dg5 solution manual ch05 1pp - flourishkh.com · the figure that is shared by the pentagon and the...

the figure that is shared by the pentagon and thequadrilateral. The sum of the measures of the threeangles at this vertex is 360°, so the unmarked anglein the quadrilateral measures 122°. In the quadrilat-eral, h � 90° � 66° � 122° � 360°, so h � 82°.

8. j � 120°, k � 38°. The hexagon is equiangular, so6j � 180° � 4 � 720°, and j � 120°. Now look at theparallel lines to see that k � 38° (AIA Conjecture).

9. The sum of the interior angle measures of thisquadrilateral is 358°, but it should be 360°. Usevertical angles to find that three of the anglemeasures in the quadrilateral are 102°, 76°, and82°. Using the CA Conjecture and the Linear PairConjecture shows that the fourth angle of thequadrilateral is 180° � 82° � 98°. Then the sumof the four interior angles of the quadrilateralwould be 76° � 102° � 82° � 98° � 358°, whichis impossible.

10. The sum of the measures of the interior angles is554°. However, the figure is a pentagon, so thesum of the measures of its interior angles shouldbe 540°. Three exterior angles of the pentagoneach measure 49°, so the interior angles at thesethree vertices each measure 131°. Another exteriorangle measures 154°, so the exterior angle at thatvertex measures 26°. Thus, the sum of the interiorangle measures would be 3 � 131° � 135° � 26° �554°, which is impossible.

11. 18 sides. The figure shows an equilateral triangle,

a regular nonagon, and a third regular polygon

meeting at point A. The sum of the measures of the

angles sharing A as a common vertex must be 360°.

The measure of each interior angle of an equilateral

triangle is 60°, and the measure of each interior

angle of a regular nonagon is � 140°. There-

fore, the measure of each interior angle of the third

regular polygon is 360° � 60° � 140° � 160°.

Because all angles of a regular polygon have the

same measure, the measure of each interior angle is

�180°(n

n� 2)�. Use this expression to find n when each

interior angle measures 160°.

�180°(n

n� 2)� � 160°

180°(n � 2) � 160°n

180° � 7�9

CHAPTER 5

LESSON 5.1

EXERCISES

1. (See table at bottom of page.)

For each given value of k (the number of sides ofthe polygon), calculate 180°(k � 2) (the sum of themeasures of the interior angles).


For each given value of k (number of sides), calcu-late 180°(k � 2) to find the sum of the measures ofthe interior angles, and then divide by n to find themeasure of each interior angle. In an equiangularpolygon, all interior angles have the same measure.

3. 122°. In the quadrilateral, the unmarked anglemeasures 90° (Linear Pair Conjecture). Then, bythe Quadrilateral Sum Conjecture, a � 72° � 76° �90° � 360°, so a � 122°.

4. 136°. The two unmarked angles in the hexagonmeasure 112° and 110° (Linear Pair Conjecture),and the sum of the interior angles of a hexagon is180° � 4 � 720° (Polygon Sum Conjecture). There-fore, 2b � 110° � 112° � 116° � 110° � 720°, so2b � 272° and b � 136°.

5. e � 108°, f � 36°. In the equiangular pentagon,5e � 540° (Pentagon Sum Conjecture), so e � 108°.The triangle is isosceles with base angles of measure72° (Linear Pair Conjecture), so f � 2 � 72° � 180°and f � 36°.

6. c � 108°, d � 106°. One of the unmarked angles isa vertical angle of the angle marked as 44°, so itsmeasure is also 44°, and the other unmarked angleof the quadrilateral measures 102° (Linear PairConjecture). Look at the triangle containing theangle with measure d. Here, d � 44° � 30° � 180°(Triangle Sum Conjecture), so d � 106°. Thenapply the Quadrilateral Sum Conjecture:c � d � 102° � 44° � 360°, so c � 108°.

7. g � 105°, h � 82°. In the pentagon, 3g � 108° �117° � 540°, so g � �

13�[540° � (117° � 108°)] �

105°. Now look at the vertex on the right side of

90 CHAPTER 5 Discovering Geometry Solutions Manual

©2015 Kendall Hunt Publishing

Lesson 5.1, Exercises 1, 2

1.

2. Number of sides of equiangular polygon 4 5 6 7 8 9 ... k 100

Measure of each angle of equiangular polygon 90° 108° 120° 128�47�° 135° 140° ... 176�

25�°

Number of sides of polygon 7 8 9 10 11 12 ... k 100

Sum of measures of angles 900° 1080° 1260° 1440° 1620° 1800° ... 180(k – 2)° 17,640°

1180(k – 2)

kQ R°

DG5_Solution_Manual_CH05_Printer.qxd 3/16/16 12:36 PM Page 90

180°n � 360° � 156°n

24°n � 360°

n � 15

Therefore, an equiangular polygon in which eachinterior angle measures 156° has 15 sides.

15. The 12th century. Compare the measures ofeach interior angle in a regular 16-gon to thosein a regular 18-gon. Each interior angle of a16-gon measures � 157.7°, while eachinterior angle of an 18-gon measures �

160°. Therefore, the plate is probably from the12th century.

16. The angles of the trapezoidmeasure 67.5° and 112.5°. 67.5°is half the value of each angle ofa regular octagon, and 112.5° ishalf the value of 360° � 135°.

17. First proof: Apply the Triangle Sum Conjecture to�QDU to obtain q � d � u � 180°, and to �ADUto obtain a � e � v � 180°. The sum of theangle measures in quadrilateral QDAU is m�Q �m�QDA � m�A � m�AUQ � q � d � e � a �u � v. Rearrange terms to obtain q � d � u � a �e � v � (q � d � u) � (a � e � v) � 180° �180° � 360°.

