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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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Solution Manual for Geometry, 1st Edition - Page 1 preview imageINSTRUCTOR’SSOLUTIONSMANUALGEXPUBLISHINGSERVICESGEOMETRYElayn Martin-GayUniversity of New Orleans
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Solution Manual for Geometry, 1st Edition - Page 2 preview image
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Solution Manual for Geometry, 1st Edition - Page 3 preview imageTable of ContentsChapter 1 A Beginning of Geometry ................................................................................. 1Chapter 2 Introduction to Reasoning and Proofs ............................................................. 30Chapter 3 Parallel and Perpendicular Lines ..................................................................... 56Chapter 4 Triangles and Congruence ............................................................................... 91Chapter 5 Special Properties of Triangles...................................................................... 120Chapter 6 Quadrilaterals................................................................................................. 156Chapter 7 Similarity ....................................................................................................... 185Chapter 8 Transformations............................................................................................. 216Chapter 9 Right Triangles and Trigonometry ................................................................ 247Chapter 10 Area ............................................................................................................. 290Chapter 11 Surface Area and Volume............................................................................ 346Chapter 12 Circles and Other Conic Sections................................................................ 404
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Solution Manual for Geometry, 1st Edition - Page 4 preview image
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Solution Manual for Geometry, 1st Edition - Page 5 preview image1Chapter 1Section 1.2 Practice1.Each figure is a square divided into foursmaller squares, with one of the smallersquares shaded with an X in it. So, predictthat the next figure is a square divided intofour smaller squares, with the upper-rightsquare shaded and marked, as sketchedbelow.2.a.The second number is the first plus 2, thethird is the second minus 3, the fourth isthe third plus 4, and the fifth is the fourthminus 5. So, predict that the sixth is thefifth plus 6, or 3.b.Each number is13of the previousnumber, so predict that the next number is1 .813.No. None of the angles are marked, so wemay not conclude that they have the samemeasure.Vocabulary & Readiness Check 1.21.defined term2.undefined term3.A statement that we prove is called atheorem.4.A statement that we accept as true but do notprove is called an axiom or a postulate.Exercise Set 1.22.Each figure is a light brown rectangle dividedinto 1 more equal-sized section than theprevious figure. A sketch of the next figure isprovided.4.Each figure is a square divided into 4different colored squares. Each figure is theprevious image rotated 90° clockwise. Asketch of the next figure is provided. Gstands for green, R stands for red, B standsfor blue, and Y stands for yellow.6.Each subsequent figure has 1 square addedon top of the existing columns, and 1 squareadded to the right. The next figure shouldhave 4 squares in the first column, 3 in thesecond column, 2 in the third column, and 1in the fourth column. A sketch of the nextfigure is provided.8.The figures are rectangles that alternatebetween tan and green. There is an X in abottom corner of each tan rectangle, and in atop corner of each green rectangle. The Xswitches from left to right in figures with thesame color. The next figure will be greenwith an X in the top right corner. A sketch ofthe next figure is provided.10.The figures are green right triangles. Thesecond figure is the first figure reflectedvertically. The third figure is the secondfigure reflected horizontally. The next figurewill be the third figure reflected vertically. Asketch of the next figure is provided.12.Each number is 2 more than the previousnumber, so predict that the next two numbersare 2 and 4.
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Solution Manual for Geometry, 1st Edition - Page 6 preview imageChapter 1:A Beginning of GeometryISM:Geometry214.Each number has a denominator that is onelarger than the denominator of the previousnumber when written as a fraction, so predictthat the next two numbers are 15 and 1 .616.Each number is 3 times the previous number,so predict that the next two numbers are 162and 486.18.Each number is half of the previous number,so predict that the next two numbers are140and1 .8020.Add 1 to the first number to get the secondnumber, and add 1 to the second number toget the third number. Subtract 1 from thethird number to get the fourth number, andsubtract 1 from the fourth number to get thefifth number. So, a pattern could indicate toadd 1 to the fifth number to get the sixthnumber, and add 1 to the sixth number to getthe second number. So, predict that the nexttwo numbers are 0 and1.22.Yes. The figure has 3 sides that areconnected, which is the requirement for beingcalled a triangle.24.No. The figure is a triangle because it has 3connected sides, but no indication is providedthat one of the angles is a right angle.26.Yes. Both angles measure 37°, so they havethe same measure.28.No. No indication is provided that thesegments are the same length.30.Yes. Both angles have the middle ray as aside.32.No. The opposite sides of the figure are notmarked parallel, so it cannot be concludedthat the figure is a parallelogram.34.The student’s number is not correct. Sincethe number that follows 14 is 1 ,6thedenominator is not doubled to get the nextdenominator. Add 2 to the denominator to getthe next denominator. Therefore, the nextnumber in the list is 1 .836.The student’s figure is not correct. In thispattern, the positions of the dot and the x arereversed to generate the next figure. Then theimage is flipped vertically to generate thenext figure. The correct figure should havethe positions of the dot and the x reversed, asshown.38.Neither axioms nor postulates are proven.They are accepted as true.Section 1.3 Practice1.a.Two other ways to nameEHHJJGareHEHJJGand line.nb.There are many correct answers. Twoother ways to name planeMare planeGEDand plane.GFDc.Points,D,EandFlie on the sameline, so they are collinear.d.Points,D,E,FandGlie on the sameplane, so they are coplanar.e.There are many correct answers. Forinstance, points,H,EandFdo not lieon the same line.2.a.The three segments areGJor,JGJQor,QJandGQor.QGb.The four rays areGJJJJGor,GQJJJG,JGJJJG,JQJJJGandQGJJJGor.QJJJJGc.The opposite rays areJGJJJGand.JQJJJG3.a.b.
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Solution Manual for Geometry, 1st Edition - Page 7 preview imageISM:GeometryChapter 1:A Beginning of Geometry3c.d.4.Points,W,ZandAare collinear andAisbetweenWand.ZSo,AWJJJJGandAZJJJGareopposite rays.Points,Y,XandAare collinear andAisbetweenYand.XSo,AYJJJGandAXJJJGareopposite rays.5.The plane intersects in.EHHJJG6.Vocabulary & Readiness Check 1.31.C. The symbolRSmeans a segment.RS2.A. The symbolRSHJJGmeans a line.RS3.D. The symbolRSJJJGmeans a ray.RS4.B. The symbolSRJJGmeans a ray.SR5.A line extends in opposite directions withoutend.6.A plane extends in two dimensions withoutend.7.A ray consists of an endpoint and all pointsof a line on one side of the endpoint.8.A segment consists of two endpoints and allthe 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 ofspace.12.Opposite rays are two rays that share anendpoint and form a line.13.A pointhas a location but no dimension.14.Space is the set of all points.Exercise Set 1.32.,NTHJJG,TNHJJG,TRHJG,RNHJJG,NRHJJGand linelare allpossible answers.4.There are many correct answers. Sampleanswers: planeTNJand planeTRJ6.Points,R,NandTlie on the same line, sothey are collinear.8.Points,R,N,TandJlie on the sameplane, so they are coplanar.10.The segments areAZor,ZAAXor,XAXYor,YXAYorYA,XZorZXandYZor.ZY12.The rays areAZJJJGorAXJJJG,AYJJJGXZJJJGor,XYJJJG,YZJJGZAJJGorZXJJJGor,ZYJJJGYAJJGor,YXJJJGand.XAJJJG14.The opposite rays with endpointXareXAJJJGandXZJJJGor.XYJJJG16.The only line isBCHJJG18.The rays are,ABJJJG,BCJJJGand.CBJJJG20.False.XPJJJGandXNJJJGare not opposite raysbecause,P,XandNare not collinear.22.True.XLJJJGandXNJJJGare opposite raysbecauseXLJJJGandXNJJJGare two rays with thesame endpoints going in opposite directionsand all three points are collinear.
