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Solution Manual for Geometry, 1st Edition

Solution Manual for Geometry, 1st Edition is packed with detailed solutions to help you grasp concepts effortlessly.

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    I NSTRUCTOR ’ S S OLUTIONS M ANUAL GEX P UBLISHING S ERVICES G EOMETRY Elayn Martin-Gay University of New Orleans

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    Table of Contents Chapter 1 A Beginning of Geometry ................................................................................. 1 Chapter 2 Introduction to Reasoning and Proofs ............................................................. 30 Chapter 3 Parallel and Perpendicular Lines ..................................................................... 56 Chapter 4 Triangles and Congruence ............................................................................... 91 Chapter 5 Special Properties of Triangles...................................................................... 120 Chapter 6 Quadrilaterals................................................................................................. 156 Chapter 7 Similarity ....................................................................................................... 185 Chapter 8 Transformations............................................................................................. 216 Chapter 9 Right Triangles and Trigonometry ................................................................ 247 Chapter 10 Area ............................................................................................................. 290 Chapter 11 Surface Area and Volume............................................................................ 346 Chapter 12 Circles and Other Conic Sections................................................................ 404

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    1 Chapter 1 Section 1.2 Practice 1. Each figure is a square divided into four smaller squares, with one of the smaller squares shaded with an X in it. So, predict that the next figure is a square divided into four smaller squares, with the upper-right square shaded and marked, as sketched below. 2. a. The second number is the first plus 2, the third is the second minus 3, the fourth is the third plus 4, and the fifth is the fourth minus 5. So, predict that the sixth is the fifth plus 6, or 3. b. Each number is 1 3 of the previous number, so predict that the next number is 1 . 81 3. No. None of the angles are marked, so we may not conclude that they have the same measure. Vocabulary & Readiness Check 1.2 1. defined term 2. undefined term 3. A statement that we prove is called a theorem. 4. A statement that we accept as true but do not prove is called an axiom or a postulate. Exercise Set 1.2 2. Each figure is a light brown rectangle divided into 1 more equal-sized section than the previous figure. A sketch of the next figure is provided. 4. Each figure is a square divided into 4 different colored squares. Each figure is the previous image rotated 90° clockwise. A sketch of the next figure is provided. G stands for green, R stands for red, B stands for blue, and Y stands for yellow. 6. Each subsequent figure has 1 square added on top of the existing columns, and 1 square added to the right. The next figure should have 4 squares in the first column, 3 in the second column, 2 in the third column, and 1 in the fourth column. A sketch of the next figure is provided. 8. The figures are rectangles that alternate between tan and green. There is an X in a bottom corner of each tan rectangle, and in a top corner of each green rectangle. The X switches from left to right in figures with the same color. The next figure will be green with an X in the top right corner. A sketch of the next figure is provided. 10. The figures are green right triangles. The second figure is the first figure reflected vertically. The third figure is the second figure reflected horizontally. The next figure will be the third figure reflected vertically. A sketch of the next figure is provided. 12. Each number is 2 more than the previous number, so predict that the next two numbers are 2 and 4.

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    Chapter 1: A Beginning of Geometry ISM: Geometry 2 14. Each number has a denominator that is one larger than the denominator of the previous number when written as a fraction, so predict that the next two numbers are 1 5 and 1 . 6 16. Each number is 3 times the previous number, so predict that the next two numbers are 162 and 486. 18. Each number is half of the previous number, so predict that the next two numbers are 1 40 and 1 . 80 20. Add 1 to the first number to get the second number, and add 1 to the second number to get the third number. Subtract 1 from the third number to get the fourth number, and subtract 1 from the fourth number to get the fifth number. So, a pattern could indicate to add 1 to the fifth number to get the sixth number, and add 1 to the sixth number to get the second number. So, predict that the next two numbers are 0 and 1. 22. Yes. The figure has 3 sides that are connected, which is the requirement for being called a triangle. 24. No. The figure is a triangle because it has 3 connected sides, but no indication is provided that one of the angles is a right angle. 26. Yes. Both angles measure 37°, so they have the same measure. 28. No. No indication is provided that the segments are the same length. 30. Yes. Both angles have the middle ray as a side. 32. No. The opposite sides of the figure are not marked parallel, so it cannot be concluded that the figure is a parallelogram. 34. The student’s number is not correct. Since the number that follows 1 4 is 1 , 6 the denominator is not doubled to get the next denominator. Add 2 to the denominator to get the next denominator. Therefore, the next number in the list is 1 . 8 36. The student’s figure is not correct. In this pattern, the positions of the dot and the x are reversed to generate the next figure. Then the image is flipped vertically to generate the next figure. The correct figure should have the positions of the dot and the x reversed, as shown. 38. Neither axioms nor postulates are proven. They are accepted as true. Section 1.3 Practice 1. a. Two other ways to name EH HJJG are HE HJJG and line . n b. There are many correct answers. Two other ways to name plane M are plane GED and plane . GFD c. Points , D , E and F lie on the same line, so they are collinear. d. Points , D , E , F and G lie on the same plane, so they are coplanar. e. There are many correct answers. For instance, points , H , E and F do not lie on the same line. 2. a. The three segments are GJ or , JG JQ or , QJ and GQ or . QG b. The four rays are GJ J JJ G or , GQ JJJG , JG JJJ G , JQ JJJ G and QG J JJG or . QJ J JJ G c. The opposite rays are JG JJJ G and . JQ JJJ G 3. a. b.

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    ISM: Geometry Chapter 1: A Beginning of Geometry 3 c. d. 4. Points , W , Z and A are collinear and A is between W and . Z So, A W JJJJ G and A Z J JJ G are opposite rays. Points , Y , X and A are collinear and A is between Y and . X So, A Y JJJ G and A X JJJG are opposite rays. 5. The plane intersects in . EH HJJG 6. Vocabulary & Readiness Check 1.3 1. C. The symbol R S means a segment . R S 2. A. The symbol R S H JJ G means a line . R S 3. D. The symbol R S J JJ G means a ray . R S 4. B. The symbol SR J JG means a ray . SR 5. A line extends in opposite directions without end. 6. A plane extends in two dimensions without end. 7. A ray consists of an endpoint and all points of a line on one side of the endpoint. 8. A segment consists of two endpoints and all the points between them. 9. Coplanar points lie on the same plane. 10. Collinear points lie on the same line. 11. A geometric figure is any nonempty subset of space. 12. Opposite rays are two rays that share an endpoint and form a line. 13. A point has a location but no dimension. 14. Space is the set of all points. Exercise Set 1.3 2. , NT H JJG , TN H JJ G , TR H JG , R N H JJG , NR HJJ G and line l are all possible answers. 4. There are many correct answers. Sample answers: plane TNJ and plane TRJ 6. Points , R , N and T lie on the same line, so they are collinear. 8. Points , R , N , T and J lie on the same plane, so they are coplanar. 10. The segments are A Z or , Z A A X or , XA XY or , YX AY or YA , XZ or Z X and YZ or . Z Y 12. The rays are A Z J JJ G or A X JJJG , A Y JJJ G XZ JJJ G or , XY J JJ G , YZ J JG Z A J JG or Z X J JJ G or , Z Y JJJ G YA JJG or , YX JJJ G and . XA J JJ G 14. The opposite rays with endpoint X are XA J JJ G and XZ J JJ G or . XY J JJ G 16. The only line is BC H JJ G 18. The rays are , A B J JJ G , BC JJJ G and . CB JJJ G 20. False. XP J JJ G and XN J JJG are not opposite rays because , P , X and N are not collinear. 22. True. XL J JJ G and XN J JJG are opposite rays because XL J JJ G and XN JJJG are two rays with the same endpoints going in opposite directions and all three points are collinear.