Second proof: Using the Triangle Sum Conjecture,(a � b � j) � (c � d � k) � (e � f � l) �(g � h � i) � 4(180°), or 720°. The four angles inthe center sum to 360°, so j � k � l � i � 360°.Subtract to get a � b � c � d � e � f � g �h � 360°.

18. The Quadrilateral Sum Conjecture says the sum ofthe angles of a quadrilateral is 360°. An equiangularquadrilateral has four congruent angles that add upto the 360°. The measure of each angle would be360° divided by 4, or 90°.

19. x = 120°. The sum of the angle measures at a vertexof the tiling must be 360°, so 90° � 60° � 90° +x = 360°, and x = 120°.

20. D. As vertex P moves from left to right: Thedistance PA decreases and then soon increases,the perimeter of the triangle decreases before itincreases, and the measure of �APB increases fora while, but then decreases.

21. The base angles of an isosceles right trianglemeasure 45°; thus, they are complementary.

45°

45°

67.5°

135°

180° � 14�16

180° �16�18

180°n � 360° � 160°n

20°n � 360°

n � 18

Therefore, the largest polygon has 18 sides.

12. a � 116°, b� 64°, c� 90°, d� 82°, e� 99°, f � 88°,g� 150°, h� 56°, j� 106°, k� 74°, m� 136°, n�118°, and p� 99°. a� 116° (Vertical Angles Conjec-ture). b� 64° (Corresponding Angles (CA) or AlternateInterior Angles (AIA) Conjecture, and Linear PairConjecture). c� 90° (CA or AIA Conjecture, andLinear Pair Conjecture). a� b� d� 98° � 360°(Quadrilateral Sum Conjecture), so d� 82°. Now lookat the quadrilateral containing angles with measures eand p. Here, p� e, and the unmarked angles measure98° and 64° (Linear Pair Conjecture), so 2e� 98° �64° � 360°. Therefore, 2e� 198° and e� 99°. Nextlook at the pentagon in the lower left. By the LinearPair Conjecture, the two unmarked angles measure 82°and 116°. The sum of the measures of the interiorangles of a pentagon is 540°, so f� 138° � 116° �82° � 116° � 540°, and f� 88°. f� g� 122° � 360°,so g� 150°. Next look at the quadrilateral that includesthe angles with measures g and h. By the Linear PairConjecture, the unmarked angles measure 64° and 90°,so g� h� 64° � 90° � 360°, and thus h� 56°.Nowlook at the quadrilateral in the lower right that includesthe angle with measure j. The unmarked angle is 90°(vertical angle with angle of measure c), so j� 90° �77° � 87° � 360°, and thus j� 106°. Either of theangles that form a linear pair with the angle ofmeasure j measures 74°, so k� 74° (AIA Conjecture orCA Conjecture). Next look at the quadrilateral in theupper right that includes the angle with measure m.Here, the unmarked angles measure 90° and 74°(Linear Pair Conjecture and Vertical Angles Conjec-ture), so m� 60° � 74° � 90° � 360°, and thus m�136°. n�m� 106° � 360°, so n� 118°. Finally, p� e,so p� 99°.

13. 17 sides. Find the value of n for which180°(n � 2) � 2700°. Solve this equation.

180°(n � 2) � 2700°

n � 2 � 15

n � 17

Therefore, if the sum of the measures of the interiorangles is 2700°, the polygon has 17 sides.

14. 15 sides. The measure of each interior angle of

an equiangular polygon is �180°(n

n� 2)�, so

solve the equation �180°(n

n� 2)� � 156°.

�180°(n

n� 2)� � 156°

180°(n � 2) � 156°n

Discovering Geometry Solutions Manual CHAPTER 5 91©2015 Kendall Hunt Publishing


is � 108°, so e � 180° � 108° � 72°. Inthe regular octagon, the measure of each interior

angle is � 135°, so f � 180° � 135° � 45°.g is the measure of one of three angles whose sumis 360°. Because one of these angles is an interiorangle of the pentagon and another is an interiorangle of the octagon, g � 108° � 135° � 360°,so g � 117°. Finally, h � e � f � g � 360°(Quadrilateral Sum Conjecture), so h � 234° �

360°, and h � 126°.

9. a � 30°, b � 30°, c � 106°, and d � 136°. First,a � 56° � 94° � 180° (Triangle Sum Conjecture),so a � 30°. Next, b � a (AIA Conjecture), sob � 30°. From the triangle on the left, b � c � 44° �180°, so c � 106°. Finally, look at the quadrilateralthat contains angles with measures 56°, 94°, and d,as well as an unmarked angle. The measure of theunmarked angle is 180° � c � 74° (LinearPair Conjecture), so d � 94° � 56° � 74° �360° (Quadrilateral Sum Conjecture), andd � 136°.

10. a � 162°, b � 83°, c � 102°, d � 39°, e � 129°,f � 51°, g � 55°, h � 97°, and k � 83°.a � 180° � 18° � 162°. To find b, look at theexterior angles of the large pentagon (which issubdivided into a triangle, a quadrilateral, and apentagon). The unmarked exterior angle at thelower right forms a linear pair with an angle thatis marked as congruent to the angle with measureh, so the measure of this exterior angle is b. TheExterior Angle Sum Conjecture says that the sumof the measures of a set of exterior angles (one ateach vertex) of any polygon is 360°. Therefore,86° � b � 18° � b � 90° � 360°, or 2b � 194° �360°, and b � 83°. Next look at the isoscelestriangle. Here, d � 39° (Isosceles Triangle Conjec-ture), and 2 � 39° � c � 180° (Triangle SumConjecture), so c � 102°. Next look at the vertex ofthe triangle with measure c. Here, 2e � c � 360°, so2e � 258° and e � 129°. Now look at the upper-right corner of the figure. Here, d � f � 90° � 180°,so f � 51°. Next look at the upper-left corner of thefigure. Here, 86° � g � 39° � 180°, so g � 55°. Nowlook at the lower-left corner. h � b � 180°, so h �97°. Finally, look at the quadrilateral. The angle inthe lower-right corner of the quadrilateral iscongruent to the angle with measure h, so itsmeasure is also 97°. By the Quadrilateral SumConjecture, k � 97° � f � 129° � 360°, or k �277° � 360°, so k � 83°.

