wheel is the incircle of the triangle and its center is the incentre of the triangle.
• We know that the point of intersection of the angle bisectors of a triangle is called the incenter of the triangle.
• Since, BG, CG, and AG intersect at the center of the circle which is the incenter of the triangle. So, BG, CG, and AG are the angle bisectors of the triangle ABC.

b. Which theorem can you use to prove that ∆BGF ≅ ∆BGE? Answer: AG, BG, and CG are the angle bisectors of the triangle. As BG is the angle bisector of ∠EBF. ∠FBG = ∠EBG—-(1) ∠GFC = ∠GEA = 90 degrees As ∠GFC forms a linear pair with ∠GFB and ∠GEA forms a linear pair with ∠GEB ∠GFB = 180 – ∠GFC ∠GEB = 180 – ∠GEA ∠GFB = ∠GEB = 90 degrees —-(2) In ΔBGF and ΔBGE ∠GFB ≅ ∠GEB ∠FBG ≅ ∠EBG BG ≅ BG

c. Find the radius of the wheel to the nearest tenth of a centimeter. Justify your answer. Answer: ΔBGF ≅ ΔBGE BF = BE the length of BE is also 3 cm BE + EA = 10 3 + EA = 10 EA = 10 – 3 EA = 7 Apply the Pythagoras theorem in ΔAEG EA² + GE² = AG² 7² + GE² = 8² GE² = 8² – 7² GE² = 64 – 49 GE² = 15 GE = √15 = 3.88 So, the length of GE is 3.88 cm. So, the radius of the wheel to the nearest tenth of cm is 3.9

Question 14. The Deer County Parks Committee plans to build a park at point P, equidistant from the three largest cities labeled X, Y, and Z. The map shown was created b the committee.

a. Which point of concurrency did the commIttee use as the location of the Park? Answer:

• That committee plans to build a park at point P such that it is at an equal distance from three big cities. We were asked to find which point of concurrency the committee used to locate the park.
• As the three big cities are vertices X, Y, and Z, and point P is equidistant to them, point P must be the point where the perpendicular bisectors of the triangle meet.
• There the committee used the point of concurrency of the perpendicular bisectors.

b. Did the committee use the best point of concurrency for the location of the park? If not, which point would be better to use? Explain. Answer:

• Give is to check whether the committee used the best point of concurrency for the location of the park.
• A point of intersection of perpendicular bisectors is the only point that is at equidistance from all three vertices.
• It is the best point for the location of the park.

6.4 The Triangle Midsegment Theorem

Midsegments of a Triangle

a. Plot midpoint D of \(\overline{A B}\) and midpoint E of \(\overline{B C}\). Draw \(\overline{D E}\), which is a midsegment of ∆ABC. Answer:

b. Compare the slope and length of \(\overline{D E}\) with the slope and length of \(\overline{A C}\). Answer:

c. Write a conjecture about the relationships between the midsegments and sides of a triangle. Test your conjecture by drawing the other midsegments of ∆ABC, dragging vertices to change ∆ABC. and noting whether the relationships hold. Answer:

a. Draw all three midsegments of ∆ABC. Answer:

b. Use the drawing to write a Conjecture about the triangle formed by the midsegments of the original triangle. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures. Answer:

Question 3. How are the midsegments of a triangle related to the sides of the triangle? Answer:

Question 4. In ∆RST. \(\overline{U V}\) is the rnidsegrnent connecting the midpoints of \(\overline{R S}\) and \(\overline{S T}\). Given UV = 12, find RT. Answer:

Lesson 6.4 The Triangle Midsegment Theorem

Monitoring progress

Use the graph of △ABC.

Question 1. In △ABC, show that midsegments \(\overline{D E}\) is parallel to \(\overline{A C}\) and that DE = \(\frac { 1 }{ 2 }\)AC.

Answer: Find the coordinates of D and E by getting the midpoints of AB, BC AB midpoint = D(\(\frac { 1 – 1 }{ 2 } \), \(\frac { 4 – 6 }{ 2 } \)) = D(0, -1) Midpoint of BC = E(\(\frac { 5 – 1 }{ 2 } \), \(\frac { 4 + 0 }{ 2 } \)) = E(2, 2) Slope of AC = \(\frac { 0 + 6 }{ 5 – 1 } \) =\(\frac { 3 }{ 2 } \) Slope of DE = \(\frac { 2 + 1 }{ 2 – 0 } \) = \(\frac { 3 }{ 2 } \) The slopes are same, so the lines are not parallel. DE = √(2 – 0)² + (2 + 1)² = √13 AC = √(5 – 1)² + (0 + 6)² = √16 + 36 = √52 So, DE = \(\frac { 1 }{ 2 }\)AC

Question 2. Find the coordinates of the endpoints of midsegments \(\overline{E F}\), Which opposite \(\overline{A B}\). show that \(\overline{E F}\) is parallel to \(\overline{A B}\) and that EF = \(\frac { 1 }{ 2 }\)AB.

