20. First, draw a segment ST� with midpoint M. Addcongruence marks.
Next, add point P so thatSP � PT. Draw SP� and PT�and add congruence marks.
Finally, extend PT� totwice its length. Mark theendpoint Q. BecausePT � TQ, mark TQ�with the samecongruence markused for PT�.
21. The first point listed must be the endpoint, A; theother point can be either B or C. So, the two namesare AB�� and AC��.
22. PM��, PN��
23. XY��, XZ��
24. The first point listed must be the endpoint ofthe ray.
25.
26.
27.
28. 10. First, list all the segments with endpoint A: AB�,AC�, AD��, AE�. Next, list the segments with endpoint B,but don’t include AB� because you already listed it:BC�, BD�, BE�. Next, list all the segments withendpoint C, but don’t include AC� or BC� becauseyou already listed them: CD��, CE�. The only segmentwith endpoint D that hasn’t been listed yet is DE�.All the segments with endpoint E have been listed.In all, there are 10 possible segments: AB�, AC�, AD�,AE�, BC�, BD�, BE�, CD��, CE�, DE�.
SM
T
C
A BD
M N
X
Y
BA
P
Q
SM
T
P
SM
T
CHAPTER 1
LESSON 1.1
EXERCISES1. Possible answers: Point: ball, corners of cube andnet; Segment: connectors of corners on cube, edgeof net, edge of wall, table markings; Plane: sides ofcube, table top, net, wall; collinear points: Intersec-tion points on one string of netting, edges of smallcubes that form edges of large cube; coplanar: eachperson’s paddle, hand and forearm, designs onshirts, designs on walls, lines on netting
2. PT���, TP���
3. The name can have any two of the point names—A,R, or T—in either order. So, the correct answer can beany two of the following: AR���, RA���, AT���, TA���, RT���, TR���.
4. Any two of the following: MA���, MS���, AS���, AM���, SA���, SM���
5. 6.
7. 8. AC� or CA�
9. PQ� or QP�
10. TR� or RT�, RI� or IR�, and TI� or IT�
11.
12.
13. mAB�� 14.3 cm 14. mCD�� � 6.7 cm
15–17. Check the length of each segment to see if itis correct. Refer to each exercise for the correctmeasurements.
18. R is the midpoint of PQ�. X is the midpoint of WY��.Y is the midpoint of XZ�. No midpoints are shownin �ABC. (Note: Even though AD�� and DC�� appearto be the same length, you cannot assume they arecongruent without the markings.)
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DEVELOPING MATHEMATICAL REASONINGThe other four tetrominoes:
Note that if two tetrominoes are congruent, they areconsidered to be the same. For example, these tetromi-noes are all the same:
COORDINATE GEOMETRY1
EXERCISES
1. x � �x1 �2x2� � �
12 �2(�6)� � 3 and y � �
y1 �2y2�
� ��7 �
215
� � 4, so the midpoint is (3, 4).
2. x � �x1 �2x2� � �
�17 �2(�1)� � �9 and y � �
y1 �2y2�
� ��8 �
211
� � 1.5, so the midpoint is (�9, 1.5).
3. x � �x1 �2x2� � �
14 �2(�3)� � 5.5 and y � �
y1 �2y2�
� �(�7)
2� 18� � 5.5, so the midpoint is (5.5, 5.5).
4. Let (x, y) represent the other endpoint; then
3 � �122� x� and 18 � �
�82� y�. Here are the
step-by-step solutions to these equations:
3 � �122� x�
6 � 12 � x Multiply both sides by 2.
�6 � x Subtract 12 from both sides.
18 � ��82� y�
36 � �8 � y Multiply both sides by 2.
44 � y Add 8 to both sides.
The endpoint is (�6, 44).
5. Yes. The coordinates of the midpoint of a segment
with endpoints (a, b) and (c, d) are found by taking
the average of the x-coordinates, �a �2c
�, and the
average of the y-coordinates, �b �2d
�. Thus, the
midpoint is ��a �2c
�, �b �2d
��.6. (3, 2) and (6, 4). To get the first point of trisection,add the coordinates of points A and B to get (9, 6),then multiply those coordinates by �
13� to get (3, 2).
To get the second point of trisection, add thecoordinates of points A and B to get (9, 6), thenmultiply those coordinates by �
23� to get (6, 4). This
only works because the coordinates of one of thepoints are (0, 0).
7. Possible method: Find the midpoint of the segment,and then find the midpoint of each half of thesegment.
29–31.
32. Yes. Change the signs of the coordinates to getP�(6, 2), Q�(5, �2), R�(4, �6).
Because the slope ��rruisne
�� between any two points is�41�, the points are collinear.