11. 1. Linear Pair Conjecture; 2. Linear Pair Conjecture;3. Linear Pair Conjecture; 4. 540°; 5. 180°, TriangleSum Conjecture; 6. 360°

3 � 180°�5

6 � 180°�8

LESSON 5.2

EXERCISES1. 360°. By the Exterior Angle Sum Conjecture, the

sum of the measures of a set of exterior angles forany polygon is 360°.

2. 72°; 60°. In an equiangular polygon, all interiorangles are congruent, so all exterior angles are alsocongruent. For a polygon with n sides, the measureof each exterior angle is �

36n0°�, so the measure of

each exterior angle of an equiangular pentagon is�3650°� � 72°, and the measure of each exterior angle

of an equilateral hexagon is �3660°� � 60°.

3. 15 sides. The measure of each exterior angle of aregular polygon is �

36n0°�. (See solution for Exercise 2.)

If �36n0°� � 24°, 360° � 24° � n, so n � 15. Thus, if

each exterior angle of a regular polygon measures24°, the polygon has 15 sides.

4. 43 sides. Use the Polygon Sum Conjecture. Find n if180°(n � 2) � 7380°.

180°(n � 2) � 7380°

180° � n � 360° � 7380°

180° � n � 7740°

n � 43

Thus, if the sum of the measures of the interiorangles of a polygon is 7380°, the polygon has43 sides.

5. a � 108°. Use the Exterior Angle Sum Conjecture.The figure shows a set of exterior angles (one ateach vertex) for a pentagon. The measure of theunmarked exterior angle is 180° � 140° � 40°(Linear Pair Conjecture). Therefore, a � 68° �84° � 60° � 40° � 360°, so a � 108°.

6. b � 45�13�°. Use the Exterior Angle Sum Conjecture.

The measure of the unmarked exterior angle is180° � 68° � 112° (Linear Pair Conjecture). Thereare three exterior angles with measure b. Therefore,3b � 69° � 43° � 112° � 360°, so 3b � 136°, andb � 45�

13�°.

7. c � 51�37�°, d � 115�

57�°. By the Equiangular Polygon

Conjecture, the measure of an interior angle of anequiangular heptagon is � 128�

47�°. Then c �

180° � 128�47�° � 51�

37�° (Linear Pair Conjecture). Now

look at the common vertex of the heptagon and thetwo quadrilaterals. The sum of the angle measuresat this vertex is 360°, so 2d � 128�

47�° � 360°. There-

fore, 2d � 231�37�°, and d � 115�

57�°.

8. e � 72°, f � 45°, g � 117°, and h � 126°. Use theEquiangular Polygon Conjecture. In the regularpentagon, the measure of each interior angle

5 � 180°�7




b.

Figure 2 is a regular dodecagon with an inscribedregular hexagon and six congruent isosceles trian-gles at the other vertex angles. By the EquiangularPolygon Conjecture, the measure of each interior

angle of an equiangular dodecagon is

�150°, so m�1 � 150°. By the Isosceles TriangleConjecture �2 � �3. By the Triangle SumConjecture m�1 � m�2 � m�3 � 180°. Bysubstitution, 150° � 2(m�2) � 180°. m�2 �m�3 � 15°.

c.

Figure 3 is a regular dodecagon with an inscribedsquare and four congruent isosceles trapezoids. Bythe Equiangular Polygon Conjecture, the measure ofeach interior angle of an equiangular dodecagon is

� 150°, so m�1 � m�2 � 150°. By

definition, the base angles of an isosceles trapezoidare congruent, so �3 � �4. By the QuadrilateralSum Conjecture m�1 � m�2 � m�3 � m�4 �360°. By substitution, 150° � 150° � 2(m�3) �360°. m�2 � m�3 � 30°.

d.

Figure 4 is a regular dodecagon with an inscribedregular hexagon, an equilateral triangle, three largecongruent isosceles triangles and six smallcongruent isosceles triangles. By the EquiangularPolygon Conjecture, the measure of each interior

angle of an equiangular dodecagon is

� 150°, so the measure of the vertex angle of thesmall triangle is 150°. By the Triangle Sum Conjec-ture, the sum of the measure of the other twoangles of the small triangle would be 30°, so, by theIsosceles Triangle Conjecture, each angle measure is

Q180(12�2)12

°R

231

Q180(12�2)12

°R

2

1

3

4

1

2

3

Q180(12�2)12

°R

12. Yes, the maximum number of obtuse exterior anglesthat a polygon can have is three. This is because themeasure of an obtuse angle is greater than 90°, andif there were four or more exterior angles that eachmeasured more than 90°, their sum would begreater than 360°. That is impossible because thesum of all the exterior angles (one at each vertex)is 360° for any polygon.

The minimum number of acute interior angles thata polygon must have is 0. It’s possible for a polygonto have no acute interior angles. A simple exampleof this situation is a rectangle, which has four rightangles.

13. First proof: Use the diagram on the left. a � w �h � 180°, c � b � z � 180°, e � d � y � 180°,and g � f � x � 180° by the Triangle Sum Conjec-ture. a � b � c � d � e � f � g � h � w �x � y � z � 720° by the addition property ofequality. Angles w, x, y and z are linear angleswhose sum is 180°. Substituting 180° for w �x � y � z and subtracting 180° from both sidesgives the sum of the interior angles of a pentagon is540°.