Answer: The coordinates of E = (\(\frac { -1 + 5 }{ 2 } \), \(\frac { 0 + 4 }{ 2 } \)) = (2, 2) The coordinates of F = (\(\frac { 1 + 5 }{ 2 } \), \(\frac { 0 + 6 }{ 2 } \)) = (3, 3) The slope of AB = \(\frac { 4 + 6 }{ -1 – 1 } \) = -5 The slope of EF = \(\frac { 3 – 2 }{ 3 – 2 } \) = 1 The slopes are different. So the lines are not parallel EF = √(3 – 2)² + (3 – 2)² = √2 AB = √(-1 – 1)² + (4 + 6)² = √4 + 100 = √104

Question 3. In Example 2, find the coordinates of F, the midpoint of \(\overline{O C}\). Show that \(\overline{F E}\) || \(\overline{O B}\) and FE = \(\frac { 1 }{ 2 }\)OB.

Answer: The coordinates of F = (\(\frac { 0 + 2p }{ 2 } \), \(\frac { 0 + 0 }{ 2 } \)) = (p, 0) The slope of OB = \(\frac { 2r – 0 }{ 2q – 0 } \) = \(\frac { r }{ q } \) The slope of FE = \(\frac { r – 0 }{ q + p – p }\) = \(\frac { r }{ q }\) The slopes are equal. So, the lines are parallel. FE = √(q + p – p)² + (r – 0)² = √q² + r² OB = √(2q – 0)² + (2r – 0)² = √4q² + 4r²  = 2√q² + r² OB = 2(FE) FE = \(\frac { 1 }{ 2 }\)OB.

Question 4. Copy the diagram in Example 3. Draw and name the third midsegment. Then find the length of \(\overline{V S}\) when the length of the third midsegment is 81 inches.

Question 5. In Example 4. if F is the midpoint of \(\overline{C B}\), what do you know about \(\overline{D F}\)?

Answer: DF is parallel to AE.

Question 6. WHAT IF? In Example 5, you jog down Peach Street to Plum Street, over Plum Street to Cherry Street. up Cherry Street to Pear Street. over Pear Street to Peach Street. and then back home up Peach Street. Do you jog more miles in Example 5? Explain.

Answer: The distance you jog = 2.25 + 1.4 + 1.3 + .65 = 5.6

Exercise 6.4 The Triangle Midsegment Theorem

Question 2. COMPLETE THE SENTENCE If \(\overline{D E}\) is the midsegment opposile \(\overline{A C}\) in ∆ABC, then \(\overline{D E}\) || \(\overline{A C}\) and DE = ________ AC by the Triangle Midsegrnent Theorem

Answer: If \(\overline{D E}\) is the midsegment opposile \(\overline{A C}\) in ∆ABC, then \(\overline{D E}\) || \(\overline{A C}\) and DE = \(\frac { 1 }{ 2 } \) AC by the Triangle Midsegrnent Theorem (Theorem 6.8).

In Exercises 3-6, use the graph of ∆ABC with midsegments \(\overline{D E}\), \(\overline{E F}\), and \(\overline{D F}\).

Question 4. Show that \(\overline{D E}\) is parallel to \(\overline{C B}\) and that DE = \(\frac{1}{2}\)CB.

Answer: The slope of DE = \(\frac { 0 + 2 }{ -2 + 4 } \) = 1 The slope of CB = \(\frac { -2 + 6 }{ 1 + 3 } \)  = 1 The slopes are same. So \(\overline{D E}\) is parallel to \(\overline{C B}\) DE = √(-2 + 4)² + (0 + 2)² = √4 + 4 = √8 CB = √(1 + 3)² + (-2 + 6)² = √16 + 16 = √32 So, DE = \(\frac{1}{2}\)CB

Question 6. Show that \(\overline{D F}\) is parallel to \(\overline{A B}\) and that DF = \(\frac{1}{2}\)AB.

Answer: The slope of DF = \(\frac { -4 + 2 }{ -1 + 4 } \) = \(\frac { -2 }{ 3 } \) The slope of AB = \(\frac { -2 – 2 }{ 1 + 5 } \) = \(\frac { -4 }{ 6 } \) = \(\frac { -2 }{ 3 } \) The slopes are equal. So, the lines are parallel. DF = √(-1 + 4)² + (-4 + 2)² = √9 + 4 = √13 AB = √(1 + 5)² + (-2 – 2)² = √36 + 16 = √52 So, DF = \(\frac{1}{2}\)AB.

In Exercises 7-10, \(\overline{D E}\) is a midsegment of ∆ABC Find the value of x.

Answer: x = 10

Explanation: DE = \(\frac { 1 }{ 2 } \)AB 5 = \(\frac { 1 }{ 2 } \)x x = 10

Answer: x = 8

Explanation: BE = EC x = 8

In Exercise 11-16, \(\overline{X J} \cong \overline{J Y}\), \(\overline{Y L} \cong \overline{L Z}\), and \(\overline{X K} \cong \overline{K Z}\). Copy and complete the statement.

Question 12. \(\overline{X Y}\) || __________

Answer: \(\overline{X Y}\) || KL

Question 14. \(\overline{J L}\) ≅ __________ ≅ __________

Answer: \(\overline{J L}\) ≅ JK ≅ KL

Question 16. \(\overline{J K}\) ≅ __________ ≅ __________

Answer: \(\overline{J K}\) ≅ JL ≅ KL

Question 18. When AC = 3y – 5 and HJ = 4y + 2, what is HB?