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    Chapter 1: A Beginning of Geometry ISM: Geometry 4 24. There are many correct answers. Sample Answer: Points , A , E and B are not collinear. 26. There are many correct answers. Sample Answer: Points , W , A and X are not collinear. 28. False. Point D lies on line , l not line . m 30. True. Point C does lie on line . m 32. True. Points , D , E and B are collinear because they all lie on the same line, line . l 34. False. The fact that the points are collinear means that many planes pass through these three points, not just one. 36. The plane intersects in . VW HJJG 38. The plane intersects in . TX HJJ G 40. Sample Answer: planes TXW and TSR 42. Sample Answer: planes UXV and VWS 44. Point G 46. Point H 48. Point E 50. Point H 52. Coplanar. All four points are on the same plane. 54. Coplanar. All four points are on the same plane. 56. Noncoplanar. All four points are not on the same plane. 58. 60. 62. 64. Sometimes. This is true if JL JJG is a ray where point K is on the ray . JL JJG This is false if KL H JJ G is a line where J is on that line between K and . L Then the two rays would have the same endpoints and be going in opposite directions. 66. Sometimes. This could be true like Exercise 52 where there are four points that all are on the same plane. This could be false like Exercise 56 where there are four points and one of them is on a different plane from the other three. 68. Never. Postulate 1.3-2 states that two distinct lines intersect at exactly one point. 70. The following times can also be represented as opposite rays on a clock: 1:35, 2:40, 3:45, 4:50, 5:55, 7:05, 8:10, 9:15, 10:20, 11:25, and 12:30. 72. Yes.

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    ISM: Geometry Chapter 1: A Beginning of Geometry 5 74. No. 76. No. 78. One line can have many planes pass through it. A single line can be the intersection of multiple planes, but there is no distinct number. 80. If two lines intersect than only one plane can contain both of those lines. Postulate 1.3-3 states that if two distinct planes intersect, then they intersect in exactly one line. This means that no two planes can contain two distinct lines. Using the diagram, the plane can be labeled , A BC which refers to a single plane. Section 1.4 Practice 1. NQ is the distance between points N and Q . Use the Ruler Postulate to find NQ . The coordinate of N is 2, − and the coordinate of Q is 13. 13 ( 2) 16 16 NQ NQ NQ = − − = = 2. a. A C and DF are marked the same, so A C DF ≅ and . A C DF = A B and DE are marked the same, so A B DE ≅ and . A B DE = BC and EF are marked the same, so BC EF ≅ and . BC EF = b. , A C DF = so if 10 AC = cm, 10 DF = cm. c. , CB FE = so if 25 CB = cm, 25 FE = cm. 3. If BC GD = , then . BC GD ≅ ( 1) ( 6) 5 5 8 3 5 5 BC GD = − − − = = = − = = Yes, , BC GD = so . BC GD ≅ 4. Use the Segment Addition Postulate and algebra to find x . (12 3) (8 2) 79 20 1 79 20 80 20 80 20 20 4 QR RS QS x x x x x x + = − + + = − = = = = Use the value of x to find QR and RS . 12 3 12(4) 3 48 3 45 QR x = − = − = − = 8 2 8(4) 2 34 RS x = + = + = 5. Use the meaning of the midpoint to find the value of x . 11 8 10 1 8 1 7 QP PR x x x x = − = − − = − =

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    Chapter 1: A Beginning of Geometry ISM: Geometry 6 Now, use the value of x to find QP and PR , which should be the same. 11 8 11(7) 8 77 8 69 QP x = − = − = − = 10 1 10(7) 1 70 1 69 PR x = − = − = − = To find QR , add. 69 69 138 QR QP PR = + = + = Vocabulary & Readiness Check 1.4 1. Two segments that have the same length are called congruent segments. 2. The real number that corresponds to a point is called the coordinate of the point. 3. Given three distinct points on a line, one of the points will always be between the other two. 4. A point that divides a segment into two congruent segments is called a midpoint. 5. If two segments are congruent, then their lengths are equal. 6. The distance between two points is the absolute value of the difference of their coordinates. 7. A line, ray, segment, or plane that intersects a segment at its midpoint is called a segment bisector. 8. In geometry, to bisect a segment means to divide it into two equal parts. Exercise Set 1.4 2. Yes; N is between Q and T because N appears on . QT 4. No; M is not between N and P because M does not appear on . NP 6. No; P is not between R and T because P is not on . R T 8. No; N is not between R and Q because N does not appear on . R Q 10. Yes; Q is between R and T because Q appears on . R T 12. AB is the distance between points A and B . Use the ruler postulate to find AB . 6 ( 8) 2 2 AB AB AB = − − − = = 14. AD is the distance between points A and D . Use the ruler postulate to find AD . 3 ( 8) 11 11 AD AD AD = − − = = 16. VX and UW are marked the same, so VX UW ≅ and , VX UW = which means since 105 VX = cm, then 105 UW = cm. 18. DE and FG are marked the same, so DE FG ≅ and , DE FG = which means since 7 DE = in., then 7 FG = in. 20. QS and R T are marked the same, so QS RT ≅ and , QS RT = which means since 33 QS = ft, then 33 RT = ft. 22. If , MP NQ = then . MP NQ ≅ 6 ( 3) 9 9 12 3 9 9 MP NQ = − − = = = − = = Yes, , MP NQ = so . MP NQ ≅ 24. If , LP MQ = then . LP MQ ≅ 6 ( 8) 14 14 12 ( 3) 15 15 LP MQ = − − = = = − − = = No, , LP MQ ≠ so LP is not congruent to . MQ 26. If 15 ST = and 40, RT = then 40 15 25. RS = − =

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    ISM: Geometry Chapter 1: A Beginning of Geometry 7 28. a. Use the Segment Addition Postulate and algebra to find x . (14 8) (9 10) 232 23 2 232 23 230 23 230 23 23 10 R S ST RT x x x x x x + = − + + = + = = = = b. Use the value of x to find RS and ST . 14 8 14(10) 8 140 8 132 RS x = − = − = − = 9 10 9(10) 10 90 10 100 ST x = + = + = + = 30. a. Use the meaning of the midpoint to find the value of y . 9 25 4 5 25 0 5 25 5 25 5 5 5 y y y y y y − = − = = = = b. Use the value of y to find QC and CR , which should be the same. 9 25 9(5) 25 45 25 20 QC y = − = − = − = 4 4(5) 20 CR y = = = To find QR, add. 20 20 40 QR QC CR = + = + = 32. PT and TQ are marked the same, so PT TQ ≅ and . PT TQ = Use this to solve for x . 6 35 2 3 4 32 4 32 4 4 8 PT TQ x x x x x = − = − = = = Use x to find . PT 6 35 6(8) 35 48 35 13 PT x = − = − = − = 34. PT and TQ are marked the same, so PT TQ ≅ and . PT TQ = Use this to solve for x . 7 24 6 2 22 PT TQ x x x = − = − = Use x to find . PT 7 24 7(22) 24 154 24 130 PT x = − = − = − = 36. Find ZX and WY . 7 1 8 8 ZX = − − = − = 5 ( 3) 8 8 WY = − − = = Because , Z X WY = . Z X WY ≅ 38. You are given that 7. AT = Because , DEC ∠ the coordinate of T must be either 7 or 7. − The midpoint is halfway between these coordinates. Take half of the distance from zero. 7 0 1 3 , 2 2 − = so the coordinate of the midpoint is either 1 3 2 or 1 3 . 2 − 40. The coordinate of G must either be 6 units greater than or 6 units less than the coordinate of P . 10 6 16 10 6 4 + = − = The possible coordinates of G are 4 and 16. 42. ED and DB are marked the same, so ED DB ≅ and . ED DB = Use this to solve for x . 4 3 8 2 4 8 2 12 2 12 2 2 6 ED DB x x x x x x = + = − − + = − − = − − − = − − =