Second proof: Use the diagram on the right. Bythe Triangle Sum Conjecture, (a � b � q) �(c � d � r) � (e � f � s) � (g � h � t) �(i � j � p) � 5(180°), or 900°. The five angles inthe center sum to 360°, so p � q � r � s � t �360°. Subtract to get a � b � c � d � e � f � g �h � i � j � 540°.

14. Regular polygons: triangle and dodecagon. Anglemeasures: 60°, 150°, and 150°. Point A issurrounded by an interior angle of a triangle andan interior angle of two dodecagons. The measureof each interior angle of an equilateral triangle is60°, while the measure of each interior angle of aregular dodecagon is � 150°.

15. a.

Figure 1 is a regular dodecagon divided intotwelve congruent isosceles triangles. The sum ofthe measures of the twelve vertex angles is 360°,

so m�1 � , or 30°. By the Isosceles Triangle

Conjecture �2 � �3. By the Triangle SumConjecture m�1 � m�2 � m�3 � 180°.By substitution, 30° � 2(m�2) � 180°. m�2 �m�3 � 75°.

360°

12

10 �180°�12

1

2

3



17. Yes. �RAC � �DCA by SAS. AD�� CR� byCPCTC.

18. Yes. �DAT � �RAT by SSS. �D � �R by CPCTC.(Draw AT� to form the two triangles.)

19. 180°

20.

A, C, D, and E are true. The lengths of the sides of�A�B�C� are three times the lengths of the sides of�ABC, so B is not true.

DEVELOPING MATHEMATICAL REASONINGEleven possible answers are shown.

EXTENSIONSBecause it is easy to construct angles with measures 60° and90° with compass and straightedge, and these angles may bebisected to obtain angles with measures 30° and 45°, youcan construct regular polygons that have any of these fourinterior angle measures or angle measures that can be foundby adding or subtracting these angles. This includesequilateral triangles (interior angle measure � 60°), squares(interior angle measure � 90°), regular hexagons (interiorangle measure � 120° � 2 � 60°), regular octagons (interiorangle measure � 135° � 90° � 45°), and dodecagons(interior angle measure � 150° � 2 � 60° � 30°). It is alsopossible (but difficult) to construct regular polygons with10, 16, 17, and 20 sides.

y

x

A

B

C

C'

B'

A'

15°. The measure of �1 is 150° � 15° � 15° �120° . By the Isosceles Triangle Conjecture �2 ��3. By the Triangle Sum Conjecture m�1 � m�2� m�3 � 180°. By substitution, 120° � 2(m�2)� 180°. m�2 � m�3 � 30°.

e.

Figure 5 is a regular dodecagon with twelveinscribed congruent rhombuses and twelvecongruent isosceles triangles. The sum of themeasures of the twelve angles at the center is 360°,so m�4 � , or 30°. By definition, opposite

angles of a rhombus are congruent so �1 � �4and m�1 � 30°. By definition, opposite angles of arhombus are congruent, so �2 � �3. By theQuadrilateral Sum Conjecture m�1 � m�2 �m�3 � m�4 � 360°. By substitution, 30° � 30°� 2(m�3) � 360°. m�2 � m�3 � 150°.

16. a.

b. A�(�9, 2), B�(3, 0), C�(�2, 5), D�(�8, 6)

c.

d. A�(�6, 7), B�(6, 5), C�(1, 10), D�(�5, 11)

e. (x, y) → (x, y � 9)

f. (x, y) → (x, y � 9)

y

x

A

B

C

D'

A'D

D''C''

B'

A''

B''C'

36012

°

13 2

4

y

x

A

B

CD'

D'C

A'

B




8. w � 1.6 cm, x � 48°, y � 42°. By the definition ofa kite, FLYE must have two distinct pairs of consec-utive congruent sides. Because FL � LY is given, wealso know that FE � EY. Therefore, �FLY and�FEY are the vertex angles of the kite, so, by theKite Diagonal Bisector Conjecture, LE� bisects FY�;thus, w � 1.6 cm. �FEY is isosceles with FE � EY,so by the Isosceles Triangle Conjecture, �RFE ��RYE; thus, x � 48°. Finally, �YRE is a righttriangle because LE� � FY� by the Kite DiagonalsConjecture. Therefore, by the Triangle Sum Conjec-ture, x � y � 90° � 180°. Substitute 48° for x toobtain y � 48° � 90° � 180°, so y � 42°.

9. 3. BN� � BN�; 4. �BEN � �BYN by SSS; 5. �1 ��2 and �3 � �4 by CPCTC; 6. BN� bisects �B,BN� bisects �N.

10. First proof: This proof uses theKite Angle Bisector Conjecture.Given: Kite BENY with vertexangles �B and �N, and diagonalsintersecting at point X.

�1 � �2 by the Kite Angle Bisector Conjecture. BySAS, �BXY � �BXE, so XY� � XE� by CPCTC.Because �YXB and �EXB form a linear pair, theyare supplementary, that is, m�YXB � m�EXB �180°. By CPCTC, �YXB � �EXB, so 2m�YXB �180°, or m�YXB � m�EXB � 90°. Hence, BN� isthe perpendicular bisector of YE�.

Second proof: This proof usesthe Converse of the PerpendicularBisector Conjecture. The definitionof a kite says that KI� � IT� andTE� � EK�. So point I is equidistantfrom points K and T. Likewise,point E is equidistant from points Kand T. Therefore, by the Converse ofthe Perpendicular Bisector Conjecture, both I and Elie on the perpendicular bisector of KT�. So diagonalIE� is the perpendicular bisector of diagonal KT�.