33. Plot point N. It will be easiest to determine thelengths if you draw a horizontal or vertical segment.To locate P and Q, mark points the same number ofunits above and below (or to the left and the rightof) point N.
Possible answer:
34.
35. Possible answer: use the ruler to drawAB� of length 8 cm. Then, use theruler to draw any segment BC� oflength 11 cm for which A, B, and Care not collinear. Connect points A and C to formAC�, which becomes the third side of �ABC.
For the second triangle, use thesame procedure to draw �DEF,but change the size of theopening between the two sides with lengths 8 cmand 11 cm.
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5. In the first figure, �S, �P, �R, �Q. The otherangles all have the same vertex, T, so none of themcan be described with just one letter. In the secondfigure, all the angles have the same vertex, D, sonone of them can be described with just one letter.
6. 90°
7. Side QC�� passes through 60° on one scale and 120°on the other. Because �AQC is larger than a 90°angle, 120° must be the correct measure.
8. 45° 9. 135° 10. 45° 11. 135°
12. One way to find m�CQB is to find m�AQC andm�AQB and subtract. m�AQC � 120° andm�AQB � 90°, so m�CQB � 30°.
13. To find m�XQY, find m�AQY and m�AQX andsubtract. m�AQY � 135° and m�AQX � 45°, som�XQY � 90°.
16. 69°. Place the center mark of the protractor onpoint A. Rotate the zero-edge of the protractor toline up with side AC��. AM�� passes through 69° onone scale and 111° on another. Because �MAC issmaller than a 90° angle, m�MAC � 69°.
17. 110° 18. 40° 19. 125° 20. 55°
21. �SML has the greater measure becausem�SML � 30° and m�BIG � 20°. The measure ofan angle is the amount of rotation from one side tothe other. It has nothing to do with the lengths ofthe angle’s sides.
AD
C
B
A
180
17016
015
014
013
012
011
010
090
8070
60 50 40 30 20 100M
C
0
1020
3040
5060
7080
90
100110
120 130 140 150 160 170180
8. a. For Figure 1, both midpoints are (5.5, 6.5). ForFigure 2, both midpoints are (16, 6.75). ForFigure 3, both midpoints are (29.75, 5.5).
b. For these figures, the midpoints of the twodiagonals are the same point. This means thediagonals intersect each other at their midpoints.
DEVELOPING MATHEMATICAL REASONINGHere is one way to organize your work to make sure youhave found all the possibilities. Keep in mind thatcongruent pentominoes are considered to be the same.
LESSON 1.2
EXERCISES1. When three letters are used to name an angle, themiddle letter must be the vertex. Also, a single lettercan name an angle only if there is no confusionabout which angle it is referring to. For example,there is only one angle with vertex E, so it can benamed �E. However, there are two angles withvertex O, so the name �O is ambiguous. �TEN,�NET, �E; �FOU, �UOF, �1; �ROU,�UOR, �2.
In Exercises 2–4, the angles may be any size, but thevertex labels should be the same as those in theanswers below.
2. 3.
4.M
L
S
A T
N
G
I
B
Pentominoes with arow of five squares
Pentominoes with a rowof four squares
Pentominoes with a row of three squares and both othersquares on the same side of that row
Pentominoes with a row of three squares and the othersquares on opposite sides of that row
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27. Don’t forget that at half past the hour, the hourhand will be halfway between 3 and 4.
28.
29.
30. Possible answer: 4:00
31.
32.
33.
34.
35. MI� and MY�� have the same congruence markings, sothey are the same length. That is, MI � MY. IC� andCK� have the same markings, so IC � CK. �M and�I have the same markings, so m�M � m�I.
36. �SEU; �EUO; MO
37. A quarter of a rotation is 90°. So,x � 90° � (15° � 21°) � 54°.
38. Half of a rotation is 180°. So,y � 180° � (41° � 37°) � 102°.
W
B OT
I
G T
I
A
N
R
T H
SA
T
H
O
908
6 8
22. Draw a ray with endpoint A.
Place the center mark of the protractor on point Aand line up the zero-edge with the ray.
Mark a point next to the 44° mark of theprotractor. (Use the scale that will give you an anglesmaller than 90°.)
Remove the protractor, and draw a ray frompoint A through the point you marked.
23. 24.
25. Align your protractor as if you were going tomeasure �A. Mark a point at the 22° mark (halfthe measure of �A). Remove the protractor, anddraw a ray from point A through the point youmarked. Add markings to show that the two anglesformed are congruent.
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You can draw a different triangle with these twoangle measures, but any two such triangles will besimilar (look alike). Triangle DEF with DE � 6 cmis one such triangle.
46.