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    Chapter 1: A Beginning of Geometry ISM: Geometry 8 Use x to find ED and DB . They should be the same. 4 6 4 10 ED x = + = + = 3 8 3(6) 8 18 8 10 DB x = − = − = − = Add ED and DB to find EB . 10 10 20 EB ED DB = + = + = 44. Mile markers for Macclenny and Tallahassee are shown, so take the absolute value of the difference to find the distance. 199 335 136 136 − = − = Divide the distance by the number of miles per hour to find the time. 136 20 2 58 58 = hours, or 2 hours and 20 20 29 minutes. 46. The highway sign gives the number of miles away each town is from the sign. If your friend is reading the sign, then your friend is the same distance away from the town as the sign. Because 80 is shown on the sign for Watertown; Watertown is 80 miles away, so your friend is incorrect. 48. a. Add GJ and HK , but because they overlap, you will need to subtract the distance that they share, which is JH . (2 3) (4 3) 5 GK GH HJ HK GK x x x GK x = − + = + − + − = b. Solve for x using 30. GK = 5 30 5 30 5 5 5 6 GK x x x x = = = = Subtract HJ from GJ to find GH . (2 3) 3 6 3 9 GH GJ HJ x x x = − = + − = + = + = Subtract HJ from HK to find JK . (4 3) 3 3 3(6) 3 18 3 15 JK HK HJ x x x = − = − − = − = − = − = 50. The mile markers are like numbers on a number line. The number at each marker is the distance from the start of the road. This means that you can subtract the number on a mile marker from the number on another mile marker and take the absolute value to get the distance between the markers, just like finding the distance between two coordinates on a number line. Section 1.5 Practice 1. a. , NPQ ∠ QPN ∠ b. QPM ∠ 2. 180 165 15 , m TQN ∠ = − ° = ° acute angle 180 45 135 , m TQM ∠ = − ° = ° obtuse angle 180 0 180 m TQR ∠ = − ° = ° , straight angle 3. a. , A D ∠ ≅ ∠ so m A m D ∠ = ∠ . , C F ∠ ≅ ∠ so . m C m F ∠ = ∠ b. 25 , m A ∠ = ° so 25 . m D ∠ = ° c. 120 , m F ∠ = ° so 120 . m C ∠ = ° 4. 360 60 6 x ° = = ° 5. Use the Angle Addition Postulate and solve for x. (4 10) (11 10) 180 15 180 12 m PQR m RQS m PQS x x x x ∠ + ∠ = ∠ − + + = = = Substitute the value of x to find . m PQR ∠ (4 10) (4 12 10) 38 m PQR x m PQR m PQR ∠ = − ° ∠ = ⋅ − ° ∠ = ° Substitute the value of x to find . m RQS ∠ (11 10) (11 12 10) 142 m RQS x m RQS m RQS ∠ = + ° ∠ = ⋅ + ° ∠ = °

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    ISM: Geometry Chapter 1: A Beginning of Geometry 9 Vocabulary & Readiness Check 1.5 1. An instrument used to measure angles in degrees is called a protractor. 2. Two angles that have the same measure are called congruent angles. 3. An angle consists of two different rays with a common endpoint. 4. The rays of an angle are also called the sides of the angle. 5. The common endpoint of the rays of an angle is called the vertex of the angle. 6. If two angles are congruent, then their measures are equal. 7. An obtuse angle has a degree measure between 90º and 180º. 8. An acute angle has a degree measure between 0º and 90º. 9. A straight angle has a degree measure equal to 180º. 10. A right angle has a degree measure of 90º. Exercise Set 1.5 2. , 1, , A BC CBA B ∠ ∠ ∠ ∠ 4. , 3, , MKL LKM K ∠ ∠ ∠ ∠ 6. 90 0 90 ; DAF ∠ = − ° = ° right angle 8. 25 0 25 ; BAC ∠ = − ° = ° acute angle 10. 90 70 20 , DAE ∠ = − ° = ° acute angle 12. Sample of an acute angle, . GHJ ∠ 14. Sample of a right angle, . QRS ∠ 16. FJH BJA ∠ ≅ ∠ 18. If 130 , m GHF ∠ = ° then 130 . m JBC ∠ = ° 20. A clock face is a circle which is divided into 12 equal slices. Each slice represents an angle measuring 360 30 . 12 ° = ° The angle measure of the clock hands at 5:00 is equal to 5 30 150 . ⋅ ° = ° 22. A clock face is a circle which is divided into 12 equal slices. Each slice represents an angle measuring 360 30 . 12 ° = ° The angle measure of the clock hands at 11:00 is equal to 1 30 30 . ⋅ ° = ° 24. 360 72 5 y ° = = ° 26. 360 30 12 x ° = = ° 28. Use the Angle Addition Postulate and solve for x . 45 (6 2) (2 1) 45 8 3 45 8 48 6 m YNZ m ZNM m YNM x x x x x ∠ + ∠ = ∠ = ° − + − = − = = = Substitute the value of x to solve for . m YNZ ∠ (6 2) (6 6 2) 34 m YNZ x m YNZ m YNZ ∠ = − ° ∠ = ⋅ − ° ∠ = ° Substitute the value of x to solve for . m ZNM ∠ (2 1) (2 6 1) 11 m ZNM x m ZNM m ZNM ∠ = − ° ∠ = ⋅ − ° ∠ = °

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    Chapter 1: A Beginning of Geometry ISM: Geometry 10 30. Use the Angle Addition Postulate and solve for x . 180 (5 3) (3 23) 180 8 20 180 8 160 20 m ABD m DBC m ABC x x x x x ∠ + ∠ = ∠ = ° − + + = + = = = Substitute the value of x to solve for . m ABD ∠ (5 3) (5 20 3) 97 m ABD x m ABD m ABD ∠ = − ° ∠ = ⋅ − ° ∠ = ° Substitute the value of x to solve for . m DBC ∠ (3 23) (3 20 23) 83 m DBC x m DBC m DBC ∠ = + ° ∠ = ⋅ + ° ∠ = ° 32. Use the Angle Addition Postulate and solve for x . 90 (8 1) ( 8) 90 9 9 90 9 81 9 m MRJ m JRK m MRK x x x x x ∠ + ∠ = ∠ = ° + + + = + = = = Substitute the value of x to solve for . m MRJ ∠ (8 1) (8 9 1) 73 m MRJ x m MRJ m MRJ ∠ = + ° ∠ = ⋅ + ° ∠ = ° Substitute the value of x to solve for . m JRK ∠ ( 8) (9 8) 17 m JRK x m JRK m JRK ∠ = + ° ∠ = + ∠ = ° 34. The measure of the angle is 90 ° and it is a right angle. 36. The measure of the angle is 88 ° and it is an acute angle. 38. 28 , m AOB m COD ∠ = ∠ = ° (3 2) , m BOC x ∠ = − ° and 6 . m AOD x ∠ = ° Use the Angle Addition Postulate and solve for x . 28 (3 2) 28 6 3 54 18 m AOB m BOC m COD m AOD x x x x ∠ + ∠ + ∠ = ∠ + − + = = = Substitute the value of x to solve for . m BOC ∠ (3 2) 52 m BOC x ∠ = − ° = ° Substitute the value of x to solve for . m AOD ∠ 6 108 m AOD x ∠ = ° = ° Check your work by adding the interior angles to be sure that their sum equals . m AOD ∠ m AOB m BOC m COD m AOD ∠ + ∠ + ∠ = ∠ 28 52 28 108 m AOD ° + ° + ° = ° = ∠ 40. The clock face is a circle with 12 equal intervals. Each interval represents 360 30 . 12 ° = ° The minute hand moves 30° every 5 minutes and the hour hand moves 30° every hour. At 8:40, the minute hand is on the 8 and the hour hand has moved 2 3 of the way from 8 to 9, so the clock hands are 2 3 of an interval apart. Since each interval represents 30°, at 8:40 the angle formed by the clock hands is 2 30 20 . 3 ⋅ ° = ° 42. The clock face is a circle with 12 equal intervals. Each interval represents 360 30 . 12 ° = ° The minute hand moves 30° every 5 minutes and the hour hand moves 30° every hour. At 4:30, the minute hand is on the 6 and the hour hand has moved 1 2 of the way from 4 to 5, so the hands are 1.5 intervals apart. Since each interval represents 30°, at 4:30 the angle formed by the clock hands is 1 30 30 45 . 2 ⋅ ° + ° = ° 44. There are two ways to sketch 135 . PVB ∠ = °