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Solution Manual for Geometry, 1st Edition - Page 8 preview imageChapter 1:A Beginning of GeometryISM:Geometry424.There are many correct answers. SampleAnswer: Points,A,EandBare notcollinear.26.There are many correct answers. SampleAnswer: Points,W,AandXare notcollinear.28.False. PointDlies on line,lnot line.m30.True. PointCdoes lie on line.m32.True. Points,D,EandBare collinearbecause they all lie on the same line, line.l34.False. The fact that the points are collinearmeans that many planes pass through thesethree points, not just one.36.The plane intersects in.VWHJJG38.The plane intersects in.TXHJJG40.Sample Answer: planesTXWandTSR42.Sample Answer: planesUXVandVWS44.PointG46.PointH48.PointE50.PointH52.Coplanar. All four points are on the sameplane.54.Coplanar. All four points are on the sameplane.56.Noncoplanar. All four points are not on thesame plane.58.60.62.64.Sometimes. This is true ifJLJJGis a ray wherepointKis on the ray.JLJJGThis is false ifKLHJJGis a line whereJis on that line betweenKand.LThen the two rays would have thesame endpoints and be going in oppositedirections.66.Sometimes. This could be true like Exercise52 where there are four points that all are onthe same plane. This could be false likeExercise 56 where there are four points andone of them is on a different plane from theother three.68.Never. Postulate 1.3-2 states that two distinctlines intersect at exactly one point.70.The following times can also be representedas 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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Solution Manual for Geometry, 1st Edition - Page 9 preview imageISM:GeometryChapter 1:A Beginning of Geometry574.No.76.No.78.One line can have many planes pass throughit. A single line can be the intersection ofmultiple planes, but there is no distinctnumber.80.If two lines intersect than only one plane cancontain both of those lines. Postulate 1.3-3states that if two distinct planes intersect,then they intersect in exactly one line. Thismeans that no two planes can contain twodistinct lines. Using the diagram, the planecan be labeled,ABCwhich refers to asingle plane.Section 1.4 Practice1.NQis the distance between pointsNandQ.Use the Ruler Postulate to findNQ. Thecoordinate ofNis2,−and the coordinate ofQis 13.13( 2)1616NQNQNQ=− −==2.a.ACandDFare marked the same, soACDF≅and.ACDF=ABandDEare marked the same, soABDE≅and.ABDE=BCandEFare marked thesame, soBCEF≅and.BCEF=b.,ACDF=so if10AC=cm,10DF=cm.c.,CBFE=so if25CB=cm,25FE=cm.3.IfBCGD=, then.BCGD≅( 1)( 6)558355BCGD=−− −===−==Yes,,BCGD=so.BCGD≅4.Use the Segment Addition Postulate andalgebra to findx.(123)(82)79201792080208020204QRRSQSxxxxxx+=−++=−====Use the value ofxto findQRandRS.12312(4)348345QRx=−=−=−=828(4)234RSx=+=+=5.Use the meaning of the midpoint to find thevalue ofx.118101817QPPRxxxx=−=−−= −=
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Solution Manual for Geometry, 1st Edition - Page 10 preview imageChapter 1:A Beginning of GeometryISM:Geometry6Now, use the value ofxto findQPandPR,which should be the same.11811(7)877869QPx=−=−=−=10110(7)170169PRx=−=−=−=To findQR, add.6969138QRQPPR=+=+=Vocabulary & Readiness Check 1.41.Two segments that have the same length arecalled congruent segments.2.The real number that corresponds to a pointis called the coordinate of the point.3.Given three distinct points on a line, one ofthe points will always be between the othertwo.4.A point that divides a segment into twocongruent segments is called a midpoint.5.If two segments are congruent, then theirlengths are equal.6.The distance between two points is theabsolute value of the difference of theircoordinates.7.A line, ray, segment, or plane that intersects asegment at its midpoint is called a segmentbisector.8.In geometry, to bisect a segment means todivide it into two equal parts.Exercise Set 1.42.Yes;Nis betweenQandTbecauseNappearson.QT4.No;Mis not betweenNandPbecauseMdoes notappear on.NP6.No;Pis not betweenRandTbecausePisnot on.RT8.No;Nis not betweenRandQbecauseNdoes not appear on.RQ10.Yes;Qis betweenRandTbecauseQappearson.RT12.ABis the distance between pointsAandB.Use the ruler postulate to findAB.6( 8)22ABABAB=−− −==14.ADis the distance between pointsAandD.Use the ruler postulate to findAD.3( 8)1111ADADAD=− −==16.VXandUWare marked the same, soVXUW≅and,VXUW=which meanssince105VX=cm, then105UW=cm.18.DEandFGare marked the same, soDEFG≅and,DEFG=which meanssince7DE=in., then7FG=in.20.QSandRTare marked the same, soQSRT≅and,QSRT=which means since33QS=ft, then33RT=ft.22.If,MPNQ=then.MPNQ≅6( 3)9912399MPNQ=− −===−==Yes,,MPNQ=so.MPNQ≅24.If,LPMQ=then.LPMQ≅6( 8)141412( 3)1515LPMQ=− −===− −==No,,LPMQ≠soLPis not congruent to.MQ26.If15ST=and40,RT=then401525.RS=−=