47. Possible answer: MS � DG means that the distancebetween M and S equals the distance between Dand G. MS�� � DG�� means that segment MS iscongruent to segment DG. The first statementequates two numbers. The second statementconcerns the congruence between two geometricfigures. However, both statements convey the sameinformation and are marked the same way ona diagram.
48. The wire has length 4.36 cm. The center of gravityis the midpoint of the wire, which is 4.36 dividedby 2, or 2.18 cm from the end of the wire.
DEVELOPING MATHEMATICAL REASONING
LESSON 1.3
EXERCISES1. Possible answer: 2. Possible answer:
O G
D
458
T E
R
708408
D E6 cm
F
708408
4 cmA B
C
12 cm
6 cm 6 cm
4.36 cm
2.18 cm 2.18 cm
39. A full rotation has measure 360°. Both anglesmarked as congruent have measure z, so2z � 360° � (87° � 135° � 74°) � 64°. Therefore,z � 32°.
40. 288°. First, measure �ACU with a protractor to findm�ACU � 72°. Because a complete rotationaround a point is 360°, the reflex measure of �ACUis 360° � 72° � 288°.
41. 242°. First, measure �QUA with a protractor to findm�QUA � 118°. Because a complete rotationaround a point is 360°, the reflex measure of �QUAis 360° � 118° � 242°.
42. They add to 360° because a complete rotationaround a point is 360°.
43. The incoming angle has measure 50°. Draw anoutgoing angle of measure 50°. The side of theangle does not pass through the pocket, so the ballwill not go in.
44. You will miss the target because the incoming angleis too big.
45. Possible answer: Using the ruler, draw AB� of length4 cm. (Any length can be used for AB�.) Using theprotractor, draw an angle with vertex A so thatm�A � 40° and AB� lies along one of the sides of�A. Using the protractor, now draw an angle withvertex B so that m�B � 70° and AB� also lies alongone of the sides of �B. If necessary, extend the tworays of each angle that do not contain AB� until theyintersect. Label their point of intersection as C.Then �ABC will meet all of the requirements.
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required for a pair of angles to be a linear pair.
13. a. an angle; measures less than 90°
b. angles; have measures that add to 90°
c. a point; divides a segment into two congruentsegments
d. a geometry tool; is used to measure the size ofangles in degrees
14. If P is not between A and B or not between Cand D, then the definition is not true. Here isa counterexample:
The following is a correct definition: If AB��� and CD���intersect at point P so that P is between A and Band P is between C and D, then �APC and �BPDare a pair of vertical angles.
15. True
16. True 17. True
18. False. Think in three dimensions. There are aninfinite number of planes that contain the givenline, and each one contains a line perpendicularthrough the given point.
19. True. Think in three dimensions.
P
BC
D
A
A
458
1358
B
3. Possible answer: 4. Note that DG�� must bea segment and MS���must be a line.
5. Note that PE� must be a segment and AR�� must bea ray.
6.
7. �A and �B are complementary, so their measuresmust add to 90°. Because m�A � 40°, m�B � 50°.
8. �C and �D are supplementary, so their measuresmust add to 180°. Because m�D � 40°,m�C � 140°.
9. B is a Zoid. A Zoid is a creaturewhose interior contains a smalltriangle surrounding a black dot.
10. A good definition places an object in a class andalso differentiates it from other objects in that class.A good definition has no counterexamples.
11. The sum of the measures of two complementaryangles is 90°, whereas the sum of the measures oftwo supplementary angles is 180°.
12. No. Supplementary angles can be unconnected, buta linear pair must share a vertex and a commonside. In the figure, �A and �B are supplementarybecause m�A � m�B � 180°, but they are not alinear pair. Note that they neither share a commonvertex nor a common side, both of which are
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You should find that the reflected ray and the raythat passes through (called the “refracted” ray) aremirror images of each other. They form congruentangles with the mirror.
28. This problem is best solved by trial and error. Hereis a possible method. Point C has to lie on TR���, sochoose that point first. Put it far enough to the leftof point S so that you have room for points A andB. Because �BAC will be a right angle, you knowpoints A and B will be on a segment perpendicularto AC�. Because �ABS will be obtuse, B needs tobe to the right of A and to the left of S. �BATneeds to be an acute angle, so place segment ABsomewhere such that �ABS is obtuse and �BATis acute. Possible answer: A(�8, 8), B(�4.5, 6.5),C(�11, 1).
29. Make a sketch showing the given information.Because D is the midpoint of AC�, AC � 6 cm.Because C is the midpoint of AB�, AB � 12 cm.
30. Make a sketch showing the given information.Because BD�� is the bisector of �ABC,m�ABD � 24°. Because BE�� is the bisectorof �ABD, m�EBD � 12°. Therefore,m�EBC � 12° � 24°, or 36°.