11. �E � �I

12. The other base is ZI�. �Q and�U are a pair of base angles. �Zand �I are a pair of base angles.

13. OW�� is the other base. �S and �Hare a pair of base angles. �O and�W are a pair of base angles.SW�� HO��.

14. Start by duplicating BN�, which will form one diag-onal of the kite. Next, draw an arc centered at N

W O

S H

Z I

Q U

E

KT

I

TK

I

E

B N

Y

E

12 X

LESSON 5.3

EXERCISES1. 64 cm. Because ABCD is a kite, AB � AD � 20 cm

and BC � DC � 12 cm. Therefore, the perimeter is2 � 12 � 2 � 20 � 64 cm.

2. x � 21°, y � 146°. By the Kite Angles Conjecture,y � 146°. Then, by the Quadrilateral Sum Conjec-ture, x � 2 � 146° � 47° � 360°, so x � 21°.

3. x � 52°, y � 128°. By the Isosceles TrapezoidConjecture, y � 128°. Also, by the Isosceles TriangleConjecture, the fourth angle of the trapezoid hasmeasure x. By the Quadrilateral Sum Conjecture,2x � 2 � 128° � 360°, so 2x � 104° and x � 52°.

4. 15 cm. The perimeter of the trapezoid is 85 cm, so2x � 18 � 37 � 85. Thus, 2x � 30 and x � 15 cm.

5. x � 72°, y � 61°. The Kite Diagonals Conjecturesays that the diagonals of a kite are perpendicular,so the two diagonals form four right triangles. Also,the diagonal that connects the vertices of thenonvertex angles divides the kite into two isoscelestriangles. Look at the small right triangle in theupper right. The smallest angle in this trianglemeasures 18° because this is one of the base anglesof the smaller isosceles triangle (Isosceles TriangleConjecture). Then x � 90° � 18° � 180° (TriangleSum Conjecture), so x � 72°. In the right triangle inthe lower right, the acute angles are 29° and 90° �29° � 61°. The 61° angle in this triangle is a baseangle of the larger isosceles triangle, so y � 61°(Isosceles Triangle Conjecture).

6. x � 99°, y � 38 cm. Notice that the figure is anisosceles trapezoid. First look at the unmarked angleat the lower-right corner of the trapezoid. By theTrapezoid Consecutive Angles Conjecture, themeasure of this angle is 180° � 81° � 99°. By theIsosceles Triangle Conjecture, this angle is congruentto x, so x � 99°. The perimeter of this isoscelestrapezoid is 164 cm, so y � 2(y � 12) � (y � 12) �164. Solve this equation.

y � 2y � 24 � y � 12 � 164

4y � 12 � 164

4y � 152

y � 38

7. w � 120°, x � 45°, y � 30°. From the hint, �PAT ��APR. Then, by CPCTC, �PAT � �APR, soy � 30°. By the Isosceles Trapezoid Conjecture,�PAR � �APT, so x � 30° � y � 45°. Becausey � 30°, this gives x � 45°. Now look at �ZAP. Bythe Vertical Angles Conjecture, m�AZP � w. Thus,by the Triangle Sum Conjecture, w � y � 30° �180°. Substitute 30° for y to obtain w � 30° �30° � 180°, so w � 120°.



two base angles of the isosceles trapezoid along theinner edge of the arch each measure 100°, and thetwo base angles along the outer edge of the archeach measure 80°.

Notice that the exercise asks you to draw the arch,not to construct it, so you may use a protractor tomeasure angles.

To draw the arch, begin by drawing a horizontalline. Using a point on the line as the center, drawtwo concentric semicircles above the line. To formhalf of the regular 18-gons inscribed in the semicir-cles, draw central angles of measure �

1890°� � 20°. On

each semicircle, draw line segments connectingconsecutive intersections of the semicircles with therays of the central angles. Then draw segmentsconnecting the two semicircles at those points.

18. The angles of the trapezoid that form half then-gon are 96°. Therefore 96° � 96° � x � 360°,which gives us 168° as the angle measure of theregular n-gon. Using the formula �168°,

n � 30. So it would be a 30-gon. Half of that wouldbe 15 cartons.

19. Because ABCD is an isosceles trapezoid, �A � �B.�AGF � �BHE by SAA. Thus, AG� � BH� byCPCTC.

20. a � 80°, b � 20°, c � 160°, d � 20°, e � 80°,f � 80°, g � 110°, h � 70°, m � 110°, and n � 100°.First a � 100° � 180° (Linear Pair Conjecture), soa � 80°. In the isosceles triangle that contains theangles with measures a and b, notice that b is themeasure of the vertex angle. Therefore, b � 2a �180° (Triangle Sum Conjecture), so b � 20°. c �b � 180° (Linear Pair Conjecture), so c � 160°.The angles with measures b and d are correspond-ing angles, so d � b (CA Conjecture), and d � 20°.Next 2e � d � 180°, so e � 80°. The angle thatforms a linear pair with the angle of measure f hasthe same measure as the angle formed bycombining the angles of measures e and d, so itsmeasure is 80° � 20° � 100°, and thus f � 180° �100° � 80°. Now look at the small right triangle inthe lower left that contains the angle with measureh. The measure of the other acute angle of thistriangle is d (Vertical Angles Conjecture), soh � d � 90° and h � 70°. Then g � 180° � h(Linear Pair Conjecture), so g � 110°. Finally, bythe CA Conjecture, m � g, so m � 110°, andsimilarly n � 100°.

(n�2)180°n

20° 80°

100°160°

with radius NE and a second arc centered at B withradius BE. Draw these two arcs large enough so thatthey will intersect twice on opposite sides of BN�.Label the two points of intersection of the arcs as Fand E. Connect the four vertices B, E, N, and F inorder, forming the kite.