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Solution Manual for Geometry, 1st Edition - Page 11 preview imageISM:GeometryChapter 1:A Beginning of Geometry728.a.Use the Segment Addition Postulate andalgebra to findx.(148)(910)2322322322323023230232310RSSTRTxxxxxx+=−++=+====b.Use the value ofxto findRSandST.14814(10)81408132RSx=−=−=−=9109(10)109010100STx=+=+=+=30.a.Use the meaning of the midpoint to findthe value ofy.92545250525525555yyyyyy−=−====b.Use the value ofyto findQCandCR,which should be the same.9259(5)25452520QCy=−=−=−=44(5)20CRy===To find QR, add.202040QRQCCR=+=+=32.PTandTQare marked the same, soPTTQ≅and.PTTQ=Use this to solveforx.63523432432448PTTQxxxxx=−=−===Usexto find.PT6356(8)35483513PTx=−=−=−=34.PTandTQare marked the same, soPTTQ≅and.PTTQ=Use this to solveforx.7246222PTTQxxx=−=−=Usexto find.PT7247(22)2415424130PTx=−=−=−=36.FindZXandWY.7188ZX= −−= −=5( 3)88WY=− −==Because,ZXWY=.ZXWY≅38.You are given that7.AT=Because,DEC∠the coordinate ofTmust be either 7or7.−The midpoint is halfway betweenthese coordinates. Take half of the distancefrom zero. 7013,22−=so the coordinate ofthe midpoint is either13 2 or13.2−40.The coordinate ofGmust either be 6 unitsgreater than or 6 units less than thecoordinate ofP.106161064+=−=The possible coordinates ofGare 4 and 16.42.EDandDBare marked the same, soEDDB≅and.EDDB=Use this to solveforx.438248212212226EDDBxxxxxx=+=−−+= −−= −−−=−−=
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Solution Manual for Geometry, 1st Edition - Page 12 preview imageChapter 1:A Beginning of GeometryISM:Geometry8Usexto findEDandDB. They should be thesame.46410EDx=+=+=383(6)818810DBx=−=−=−=AddEDandDBto findEB.101020EBEDDB=+=+=44.Mile markers for Macclenny and Tallahasseeare shown, so take the absolute value of thedifference to find the distance.199335136136−= −=Divide the distance by the number of milesper hour to find the time.1362025858=hours, or 2 hours and2020 29 minutes.46.The highway sign gives the number of milesaway each town is from the sign. If yourfriend is reading the sign, then your friend isthe same distance away from the town as thesign. Because 80 is shown on the sign forWatertown; Watertown is 80 miles away, soyour friend is incorrect.48.a.AddGJandHK, but because theyoverlap, you will need to subtract thedistance that they share, which isJH.(23)(43)5GKGHHJHKGKxxxGKx=−+=+−+−=b.Solve forxusing30.GK=5305305556GKxxxx====SubtractHJfromGJto findGH.(23)3639GHGJHJxxx=−=+−=+=+=SubtractHJfromHKto findJK.(43)333(6)318315JKHKHJxxx=−=−−=−=−=−=50.The mile markers are like numbers on anumber line. The number at each marker isthe distance from the start of the road. Thismeans that you can subtract the number on amile marker from the number on another milemarker and take the absolute value to get thedistance between the markers, just likefinding the distance between two coordinateson a number line.Section 1.5 Practice1.a.,NPQ∠QPN∠b.QPM∠2.18016515 ,mTQN∠=−° =°acute angle18045135 ,mTQM∠=−° =°obtuse angle1800180mTQR∠=−° =°, straight angle3.a.,AD∠≅ ∠somAmD∠=∠.,CF∠≅ ∠so.mCmF∠=∠b.25 ,mA∠=°so25 .mD∠=°c.120 ,mF∠=°so120 .mC∠=°4.360606x°==°5.Use the Angle Addition Postulate and solveforx.(410)(1110)1801518012mPQRmRQSmPQSxxxx∠+∠=∠−++===Substitute the value ofxto find.mPQR∠(410)(4 1210)38mPQRxmPQRmPQR∠=−°∠=⋅−°∠=°Substitute the value ofxto find.mRQS∠(1110)(11 1210)142mRQSxmRQSmRQS∠=+°∠=⋅+°∠=°
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Solution Manual for Geometry, 1st Edition - Page 13 preview imageISM:GeometryChapter 1:A Beginning of Geometry9Vocabulary & Readiness Check 1.51.An instrument used to measure angles indegrees is called a protractor.2.Two angles that have the same measure arecalled congruent angles.3.An angle consists of two different rays with acommon endpoint.4.The rays of an angle are also called the sidesof the angle.5.The common endpoint of the rays of an angleis called the vertex of the angle.6.If two angles are congruent, then theirmeasures are equal.7.An obtuse angle has a degree measurebetween 90º and 180º.8.An acute angle has a degree measure between0º and 90º.9.A straight angle has a degree measure equalto 180º.10.A right angle has a degree measure of 90º.Exercise Set 1.52.,1,,ABCCBAB∠∠∠∠4.,3,,MKLLKMK∠∠∠∠6.90090 ;DAF∠=−° =°right angle8.25025 ;BAC∠=−° =°acute angle10.907020 ,DAE∠=−° =°acute angle12.Sample of an acute angle,.GHJ∠14.Sample of a right angle,.QRS∠16.FJHBJA∠≅ ∠18.If130 ,mGHF∠=°then130 .mJBC∠=°20.A clock face is a circle which is divided into12 equal slices. Each slice represents anangle measuring 36030 .12° =°The anglemeasure of the clock hands at 5:00 is equal to5 30150 .⋅° =°22.A clock face is a circle which is divided into12 equal slices. Each slice represents anangle measuring 36030 .12° =°The anglemeasure of the clock hands at 11:00 is equalto 1 3030 .⋅° =°24.360725y°==°26.3603012x°==°28.Use the Angle Addition Postulate and solveforx.45(62)(21)4583458486mYNZmZNMmYNMxxxxx∠+∠=∠=°−+−=−===Substitute the value ofxto solvefor.mYNZ∠(62)(6 62)34mYNZxmYNZmYNZ∠=−°∠=⋅−°∠=°Substitute the value ofxto solvefor.mZNM∠(21)(2 61)11mZNMxmZNMmZNM∠=−°∠=⋅−°∠=°
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Solution Manual for Geometry, 1st Edition - Page 14 preview imageChapter 1:A Beginning of GeometryISM:Geometry1030.Use the Angle Addition Postulate and solveforx.180(53)(323)180820180816020mABDmDBCmABCxxxxx∠+∠=∠=°−++=+===Substitute the value ofxto solvefor.mABD∠(53)(5 203)97mABDxmABDmABD∠=−°∠=⋅−°∠=°Substitute the value ofxto solvefor.mDBC∠(323)(3 2023)83mDBCxmDBCmDBC∠=+°∠=⋅+°∠=°32.Use the Angle Addition Postulate and solveforx.90(81)(8)9099909819mMRJmJRKmMRKxxxxx∠+∠=∠=°+++=+===Substitute the value ofxto solvefor.mMRJ∠(81)(8 91)73mMRJxmMRJmMRJ∠=+°∠=⋅+°∠=°Substitute the value ofxto solvefor.mJRK∠(8)(98)17mJRKxmJRKmJRK∠=+°∠=+∠=°34.The measure of the angle is 90°and it is aright angle.36.The measure of the angle is 88°and it is anacute angle.38.28 ,mAOBmCOD∠=∠=°(32) ,mBOCx∠=−°and6.mAODx∠=°Use the Angle Addition Postulate and solveforx.28(32)28635418mAOBmBOCmCODmAODxxxx∠+∠+∠=∠+−+===Substitute the value ofxto solvefor.mBOC∠(32)52mBOCx∠=−° =°Substitute the value ofxto solvefor.mAOD∠6108mAODx∠=° =°Check your work by