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    ISM: Geometry Chapter 1: A Beginning of Geometry 11 Section 1.6 Practice 1. a. True. 8 ∠ and 5 ∠ are two adjacent angles, and their noncommon sides form opposite rays. b. True. 8 ∠ and 6 ∠ have sides that form opposite rays. c. False. 7 ∠ and 5 ∠ are not adjacent angles. d. False. 7 ∠ and 6 ∠ do not have sides that form opposite rays. 2. Find any angle pairs with measures that have a sum of 90°. 8 ∠ and 7 ∠ are complementary. 7 ∠ and 6 ∠ are complementary. 6 ∠ and 5 ∠ are complementary. 8 ∠ and 5 ∠ are complementary. 3. Find any angle pairs with measures that have a sum of 180°. 5 ∠ and 6 ∠ are supplementary. 6 ∠ and 7 ∠ are supplementary. 7 ∠ and 8 ∠ are supplementary. 8 ∠ and 5 ∠ are supplementary. 4. a. Since ∠ M and ∠ C are supplementary, the sum of their measures is 180 . ° 180 16 180 164 m M m C m M m M ∠ + ∠ = ° ∠ + ° = ° ∠ = ° b. Since ∠ N and ∠ C are complementary, the sum of their measures is 90 . ° 90 16 90 74 ∠ + ∠ = ° ∠ + ° = ° ∠ = ° m N m C m N m N 5. a. Since , m LKM m MKN ∠ = ∠ ( ) . 72 2 36 ∠ = ∠ + ∠ ∠ = ∠ + ∠ ° = ∠ ° = ∠ m LKN m LKM m MKN m LKN m LKM m LKM m LKM m LKM b. Since , m LKM m MKN ∠ = ∠ 36 . m MKN ∠ = ° 6. Since HL J JJ G bisects , KHJ ∠ . m KHL m LHJ ∠ = ∠ Solve for x . ( ) . 3 5 46 2 46 23 ∠ = ∠ ° = − ° − = − = m KHL m LHJ x x x x Use substitution to find the measure of each angle. ( ) 3 3 23 69 m KHL x ∠ = ° = ° = ° ( ) ( ) ( ) 5 46 5 23 46 115 46 69 m LHJ x ∠ = − ° = ⋅ − ° = − ° = ° 7. DEA ∠ and B EA ∠ form a linear pair, so the sum of their measures is 180 . ° ( ) ( ) 180 5 6 2 46 180 7 40 180 7 140 20 m DEA m BEA x x x x x ∠ + ∠ = ° − ° + + ° = ° + = = = DEC ∠ and BEC ∠ form a linear pair. ( ) ( ) 180 2 56 6 4 180 8 60 180 8 120 15 m DEC m BEC y y y y y ∠ + ∠ = ° + ° + + ° = ° + = = = Use substitution to find the measure of each angle. ( ) ( ) ( ) 5 6 5 20 6 100 6 94 m DEA x ∠ = − ° = ⋅ − ° = − ° = ° ( ) ( ) ( ) 2 46 2 20 46 40 46 86 m BEA x ∠ = + ° = ⋅ + ° = + ° = °

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    Chapter 1: A Beginning of Geometry ISM: Geometry 12 ( ) ( ) ( ) 2 56 2 15 56 30 56 86 m DEC y ∠ = + ° = ⋅ + ° = + ° = ° ( ) ( ) ( ) 6 4 6 15 4 90 4 94 m BEC y ∠ = + ° = ⋅ + ° = + ° = ° Vocabulary & Readiness Check 1.6 1. Two angles are complementary angles if their measures have a sum of 90°. 2. Two angles are supplementary angles if their measures have a sum of 180°. 3. Two angles are vertical angles if their sides form opposite rays. 4. Two angles are adjacent angles if they share a common side and a common vertex, but have no interior points in common. 5. Two adjacent angles form a linear pair if their noncommon sides are opposite rays. 6. A ray that divides an angle into two adjacent congruent angles is called an angle bisector. Exercise Set 1.6 2. 3 ∠ and 1 ∠ are vertical angles. 4. True. 2 ∠ and 1 ∠ are adjacent angles, and their noncommon sides form opposite rays. 6. False. 4 ∠ and 2 ∠ are not adjacent angles. 8. False. 3 ∠ and 5 ∠ do not have sides that form opposite rays. 10. False. 1 ∠ and 2 ∠ do not have measures that add up to 180°. 12. COE ∠ shares side OE and is also a right angle. 14. COD ∠ and EOD ∠ form right angle . COE ∠ 16. A OD ∠ is vertical to . BOC ∠ 18. False. 1 4 90 m m ∠ + ∠ ≠ ° 20. True. 1 3 90 m m ∠ + ∠ = ° 22. True. 3 4 90 m m ∠ + ∠ = ° 24. False. 3 4 180 m m ∠ + ∠ ≠ ° 26. True. 4 1 180 m m ∠ + ∠ = ° 28. The conclusion cannot be made from the given information. There is no indication that A C J JJG is an angle bisector. 30. The conclusion can be made from the given information, assuming that JD is a line segment. Then JCA ∠ and A CD ∠ form a linear pair, which means the angles are supplementary, and the sum of their measures is 180°. 32. The conclusion cannot be made from the given information. There is no indication that . JC DC ≅ 34. The conclusion can be made from the given information, assuming DE JJJG and JF JJJ G are straight rays. Then, E AF ∠ and JAD ∠ have sides that form opposite rays. 36. A CD ∠ and A CF ∠ are a linear pair, so they are supplementary, and it is given that the sum of their measures is 180°. So, 180 2 90 . ∠ = ° ÷ = ° m ACD m ACD m ACB m BCD ∠ − ∠ = ∠ . Therefore, 90 65 25 m BCD ° − ° = ∠ = ° 38. From Exercise 36, 25 , m BCD ∠ = ° and it is given that . m BCD m ECF ∠ = ∠ , m ECF m ACF m ACE ∠ + ∠ = ∠ so 25 90 115 . m ACE ° + ° = ∠ = ° 40. Supplement: 180 70 110 ° − ° = ° Complement: 90 70 20 ° − ° = ° 42. Supplement: 180 1 179 ° − ° = ° Complement: 90 1 89 ° − ° = ° 44. Supplement: 180 175 5 ° − ° = ° Complement: There is no compliment because 175 90 . ° > °