Only one kite is possible because three sidesdetermine a triangle. (Note: The two points labeledE and F are interchangeable, so the figure can belabeled in two ways, but there is only onepossible figure.)

15. Duplicate WI�. Duplicate �I and �W, using theendpoints of WI� as their vertices. Duplicate IS�along the side of �I that does not lie on WI�.Construct a line through point S parallel to WI�.Label the point where this line intersects the side of�W that does not lie along WI� as H. WISH is therequired trapezoid.

16. Possible construction: Duplicate NE�. Use congruentcorresponding angles to construct a line parallel toNE�, and duplicate the length BO along this line.Connect the vertices B, O, N, and E to forma trapezoid.

Infinitely many trapezoids can be drawn. Noticethat none of the angles of the trapezoid, nor thelengths of either of the nonparallel sides, are given.In the construction shown above, the angle that wasused to construct the parallel lines is arbitrary as isthe position of point O along one of the rays ofthat angle.

17. 80°, 80°, 100°, 100°. Look at one of the commonvertices where two of the trapezoids and the 18-gonmeet. The sum of the three angles that meet at thisvertex must be 360°. Let x represent the measure ofone of the base angles of the trapezoid along theinner edge of the arch. Each interior angle of aregular 18-gon measures � �

160°. Therefore, 2x � 160° � 360°, so x � 100°. The

(18 � 2) � 180°��18

16 � 180°�18

B O

NE

S H

WI

F

E

BN




Therefore, the perimeter of the parallelogram is2 � 17 � 2 � 23 � 34 � 46 � 80.

6. e � 63°, f � 78°. By the definition of a parallelo-gram, both pairs of opposite sides are parallel, soe � 63° by the AIA Conjecture, and f � 78° by theParallelogram Opposite Angles Conjecture.

7. Possible construction: First, duplicate LA�; then,duplicate �L with one of its sides along LA��. Markoff an arc of length AS along the other side of �L,and label the intersection point of this arc with theray that forms the second side of the angle as T.Draw an arc of length AS centered at A and an arcof length LA centered at T. Label the intersectionmark of these two arcs as S. Connect the vertices inorder to form parallelogram LAST. Notice that youused two pairs of parallel sides in this construction.

8. Possible construction: Recall that the diagonals ofa parallelogram bisect each other (ParallelogramDiagonals Conjecture). First duplicate DO��, whichwill form the longer diagonal of the parallelogram.Bisect DO�� to locate its midpoint. Also, bisect thegiven segment PR�, and then open your compass toa radius of �

12�PR. Draw arcs (or a complete circle)

with centers at the midpoint of DO�� and radius �12�PR.

Also, draw arcs with centers at D and O and withlengths DR. Label as R the point where the arccentered at D intersects one of the arcs (or thecircle) with radius �

12�PR. Label as P the point where

the arc centered at O intersects one of the arcs (orthe circle) with radius �

12�PR. Connect the vertices D,

R, O, and P to form the parallelogram.

9. (b � a, c). Use the definition of a parallelogramand slopes. By the definition of a parallelogram,MA�� PR�; because PR� is a horizontal segment (andhas a slope of 0), so is MA��, and therefore, they-coordinate of M is c. Also, by the definition of aparallelogram, PM�� RA�, so PM�� and RA� have equalslopes. The slope of RA� is �b

c ��0a� � �b �

ca�. The coor-

dinates of P are (0, 0), so if you let x represent thex-coordinate of M, then the slope of PM�� is �x

c��00� � �x

c�.

Equating the slopes of RA� and PM�� gives �b �ca� � �x

c�.

P RD

P

O

R

T S

AL

21. a.

b. A�(8, 2), B�(4, �2), C�(0, 2), D�(4, 10)

c. (x, y) → (�x, y)

d

e. A�(2, �8), B�(�2, �4), C�(2, 0), D�(10, �4)

f. (x, y) → (y, �x)

g. (x, y) → (y, x)

LESSON 5.4

EXERCISES1. c � 34 cm, d � 27 cm. Use the Parallelogram

Opposite Sides Conjecture.

2. a � 132°, b � 48°. By the Parallelogram Consecu-tive Angles Conjecture, a � 180° � 48° � 132°,and by the Parallelogram Opposite AnglesConjecture, b � 48°.

3. g � 16 in., h � 14 in. Use the ParallelogramOpposite Sides Conjecture to find g and theParallelogram Diagonals Conjecture to find h.

4. 63 m. Use the Parallelogram Diagonals Conjectureto find VN � �

12�(VF) � 18 m and NI � �

12�(EI) �

21 m, and use the Parallelogram Opposite SidesConjecture to find that VI � EF � 24 m. Thenthe perimeter of �NVI � 18 � 21 � 24 � 63 m.

5. 80. By the Parallelogram Opposite Sides Conjecture,x � 3 � 17, so x � 20, and therefore, x � 3 � 23.

y

x

A

B

C

D'

A'

D

B'

C'

D''

C''

B''

A''

y

x

A

B

C

D'

A'

D

B'

C'



figure, one of which has angle measure x. The othertwo angles measure 154° (Kite Angles Conjecture)and 180° � 78° � 102° (Trapezoid ConsecutiveAngles Conjecture). Therefore, x � 154° � 102° �360°, and x � 104°. Now look at the three anglesthat share a common vertex on the right side of thefigure, one of which has angle measure y. The othertwo angles measure 160° (Kite Angles Conjecture)and 102° (Isosceles Trapezoid Conjecture), so y �102° � 160° � 360°, and y � 98°.

16. a � 84°, b � 96°. The arch is semicircular, sothe complete regular polygon on which the archis constructed has 30 sides, and each of its interiorangles measures � 168°. Look at acommon vertex along the inside edge of the archwhere a base angle from each of two isosceles trape-zoids meets an interior angle of the 30-gon. Here, 2b� 168° � 360°, so b � 96°. By the Trapezoid Consec-utive Angles Conjecture, a � b � 180°, so a � 84°.