B
A
E
D
C
24°
A D C B
3 cm
y
x–12 –6
A
B
C
T
S
R
6
B
A
Mirror
Reflected�segment
468
468
20. False. Possible counterexample:
21. True. (Parallel lines must be in the same plane.)
22. False. Possible counterexample:
23. False. Possible counterexample:
24. False. Possible counterexample:
25. Point P must be on the line through T and S. Theslope of TS��� is �
69�, or �
23�. One way to find another
point on the line is to start at point T and move up2 units and right 3 units. This would give youpoint (�3, 0). You can continue moving up andover to find more possible points. Possible answers:(�3, 0), (0, 2), or (6, 6).
26. For QR��� � TS���, the lines must have the same slope.The slope of TS��� is �
69�, or �
23�. One way to find
point Q so that the slope of QR��� is �23� is to start at
point R and move up 2 units and right 3 units. Youwill be at point (2, �3). You can continue movingup and over to find more possible points. Possibleanswers: (2, �3) or (5, �1).
27. Just as in pool, the measure of the incoming angleis equal to the measure of the outgoing angle. Theincoming angle has measure 46°, so draw anoutgoing angle with measure 46°.
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4. The polygon has eight sides so it is an octagon.
5. The polygon has six sides so it is a hexagon.
6. The polygon has four sides so it is a quadrilateral.
7. The polygon has seven sides so it is a heptagon.
8. To name the figure, start with any vertex, and listthe letters in order as you move clockwise orcounterclockwise around the figure. A few possiblenames are FIVER, VERFI, REVIF, IVERF.
9. Possible name: quadrilateral FOUR
10. Possible name: equilateral quadrilateral BLOC
11. a. a polygon; has eight sides
b. a polygon; has at least one diagonal outside thepolygon
c. a polygon; has 20 sides
d. a polygon; has all sides of equal length
12. Consecutive angles are angles whose vertices areendpoints of the same side. For example, �C and�Y are consecutive angles. Consecutive sides aresides that share a vertex. For example, CY� and YN�are consecutive sides.
13. Nine. A concave hexagon is a hexagon with at leastone diagonal outside the polygon. The hexagon onthe left below is a convex hexagon. On the right,the hexagon’s nine diagonals have been drawn.
14. Make a sketch. The diagonals are AC�, AD�, BD�,BE�, and CE�.
15. When stating that two triangles are congruent, youneed to list the vertices in the order that shows thecorrespondences. Because T corresponds to E,I corresponds to A, and N corresponds to R,�EAR � �TIN.
16. �OLD � �WEN
17. a. a � 44; b � 58; c � 34. Match up correspondingsides of the two congruent pentagons, and applythe part of the definition of congruent polygons
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x � 17 Divide both sidesby 4.
If x � 17, then x � 3 � 14 and x � 8 � 25.Therefore, AB � 14 m and CD � 25 m.
24. Complementary angles: �AOS and �SOC; verticalangles: �OCT and �ECR, or �TCE and �RCO
25.
26. Possible answer:
27. All are possible except two points.
PERFORMANCE TASKThere can be 1, 2, 3, 4, or infinite intersections of pointsif one or more sides overlap or the quadrilaterals arecongruent.
LESSON 1.5
EXERCISES1. Figure D. An equilateral triangle has three sides ofequal length.
2. Figure A. A scalene right triangle has three sides ofdifferent lengths and one right angle.
3. Figure C. An isosceles right triangle has two sides ofthe same length and one right angle.
0 pts.
3 pts. 4 pts.
1 pt.
5 pts. 6 pts.
20
5
E
A B
C DF
that says that corresponding sides of congruentpolygons are congruent:
OW�� corresponds to HI�, so a � 44.
ER� corresponds to NK�, so b � 58.
PR� corresponds to TK�, so c � 34.
b. Because �P and �T correspond, m�T � m�P� 87°. Because �W and �I correspond, m�I� m�W � 165°.
18. PA� � FI� and �IVE � �ANC.
19. Many hexagons can be drawn that satisfy therequirements. The figure shows one such hexagon.
20. An equilateral concave polygon has five sides ofequal length and at least one diagonal outside thepentagon. The figure shows one such pentagon.
An equiangular convex polygon has five angles ofequal measure. The figure shows one such pentagon.
21. 84 in. A regular dodecagon has 12 congruent sides.If the length of each side is 7 in., the perimeter willbe (12)(7 in.) � 84 in.
22. 5.25 cm. An equilateral octagon has eight congruentsides. If the perimeter is 42 cm, the length of eachside will be �
428cm� � 5.25 cm.