Answer: HB = 13

Explanation: AC = 0.5(HJ) 3y – 5 = 0.5(4y + 2) 3y – 5 = 2y + 1 3y – 2y = 1 + 5 y = 6 HB = 0.5(HJ) = 0.5(4(6) + 2) = 0.5(26) = 13

Answer: BC = \(\frac { 1 }{ 2 } \)DE 10 = \(\frac { 12 }{ 2 } \)(5) According to triangle midsegment theorem, AD = DB and AE = EC.

Question 22. PROVING A THEOREM Use the figure from Example 2 to prose the Triangle Midsegment Theorem (Theorem 6.8) for midsegment \(\overline{D F}\), where F is the midpoint of \(\overline{O C}\). Answer:

Question 24. THOUGHT PROVOKING Find a real-life object that uses midsegments as part of its structure. Print a photograph of the object and identify the midsegments of one of the triangles in the structure. Answer:

• We have to look for an object which has midsegments in our surroundings. Then identify the midsegment in one of the triangles present in it.
• When we look around, we will see that many structures like swings, rooftops of house, bridges etc, use midsegments in their design.

Answer: Because the yellow triangle vertices are located at the midpoints of the red triangle sides.

Find a counter example to show that the conjecture is false.

Question 28. The difference of two numbers is always less than the greater number.

Answer: When we subtract -3 from 4 i.e 4 -(-3) = 7. The number is greater than the individual integers. So, the subtraction of a negative number from a positive number gives an answer greater than both numbers.

6.5 Indirect Proof and Inequalities in One Triangle

Comparing Angle Measures and Side Lengths

Work with a partner: Use dynamic geometry software. Draw any scalene ∆ABC

b. Order the side lengths. Order the angle measures. What do you observe? Answer:

c. Drug the vertices of ∆ABC to form new triangles. Record the side lengths and angle measures in a table. Write a conjecture about your findings. Answer:

A Relationship of the Side Lengths of a Triangle

a. Find the side lengths of the triangle. Answer:

c. Drag the vertices of ∆ABC to form new triangles and repeat parts (a) and (b). Organize your results in a table. Write a conjecture about your findings. Answer:

Question 3. How are the sides related to the angles of a triangle? How are any two sides of a triangle related to the third side? Answer: The relation between the sides and the angles of the triangle is that in any triangle the largest side and the largest angle of the triangle are always opposite to each other and the shortest side and the shortest angle are always opposite to each other. The relation of the two sides of a triangle to the third side is that the sum of the lengths of any two sides of a triangle is greater than the length of its third side.

Question 4. Is it possible for a triangle to have side lengths of 3, 4, and 10? Explain. Answer: Given side lengths are 3, 4 and 10. 3 + 4 > 10 7 > 10 So, it is not possible to construct a triangle.

Lesson 6.5 Indirect Proof and Inequalities in One Triangle

Question 1. Write in indirect proof that a scalene triangle cannot have two congruent angles. Answer:

• Let us consider the contradiction that a scalene triangle has two congruent angles and no congruent sides. This means that if two angles of a scalene triangle are congruent, then the two sides opposite two these congruent angles also congruent. But this is a contradiction to the above assumption that the triangle has no congruent sides.
• So, we can conclude that a scalene triangle cannot have two congruent angles.

Answer: The sides from the largest to smallest are PQ, QR, PR The angles opposite to sides are ∠R, ∠P, ∠Q The angles from smallest to the largest are ∠Q, ∠P, ∠R

Answer: The angles from shortest to longest are ∠R, ∠T, ∠S

Question 4. A triangle has one side of length 12 inches and another side of length 20 inches. Describe the possible lengths of the third side.

Answer: The length of the third side must be greater than 8 and lesser than 32.

Explanation: Let x represent the length of the third side. Draw diagrams to help visualize the small and large values of x. Then use the Triangle Inequality Theorem to write and solve inequalities. The smallest value of x is x + 12 > 20. So, x > 8 The largest value of x is 12 + 20 > x 32 > x So, 32 > x or 32 < x The length of the third side must be greater than 8 and lesser than 32.

Decide Whether it is possible to construct a triangle with the given side lengths. Explain your reasoning.

Question 5. 4 ft, 9 ft, 10 ft

Answer: It is not possible to construct a triangle.

Explanation: According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side. So, 9 + 4 = 13 13 is not greater than 10. So, it is not possible to construct a triangle.

Question 6. 8 ft, 9 ft, 18 ft

Answer: It is possible to construct a triangle.

Explanation: According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side. 8 + 9 = 17 > 18 So, it is possible to construct a triangle.

Question 7. 5 cm, 7 cm, 12 cm

Explanation: According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle is greater than the length of the third side. 7 + 5 = 12 12 is not greater than 12 so, it is not possible to construct a triangle.

Exercise 6.5 Indirect Proof and Inequalities in One Triangle

Question 2. WRITING How can you tell which side of a triangle is the longest Irom the angle measures of the triangle? How can you tell which side is the shortest?

Answer: The angle which is opposite to the longest side is the longest angle.

In Exercises 3-6, write the first step in an indirect proof of the statement.

Question 4. If x and y are odd integers. then xy is odd.