adding the interiorangles to be sure that their sum equals.mAOD∠mAOBmBOCmCODmAOD∠+∠+∠=∠285228108mAOD° +° +° =° =∠40.The clock face is a circle with 12 equalintervals. Each interval represents36030 .12° =°The minute hand moves 30°every 5 minutes and the hour hand moves 30°every hour. At 8:40, the minute hand is onthe 8 and the hour hand has moved 23 of theway from 8 to 9, so the clock hands are 23 ofan interval apart. Since each intervalrepresents 30°, at 8:40 the angle formed bythe clock hands is 2 3020 .3⋅° =°42.The clock face is a circle with 12 equalintervals. Each interval represents36030 .12° =°The minute hand moves 30°every 5 minutes and the hour hand moves 30°every hour. At 4:30, the minute hand is onthe 6 and the hour hand has moved 12 of theway from 4 to 5, so the hands are 1.5intervals apart. Since each interval represents30°, at 4:30 the angle formed by the clockhands is 1 303045 .2⋅° +° =°44.There are two ways to sketch135 .PVB∠=°
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Solution Manual for Geometry, 1st Edition - Page 15 preview imageISM:GeometryChapter 1:A Beginning of Geometry11Section 1.6 Practice1.a.True.8∠and5∠are two adjacentangles, and their noncommon sides formopposite rays.b.True.8∠and6∠have sides that formopposite rays.c.False.7∠and5∠are not adjacentangles.d.False.7∠and6∠do not have sides thatform opposite rays.2.Find any angle pairs with measures that havea sum of 90°.8∠and7∠are complementary.7∠and6∠are complementary.6∠and5∠are complementary.8∠and5∠are complementary.3.Find any angle pairs with measures that havea sum of 180°.5∠and6∠are supplementary.6∠and7∠are supplementary.7∠and8∠are supplementary.8∠and5∠are supplementary.4.a.Since∠Mand∠Care supplementary,the sum of their measures is 180 .°18016180164mMmCmMmM∠+∠=°∠+° =°∠=°b.Since∠Nand∠Care complementary,the sum of their measures is 90 .°90169074∠+∠=°∠+° =°∠=°mNmCmNmN5.a.Since,mLKMmMKN∠=∠().72236∠=∠+∠∠=∠+∠° =∠° =∠mLKNmLKMmMKNmLKNmLKMmLKMmLKMmLKMb.Since,mLKMmMKN∠=∠36 .mMKN∠=°6.SinceHLJJJGbisects,KHJ∠.mKHLmLHJ∠=∠Solve forx.().354624623∠=∠° =−°−= −=mKHLmLHJxxxxUse substitution to find the measure of eachangle.()33 2369mKHLx∠=°=°=°()()()5465 23461154669mLHJx∠=−°=⋅−°=−°=°7.DEA∠andBEA∠form a linear pair, so thesum of their measures is 180 .°()()18056246180740180714020mDEAmBEAxxxxx∠+∠=°−° ++° =°+===DEC∠andBEC∠form a linear pair.()()18025664180860180812015mDECmBECyyyyy∠+∠=°+° ++° =°+===Use substitution to find the measure of eachangle.()()()565 206100694mDEAx∠=−°=⋅−°=−°=°()()()2462 2046404686mBEAx∠=+°=⋅+°=+°=°
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Solution Manual for Geometry, 1st Edition - Page 16 preview imageChapter 1:A Beginning of GeometryISM:Geometry12()()()2562 1556305686mDECy∠=+°=⋅+°=+°=°()()()646 15490494mBECy∠=+°=⋅+°=+°=°Vocabulary & Readiness Check 1.61.Two angles are complementary angles if theirmeasures have a sum of 90°.2.Two angles are supplementary angles if theirmeasures have a sum of 180°.3.Two angles are vertical angles if their sidesform opposite rays.4.Two angles are adjacent angles if they sharea common side and a common vertex, buthave no interior points in common.5.Two adjacent angles form a linear pair iftheir noncommon sides are opposite rays.6.A ray that divides an angle into two adjacentcongruent angles is called an angle bisector.Exercise Set 1.62.3∠and1∠are vertical angles.4.True.2∠and1∠are adjacent angles, andtheir noncommon sides form opposite rays.6.False.4∠and2∠are not adjacent angles.8.False.3∠and5∠do not have sides thatform opposite rays.10.False.1∠and2∠do not have measuresthat add up to 180°.12.COE∠shares sideOEand is also a rightangle.14.COD∠andEOD∠form right angle.COE∠16.AOD∠is vertical to.BOC∠18.False.1490mm∠ +∠≠°20.True.1390mm∠ +∠=°22.True.3490mm∠+∠=°24.False.34180mm∠+∠≠°26.True.41180mm∠+∠ =°28.The conclusion cannot be made from thegiven information. There is no indication thatACJJJGis an angle bisector.30.The conclusion can be made from the giveninformation, assuming thatJDis a linesegment. ThenJCA∠andACD∠form alinear pair, which means the angles aresupplementary, and the sum of their measuresis 180°.32.The conclusion cannot be made from thegiven information. There is no indication that.JCDC≅34.The conclusion can be made from the giveninformation, assumingDEJJJGandJFJJJGarestraight rays. Then,EAF∠andJAD∠havesides that form opposite rays.36.ACD∠andACF∠are a linear pair, so theyare supplementary, and it is given that thesum of their measures is 180°. So,180290 .∠=° ÷=°mACDmACDmACBmBCD∠−∠=∠. Therefore,906525mBCD° −° =∠=°38.From Exercise 36,25 ,mBCD∠=°and it isgiven that.mBCDmECF∠=∠,mECFmACFmACE∠+∠=∠so2590115 .mACE° +° =∠=°40.Supplement: 18070110° −° =°Complement: 907020° −° =°42.Supplement: 1801179° − ° =°Complement: 90189° − ° =°44.Supplement: 1801755° −° =°Complement: There is no complimentbecause 17590 .° >°
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Solution Manual for Geometry, 1st Edition - Page 17 preview imageISM:GeometryChapter 1:A Beginning of Geometry1346.29295858mGJHmDJGmDJHmDJH∠+∠=∠° +° =°∠=°48.36mSVTmRVSmSVT∠=∠∠=°50.a.()()859135138mAQBmGQBxxxx∠=∠−° =−°−=−=b.()()858 8559mAQBx∠=−°⋅−° =°c.()()9139 81359mBQGx∠=−°⋅−° =°d.5959118mAQGmAQBmBQG∠=∠+∠=° +°=°52.Solve forx.()41224244124846015mABCmABDmCBDxxxx∠=∠+∠−° =° +°−===Solve for∠mABCusing the value ofx.()()()4124 1512601248mABCx∠=−°=⋅−°=−°=°54.Solve forx.()()3206162031636312mABDmCBDxxxxx∠=∠+° =−°=−==Solve for∠mABCusing the value ofx.()()()()()320616949 1241084112∠=∠+∠=+° +−°=+°=⋅+°=+°=°mABCmABDmCBDxxx56.a.Since∠EFGand∠GFHform a linearpair, they aresupplementary, and the sumof their measures is 180 .°()()180221415180636180614424∠+∠=°+° ++° =°+===mEFGmGFHnnnnnb.()()()2212 2421482169mEFGn∠=+°=⋅+°=+°=°()()()4154 24159615111mGFHn∠=+°=⋅+°=+°=°c.18069111180mEFGmGFH∠+∠=°° +° =°58.Findx.()()1802474180918020mFGJmHGJxxxx∠+∠=°+° +−° =°° =°=Substitute the value ofxto find∠mFGJand.mHGJ∠()()()242 20440444mFGJx∠=+°=⋅+°=+°=°()()()747 2041404136mHGJx∠=−°=⋅−°=−°=°Findy.()()180346216180530180515030mFGKmHGKyyyyy∠+∠=°+° +−° =°+===