23. AB � 14 m, CD � 25 m. Use the information givenin the figure to write an equation. Notice from thefigure markings that AB � BC, so AB � x � 3; andCD � AE, so CD � x � 8. The perimeter of pentagonABCDE is 94 m, so 2(x � 3) � 2(x � 8) � 16 � 94.Solve this equation to find the value of x.
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12. There is an infinite number of possible answers.Here are a few possibilities: (�1, �1), (�1, 0),(�4, 3).
13. There are six possible answers: (3, �3), (3, 4),(�11, 4), (�11, �3), (�0.5, 0.5), or (�7.5, 0.5).The triangle must have one right angle and twocongruent sides. To find the third vertex of anisosceles right triangle with right angle at vertexM, draw a horizontal line through M. BecauseME�� is 7 units long, count over 7 units in eitherdirection to locate point O. You will get (3, 4)or (�11, 4).
You can use a similar method to find the thirdvertex if the right angle is at E. You will get (3, �3)or (�11, �3).
If the right angle is at O, then MO�� and EO� will bethe congruent sides. Visually, you can see that vertexO must be located on the horizontal line y � 0.5(that is, the horizontal line through the midpoint ofME��). For O to be a right angle, the slopes of MO��and EO� must be opposite reciprocals. The twopoints that satisfy these criteria are (�0.5, 0.5)or (�7.5, 0.5).
14. Possible answers: (�3, 1), (�1, �9), (2, 2), (4, �8),(0, �1), or (1, �6). If the right angle is at C, thenthe slope of CR� will be �5, the opposite reciprocalof the slope of CL�. Draw a line through point Cwith slope �5 and locate point R so that CR � CL.You can use similar reasoning to locate point R ifthe right angle is at L. If the right angle is at R, theproblem is a little trickier. You can see that in orderfor RC to equal RL, point R will have to be on theperpendicular line through the midpoint of CL�. Themidpoint of CL� is (0.5, �3.5). Draw a line withslope �5 through this point. Then try to locate apoint R so that the slopes of RC� and RL� are oppo-site reciprocals.
15. Possible answer: Use your ruler to draw linesegment AB� of length 9 cm. Use your protractor todraw an angle with vertex A so that m�A � 45°and AB� lies along one of the sides of �A. Pick anypoint on the other side of �A, and label it C.Finally, draw segments AC� and BC�. The triangle willbe different depending on which point you choosefor C. Two sample triangles are shown.
O OM
E
–4–8 2
2
–2
–4
x
y
–6 –2
4
4. Figure B. An isosceles obtuse triangle has two sidesof the same length and one obtuse angle.
5. 6.
7. 8. Possible answer:
9. Possible answer:
10. Possible answer: Use the ruler to draw AB� of anylength. (In the figure, the length of AB� is 4 cm.)Then draw BC� so that BC � AB, and A, B, and Care not collinear. (Any angle measure can be usedfor �B. In the figure, an 80° angle was used.)Connect points A and C to form the third side of�ABC. Notice that because AB� and BC� are thecongruent sides of this isosceles triangle, AC� is thebase, as required.
11. Possible answer: Use the protractor to draw anyobtuse angle, and label its vertex as P. (In the figure,this angle measure was chosen to be 120°.) Alongthe rays of �P, measure equal distances to form PA�and PZ�. (Any length can be used for these sides, aslong as the two segments have the same length. Inthe figure, this length was chosen to be 4 cm.)Connect points A and Z to form the third side ofthe triangle. Notice that because PA� and PZ� are thetwo congruent sides of isosceles triangle ZAP, thebase angles are �A and �Z, as required.
DG5_Solution_Manual_CH01_Printer.qxd 3/16/16 10:45 AM Page 16
look like a kite, but the markings indicate it isa kite.)
Third figure: �I and �J are right angles.
Fourth figure: QP� � MN�� and QM�� � PN�, or MNPQis a parallelogram.
2. Figure B. A trapezoid is a quadrilateral with exactlyone pair of parallel sides.
3. Figure D. A rhombus is a quadrilateral with fourcongruent sides.
4. Figure F. A rectangle is a parallelogram with fourcongruent angles.
5. Figure C. A kite is a quadrilateral with two distinctpairs of consecutive congruent sides.
6. Figures A, D, and F. A parallelogram is a quadrilat-eral with two pairs of parallel sides.
7.
8.
9. 10.
11. To have two outside diagonals, the hexagon must beconcave with two “dents.”
12. A regular quadrilateral has four congruent sides andfour congruent angles. It is a square.
13. There are three possibilities for the other twovertices: (2, 6) and (�2, 4); (6, �2) and (2, �4);(3, �1) and (1, 3). In two cases, the given verticesare the endpoints of a side. In the third case, theyare the endpoints of a diagonal.