Answer: Assume x = 1, y = 3 xy = 3

Question 6. In ∆JKL, if M is the midpoint of \(\overline{K L}\), then \(\overline{J M}\) is a median.

Answer: Assume temporarily JM is a median.

In Exercises 7 and 8, determine which two statements contradict each other. Explain your reasoning.

Question 8. (A) Both ∠X and ∠Y have measures greater than 20°. (B) Both ∠X and ∠Y have measures less than 30°. (C) m∠X + m∠Y = 62° Answer: B and C. If ∠X and ∠Y are less than 30, then  m∠X + m∠Y is not equal to 62.

In Exercises 9 and 10, use a ruler and protractor to draw the given type of triangle. Mark the largest angle and longest side in red and the smallest angle and shortest side in blue. What do you notice?

Question 10. right scalene

In Exercises 11 and 12, list the angles of the given triangle from smallest to largest.

In Exercises 13-16, list the sides of the given triangle from shortest to longest.

In Exercises 17-20, describe the possible lengths of the third side of the triangle given the lengths of the other to sides.

Question 18. 12 feet, 18 feet

Answer: The possible lengths of the third side are greater than 6 and less than 30 ft

Explanation: x + 12 > 18 x > 6 12 + 18 > x 30 > x or x < 30 The possible lengths of the third side are greater than 6 and less than 30 ft

Question 20. 25 meters, 25 meters

Answer: The possible lengths of the third side are greater than 0 and less than 50 m.

Explanation: x + 25 > 25 x > 0 25 + 25 > x 50 > x or x < 50 The possible lengths of the third side are greater than 0 and less than 50 m.

In Exercises 21-23, is it possible to construct a triangle with the given side lengths? If not, explain why not.

Question 22. 3, 6, 9

Explanation: 3 + 6 = 9 → 9 = 9 6 + 9 = 15 → 15 < 9 9 + 3 = 12 → 12 < 9 No, the sum of any two side lengths of a triangle is less than the length of the third side.

Question 24. 35, 120, 125

Answer: 35 + 120 = 155 → 155 < 125 So, No.

Answer: Number 1 is opposite to 90 degrees and 2 is opposite to 60 degrees and the remaining is √3 on the triangle.

Question 28. REASONING Your class has fewer than 30 students. The teacher divides your class into two groups. The first group has 15 students. Use indirect reasoning to show that the second group must have fewer than 15 students.

Answer: The assumption that N2 ≥ 15 is false.

Explanation: The number of students in a class N < 30 The number of students in the first group N1 = 15 The number of students in the second group N2 N = N1 + N2 Assume temporarily that the number of student in the second group has 15 or more students N2 ≥ 15 N1 + N2 ≥ 15 + 15 30 ≥ 30 So, the assumption that N2 ≥ 15 is false.

Explanation: ∠R = 180 – (65 + 56) = 180 – 121 = 59, which indicates that ∆RST i isosceles. By the trinagle longer side theorem, ST is 7

Question 32. PROOF Write an indirect proof of the statement “In ∆QRS, if m∠Q + m∠R = 90°, then m∠S = 90°.” Answer:

Question 34. CRITICAL THINKING Is it possible to decide if three side lengths form a triangle without checking all three inequalities shown in the Triangle Inequality Theorem (Theorem 6. 11)? Explain your reasoning. Answer: Given, Is it possible to decide if three side lengths form a triangle without checking all three inequalities shown in the Triangle Inequality Theorem? Yes, it is possible to decide if three side lengths form a triangle without checking all three inequalities shown in the Triangle Inequality Theorem. If the sum of the lengths of the two shortest sides is greater than the length of the longest side. then the other two inequalities will also be true.

b. How is your answer to part (a) affected if you know that m∠2 < m∠1 and m∠2 < m∠3? Answer: ∠WXY is smaller than ∠YXZ, ∠YXZ is smaller than ∠WYX, ∠WYX is smaller than ∠XZY and ∠XZY is smaller than ∠XYZ.

Question 38. MATHEMATICAL CONNECTIONS In ∆DEF, m∠D = (x + 25)°. m∠E = (2x – 4)°, and in m∠F = 63°. List the side lengths and angle measures of the triangle in order from least to greatest.

Answer: sum of angles is 180 degrees x + 25 + 2x – 4 + 63 = 180 3x + 84 = 180 3x = 96 x = 32 m∠D = 32 + 25 = 57 m∠E = 2(32) – 4 = 60 m∠F = 63 The angles from least to greatest are m∠D, m∠E, m∠F and the side lengths are EF, DF, DE.

Question 39. ANALYZING RELATIONSHIPS Another triangle inequality relationship is given by the Exterior Angle Inequality Theorem. It states:

The measures of an exterior angle of a triangle is greater than the measure of either of the nonadjacent interior angles.

MATHEMATICAL CONNECTIONS In Exercises 40 and 41, describe the possible values of x.