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    ISM: Geometry Chapter 1: A Beginning of Geometry 13 46. 29 29 58 58 m GJH m DJG m DJH m DJH ∠ + ∠ = ∠ ° + ° = ° ∠ = ° 48. 36 m SVT m RVS m SVT ∠ = ∠ ∠ = ° 50. a. ( ) ( ) 8 5 9 13 5 13 8 m AQB m GQB x x x x ∠ = ∠ − ° = − ° − = − = b. ( ) ( ) 8 5 8 8 5 59 m AQB x ∠ = − ° ⋅ − ° = ° c. ( ) ( ) 9 13 9 8 13 59 m BQG x ∠ = − ° ⋅ − ° = ° d. 59 59 118 m AQG m AQB m BQG ∠ = ∠ + ∠ = ° + ° = ° 52. Solve for x . ( ) 4 12 24 24 4 12 48 4 60 15 m ABC m ABD m CBD x x x x ∠ = ∠ + ∠ − ° = ° + ° − = = = Solve for ∠ m ABC using the value of x . ( ) ( ) ( ) 4 12 4 15 12 60 12 48 m ABC x ∠ = − ° = ⋅ − ° = − ° = ° 54. Solve for x . ( ) ( ) 3 20 6 16 20 3 16 36 3 12 m ABD m CBD x x x x x ∠ = ∠ + ° = − ° = − = = Solve for ∠ m ABC using the value of x . ( ) ( ) ( ) ( ) ( ) 3 20 6 16 9 4 9 12 4 108 4 112 ∠ = ∠ + ∠ = + ° + − ° = + ° = ⋅ + ° = + ° = ° m ABC m ABD m CBD x x x 56. a. Since ∠ EFG and ∠ GFH form a linear pair, they are supplementary, and the sum of their measures is 180 . ° ( ) ( ) 180 2 21 4 15 180 6 36 180 6 144 24 ∠ + ∠ = ° + ° + + ° = ° + = = = m EFG m GFH n n n n n b. ( ) ( ) ( ) 2 21 2 24 21 48 21 69 m EFG n ∠ = + ° = ⋅ + ° = + ° = ° ( ) ( ) ( ) 4 15 4 24 15 96 15 111 m GFH n ∠ = + ° = ⋅ + ° = + ° = ° c. 180 69 111 180 m EFG m GFH ∠ + ∠ = ° ° + ° = ° 58. Find x . ( ) ( ) 180 2 4 7 4 180 9 180 20 m FGJ m HGJ x x x x ∠ + ∠ = ° + ° + − ° = ° ° = ° = Substitute the value of x to find ∠ m FGJ and . m HGJ ∠ ( ) ( ) ( ) 2 4 2 20 4 40 4 44 m FGJ x ∠ = + ° = ⋅ + ° = + ° = ° ( ) ( ) ( ) 7 4 7 20 4 140 4 136 m HGJ x ∠ = − ° = ⋅ − ° = − ° = ° Find y . ( ) ( ) 180 3 46 2 16 180 5 30 180 5 150 30 m FGK m HGK y y y y y ∠ + ∠ = ° + ° + − ° = ° + = = =

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    Chapter 1: A Beginning of Geometry ISM: Geometry 14 Substitute the value of y to find ∠ m FGK and . m HGK ∠ ( ) ( ) ( ) 3 46 3 30 46 90 46 136 m FGK y ∠ = + ° = ⋅ + ° = + ° = ° ( ) ( ) ( ) 2 16 2 30 16 60 16 44 m HGK y ∠ = − ° = ⋅ − ° = − ° = ° 60. Find x . ( ) ( ) 180 3 2 26 180 4 24 180 4 204 51 m LWT m TWZ x x x x x ∠ + ∠ = ° + ° + − ° = ° − = = = Substitute the value of x to find ∠ m LWT and . m TWZ ∠ ( ) ( ) ( ) 3 2 3 51 2 153 2 155 m LWT x ∠ = + ° = ⋅ + ° = + ° = ° ( ) ( ) 26 51 26 25 m TWZ x ∠ = − ° = − ° = ° Find y . ( ) ( ) 180 2 1 12 11 180 14 12 180 14 168 12 m LWV m VWZ y y y y y ∠ + ∠ = ° + ° + + ° = ° + = = = Substitute the value of y to find ∠ m LWV and . m VWZ ∠ ( ) ( ) ( ) 2 1 2 12 1 24 1 25 m LWV y ∠ = + ° = ⋅ + ° = + ° = ° ( ) ( ) ( ) 12 11 12 12 11 144 11 155 m VWZ y ∠ = + ° = ⋅ + ° = + ° = ° 62. Let the angles be A ∠ and , B ∠ where ( ) 20 . m B m A ∠ = ∠ − ° Since A ∠ and B ∠ are complementary, the sum of their measures is 90 . ° Find . m A ∠ ( ) ( ) ( ) 90 20 90 2 20 90 2 110 55 m A m B m A m A m A m A m A ∠ + ∠ = ° ∠ + ∠ − = ° ∠ − = ° ∠ = ° ∠ = ° Find . m B ∠ 90 55 90 35 m A m B m B m B ∠ + ∠ = ° ° + ∠ = ° ∠ = ° 64. The first student is correct. The correct equation is 1 . 2 m ABX m ABC ∠ = ∠ The angle bisector splits A BC ∠ in half to make ABX ∠ and . CBX ∠ The second student’s equation states that one of the halves is twice as big as . A BC ∠ 66. For this to occur, m A m B ∠ = ∠ and 180 m A m B ∠ + ∠ = ° ( ) 180 180 2 180 90 , 90 m A m B m A m B m A m A m A m A m B ∠ = ∠ ∠ + ∠ = ° ∠ + ∠ = ° ∠ = ° ∠ = ° ∠ = ° The only instance where vertical angles are supplementary occurs when the angles are right angles. 68. Two angles are supplementary if the sum of their measures is 180 . ° All of the acute angles in the figure are supplementary to . JQM ∠ 70. Two adjacent angles form a linear pair if their noncommon sides are opposite rays. , LMN ∠ , QMP ∠ and NMP ∠ form linear pairs with . LMQ ∠

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    ISM: Geometry Chapter 1: A Beginning of Geometry 15 Section 1.7 Practice 1. Use the Midpoint Formula , 2 a b M + = where a = –9 and b = 4. 2 9 4 2 5 2 2.5 + = − + = − = = − a b M 2. Let ( ) ( ) 1 1 , 5, 2 , x y P = − and ( ) ( ) 2 2 , 8, 6 . x y Q = − x -coordinate of M : 2 5 8 2 13 2 6.5 + = + = = = a b M y -coordinate of M : ( ) 2 2 6 2 8 2 4 + = − + − = − = = − a b M The midpoint of segment PQ is ( ) 6.5, 4 . M − 3. Let the coordinates of B be ( ) 2 2 , . x y ( ) 2 2 3 5 4, 9 , 2 2 x y − + − + ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ 2 : x 2 2 2 3 4 , 2 8 3 11 x x x − + = = − + = and 2 : y 2 2 2 5 9 2 18 5 13 y y y − + − = − = − + − = The coordinates of B are ( ) 11, 13 . − 4. Let the coordinates of P be ( ) 1 1 , x y and the coordinates of Q be ( ) 2 2 , . x y Use the Distance Formula. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 3 3 7 1 4 1 16 17 4.1 d x x y y = − + − ⎡ ⎤ = − − − + − ⎣ ⎦ = + − = + = ≈ Vocabulary & Readiness Check 1.7 1. The midpoint of a line segment is a point exactly halfway between the two endpoints of the line segment. 2. The Distance Formula is ( ) ( ) 2 2 2 1 2 1 . d x x y y = − + − 3. The Midpoint Formula for a segment on the coordinate plane is 1 2 1 2 , . 2 2 x x y y M + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 4. The Midpoint Formula for a segment on a number line is . 2 a b M + = Exercise Set 1.7 2. 9 6 3 1.5 2 2 2 + − + − = = = = − a b M 4. ( ) 8 12 20 10 2 2 2 − + − + − = = = = − a b M 6. ( ) 1 2 1 2 , 2 2 3 7 9 11 , 2 2 10 20 , 2 2 5,10 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = x x y y M