E U
L
Q
A
T
R
H
G
I
N
BF
E
D I
Z O
16. Possible answer: Use your ruler to draw line segmentAB� of length 10 cm. On a sheet of patty paper, useyour protractor to draw an angle with vertex C sothat m�C � 40°, and extend sides of the angle.Move the patty paper with �C over the drawing ofAB� until the points A and B touch the sides of �C.Trace over AB� to create the triangle on the pattypaper. The triangle will be different depending onwhere A and B touch the sides of �C. Two sampletriangles are shown.
17. True 18. True
19. False. A diagonal connects nonconsecutive vertices.
20. False. An angle bisector divides an angle into twocongruent angles.
21. True
22. Find the midpoint of each rod. All the midpoints lieon the same line; place the edge of a ruler underthis line.
23. Possible answer: 24. Possible answer:
LESSON 1.6
EXERCISES1. First figure: AD�� � BC� and �D is a right angle.
Second figure: EH� � HG�� and EF� � FG�, or EFGHis a kite. (Note: EFGH is drawn so that it doesn’t
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hypotenuses, and the common side becomes adiagonal of the rectangle.
b. Isosceles right triangles. Two congruent isoscelesright triangles must be used in order to get rightangles and four congruent sides for the square.The isosceles right triangles are joined by theirhypotenuses, and the common side becomes adiagonal of the square.
20. Possible answer:
21. Possible answer:
22. Use your compass to draw a circle. Because aregular pentagon has five equal sides, you want todivide the circle into five equal pieces. Draw five 72°angles (360° � 5 � 72°) with vertices at the centerof the circle as shown below. Then connect thepoints where the sides of the angles intersectthe circle.
23. Possible answer: draw two sides of the octagon byconnecting A, B, and C. Get two more sides byreflecting these sides over the x-axis. Get theremaining sides by reflecting those four sides overthe y-axis. The octagon is not regular because itsangles are not all the same size.
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5 cm5 cm
14. 90 cm. Let l and w representthe length and the width ofone of the small rectangles.Then, as the diagram at rightillustrates, the perimeter ofthe large rectangle is 4l � 5w.So, 4l � 5w � 198. You can also seefrom the diagram that 2l � 3w.
Now do a little algebra.
4l � 5w � 198 The equation for the perimeterof the large rectangle.
2(2l) � 5w � 198 Rewrite 4l as 2(2l).
2(3w) � 5w � 198 Substitute 3w for 2l.
11w � 198 Simplify.
w � 18 Solve for w.
To find l, go back to the equation 2l � 3w.
2l � 3w
2l � 3(18) Substitute 18 for w.
2l � 54 Multiply.
l � 27 Solve for l.
So, each small rectangle has length 27 cm and width18 cm. Therefore, the perimeter of a small rectangleis 2(27 � 18) � 90 cm.
15. (5, 6). Here is one way to reason through thisproblem: Point P is 5 units right of point C and1 unit up. Point E is 5 units right of point Aand 1 unit up. So, point T could be 5 units rightand 1 unit up from point R. Counting squares,this gives one possibility for T as (5, 6).
16. S(9, 0), I(4, �2). Use reasoning similar to thatdescribed in the solution to Exercise 15.
17. S(�3, 0), I(�1, �5)
18. A(7, 6), N(5, 9) or A(�5, �2), N(�7, 1)
19. a. Right triangles. Two congruent right trianglesmust be used in order to get right angles for therectangle. The right triangles are joined by their
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11. mPQX is the measure of the central angle whose sidespass through points P and Q. So, mPQX� 110°.Because PQX and PRQXmake up a full circle, and afull circle has measure 360°, mPRQX� 360° � 110°,or 250°.
12. To make the 65° arc, first draw a circle. Then draw a65° angle with its vertex at the center of the circle.The minor arc with endpoints at the intersectionpoints has measure 65°. To make the 215° arc, firstdraw a circle. Then draw a 145° angle with its vertexat the center of the circle. The major arc withendpoints at the intersection points has measure360° � 145°, or 215°.
13. Possible answers: concentric rings on cross sectionsof trees (annual rings), bull’s-eye or target, ripplesfrom a rock falling into a pond
14.
15. Every triangle can have an inscribed and circum-scribed circle.
16. All circumscribed rectangles are squares with sidelengths equal to the length of the diameter of thecircle. All inscribed rectangles are not squares. Sidesof equal lengths would be diameters and would gothrough the center, so they would not form asquare. Yes.