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Solution Manual for Geometry, 1st Edition - Page 18 preview imageChapter 1:A Beginning of GeometryISM:Geometry14Substitute the value ofyto find∠mFGKand.mHGK∠()()()3463 30469046136mFGKy∠=+°=⋅+°=+°=°()()()2162 3016601644mHGKy∠=−°=⋅−°=−°=°60.Findx.()()1803226180424180420451mLWTmTWZxxxxx∠+∠=°+° +−° =°−===Substitute the value ofxto find∠mLWTand.mTWZ∠()()()323 5121532155mLWTx∠=+°=⋅+°=+°=°()()26512625mTWZx∠=−°=−°=°Findy.()()18021121118014121801416812mLWVmVWZyyyyy∠+∠=°+° ++° =°+===Substitute the value ofyto find∠mLWVand.mVWZ∠()()()212 12124125mLWVy∠=+°=⋅+°=+°=°()()()121112 121114411155mVWZy∠=+°=⋅+°=+°=°62.Let the angles beA∠and,B∠where()20.mBmA∠=∠−°SinceA∠andB∠are complementary, the sum of theirmeasures is 90 .°Find.mA∠()()()90209022090211055mAmBmAmAmAmAmA∠+∠=°∠+∠−=°∠−=°∠=°∠=°Find.mB∠90559035mAmBmBmB∠+∠=°° +∠=°∠=°64.The first student is correct. The correctequation is1.2mABXmABC∠=∠The anglebisector splitsABC∠in half to makeABX∠and.CBX∠The second student’sequation states that one of the halves is twiceas big as.ABC∠66.For this to occur,mAmB∠=∠and180mAmB∠+∠=°()180180218090 ,90mAmBmAmBmAmAmAmAmB∠=∠∠+∠=°∠+∠=°∠=°∠=°∠=°The only instance where vertical angles aresupplementary occurs when the angles areright angles.68.Two angles are supplementary if the sum oftheir measures is 180 .°All of the acuteangles in the figure are supplementary to.JQM∠70.Two adjacent angles form a linear pair iftheir noncommon sides are opposite rays.,LMN∠,QMP∠andNMP∠form linearpairs with.LMQ∠
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Solution Manual for Geometry, 1st Edition - Page 19 preview imageISM:GeometryChapter 1:A Beginning of Geometry15Section 1.7 Practice1.Use the Midpoint Formula,2abM+=wherea= –9 andb= 4.2942522.5+=−+=−== −abM2.Let()()11,5,2 ,xyP=−and()()22,8,6 .xyQ=−x-coordinate ofM:25821326.5+=+===abMy-coordinate ofM:()2262824+=−+ −=−== −abMThe midpoint of segmentPQis()6.5,4 .M−3.Let the coordinates ofBbe()22,.xy()22354,9,22xy−+−+⎛⎞−=⎜⎟⎝⎠2:x22234,28311xxx−+== −+=and2:y22259218513yyy−+−=−= −+−=The coordinates ofBare()11,13 .−4.Let the coordinates ofPbe()11,xyand thecoordinates ofQbe()22,.xyUse theDistance Formula.()()()()( )()2221212222233714116174.1dxxyy=−+−⎡⎤=−− −+−⎣⎦=+ −=+=≈Vocabulary & Readiness Check 1.71.The midpoint of a line segment is a pointexactly halfway between the two endpointsof the line segment.2.The Distance Formula is()()222121.dxxyy=−+−3.The Midpoint Formula for a segment on thecoordinate plane is1212,.22xxyyM++⎛⎞=⎜⎟⎝⎠4.The Midpoint Formula for a segment on anumber line is.2abM+=Exercise Set 1.72.9631.5222+−+−==== −abM4.()8122010222−+ −+−==== −abM6.()1212,2237911,221020,225,10++⎛⎞=⎜⎟⎝⎠++⎛⎞=⎜⎟⎝⎠⎛⎞=⎜⎟⎝⎠=xxyyM
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Solution Manual for Geometry, 1st Edition - Page 20 preview imageChapter 1:A Beginning of GeometryISM:Geometry168.()()1212,224836 ,22312,221.5,6++⎛⎞=⎜⎟⎝⎠⎛⎞−+ −−+=⎜⎟⎜⎟⎝⎠−⎛⎞=⎜⎟⎝⎠=−xxyyM10.()1212,2222515,22020,220,10++⎛⎞=⎜⎟⎝⎠−++⎛⎞=⎜⎟⎝⎠⎛⎞=⎜⎟⎝⎠=xxyyM12.1212,22xxyyM++⎛⎞=⎜⎟⎝⎠2274551515,2243515,2243,103021,5 10M⎛⎞⎛⎞⎛⎞−+−+−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎜⎟=⎜⎟⎜⎟⎝⎠⎛⎞⎛⎞⎛⎞−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎜⎟=⎜⎟⎜⎟⎝⎠⎛⎞=−⎜⎟⎝⎠⎛⎞=−⎜⎟⎝⎠14.Let the coordinates ofSbe()22,.xy()225155,8,22xy+−+⎛⎞−=⎜⎟⎝⎠2:x22255,21055xxx+==+=and2:y222158216151yyy−+−=−= −+−=The coordinates ofSare()5,1 .−16.Let the coordinates ofSbe()22,.xy()22285,8,22xy−++⎛⎞−=⎜⎟⎝⎠2:x22225,210212xxx−+== −+=and2:y22288216824yyy+−=−=+−=The coordinates ofSare()12,24 .−18.()()()()()()22212122228101414180324032418=−+−=−−+−=−+=+==dxxyy20.()()()()()()22212122220012309081819=−+−=−+−=+=+==dxxyy22.()()()()()()222121222251212127244957662525=−+−⎡⎤=−+− −⎣⎦=−+=+==dxxyy24.()()()()()()2221212222812186201240014454443423.3=−+−=−−+−=−+=+==≈dxxyy
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Solution Manual for Geometry, 1st Edition - Page 21 preview imageISM:GeometryChapter 1:A Beginning of Geometry1726.Brookline has the coordinates()8, 2 , andCharleston has the coordinates()4,5 .()()()()()( )2221212222485243169255=−+−=−+−=−+=+==dxxyy28.Everett has the coordinates()3,5 ,−andFairfield has the coordinates()8,3 .−−()()()()()()22212122228335582564899.3=−+−⎡⎤=−− −+ −−⎣⎦=−+ −=+=≈dxxyy30.Platform D has coordinates()20, 20 , andPlatform E has coordinates()30,15 .−()()()()()()22212122223020152010351001225132536.4 m=−+−=−+ −−=+ −=+=≈dxxyy32.a.()()()()( )()2221212222537186643610010=−+−⎡⎤⎡⎤=− −+−− −⎣⎦⎣⎦=+ −=+==dxxyyb.()()1212,221735 ,2228,221,4++⎛⎞=⎜⎟⎝⎠⎛⎞− + −−+=⎜⎟⎜⎟⎝⎠−⎛⎞=⎜⎟⎝⎠=−xxyyM34.a.()()()()()()22212122221432552525507.1=−+−⎡⎤⎡⎤=− −+− −⎣⎦⎣⎦=+=+=≈dxxyyb.()1212,224123,2231,221.5, 0.5++⎛⎞=⎜⎟⎝⎠−+−+⎛⎞=⎜⎟⎝⎠−⎛⎞=⎜⎟⎝⎠=−xxyyM36.a.()()()()()()22212122223553224482.8=−+−⎡⎤⎡⎤=−− −+−− −⎣⎦⎣⎦=+ −=+=≈dxxyyb.()()()1212,225335,2288,224,4++⎛⎞=⎜⎟⎝⎠⎛⎞−+ −−+ −=⎜⎟⎜⎟⎝⎠−−⎛⎞=⎜⎟⎝⎠=−−xxyyM