Explanation: JK + JL > KL x + 11 + 5x – 9 > 2x +10 6x + 2 > 2x + 10 4x > 8 x > 2 JK + KL > JL x + 11 + 2x + 10 > 5x – 9 3x + 21 > 5x – 9 30 > 2x 15 > x JL + KL > JK 5x – 9 + 2x + 10 > x + 11 7x + 1 > x + 11 6x > 10 x > \(\frac { 5 }{ 3 } \) The possible values of x are x > \(\frac { 5 }{ 3 } \) and x < 15

Question 44. USING STRUCTURE The length of the base of an isosceles triangle is l. Describe the possible lengths for each leg. Explain our reasoning. Answer: If other lengths are x and x Then, x + x > l 2x > l x > l/2 x + l > x The possible lengths of each leg are l are l/2.

Name the indicated angle between the pair of sides given.

Question 50. \(\overline{A E}\) and \(\overline{B E}\) Answer: The included angle is \(\overline{A E}\) and \(\overline{B E}\) is ∠AEB

Question 52. \(\overline{A D}\) and \(\overline{D C}\) Answer: The included angle is \(\overline{A D}\) and \(\overline{D C}\) is ∠ADC

6.6 Inequalities in Two Triangles

Comparing Measures in Triangles

a. Draw ∆ABC, as shown below. Answer:

b. Draw the circle with center C(3, 3) through the point A(1, 3). Answer:

d. Which two sides of ∆ABC are congruent to two sides of ∆DBC? Justify your answer. Answer:

e. Compare the lengths of \(\overline{A B}\) and \(\overline{D B}\). Then compare the measures of ∠ACB and ∠DCB. Are the results what you expected? Explain. Answer:

g. Look for a pattern of the measures in our table. Then write a conjecture that summarizes your observations. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures. Answer:

Question 3. Explain how you can use the hinge shown at the left to model the concept described in Question 2. Answer:

Lesson 6.6 Inequalities in Two Triangles

Use the diagram

Question 1. If PR = PS and m∠QPR > m∠QPS, which is longer, \(\overline{S Q}\) or \(\overline{R Q}\)?

Answer: Given that, PR = PS and PQ ≅ PQ by the reflexive property of congruence. m∠QPR > m∠QPS. So, two sides of △PSQ are congruent to two sides of △PRQ and included angle in △PRQ is larger. By the hinge theorem RQ > SQ.

Question 2. If PR = PS and RQ < SQ, which is larger, ∠RPQ or ∠SPQ?

Answer: Given that, PR = PS, RQ < SQ PQ ≅ PQ by the reflexive property of congruence theorem. So, two sides of △PSQ are congruent to two sides of △PRQ and third side of △PSQ is longer. By the converse of the hinge theorem, ∠SPQ > ∠RPQ.

Question 3. Write a temporary assumption you can make to prove the Hinge Theorem indirectly. What two cases does that assumption lead to?

Answer: Hinge theorem states that if two sides of the triangle are congruent to two sides of another triangle and the included angle of the first is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second triangle.

Question 4. WHAT IF? In Example 5, Group C leaves camp and travels 2 miles due north. then turns 40° towards east and travels 1.2 miles. Compare the distances from camp for all three groups.

Exercise 6.6 Inequalities in Two Triangles

Question 2. COMPLETE THE SENTENCE In ∆ABC and ∆DEF, \(\overline{A B} \cong \overline{D E}\), \(\overline{B C} \cong \overline{E F}\). and AC < DF. So m∠_______ > m ∠ _______ by the Converse of the Hinge Theorem

Answer: In ∆ABC and ∆DEF, \(\overline{A B} \cong \overline{D E}\), \(\overline{B C} \cong \overline{E F}\). and AC < DF. So m∠E > m ∠B by the Converse of the Hinge Theorem

In Exercises 3-6, copy and complete the statement with <, >, or = Explain your reasoning.

Answer: m∠1 < m∠2 by the converse of the hinge theorem.

Answer: m∠1 > m∠2 by the converse of the hinge theorem.

In Exercises 7-10. copy and complete the statement with <, >, or =. Explain your reasoning.

Answer: MN < KL by the converse of the hinge theorem.

Answer: AC > Dc by the converse of the hinge theorem.

PROOF In Exercises 11 and 12, write a proof.

Answer: Given that, AD ≅ BC AC ≅ AC by the reflexive property of congruence theorem DC < AB is given So, m∠BCA > m∠DAC by the hinge theorem.

In Exercises 13 and 14, you and your friend leave on different flights from the same airport. Determine which flight is farther from the airport. Explain your reasoning.

Question 14. Your flight: Flies 21omiles due south, then turns 70° toward west and flies 80 miles. Friend’s flight: Flies 80 miles due north, then turns 50° toward east and flies 210 miles.

Answer: My friend flight is farther from the airport.

Answer: (D) 35°

Explanation: KL ≅ KJ is given KM ≅ KM by the reflexive property of congruence LM > MJ So, m∠LKM > m∠MKJ 25 > 35

Question 18. ABSTRACT REASONING In ∆EFG, the bisector of ∠F intersects the bisector of ∠G at point H. Explain why \(\overline{F G}\) must be longer than \(\overline{F H}\) or \(\overline{H G}\).

Answer: The sum of interior angles of a triangle is 180 degrees The sum of angles F and G cannot be larger than 180 degrees ∠HFG is half of the measurement of ∠F. ∠HGF is half of the measurement of ∠G. thus, ∠HFG + ∠HGF < 90° ∠FHG = 180° – ∠HGF – ∠HFG > 90°

MATHEMATICAL CONNECTIONS In Exercises 20 and 21, write and solve an inequality for the possible values of x.