17. No. The congruent angles and the common side donot correspond. Notice that the common side of thetwo triangles, WY��, is opposite the 83° angle in�WYZ, but is opposite the 58° angle in �XYW.

18. The section is an isosceles trapezoid.

19. Parallelogram. Possible construction: Construct asegment, AC�, of any length, and bisect it to find itsmidpoint, M. Construct any line that passes throughM; call this line p. Open your compass to a radiusthat is not equal to �

12�AC. Place your compass point at

M, and use this new radius to mark off arcs on p oneither side of AC��. Label the two points where thesearcs intersect p as B and D. Connect the vertices A, B,C, and D to form a parallelogram. Notice that M isthe midpoint of both diagonals, AC� and BD�, so thediagonals bisect each other as required.

To explain why the quadrilateral you haveconstructed is a parallelogram, look at the figure withthe construction marks removed. The marks on thefigure show that the diagonals bisect each other.

A

B

D1

4

23

M

C

A

B

D

M

C

p

(30 � 2) � 180°��30

Therefore, x � b � a and the coordinates of M are(b � a, c).

10. 4. Given; 5. Lemma just proved; 6. CPCTC

11. 1. Given; 2. Definition of parallelogram; 4. �EAL ��NLA; 5. AE� � LN�; 6. �AET � �LNT; 8. AT� �LT�; 9. EN� and LA� bisect each other

12. The parallelogram linkage is used for the tool box sothat the drawers remain parallel to each other (andto the ground) and the contents cannot fall out.

13. a � 135°, b � 90°. a is the measure of an interior

angle of a regular octagon, so a � � 135°

(Equiangular Polygon Conjecture), and b is the

measure of an interior angle of a square, so b � 90°.

14. a � 120°, b � 108°, c � 90°, d � 42°, and e � 69°.

Start by applying the Equiangular Polygon Conjec-

ture to the three equiangular polygons in the figure.

Recall that the Equiangular Polygon Conjecture

says that the measure of each interior angle of an

equiangular n-gon is �(n �

n2)180°�. a is the measure

of an interior angle of an equiangular hexagon, so

a � � 120°

b is the measure of an interior angle of an equian-gular pentagon, so

b � � 108°

c is the measure of an interior angle of an equian-gular quadrilateral, so

c � � 90°

The four polygons share a common vertex, so a �b � c � d � 360°, and d � 360° � 120° � 108° �90° � 42°. Now look at the isosceles triangle. Here,d is the measure of the vertex angle, so d � 2e �180° (Triangle Sum Conjecture). Therefore,2e � 138°, and e � 69°.

15. x � 104°, y � 98°. The quadrilaterals on the leftand right sides are kites, and the nonvertex anglesof a kite are congruent (Kite Angles Conjecture).The quadrilateral at the bottom of the figure is anisosceles trapezoid. Base angles are congruent(Isosceles Trapezoid Conjecture), and consecutiveangles are supplementary (Trapezoid ConsecutiveAngles Conjecture). Look at the three angles thatshare a common vertex on the left side of the

180° � 2�4

180° � 3�5

180° � 4�6

180° � 6�8



DG5_Solution_Manual_CH05_Printer.qxd 3/16/16 12:27 PM 8

positive x-axis. In order for opposite sides of theparallelogram to be parallel, the side opposite theside that lies on the x-axis must also lie on ahorizontal line (slope � 0), so the y-coordinates ofits endpoints will be equal.

The Parallelogram Diagonals Conjecture states thatthe diagonals of a parallelogram bisect each other,so that is what you need to show for this figure.Two segments bisect each other if they have thesame midpoint, so if you can show that the twodiagonals of the parallelogram have the samemidpoint, you will have verified the conjecture.

The midpoint of the diagonal with endpoints

(0, 0) and (d, c) is ��0 �2d

�, �0 �2c

�� d2�, �2

c��, and the

midpoint of the diagonal with endpoints (a, 0) and

(d, c) is ��a �2b

�, �c �20

�� a �2b

�, �2c

��. Now look at the

slopes of the opposite sides of the parallelogram

that are not horizontal. The slope of the side with

endpoints (0, 0) and (b, c) is �bc �

�00� � �b

c�, and the

slope of the side with endpoints (a, 0) and (d, c)

is �dc �

�0a� � �d �

ca�. Because opposite sides of a parallel-

ogram are parallel (definition of a parallelogram)

and parallel lines have the same slope, the two

slopes you have found must be equal. Therefore,

�bc

� � �d �ca�, so b � d � a, and d � a � b. By

substituting a � b for d in the midpoint coordi-

nates ��d2�, �2

c��, you can see that both diagonals have

the same midpoint, ��a �2b

�, �2c

��. Therefore, the diago-nals of the parallelogram bisect each other.

DEVELOPING MATHEMATICAL REASONINGBeginner Puzzle: green left, red up, red right, reddown

Intermediate Puzzle: orange down, orange right,red up, red right, red down

Advanced Puzzle: orange down, orange left, red up,red right, yellow down, yellow left, red down

LESSON 5.5

EXERCISES1. Sometimes true. It is only true if the parallelogram

is a rectangle.

true

false

x

y

(0, 0)

(b, c) (d, c)

(a, 0)

Because MB�� MD�� and MA�� MC�� (diagonalsbisect each other, by construction) and also �BMC� �DMA (vertical angles), �BMC � �DMA bySAS. Then �1 � �2 by CPCTC, and because �1and �2 are alternate interior angles, BC� � AD�� bythe Converse of the Parallel Lines Conjecture. Like-wise, using the same sides and vertical angles AMBand CMD, �AMB � �CMD by SAS. Then �3 ��4 by CPCTC, and because �3 and �4 are alter-nate interior angles, AB� � DC��. Therefore, both pairsof opposite sides are parallel, and ABCD is aparallelogram by the definition of a parallelogram.