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Solution Manual for Geometry, 1st Edition - Page 22 preview imageChapter 1:A Beginning of GeometryISM:Geometry1838.Point A has coordinates()0,3 , and Point Bhas coordinates()2,2 .−−a.()()()()()()2221212222202325425295.4=−+−=−−+ −−=−+ −=+=≈dxxyyb.()()()1212,220232,2221,221,0.5++⎛⎞=⎜⎟⎝⎠⎛⎞+ −+ −=⎜⎟⎜⎟⎝⎠−⎛⎞=⎜⎟⎝⎠=−xxyyM40.Central Station has coordinates()0,3 , andSouth Station has coordinates()0,4 .−()()()()()()2221212222004307049497 miles=−+−=−+ −−=+ −=+==dxxyy42.Cedar Station has coordinates()3,1 ,−andCity Plaza Station has coordinates()0, 0 .()()()()( )()222121222203013191103.2 miles=−+−⎡⎤=− −+−⎣⎦=+ −=+=≈dxxyy44.Sycamore Station has coordinates()4,6 ,−and Cedar Station has coordinates()3,1 .−()()()()()()22212122223416774949989.9 miles=−+−⎡⎤=−−+− −⎣⎦=−+=+=≈dxxyy46.a.If PointPis the midpoint, let thecoordinates ofRbe()22,.xy()22244, 6,22xy++⎛⎞−=⎜⎟⎝⎠2:x22224,28210xxx+−=−=+−=and2:y2224621248yyy+==+=IfPis the midpoint, the coordinates ofRare()10,8 .−If PointQis the midpoint, let thecoordinates ofRbe()22,.xy()22462, 4,22xy−++⎛⎞=⎜⎟⎝⎠2:x22242,2448xxx−+== −+=and2:y222642862yyy+==+=IfQis the midpoint, the coordinates ofRare()8, 2 .If PointRis the midpoint, find thecoordinates ofR.()1212,224264,222 10,221,5++⎛⎞=⎜⎟⎝⎠−++⎛⎞=⎜⎟⎝⎠−⎛⎞=⎜⎟⎝⎠=−xxyyMIfRis the midpoint, its coordinates are()1,5 .−
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Solution Manual for Geometry, 1st Edition - Page 23 preview imageISM:GeometryChapter 1:A Beginning of Geometry19b.Find the possible lengths of.RQIfPis the midpoint:()()()()()()22212122222104812414416160=−+−⎡⎤=− −+−⎣⎦=+ −=+=dxxyyIfQis the midpoint:()()()()()()222121222228426236440=−+−=−+−=−+=+=dxxyyIfRis the midpoint:()()()()( )()22212122222145319110=−+−⎡⎤=− −+−⎣⎦=+ −=+=dxxyyIn order forRQto have a length of160,Pmust be the midpoint. So,Rmust have the coordinates()10,8 .−48.The terms inside the distance formula()()222121=−+−dxxyyare squared,which always results in a positive answer.Then they are added, and the sum of twopositive numbers is always positive. Thesquare root of a positive number is alwayspositive.50.Your friend substituted the values in theDistance Formula incorrectly. Eachsubtraction is written with the wrong partner.It should say the following.()()()()222121223185=−+−=−+−dxxyySection 1.8 Practice1.RSmust be twice as long asXYbelow.Step 1:Draw a ray with endpointR.Step 2:Open the compass to the length of.XYWiththe same compass setting, put the compasspoint onRand draw an arc that intersects theray.Step 3:Without changing the compass setting, putthe compass point on the intersection of theray and the arc and draw an arc that intersectsthe ray on the opposite side of theintersection asR. Label the point ofintersectionS.2.Q∠must be congruent toB∠below.Step 1:Draw a ray with endpointQ.Step 2:With the compass point on vertexB, draw anarc that intersects the sides of.B∠Step 3:With the same compass setting, put thecompass point on pointQ.Draw an arc thatintersects the ray.Step 4:Open the compass to the distance betweenthe points of intersection between the sides ofB∠and the arc from Step 2. Keeping thesame compass setting, put the compass pointon the intersection from Step 3. Draw an arcthat intersects the arc.Step 5:Draw a ray from pointQthrough theintersection of the arcs.
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Solution Manual for Geometry, 1st Edition - Page 24 preview imageChapter 1:A Beginning of GeometryISM:Geometry203.Draw.STStep 1:Put the compass point on pointSand drawarcs above and below the segment. Be surethat the opening is greater than 1.2STStep 2:With the same compass setting, put thecompass point on pointTand draw arcsabove and below the segment that intersectthe arcs from Step 1.Step 3:Draw a line that passes through the pointswhere the arcs intersect.4.Draw an obtuse angle,.Y∠Step 1:Put the compass point on vertexY. Draw arcsthat intersect the sides of the angle. Label theintersectionsXandZ.Step 2:Put the compass point on pointXand draw anarc inside the angle. Then put the compasspoint on pointZwithout changing thecompass setting, and draw an arc thatintersects the last arc at a pointP.Step 3:Draw a ray from vertexYthrough pointP.Exercise Set 1.82.VWmust be twice as long asABbelow.Step 1:Draw a ray with endpointV.Step 2:Open the compass to the length of.ABWiththe same compass setting, put the compasspoint onVand draw an arc that intersectsthe ray.Step 3:Without changing the compass setting, putthe compass point on the intersection of theray and the arc and draw an arc that intersectsthe ray on the opposite side of theintersection asV.Label the point ofintersectionW.4.The length ofQJmust be the difference ofthe lengths ofTRandPSbelow.Step 1:Draw a ray with endpointQ.Step 2:Open the compass to the length of.TRPutthe compass point onQand draw an arc thatintersects the ray.Step 3:Open the compass to the length of.PSPutthe compass point on the intersection of theray and the arc and draw an arc that intersectsthe ray between the intersection and pointQ.Label this pointJ.6.mF∠must be twice the measure of.C∠Step 1:Draw a ray with endpointF.Step 2:With the compass point on vertexC, draw anarc that intersects the sides of.C∠Step 3:With the same compass setting, put thecompass point on pointF.Draw a large arcthat intersects the ray.Step 4:Open the compass to the distance betweenthe points of intersection between the sides ofC∠and the arc from Step 2. Keeping thesame compass setting, put the compass pointon the intersection from Step 3. Draw an arcthat intersects the arc above the ray.