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    Chapter 1: A Beginning of Geometry ISM: Geometry 16 8. ( ) ( ) 1 2 1 2 , 2 2 4 8 3 6 , 2 2 3 12 , 2 2 1.5, 6 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ − + − − + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − x x y y M 10. ( ) 1 2 1 2 , 2 2 2 2 5 15 , 2 2 0 20 , 2 2 0,10 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = x x y y M 12. 1 2 1 2 , 2 2 x x y y M + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 2 2 7 4 5 5 15 15 , 2 2 4 3 5 15 , 2 2 4 3 , 10 30 2 1 , 5 10 M ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − + − + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ 14. Let the coordinates of S be ( ) 2 2 , . x y ( ) 2 2 5 15 5, 8 , 2 2 x y + − + ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ 2 : x 2 2 2 5 5 , 2 10 5 5 x x x + = = + = and 2 : y 2 2 2 15 8 2 16 15 1 y y y − + − = − = − + − = The coordinates of S are ( ) 5, 1 . − 16. Let the coordinates of S be ( ) 2 2 , . x y ( ) 2 2 2 8 5, 8 , 2 2 x y − + + ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ 2 : x 2 2 2 2 5 , 2 10 2 12 x x x − + = = − + = and 2 : y 2 2 2 8 8 2 16 8 24 y y y + − = − = + − = The coordinates of S are ( ) 12, 24 . − 18. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 8 10 14 14 18 0 324 0 324 18 = − + − = − − + − = − + = + = = d x x y y 20. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 0 0 12 3 0 9 0 81 81 9 = − + − = − + − = + = + = = d x x y y 22. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 5 12 12 12 7 24 49 576 625 25 = − + − ⎡ ⎤ = − + − − ⎣ ⎦ = − + = + = = d x x y y 24. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 8 12 18 6 20 12 400 144 544 4 34 23.3 = − + − = − − + − = − + = + = = ≈ d x x y y

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    ISM: Geometry Chapter 1: A Beginning of Geometry 17 26. Brookline has the coordinates ( ) 8, 2 , and Charleston has the coordinates ( ) 4,5 . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 4 8 5 2 4 3 16 9 25 5 = − + − = − + − = − + = + = = d x x y y 28. Everett has the coordinates ( ) 3,5 , − and Fairfield has the coordinates ( ) 8, 3 . − − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 8 3 3 5 5 8 25 64 89 9.3 = − + − ⎡ ⎤ = − − − + − − ⎣ ⎦ = − + − = + = ≈ d x x y y 30. Platform D has coordinates ( ) 20, 20 , and Platform E has coordinates ( ) 30, 15 . − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 30 20 15 20 10 35 100 1225 1325 36.4 m = − + − = − + − − = + − = + = ≈ d x x y y 32. a. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 5 3 7 1 8 6 64 36 100 10 = − + − ⎡ ⎤ ⎡ ⎤ = − − + − − − ⎣ ⎦ ⎣ ⎦ = + − = + = = d x x y y b. ( ) ( ) 1 2 1 2 , 2 2 1 7 3 5 , 2 2 2 8 , 2 2 1, 4 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ − + − − + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − x x y y M 34. a. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 1 4 3 2 5 5 25 25 50 7.1 = − + − ⎡ ⎤ ⎡ ⎤ = − − + − − ⎣ ⎦ ⎣ ⎦ = + = + = ≈ d x x y y b. ( ) 1 2 1 2 , 2 2 4 1 2 3 , 2 2 3 1 , 2 2 1.5, 0.5 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + − + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − x x y y M 36. a. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 3 5 5 3 2 2 4 4 8 2.8 = − + − ⎡ ⎤ ⎡ ⎤ = − − − + − − − ⎣ ⎦ ⎣ ⎦ = + − = + = ≈ d x x y y b. ( ) ( ) ( ) 1 2 1 2 , 2 2 5 3 3 5 , 2 2 8 8 , 2 2 4, 4 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ − + − − + − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ − − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − − x x y y M

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    Chapter 1: A Beginning of Geometry ISM: Geometry 18 38. Point A has coordinates ( ) 0,3 , and Point B has coordinates ( ) 2, 2 . − − a. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 0 2 3 2 5 4 25 29 5.4 = − + − = − − + − − = − + − = + = ≈ d x x y y b. ( ) ( ) ( ) 1 2 1 2 , 2 2 0 2 3 2 , 2 2 2 1 , 2 2 1,0.5 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ + − + − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − x x y y M 40. Central Station has coordinates ( ) 0,3 , and South Station has coordinates ( ) 0, 4 . − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 0 0 4 3 0 7 0 49 49 7 miles = − + − = − + − − = + − = + = = d x x y y 42. Cedar Station has coordinates ( ) 3,1 , − and City Plaza Station has coordinates ( ) 0, 0 . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 0 3 0 1 3 1 9 1 10 3.2 miles = − + − ⎡ ⎤ = − − + − ⎣ ⎦ = + − = + = ≈ d x x y y 44. Sycamore Station has coordinates ( ) 4, 6 , − and Cedar Station has coordinates ( ) 3,1 . − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 3 4 1 6 7 7 49 49 98 9.9 miles = − + − ⎡ ⎤ = − − + − − ⎣ ⎦ = − + = + = ≈ d x x y y 46. a. If Point P is the midpoint, let the coordinates of R be ( ) 2 2 , . x y ( ) 2 2 2 4 4, 6 , 2 2 x y + + ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ 2 : x 2 2 2 2 4 , 2 8 2 10 x x x + − = − = + − = and 2 : y 2 2 2 4 6 2 12 4 8 y y y + = = + = If P is the midpoint, the coordinates of R are ( ) 10,8 . − If Point Q is the midpoint, let the coordinates of R be ( ) 2 2 , . x y ( ) 2 2 4 6 2, 4 , 2 2 x y − + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 2 : x 2 2 2 4 2 , 2 4 4 8 x x x − + = = − + = and 2 : y 2 2 2 6 4 2 8 6 2 y y y + = = + = If Q is the midpoint, the coordinates of R are ( ) 8, 2 . If Point R is the midpoint, find the coordinates of R . ( ) 1 2 1 2 , 2 2 4 2 6 4 , 2 2 2 10 , 2 2 1,5 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − x x y y M If R is the midpoint, its coordinates are ( ) 1,5 . −

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    ISM: Geometry Chapter 1: A Beginning of Geometry 19 b. Find the possible lengths of . R Q If P is the midpoint: ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 10 4 8 12 4 144 16 160 = − + − ⎡ ⎤ = − − + − ⎣ ⎦ = + − = + = d x x y y If Q is the midpoint: ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 8 4 2 6 2 36 4 40 = − + − = − + − = − + = + = d x x y y If R is the midpoint: ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 1 4 5 3 1 9 1 10 = − + − ⎡ ⎤ = − − + − ⎣ ⎦ = + − = + = d x x y y In order for R Q to have a length of 160, P must be the midpoint. So, R must have the coordinates ( ) 10,8 . − 48. The terms inside the distance formula ( ) ( ) 2 2 2 1 2 1 = − + − d x x y y are squared, which always results in a positive answer. Then they are added, and the sum of two positive numbers is always positive. The square root of a positive number is always positive. 50. Your friend substituted the values in the Distance Formula incorrectly. Each subtraction is written with the wrong partner. It should say the following. ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 3 1 8 5 = − + − = − + − d x x y y Section 1.8 Practice 1. R S must be twice as long as XY below. Step 1: Draw a ray with endpoint R . Step 2: Open the compass to the length of . XY With the same compass setting, put the compass point on R and draw an arc that intersects the ray. Step 3: Without changing the compass setting, put the compass point on the intersection of the ray and the arc and draw an arc that intersects the ray on the opposite side of the intersection as R . Label the point of intersection S . 2. Q ∠ must be congruent to B ∠ below. Step 1: Draw a ray with endpoint Q . Step 2: With the compass point on vertex B , draw an arc that intersects the sides of . B ∠ Step 3: With the same compass setting, put the compass point on point Q. Draw an arc that intersects the ray. Step 4: Open the compass to the distance between the points of intersection between the sides of B ∠ and the arc from Step 2. Keeping the same compass setting, put the compass point on the intersection from Step 3. Draw an arc that intersects the arc. Step 5: Draw a ray from point Q through the intersection of the arcs.