Answer: BC ≅ CD AC ≅ AC by the reflexive property of congruence theorem AB < AD So, m∠ACB < m∠ACD by the converse of the hinge theorem.

Question 24. THOUGHT PROVOKING The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, state an inequality involving the sum of the angles of a triangle. Find a formula for the area of a triangle in spherical geometry.

Maintaining Mathematical proficiency

Find the value of x.

Answer: As two sides are congruent, two angles opposite to the congruent sides are also congruent by the converse of the isosceles trinagle theorem. 36 + x + x = 180 36 + 2x = 180 2x = 144 x = 72

Relationships Within Triangles Chapter Review

Answer: DC = 20

Explanation: AD = CD 20 = 7x – 15 7x = 35 x = 5 DC = 7(5) – 15 = 20

Answer: RS = 23

Explanation: PS = SR 6x + 5 = 9x – 4 9x – 6x = 9 3x = 9 x = 3 RS = 9(3) – 4 = 23

Answer: m∠JFH = 47°

Explanation: m∠JFH = m∠JFG So, m∠JFH = 47°

Question 4. T(- 6, – 5), U(0, – 1), V(0, – 5)

Answer: The circumcenter is (-3, -3)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let T(- 6, – 5), U(0, – 1), V(0, – 5) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then PT = PU = PV PT² = PU² = PV² PT² = PU² (x + 6)² + (y + 5)² = (x – 0)² + (y + 1)² x² + 12x + 36 + y² + 10y + 25 = x² + y² + 2y + 1 12x + 8y + 61 = 1 12x + 8y + 60 = 0 3x + 2y + 15 = 0 PU² = PV² (x – 0)² + (y + 1)² = (x – 0)² + (y + 5)² x² + y² + 2y + 1 = x² + y² + 10y + 25 8y + 24 = 0 8y = -24 y = -3 Substitute y = -3 in 3x + 2y + 15 = 0 3x + 2(-3) + 15 = 0 3x – 6 + 15 = 0 3x = -9 x = -3 The circumcenter is (-3, -3)

Question 5. X(- 2, 1), Y(2, – 3), Z(6, – 3)

Answer: The circumcenter is (4, 3)

Explanation: Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let X(- 2, 1), Y(2, – 3), Z(6, – 3) be the vertices of the given triangle and let P(x, y) be the circumcentre of this triangle. Then PX = PY = PZ PX² = PY² = PZ² PX² = PY² (x + 2)² + (y – 1)² = (x – 2)² + (y + 3)² x² + 4x + 4 + y² – 2y + 1 = x² – 4x + 4 + y² + 6y + 9 8x – 8y + 5 = 13 8x – 8y – 8 = 0 x – y = 1 PY² = PZ² (x – 2)² + (y + 3)² = (x – 6)² + (y + 3)² x² – 4x + 4 + y² + 6y + 9 = x² – 12x + 36 + y² + 6y + 9 8x = 32 x = 4 Substitue x = 4 in x – y = 1 4 – y = 1 -y = -3 y = 3 The circumcenter is (4, 3)

Question 6. Point D is the incenter of ∆LMN. Find the value of x.

Answer: x = 13

Question 7. A(- 10, 3), B(- 4, 5), C(- 4, 1)

Answer: The centroid is (-6, 3)

Explanation: The centroid = (\(\frac { -10 – 4 – 4 }{ 3 } \), \(\frac { 3 + 5 + 1 }{ 3 } \)) = (-6, 3)

Question 8. D(2, – 8), E(2, – 2), F(8, – 2)

Answer: The centroid is (4, -4)

Explanation: The centroid = (\(\frac { 2 + 2 + 8 }{ 3 } \), \(\frac { -8 – 2 – 2 }{ 3 } \)) = (4, -4)

Question 9. G(1, 6), H(5, 6), J(3, 1)

Answer: The orthocenter is (3, 5.2) The orthocenter lies inside the triangle.

Explanation: The slope of the line HJ = \(\frac { 1 – 6 }{ 3 – 5} \) = \(\frac { 5 }{ 2 } \) The slope of the perpendicular line = \(\frac { -2 }{ 5 } \) The perpendicular line is (y – 6) = \(\frac { -2 }{ 5 } \)(x – 1) 5(y – 6) = -2(x – 1) 5y – 30 = -2x + 2 2x + 5y – 32 = 0 — (i) The slope of GJ = \(\frac { 1 – 6 }{ 3 – 1 } \) = \(\frac { -5 }{ 2 } \) The slope of the perpendicular line = \(\frac { 2 }{ 5 } \) The equation of perpendicular line (y – 6) = \(\frac { 2 }{ 5 } \)(x – 5) 5(y – 6) = 2(x – 5) 5y – 30 = 2x – 10 2x – 5y + 20 = 0 — (ii) Equate both equations 2x + 5y – 32 = 2x – 5y + 20 10y = 52 y = 5.2 Substitute y = 5.2 in (i) 2x + 5(5.2) – 32 = 0 2x + 26 – 32 = 0 2x = 6 x = 3 The orthocenter is (3, 5.2) The orthocenter lies inside the triangle.