A B
C
P
2158
1458
C
E
D
658
B
A
mAB 658 mCDE 2158
The octagon may be shaped very differently, for exampleby replacing the vertex at (�4, 4) with one at (�1, 1).
DEVELOPING MATHEMATICAL REASONING
LESSON 1.7
EXERCISES1. Answers will vary. Sample answers: The green areaon the irrigation photo is a circle, and the water is aradius. A path on the far side of the circle appearsto be tangent to the circle. The wood bridge is anarc of a circle, and the horizontal support beamunder the bridge is a chord.
2. Three of the following: AB�, BD�, EC�, EF�
3. EC�
4. AP�, EP�, FP�, BP�, CP�
5. Five of the following: EFX, AEX, ABX, BCX, CDX, DFX, EBX,EDX, FCX, ACX, DBX, AFX, ADX, BFX
6. EDCX (or EFCX) and EBCX (or EACX)
7. Two of the following: ECDX, ECFX, EDBX, EDAX, CBFX,CBDX, CDAX, CDBX, FEDX, FCAX, BCAX. Each of the abovecan be named in several other ways. For example,ECDXcan also be called DCEX, EADX, DAEX, EBDX, orDBEX.
8. FG���, HB���
9. Either F or B
10. Possible answers: cars, trains, motorcycles; washingmachines, dishwashers, vacuum cleaners; compactdisc players, record players, car racing, Ferris wheel
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26. Possible answer:
27. An equilateral triangle has three equal sides andthree equal angles, so it cannot have one rightangle. Therefore, it is not possible to draw atriangle that is both equilateral and right.
28. Possible answer:
29. Each side must have length �12a
3� 6b�, or 4a � 2b.
30. Possible answer: 31. Possible answer:
32. Possible answer:
DEVELOPING MATHEMATICAL REASONINGThe reasoning used for solving these puzzles will vary.Here is a possible approach to solving the first puzzle.Focus on the three squares of nine boxes at the top. Theupper left square has 1 and 9 in the middle row, andthe upper middle square has 9 and 1 in the bottom row.The top right square also needs a 1 and a 9, and bothdigits must fall in the top row to avoid repetition.Because the right column contains a 1, the 9 must go inthe top right box, leaving the 1 in the adjacent box.
The upper right square is now missing 3, 6, and 7. The 3must go in the middle row because there is already a 3 inthe bottom row of the upper left square. Because there isa 6 in the second column from the right, the 6 must goin the lower right square of the upper right box, and the7 must be adjacent to it. Note: In the completed puzzlesbelow, the bold numbers are the ones that are given inpuzzles in the textbook.
E
I
TK
120°
508 508708 708 E2p
2p
2p 2p
I U
Q558
4a + 2b
T
R
G
12081208 1208608
1208608 608
608
17. Equilateral quadrilateral (The figure is actually arhombus, but students have not yet learned theproperties needed to conclude that the sidesare parallel.)
18. From the way the hexagon was constructed, youknow that the length of each side is s, the same asthe radius of the circle. Therefore, the triangles areequilateral. The perimeter of the hexagon is 6s andthe diameter of the circle is 2s, so the ratio is 6s to2s, or 3 to 1.
19.
20. About 0.986°. Divide 360° by 365.25 days:�365.
32650°days� � 0.986° per day.
21. 15°. Divide 360° by 24 hours:�243h6o0u°rs� � 15° per hour.
22.
23. A scalene triangle has no congruent sides and anisosceles triangle has at least two congruent sides.Therefore, it is not possible to draw a triangle thatis both isosceles and scalene.
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7. Possible answer:
8. Follow Steps 1–4 below. The 3 m-by-4 m face isbiggest, so it should be on the “bottom.”
9. There are 3 ? 4, or 12, boxes in the base layer, andthere are 5 layers, so there are 12 ? 5, or 60, boxes.
10. 11.
12.
13. B, D. Net A, when folded, would have too manysides; in net C the two top squares would coincide;net E would be missing a side; and net F wouldhave too many sides in the horizontal direction andwould be missing one in the vertical direction. Onlynets B and D would fold into a cube.
EXTENSIONSA. Here is one way to divide the solids.
Solids with all straight edges: prism, pyramid
Solids with curved or no edges: cylinder, cone,sphere, hemisphere
B. To draw a cube truncated to edge midpoints, firstdraw a cube. Then mark the midpoints of each sideand connect the midpoints surrounding each face tomake triangular wedges at each corner. Finally, erasethe corners of the cube (you will erase all the cube’slines) to make the truncated cube. This is also calleda cuboctahedron.
Draw the truncated tetrahedron in a similar way.A tetrahedron truncated to edge midpoints is anoctahedron.