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    Chapter 1: A Beginning of Geometry ISM: Geometry 20 3. Draw . ST Step 1: Put the compass point on point S and draw arcs above and below the segment. Be sure that the opening is greater than 1 . 2 ST Step 2: With the same compass setting, put the compass point on point T and draw arcs above and below the segment that intersect the arcs from Step 1. Step 3: Draw a line that passes through the points where the arcs intersect. 4. Draw an obtuse angle, . Y ∠ Step 1: Put the compass point on vertex Y . Draw arcs that intersect the sides of the angle. Label the intersections X and Z . Step 2: Put the compass point on point X and draw an arc inside the angle. Then put the compass point on point Z without changing the compass setting, and draw an arc that intersects the last arc at a point P . Step 3: Draw a ray from vertex Y through point P . Exercise Set 1.8 2. VW must be twice as long as A B below. Step 1: Draw a ray with endpoint V . Step 2: Open the compass to the length of . A B With the same compass setting, put the compass point on V and draw an arc that intersects the ray. Step 3: Without changing the compass setting, put the compass point on the intersection of the ray and the arc and draw an arc that intersects the ray on the opposite side of the intersection as V. Label the point of intersection W . 4. The length of QJ must be the difference of the lengths of TR and PS below. Step 1: Draw a ray with endpoint Q . Step 2: Open the compass to the length of . TR Put the compass point on Q and draw an arc that intersects the ray. Step 3: Open the compass to the length of . PS Put the compass point on the intersection of the ray and the arc and draw an arc that intersects the ray between the intersection and point Q . Label this point J . 6. m F ∠ must be twice the measure of . C ∠ Step 1: Draw a ray with endpoint F . Step 2: With the compass point on vertex C , draw an arc that intersects the sides of . C ∠ Step 3: With the same compass setting, put the compass point on point F. Draw a large arc that intersects the ray. Step 4: Open the compass to the distance between the points of intersection between the sides of C ∠ and the arc from Step 2. Keeping the same compass setting, put the compass point on the intersection from Step 3. Draw an arc that intersects the arc above the ray.

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    ISM: Geometry Chapter 1: A Beginning of Geometry 21 Step 5: Keeping the same compass setting, put the compass point on the intersection from Step 4. Draw an arc that intersects the large arc above the ray. Step 6: Draw a ray from point F through the intersection from Step 5. 8. Construct the perpendicular bisector of TR below. Step 1: Put the compass point on point T and draw arcs above and below the segment. Be sure that the opening is greater than 1 . 2 TR Step 2: With the same compass setting, put the compass point on point R and draw arcs above and below the segment intersecting the arcs from Step 1. Step 3: Draw a line that passes through the points where the arcs intersect. 10. Draw an obtuse angle, . XQZ ∠ Step 1: Put the compass point on vertex Q . Draw arcs that intersect the sides of the angle. Step 2: Put the compass point on one of the intersections and draw an arc inside the angle. Then put the compass point on the other intersection without changing the compass setting, and draw an arc that intersects the last arc. Step 3: Draw a ray from vertex Y to the point where the two arcs you drew in Step 2 intersect. 12. Draw a right angle, . PSQ ∠ Step 1: Put the compass point on vertex S . Draw arcs that intersect the sides of the angle. Step 2: Put the compass point on the intersection of an arc and the side of the angle, and draw an arc inside the angle. Without changing the compass setting, put the compass point on the other intersection to draw an arc that intersects the last arc. Label this point T . Step 3: Draw a ray from vertex S to point T .

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    Chapter 1: A Beginning of Geometry ISM: Geometry 22 14. Construct the angle bisector of . A ∠ You now have two angles with measures 1 2 . m A ∠ Pick one of the smaller angles and do the construction of the angle bisector for that angle. The measures of the two angles are half of the original angle, or 1 4 . m A ∠ 16. There is only one point at which a segment can be cut in half, so a segment can only have one bisector. 18. In space, there is a third axis so there are many angles at which a perpendicular bisector can bisect the segment, so there are an infinite number of perpendicular bisectors for this segment in space. 20. Draw angles 1 and 2. Construct an angle congruent to 2 ∠ that shares a side with 1 ∠ so that the other side of the angle is in the interior of 1. ∠ Step 1: Put the compass point on the vertex of 2. ∠ Draw arcs that intersect each side of the angle. Label the points A and B . Step 2: With the same compass setting, put the compass point on the vertex of 1 ∠ and draw and arc that intersects the side of angle 1. Label this point D . Step 3: Set the compass to be the length between point A and point B . Put the compass on point D , and draw an arc that intersects the arc from the previous step. Label the point of intersection E . Step 4: Draw a ray from the vertex of 1 ∠ through point E . Label the angle enclosed by this ray and the leg of 1, ∠ . C ∠ 22. Draw ∆ A BC with acute angles. Step 1: Put the compass point on vertex A , and draw and arc that intersects both sides of the triangle that meet at A . Label the points D and . E Step 2: Open the compass a little more than half the distance between D and . E Put the compass point on point D and draw and arc. With the same compass setting, draw an arc using the point E that intersects the other arc. Step 3: Draw the ray with endpoint A through the intersection of the two arcs. Step 4: Repeat Steps 1−3 with the other vertices of the triangle to construct angle bisectors for B ∠ and . C ∠ The angle bisectors of a triangle intersect at a point inside the triangle.

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    ISM: Geometry Chapter 1: A Beginning of Geometry 23 24. Draw segment A B that is 5 cm long. Step 1: Set the compass length to the segment that is 5 cm long. Put the compass point on point A and draw an arc above . A B Step 2: Set the compass length to the segment that is 2 cm long. Put the compass point on point B and draw an arc that intersects the previous arc. Label the intersection C . Step 3: Use a straightedge to draw . A BC ∆ The lengths of A C and BC are 5 cm and 2 cm respectively. 26. Use the segments below and draw segment A B that is 4 cm long. Step 1: Open the compass to the length of the segment that is 2 cm long. Put the compass point at point A and draw a circle. Step 2: With the same compass setting, put the compass point on point B and draw a circle. The only place where the circles intersect is on . A B So, it is not possible to construct a triangle with the given side measures. 28. Use the segments shown below and draw segment A B that is 4 cm long. Step 1: Open the compass to the length of the segment that is 4 cm long. Put the compass point at point A and draw an arc above . A B Step 2: With the same compass setting, put the compass point on point B and draw an arc that intersects the previous arc. Label the intersection C . Step 3: Use a straightedge to draw . A BC ∆ The lengths of A C and BC are both 4 cm. 30. The answer is A. In order for the half circles to all line up correctly, the line at the bottom must be drawn first. Count the number of half circles. There are five. So to construct this, you would use a straight edge to draw the segment and then a compass to draw five half circles. Chapter 1 Review 1. Each new image has a square added inside the smaller square, rotated by 45 . ° 2. Each shape inside the circle has one more side added to it from the previous one. 3. Each fraction is being multiplied by 1 . 2 1 1 , 48 96 4. Each number is decreasing by 5. 20, 15 5. Answers may vary. Sample Answer: l and m 6. H JJ G QR