Question 10. K(-8, 5), L(- 6, 3), M(0, 5)

Answer: The othrocenter is (-6, -1) The orthocenter lies outside of the triangle

Explanation: The slope of LM = \(\frac { 5 – 3}{ 0 + 6} \) = \(\frac { 1 }{ 3 } \) The slope of the perpendicular line = -3 The perpendicular line is (y – 5) = -3(x + 8) y – 5 = -3x – 24 3x + y + 19 = 0 —- (ii) The slope of KL = \(\frac { 3 – 5 }{ -6 + 8 } \) = -1 The slope of the perpendicular line = \(\frac { 1 }{ 2 } \) The equation of perpendicular line (y – 5) = \(\frac { 1 }{ 2 } \)(x – 0) 2y – 10 = x —- (ii) Substitute (ii) in (i) 3(2y – 10) + y + 19 = 0 6y – 30 + y + 19 = 0 7y – 11 = 0 y = \(\frac { 11 }{ 7 } \) x = -6 The othrocenter is (-6, -1) The orthocenter lies outside of the triangle

Find the coordinates of the vertices of the midsegment triangle for the triangle with the given vertices.

Question 11. A(- 6, 8), B(- 6, 4), C(0, 4)

Answer: The midsegment of AB = (-6, 6) The midsegment of BC = (-3, 4) The midsegment of AC = (-3, 6)

Explanation: The midsegment of AB = (\(\frac { -6 – 6 }{ 2 } \), \(\frac { 8 + 4 }{ 2 } \)) = (-6, 6) The midsegment of BC = (\(\frac { -6 + 0 }{ 2 } \), \(\frac { 4 + 4 }{ 2 } \)) = (-3, 4) The midsegment of AC = (\(\frac { -6 + 0 }{ 2 } \), \(\frac { 8 + 4 }{ 2 } \)) = (-3, 6)

Question 12. D(- 3, 1), E(3, 5), F(1, – 5)

Answer: The midsegment of DE = (0, 3) The midsegment of EF = (2, 0) The midsegment of DF = (-1, -2)

Explanation: The midsegment of DE = (\(\frac { -3 + 3 }{ 2 } \), \(\frac { 1 + 5 }{ 2 } \)) = (0, 3) The midsegment of EF = (\(\frac { 3 + 1 }{ 2 } \), \(\frac { 5 – 5 }{ 2 } \)) = (2, 0) The midsegment of DF = (\(\frac { -3 + 1 }{ 2 } \), \(\frac { 1 – 5 }{ 2 } \)) = (-1, -2)

Describe the possible lengths of the third side of the triangle given the lengths of the other two sides.

Question 13. 4 inches, 8 inches

Answer: 4 < x < 12

Explanation: 4 + 8 > x 12 > x 4 + x > 8 x > 4 8 + x > 4 x > -4 4 < x < 12

Question 14. 6 meters, 9 meters

Answer: 3 < x < 15

Explanation: 6 + 9 > x 15 > x 6 + x > 9 x > 3 9 + x > 6 x > -3 3 < x < 15

Question 15. 11 feet, 18 feet

Answer: 7 < x < 29

Explanation: 11 + 18 > x 29 > x 11 + x > 18 x > 7 18 + x > 11 x > -7 7 < x < 29

Question 16. Write an indirect proof 0f the statement “In ∆XYZ, if XY = 4 and XZ = 8. then YZ > 4.”

Answer: 4 + 8 > x 12 > x 4 + x > 8 x > 4 8 + x > 4 x > -4 So, 12 > x > 4 Therefore, YZ > 4

Use the diagram.

Question 17. If RQ = RS and m∠QRT > m∠SRT, then how does \(\overline{Q T}\) Compare to \(\overline{S T}\)?

Answer: TQ > ST

Explanation: Given that, RQ = RS and m∠QRT > m∠SRT RT = RT by the reflexive property of congruence theorem TQ > ST by the converse of the hinge theorem.

Question 18. If RQ = RS and QT > ST, then how does ∠QRT compare to ∠SRT?

Answer: Given that, RQ = RS and QT > ST RT = RT by the reflexive property of congruence theorem So, m∠QRT > m∠SRT by the converse of the hinge theorem.

Relationships Within Triangles Chapter Test

In Exercise 1 and 2, \(\overline{M N}\) is a midsegment of ∆JKL. Find the value of x.

Answer: x = 6

Explanation: KL = 0.5(MN) 12 = 0.5x x = 6

Explanation: JM = ML x = 9

Find the indicated measure. Identify the theorem you use.

Answer: ST = 17

Explanation: RS = ST by the perpendicular bisector theorem 3x + 8 = 7x – 4 7x – 3x = 12 4x = 12 x = 3 ST = 7(3) – 4 = 21 – 4 = 17

Answer: WY = 32

Explanation: 6x + 2 = 9x – 13 by the angle bisector theorem 3x = 15 x = 5 WY = 6(5) + 2 = 32

Answer: BW = 20

Explanation: WC = BW = AW by the incenter theorem BW = 20

Copy and complete the statement with <, >, or =.

Question 6. AB _____ CB

Answer: AB > CB

Question 7. m∠1 _____ m∠2

Answer: m∠1 < m∠2 by the converse of the hinge theorem

Question 8. m∠MNP ________ m∠NPM

Answer: m∠MNP < m∠NPM

Question 9. Find the coordinates of the circumcenter, orthocenter, and centroid of the triangle with vertices A(0, – 2), B(4, – 2), and C(0, 6).