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Solution Manual for Geometry, 1st Edition - Page 25 preview imageISM:GeometryChapter 1:A Beginning of Geometry21Step 5:Keeping the same compass setting, putthe compass point on the intersection fromStep 4. Draw an arc that intersects the largearc above the ray.Step 6:Draw a ray from pointFthrough theintersection from Step 5.8.Construct the perpendicular bisector ofTRbelow.Step 1:Put the compass point on pointTand drawarcs above and below the segment. Be surethat the opening is greater than 1.2TRStep 2:With the same compass setting, put thecompass point on pointRand draw arcsabove and below the segment intersecting thearcs from Step 1.Step 3:Draw a line that passes through the pointswhere the arcs intersect.10.Draw an obtuse angle,.XQZ∠Step 1:Put the compass point on vertexQ. Draw arcsthat intersect the sides of the angle.Step 2:Put the compass point on one of theintersections and draw an arc inside theangle. Then put the compass point on theother intersection without changing thecompass setting, and draw an arc thatintersects the last arc.Step 3:Draw a ray from vertexYto the point wherethe two arcs you drew in Step 2intersect.12.Draw a right angle,.PSQ∠Step 1:Put the compass point on vertexS. Draw arcsthat intersect the sides of the angle.Step 2:Put the compass point on the intersection ofan arc and the side of the angle, and draw anarc inside the angle. Without changing thecompass setting, put the compass point on theother intersection to draw an arc thatintersects the last arc. Label this pointT.Step 3:Draw a ray from vertexSto pointT.
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Solution Manual for Geometry, 1st Edition - Page 26 preview imageChapter 1:A Beginning of GeometryISM:Geometry2214.Construct the angle bisector of.A∠Younow have two angles with measures 12.mA∠Pick one of the smaller angles and do theconstruction of the angle bisector for thatangle. The measures of the two angles arehalf of the original angle, or 14.mA∠16.There is only one point at which a segmentcan be cut in half, so a segment can only haveone bisector.18.In space, there is a third axis so there aremany angles at which a perpendicularbisector can bisect the segment, so there arean infinite number of perpendicular bisectorsfor this segment in space.20.Draw angles 1 and 2. Construct an anglecongruent to2∠that shares a side with1∠so that the other side of the angle is in theinterior of1.∠Step 1:Put the compass point on the vertex of2.∠Draw arcs that intersect each side of theangle. Label the pointsAandB.Step 2:With the same compass setting, put thecompass point on the vertex of1∠and drawand arc that intersects the side of angle 1.Label this pointD.Step 3:Set the compass to be the length betweenpointAand pointB. Put the compass on pointD, and draw an arc that intersects the arcfrom the previous step. Label the point ofintersectionE.Step 4:Draw a ray from the vertex of1∠throughpointE. Label the angle enclosed by this rayand the leg of1,∠.C∠22.Draw∆ABCwith acute angles.Step 1:Put the compass point on vertexA, and drawand arc that intersects both sides of the trianglethat meet atA. Label the pointsDand.EStep 2:Open the compass a little more than half thedistance betweenDand.EPut the compasspoint on pointDand draw and arc. With thesame compass setting, draw an arc using thepointEthat intersects the other arc.Step 3:Draw the ray with endpointAthrough theintersection of the two arcs.Step 4:Repeat Steps 1−3 with the other vertices ofthe triangle to construct angle bisectors forB∠and.C∠The angle bisectors of a triangle intersect at apoint inside the triangle.
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Solution Manual for Geometry, 1st Edition - Page 27 preview imageISM:GeometryChapter 1:A Beginning of Geometry2324.Draw segmentABthat is 5 cm long.Step 1:Set the compass length to the segment that is5 cm long. Put the compass point on pointAand draw an arc above.ABStep 2:Set the compass length to the segment that is2 cm long. Put the compass point on pointBand draw an arc that intersects the previousarc. Label the intersectionC.Step 3:Use a straightedge to draw.ABC∆Thelengths ofACandBCare 5 cm and 2 cmrespectively.26.Use the segments below and draw segmentABthat is 4 cm long.Step 1:Open the compass to the length of thesegment that is 2 cm long. Put the compasspoint at pointAand draw a circle.Step 2:With the same compass setting, put thecompass point on pointBand draw a circle.The only place where the circles intersect ison.ABSo, it is not possible to construct atriangle with the given side measures.28.Use the segments shown below and drawsegmentABthat is 4 cm long.Step 1:Open the compass to the length of thesegment that is 4 cm long. Put the compasspoint at pointAand draw an arc above.ABStep 2:With the same compass setting, put thecompass point on pointBand draw an arcthat intersects the previous arc. Label theintersectionC.Step 3:Use a straightedge to draw.ABC∆Thelengths ofACandBCare both 4 cm.30.The answer is A. In order for the half circlesto all line up correctly, the line at the bottommust be drawn first. Count the number of halfcircles. There are five. So to construct this,you would use a straight edge to draw thesegment and then a compass to draw five halfcircles.Chapter 1 Review1.Each new image has a square added insidethe smaller square, rotated by 45 .°2.Each shape inside the circle has one moreside added to it from the previous one.3.Each fraction is being multiplied by 1 .211,48964.Each number is decreasing by 5.20, 155.Answers may vary. Sample Answer:landm6.HJJGQR