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    Chapter 1: A Beginning of Geometry ISM: Geometry 24 7. Answers may vary. Sample Answer: A , B , C 8. T 9. True; Postulate 1.3-1 states, “Through any two points, there is exactly one line.” This means that two points are always collinear. 10. False; JJJJ G LM has endpoint L and JJJG ML has endpoint M . Thus, since the rays have two different endpoints they cannot be the same ray. 11. 2 5 7, 2 5 3 − − = − − + = , so two possible coordinates are 7 − and 3. 12. Use the Midpoint Formula. 2 3 1 2 2 − + = 13. 3 5 4 10 15 = + = − = AB BC m m m 14. Use the Segment Addition Postulate to find the value of a . 8 50 21 + + = = a a a Substitute the value a to find and . XY YZ 21 = = XY a ; 8 29 = + = YZ a 15. This is an acute angle. 16. This is a right angle. 17. Use the Angle Addition Postulate to find . m PQR ∠ 61 25 36 ∠ + ∠ = ∠ ∠ = ∠ − ∠ = ° − ° = ° m PQR m MQP m MQR m PQR m MQR m MQP 18. 2 8 22 14 ∠ = ∠ + = + = m NQM m PQR x x x 19. ∠ ADB (or ) BDA ∠ and ∠ BDC (or ). CDB ∠ 20. Answers may vary. Sample Answer: ∠ A DC and ∠ CDF 21. Answers may vary. Sample Answer: ∠ A DC and ∠ E DF 22. Answers may vary. Sample Answer: ∠ ADB and ∠ B DF 23. Since the angles form a linear pair, the sum of their measures is 180 . ° (3 31) (2 6) 180 5 25 180 5 155 31 + + − = + = = = x x x x x 24. Since the angles are complementary, the sum of their measures is 90 . ° 3 (4 15) 90 7 15 90 7 105 15 + − = − = = = x x x x x 25. Use the Distance Formula. 2 2 (0 ( 1)) (4 5) 1.4 − − + − ≈ 26. Use the Distance Formula. 2 2 (6 ( 1)) (2 ( 1)) 7.6 − − + − − ≈ 27. Use the Midpoint Formula to find the x -coordinate. 3 3 0 0 2 2 − + = = Use the Midpoint Formula to find the y -coordinate. 2 ( 2) 0 0 2 2 + − = = So the coordinates of the midpoint are ( ) 0, 0 . 28. Use the Distance Formula. 2 2 ( 3 3) (2 ( 2)) 7.2 − − + − − ≈

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    ISM: Geometry Chapter 1: A Beginning of Geometry 25 29. Solve the Midpoint Formula for 2 . x 1 2 1 2 2 1 2 2 2 + = + = = − x x x x x m x x m x m x Now use the x values you have been given to solve for 2 . x 2 1 2 2( 1) ( 8) 6 = − = − − − = x x m x Solve the Midpoint Formula for 2 . y 1 2 1 2 2 1 2 2 2 + = + = = − y y y y y m y y m y m y Now use the y values you have been given to solve for 2 . y 2 1 2 2(1) 4 2 = − = − = − y y m y So the coordinates of the endpoint K are ( ) 6, 2 . − 30. Solve the Midpoint Formula for 2 . x 1 2 1 2 2 1 2 2 2 + = + = = − x x x x x m x x m x m x Now use the x values you have been given to solve for 2 . x 2 1 2 2(5) 9 1 = − = − = x x m x Solve the Midpoint Formula for 2 . y 1 2 1 2 2 1 2 2 2 + = + = = − y y y y y m y y m y m y Now use the y values you have been given to solve for 2 . y 2 1 2 2( 2) ( 5) 1 = − = − − − = y y m y So the coordinate of the endpoint K is ( ) 1, 1 . 31. Holding your protractor on the paper, draw a straight line along the bottom of the protractor, shown below as . A B JJJ G Then mark a point C at 73 ° and draw a straight line from A to that point C . Step 1: Use a straight edge to draw a ray. Put the compass point on the vertex of the angle and draw an arc that intersects both sides of the angle. Then put the compass point on the endpoint of the drawn ray and draw an arc using the same compass setting. Step 2: Set the compass to the distance between the two intersection points on the angle. Then put the compass point on the intersection of the ray and the arc from Step 1, and draw an arc using the same setting that intersects the arc from Step 1. Step 3: Use a straight edge to draw a ray from the endpoint of the ray from Step 1 through the intersection from Step 2. 32. Holding your protractor on the paper, draw a straight line along the bottom of the protractor, shown below as . A B JJJ G Then mark a point C at 60 ° and draw a straight line from A to that point C .

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    Chapter 1: A Beginning of Geometry ISM: Geometry 26 33. Step 1: Put the compass point on point L and draw a long arc. Be sure the opening is greater than 1 . 2 LM With the same compass setting, put the compass point on point M and draw another long arc. Step 2: Draw a straight line between the two arc intersections. Now this new line is the perpendicular bisector. 34. a. Sketch the angle. Step 1: Use a straight edge to draw a ray. Put the compass point on the vertex of the angle and draw an arc that intersects both sides of the angle. Then put the compass point on the endpoint of the drawn ray and draw an arc using the same compass setting. Step 2: Set the compass to the distance between the two intersection points on the angle. Then put the compass point on the intersection of the ray and the arc from Step 1, and draw an arc using the same setting that intersects the arc from Step 1. Step 3: Use a straight edge to draw a ray from the endpoint of the ray from Step 1 through the intersection from Step 2. b. Step 1: Start with the angle constructed in part a. Put the compass point on an intersection of the drawn arc and a side. Draw an arc inside the angle. Then move the compass point to the intersection of the drawn arc and the other side and draw another arc inside the angle. Step 2: Draw a ray from vertex B through the intersection of the arcs from Step 1. 35. Yes; All four points are located on plane AEF . 36. Yes; All four points are located on plane DCE . 37. No; Sample Answer: Points H, G, and F are in plane HGF and point B is not in that plane. 38. No; Sample Answer: Points A, E, and B are in plane AEB and point C is not in that plane. 39. a. H JJ G BF b. H JJG EH 40. Answers may vary. Sample Answer: , PR HJJ G H JJG R Q 41. , PQ J JJG J JJ G PB 42. Answers may vary. Sample Answer: , ∠ ∠ CPQ CPB 43. Answers may vary. Sample Answer: D, P, C 44. Answers may vary. Sample Answer: JJJ G PA and J JJ G PR 45. Answers may vary. Sample Answer: , , AP PR RQ 46. Answers may vary. Sample Answer: ∠ A PC and ∠ CPR 47. Answers may vary. Sample Answer: ∠ APD and ∠ CPR 48. a. Since the angles form a linear pair, the sum of their measures is 180 . ° 180 2 3 180 5 180 36 ∠ + ∠ = + = = = m ABC m CBD x x x x

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    ISM: Geometry Chapter 1: A Beginning of Geometry 27 b. 2 2(36) 72 ∠ = = = m ABC x So, ∠ A BC is acute. 3 3(36) 108 ∠ = = = m CBD x So, ∠ CBD is obtuse. 49. 1 ( 4) 3 − − − = 50. 3 0 3 − = 51. 6 3 3 − = 52. 6 ( 1) 7 − − = 53. Use the fact that A C is congruent to CD to solve for x. 4 5 3 8 3 + = + = x x x Solve for . A C 4(3) 5 17 = + = AC Solve for . CD 3(3) 8 17 = + = CD Use the Segment Addition Postulate to find . A D 17 17 34 + = + = A C CD AD 54. Use the Angle Addition Postulate to find . m FCB ∠ 130 95 35 ∠ + ∠ = ∠ ∠ = ∠ − ∠ = ° − ° = ° m FCB m BCD m FCD m FCB m FCD m BCD 55. Use the Angle Addition Postulate. 180 180 ∠ + ∠ + ∠ = − ∠ − ∠ = ∠ m FCA m FCE m ECD m FCA m ECD m FCE Substitute ∠ m FCA for . m ECD ∠ 180 − ∠ − ∠ = ∠ m FCA m FCA m FCE Solve for . m FCE ∠ 180 50 50 80 m FCE ∠ = ° − ° − ° = ° 56. No; Q is not always the midpoint. It depends on the situation. If all three points are distinct and collinear then the answer is yes, Q is always the midpoint. If the three points are not distinct or they are not collinear then it does not mean that Q is the midpoint. 57. No; C, F, and G are collinear, so HJJ G A B must intersect H JJG GF at C in plane M . Chapter 1 Test 1. Each new number has 6 being added to it. 31, 37 2. Each new image has each section inside being cut in half. The number of sections doubles each time. 3. B 4. C 5. C 6. C 7. A; Since J JJJ G MN bisects ∠ EMG , the measure of ∠ EMG is twice the measure of . EMN ∠ 8. ∠ E FD and or 2 AFC ∠ ∠ are vertical angles. 9. ∠ E FD and 3 or 1 ∠ ∠ form a linear pair. 10. The complementary angles are and . DEF FEA ∠ ∠ 11. True 12. False 13. True 14. True

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