Answer: Circumcenter is (1.5, -0.5) orthoceter is (0, -2) centroid is (1.3, 0.7)

Question 10. Write an indirect proof of the Corollary to the Base Angles Theorem (Corollary 5.2): If ∆PQR is equilateral, then it is equiangular.

Answer: Assume for contradiction that there is an equilateral triangle namely △ABC. So, AB = BC = AC. According to the base angles theorem, if two sides of a triangle are congruent, then the angles opposite them are congruent. Since AB = AC, the base angle theorem tells that m∠B = m∠C Since AB = BC, the base angle theorem tells that m∠A = m∠C By the transitive property of equality m∠A = m∠B = m∠C, which contradicts the assumption that triangle ABC is equiangular. Thus, the corollary is true.

Answer: The area of ∆DEF = A As HG is the midsegment of ∆DEF. Area of ∆GEH = A/2

Question 12. Two hikers start at a visitor center. The first hikes 4 miles hikes due west, then turns 40° toward south and hikes 1.8 miles. The second hikes 4 miles due east, then turns 52° toward north and hikes 1 .8 miles. Which hiker is farther from the visitor center? Explain how you know.

Answer: The second hiker is farther away from the visitor center as all things being equal cos 128 < cos 140 from distance² = 4² + 1.8² — 2 x 4 x 1.8 cos (angle turned)

In Exercises 13-15, use the map.

Question 13. Describe the possible lengths of Pine Avenue.

Answer: The possible lengths of pine avenue are between 2 and 16.

Explanation: 7 + 9 > x 16 > x 7 + x > 9 x > 2 9 + x > 7 x > -2 2 < x < 16 The possible lengths of pine avenue are between 2 and 16.

Question 14. You ride your bike along a trail that represents the shortest distance from the beach to Main Street. You end up exactly halfway between your house and the movie theatre. How long is Pine Avenue? Explain.

Answer: halfway between your house and the movie theatre = 7/2 = 3.5 Pine Avenue is 3.5 mi

Question 15. A market is the same distance from your house, the movie theater, and the beach. Copy the map and locate the market.

Answer: Centroid is the market.

Relationships Within Triangles Cumulative Assessment

Answer: Definition of perpendicular bisector is used to prove the Converse of the Perpendicular Bisector Theorem.

Answer: (B) (3, 5)

Explanation: The coordinates of L(3, 8), M (1, 5), N(5, 2) Centroid = (\(\frac { 3 + 1 + 5 }{ 3 } \), \(\frac { 8 + 5 + 2 }{ 3 } \)) = (3, 5)

Answer: Draw a perpendicular bisector for AB Draw a perpendicular bisector for BC Extend those bisectors to meet at one point. By taking the point of intersection of perpendicular bisectors as radius, draw a circle that connects three vertices of the triangle.

Answer: T(0, 7), U(2, 4), V(-1, 5)

Explanation: The coordinates of Q(-3, 8), R(3, 6), S(1, 2) The midpoint of QR = T = (\(\frac { -3 +3 }{ 2 } \), \(\frac { 8 + 6 }{ 2 } \)) = (0, 7) Midpoint of RS = U = (\(\frac { 3 + 1 }{ 2 } \), \(\frac { 6 + 2 }{ 2 } \)) = (2, 4) Midpoint of SR = V = (\(\frac { 1 – 3 }{ 2 } \), \(\frac { 2 + 8 }{ 2 } \)) = (-1, 5)

b. Show that each midsegment joining the midpoints of two sides is parallel to the third side and is equal to half the length of the third side. Answer: The slope of RS = \(\frac { 2 – 6 }{ 1 – 3 } \) = 2 The slope of TV = \(\frac { 5 – 7 }{ -1 – 0 } \) = 2 As the slopes are same. The lines are parallel. TV = √(-1 – 0)² + (5 – 7)² = √1 + 4 = √5 RS = √(1 – 3)² + (2 – 6)² = √4 + 16 = √20 TV = 0.5(RS)

Question 7. A triangle has vertices X(- 2, 2), Y(1, 4), and Z(2, – 2). Your friend claims that a translation of (x, y) → (x + 2, y – 3) and a translation by a scale factor of 3 will produce a similarity transformation. Do you support our friend’s claim? Explain our reasoning.

Answer: A triangle has vertices X(- 2, 2), Y(1, 4), and Z(2, – 2). The slope of XY m1 = 1-(-2)/4-2 = 3/2 The slope of the side AB, m2 = 3-0/(1-(-1) m2 = 3/2 = 3/2 The slope of the side YZ, m3 = 2-1/(-2-4) = -1/6 The slope of the side BC m4 = 4-3/-5-1 = -1/6 The slope of the side XZ m5 = 4-(-2)/(-2-2) = 4/-4 = -1 The slope of the side AC m6 = 4-0/-5-(-1) = 4/-4 = -1 m1 = m2 m3 = m4 m5 = m6 XY || AB, YZ || BC and XZ || AC Hence ΔABC is a similar transformation of the triangle ΔXYZ.

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