C. Answers will vary. D. Results will vary.
LESSON 1.9
EXERCISES1. Rigid. Reflected, but the size and shape do not change.
2. Nonrigid. The shape changes.
3. Nonrigid. The size changes.
4. Possible answer: a boat moving across the water
5. Possible answer: a Ferris wheel
6. 7.
8.
,
P
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21. All statements are true. A reflection is a rigid transfor-mation so the triangles are congruent. Therefore, RE�� R�E�� and �S � �S�. A reflection maps each pointof the original figure onto an image of the figure onthe opposite side of a line of reflection, so, sincepoints R, O, and S are collinear, points R�, O�, and S�are collinear, By definition, the line of reflection is theperpendicular bisector of the segment joining a pointin the original figure with its image, so the points areequidistant from the line of reflection.
22. All statements are true. A translation is a rigidtransformation so the triangles are congruent.Therefore, AB� � A�B�� and �C � �C�. A transla-tion is a rigid transformation that moves each pointof a figure by the same translation vector so, sincepoint M is the midpoint of AC� then point M� is themidpoint of A�C��. By definition, each point in theimage is equidistant from the point that corre-sponds to it in the original figure, BB�� � MM� �.
23. Statements a, b, and c are true, statement d is notalways true. A rotation is a rigid transformation sothe triangles are congruent. By definition, each pointin the original figure is moved by the same anglemeasure in the same direction along a circular pathabout a fixed point. Since �DEF � �D�E�F�, DF� �D�F��� and �F � �F�. Under a rotation, if EG� is thebisector of �E, then if E�G� is the bisector of �E�,and �DEG � �D�E�G�. Part d is not true of a rota-tion. Unlike a translation, in a rotation each point inthe original figure is moved by the same anglemeasure, not necessarily the same distance.
24. Notice that each figure is made up of a capital letterattached to its “mirror image” (reflection across avertical line). Because the letters are in alphabeticalorder starting with C, the next two figures will usethe letters H and I.
25. The capital letters are in alphabetical order, andtheir position moves clockwise through the foursmall squares. Also, the placement of the letter isrotated 90° clockwise each time. Therefore, the nexttwo figures will have F placed like B, and G placedlike C. The circle that is divided into four parts isalways opposite the letter, and the quarter that isshaded moves clockwise through four positions, sothe circle for F is the same as that for B, and thecircle for G is the same as that for C.
26. Possible answer: equiangular quadrilateral QUAD isa rectangle that is not a square and QU and QD areadjacent sides.
, or
14. triangle with vertices (1, 0), (1, 4) (–2, 4)
15. Fold endpoints together to get the perpendicularbisector of the segment.
16. Fold so that the rays of the angle coincide to get theangle bisector.
17. Fold through the midpoint of one side and thevertex of the opposite angle so that sides and anglesare together.
18.
19. Construct a segment that connects two correspon-ding points. Construct the perpendicular bisector ofthat segment.
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44. Carefully trace the image on the other side of theline of reflection. A reflection is a rigid transforma-tion that produces a mirror image, so the figures arecongruent.
45. Copy the figure onto patty paper. Construct aparallel line through a corresponding point on thefigure. Duplicate the translation vector on that line.Place a second piece of patty paper on top of thefirst and make a copy of the figure and the transla-tion vectors. Using the translation vector as a guide,translate the second copy matching correspondingparts of the figure. A translation is a rigid transfor-mation that preserves size and shape, so the imageis congruent to the original.
46. Extend the line to create the 180° angle on the orig-inal figure. Mark the intersection of the figure andthe initial side of the angle. Using a patty papercopy of the figure, place a pencil point at the centerand rotate the figure so that the intersection coin-cides with the terminal side of the 180° angle. Copythe figure. A rotation is a rigid transformation thatpreserves size and shape, so the image is congruentto the original.
47. One line of reflection. There are no other lines ofreflection in which the original figure coincides withthe image.
B'
B
36.
37. Here is one possible method. Draw a circle andone diameter. Draw another diameter perpendicularto the first. Draw two more diameters so that eight45° angles are formed. Draw the regular octagonformed by connecting the endpoints of thediameters.
38. 114°
39. The diagram below illustrates the given information.3z � 12, so z � 4 cm. Therefore, AB � 4z � 16 cm.
40. The diagram below illustrates the giveninformation. Because BE�� is the bisector of�DBC, m�DBC � 64°. Because BD�� is thebisector of �ABC, m�ABD � 64°. So,m�EBA � m�DBE � m�ABD � 32° � 64° � 96°.
41. �31620°� � 30°
42. If the triangle is rotated 90° clockwise, AB� will bea vertical segment. Point B will have the samelocation as it does now, and point A will beat (2, 3).