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Solution Manual for Geometry, 1st Edition - Page 28 preview imageChapter 1:A Beginning of GeometryISM:Geometry247.Answers may vary. Sample Answer:A,B,C8.T9.True; Postulate 1.3-1 states, “Through anytwo points, there is exactly one line.” Thismeans that two points are always collinear.10.False;JJJJGLMhas endpointLandJJJGMLhasendpointM. Thus, since the rays have twodifferent endpoints they cannot be thesame ray.11.257,253−−= −−+=, so two possiblecoordinates are7−and 3.12.Use the Midpoint Formula.23122−+=13.3541015=+=−=ABBCmmm14.Use the Segment Addition Postulate to findthe value ofa.85021++==aaaSubstitute the valueato findand.XYYZ21==XYa;829=+=YZa15.This is an acute angle.16.This is a right angle.17.Use the Angle Addition Postulate to find.mPQR∠612536∠+∠=∠∠=∠−∠=° −°=°mPQRmMQPmMQRmPQRmMQRmMQP18.282214∠=∠+=+=mNQMmPQRxxx19.∠ADB(or)BDA∠and∠BDC(or).CDB∠20.Answers may vary. Sample Answer:∠ADCand∠CDF21.Answers may vary. Sample Answer:∠ADCand∠EDF22.Answers may vary. Sample Answer:∠ADBand∠BDF23.Since the angles form a linear pair, the sumof their measures is 180 .°(331)(26)180525180515531++−=+===xxxxx24.Since the angles are complementary, the sumof their measures is 90 .°3(415)9071590710515+−=−===xxxxx25.Use the Distance Formula.22(0( 1))(45)1.4− −+−≈26.Use the Distance Formula.22(6( 1))(2( 1))7.6− −+− −≈27.Use the Midpoint Formula to find thex-coordinate.330022−+==Use the Midpoint Formula to find they-coordinate.2( 2)0022+ −==So the coordinates of the midpoint are()0, 0 .28.Use the Distance Formula.22( 33)(2( 2))7.2−−+− −≈
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Solution Manual for Geometry, 1st Edition - Page 29 preview imageISM:GeometryChapter 1:A Beginning of Geometry2529.Solve the Midpoint Formula for2.x121221222+=+==−xxxxxmxxmxmxNow use thexvalues you have been given tosolve for2.x2122( 1)( 8)6=−=−− −=xxmxSolve the Midpoint Formula for2.y121221222+=+==−yyyyymyymymyNow use theyvalues you have been given tosolve for2.y2122(1)42=−=−= −yymySo the coordinates of the endpointKare()6,2 .−30.Solve the Midpoint Formula for2.x121221222+=+==−xxxxxmxxmxmxNow use thexvalues you have been given tosolve for2.x2122(5)91=−=−=xxmxSolve the Midpoint Formula for2.y121221222+=+==−yyyyymyymymyNow use theyvalues you have been given tosolve for2.y2122( 2)( 5)1=−=−− −=yymySo the coordinate of the endpointKis()1, 1 .31.Holding your protractor on the paper, draw astraight line along the bottom of theprotractor, shown below as.ABJJJGThen mark apointCat 73°and draw a straight line fromAto that pointC.Step 1:Use a straight edge to draw a ray. Put thecompass point on the vertex of the angle anddraw an arc that intersects both sides of theangle. Then put the compass point on theendpoint of the drawn ray and draw an arcusing the same compass setting.Step 2:Set the compass to the distance between thetwo intersection points on the angle. Then putthe compass point on the intersection of theray and the arc from Step 1, and draw an arcusing the same setting that intersects the arcfrom Step 1.Step 3:Use a straight edge to draw a ray from theendpoint of the ray from Step 1 through theintersection from Step 2.32.Holding your protractor on the paper, draw astraight line along the bottom of theprotractor, shown below as.ABJJJGThen mark apointCat 60°and draw a straight line fromAto that pointC.
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Solution Manual for Geometry, 1st Edition - Page 30 preview imageChapter 1:A Beginning of GeometryISM:Geometry2633.Step 1:Put the compass point on pointLand draw along arc. Be sure the opening is greater than1.2LMWith the same compass setting, putthe compass point on pointMand drawanother long arc.Step 2:Draw a straight line between the two arcintersections. Now this new line is theperpendicular bisector.34.a.Sketch the angle.Step 1:Use a straight edge to draw a ray. Put thecompass point on the vertex of the angleand draw an arc that intersects both sidesof the angle. Then put the compass pointon the endpoint of the drawn ray anddraw an arc using the same compasssetting.Step 2:Set the compass to the distance betweenthe two intersection points on the angle.Then put the compass point on theintersection of the ray and the arc fromStep 1, and draw an arc using the samesetting that intersects the arc from Step 1.Step 3:Use a straight edge to draw a ray from theendpoint of the ray from Step 1 throughthe intersection from Step 2.b.Step 1:Start with the angle constructed in part a.Put the compass point on an intersectionof the drawn arc and a side. Draw an arcinside the angle. Then move the compasspoint to the intersection of the drawn arcand the other side and draw another arcinside the angle.Step 2:Draw a ray from vertexBthrough theintersection of the arcs from Step 1.35.Yes; All four points are located on planeAEF.36.Yes; All four points are located on planeDCE.37.No; Sample Answer: PointsH, G,andFarein planeHGFand pointBis not in that plane.38.No; Sample Answer: PointsA, E,andBarein planeAEBand pointCis not in that plane.39.a.HJJGBFb.HJJGEH40.Answers may vary. Sample Answer:,PRHJJGHJJGRQ41.,PQJJJGJJJGPB42.Answers may vary. Sample Answer:,∠∠CPQCPB43.Answers may vary. Sample Answer:D, P, C44.Answers may vary. Sample Answer:JJJGPAandJJJGPR45.Answers may vary. Sample Answer:,,AP PR RQ46.Answers may vary. Sample Answer:∠APCand∠CPR47.Answers may vary. Sample Answer:∠APDand∠CPR48.a.Since the angles form a linear pair, thesum of their measures is 180 .°18023180518036∠+∠=+===mABCmCBDxxxx
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Solution Manual for Geometry, 1st Edition - Page 31 preview imageISM:GeometryChapter 1:A Beginning of Geometry27b.22(36)72∠===mABCxSo,∠ABCis acute.33(36)108∠===mCBDxSo,∠CBDis obtuse.49.1( 4)3− − −=50.303−=51.633−=52.6( 1)7− −=53.Use the fact thatACis congruent toCDtosolve forx.45383+=+=xxxSolve for.AC4(3)517=+=ACSolve for.CD3(3)817=+=CDUse the Segment Addition Postulate to find.AD171734+=+=ACCDAD54.Use the Angle Addition Postulate to find.mFCB∠1309535∠+∠=∠∠=∠−∠=° −°=°mFCBmBCDmFCDmFCBmFCDmBCD55.Use the Angle Addition Postulate.180180∠+∠+∠=−∠−∠=∠mFCAmFCEmECDmFCAmECDmFCESubstitute∠mFCAfor.mECD∠180−∠−∠=∠mFCAmFCAmFCESolve for.mFCE∠180505080mFCE∠=° −° −°=°56.No;Qis not always the midpoint. It dependson the situation. If all three points are distinctand collinear then the answer is yes,Qisalways the midpoint. If the three points arenot distinct or they are not collinear then itdoes not mean thatQis the midpoint.57.No;C, F,andGare collinear, soHJJGABmustintersectHJJGGFatCin planeM.Chapter 1 Test1.Each new number has 6 being added to it.31, 372.Each new image has each section insidebeing cut in half. The number of sectionsdoubles each time.3.B4.C5.C6.C7.A;SinceJJJJGMNbisects∠EMG, the measureof∠EMGis twice the measure of.EMN∠8.∠EFDandor2AFC∠∠are verticalangles.9.∠EFDand3 or1∠∠form a linear pair.10.The complementary angles areand.DEFFEA∠∠11.True12.False13.True14.True
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Geometry by Elayn Martin-Gay

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