20. Kite, dart, or rhombus. Construct two intersectingcircles.

The figure is a kite or a dart because all radii of agiven circle are congruent, which gives the two pairsof consecutive congruent sides. It is only a rhombusif the two circles have the same radius, so thequadrilateral has four congruent sides. Here, it is akite. If the center of the smaller circle were insidethe larger circle, the figure would be a dart.

EXTENSIONSA. Yes, the converses of the parallelogram conjectures

are true.

Converse of the Parallelogram Opposite AnglesConjecture: If both pairs of opposite angles of aquadrilateral are congruent, then the quadrilateralis a parallelogram.

Converse of the Parallelogram Consecutive AnglesConjecture: If consecutive angles of a quadrilateralare supplementary, then the quadrilateral is aparallelogram.

Converse of the Parallelogram Opposite SidesConjecture: If both pairs of opposite sides of aquadrilateral are congruent, then the quadrilateralis a parallelogram.

Converse of the Parallelogram DiagonalsConjecture: If the diagonals of a quadrilateral bisecteach other, then the quadrilateral is a parallelogram.

Investigations will vary.

B. A coordinate geometry verification of the Parallelo-gram Diagonals Conjecture is given here. Otherparallelogram conjectures could also be verified.Place a parallelogram on a coordinate system withone vertex at the origin and another along the



13. x � 45°, y � 90°. A diagonal divides a square intotwo isosceles right triangles, so x � 45°. Thediagonals of a square are perpendicular (SquareDiagonals Conjecture), so y � 90°.

14. DIAM is not a rhombus because it is not equilateraland opposite sides are not parallel. You can useslopes to determine that opposite sides of thisquadrilateral are not parallel, so it is not evena parallelogram.

15. BOXY is a rectangle because its adjacent sides areperpendicular.

16. Yes. TILE is a rhombus, and every rhombus isa parallelogram.

17. Possible construction: You knowby the Square DiagonalsConjecture that the diagonalsof a square are perpendicularand congruent. Duplicate LV�.Construct its perpendicularbisector. From the intersectionpoint of the diagonals, set the radius of yourcompass at �

12�(LV ). Draw a circle with this center

and radius. Label the two points where the circleintersects the perpendicular bisector of LV� as Eand O. Connect the points L, O, V, and E to formthe square.

18. One possible construction: Duplicate �B. Bisect it.Mark off distance BK along the bisector. At K dupli-cate angles of measure �

12�m�B on either side of BK�.

Label as A and E the intersections of the rays of �Bwith the rays of the new angles which are not BK�.

Another possible construction: Duplicate �B andthen bisect it. Mark off the length BK on the anglebisector. Then construct the perpendicular bisectorof BK�. Label the points where this perpendicularintersects the sides of �B as E and K. Connect thepoints B, A, K, and E to form the rhombus.

19. Possible construction: DuplicatePS� and construct a perpendi-cular to PS� through S. Openyour compass to radius PEand draw an arc intersectingthe perpendicular line that

I E

P S

A

B

K

E

A

B E

K

E V

L O

2. Always true. By the definition of rectangle, all theangles are congruent. By the Quadrilateral SumConjecture and division, each angle measures 90°,so any two angles of a rectangle, including consecu-tive angles, are supplementary.

3. Always true by the Rectangle Diagonals Conjecture

4. Sometimes true; it is only true if the rectangle is asquare.

5. Always true by the Square Diagonals Conjecture

6. Sometimes true. It is true only if the rhombus isequiangular.

7. Always true. All squares fit the definition of rectangle.

8. Always true. All sides of a square are congruent andform right angles, so the sides become the legs ofthe isosceles right triangle, and the diagonal is thehypotenuse.

9. Always true by the Parallelogram Opposite AnglesConjecture

10. Sometimes true. It is true only if the parallelogramis a rectangle. Consecutive angles of a parallelogramare always supplementary, but are only congruent ifthey are right angles.

11. 20. By the Rectangle Diagonals Conjecture, the diag-onals of a rectangle bisect each other and arecongruent. By the “bisect each other” part, KC �CR � 10, so KR � 20, and by the “are congruent”part, WE � KR, so WE � 20.

12. 37°. By the Parallelogram Consecutive AnglesConjecture, �P and �PAR are supplementary, so48° � (y � 95°) � 180°. Then y � 95° � 132°,so y � 37°.

true

false

true

false

true

false




so e � 36°. (Or, by the Corresponding Angles (CA)Conjecture, e � b � 36°.) Now look at the quadri-lateral that contains the angle with measure f. Thisis a parallelogram because both pairs of oppositesides are parallel (from the given pairs of parallellines). Notice that the angle that forms a linear pairwith the angle of measure e is the angle oppositethe angle of measure f in the parallelogram. There-fore, by the Parallelogram Opposite Angles Conjec-ture and the Linear Pair Conjecture, f � 180° � e� 144°. Now look at the isosceles triangle in whichg is the measure of a base angle. In this triangle, thevertex angle measures 144° (Vertical AnglesConjecture), so 2g � 144° � 180°; then 2g � 36°,and g � 18°. Along the upper horizontal line, thereare three marked angles, all of measure h, and oneunmarked angle. The unmarked angle is the alter-nate interior angle of the supplement of the angleof measure f, so its measure is 180° � 144° � 36°.Then 3h � 36° � 180°, so 3h � 144°, andh � 48°. Next, j is an alternate interior angle ofone of the angles of measure h, so j � 48°. Finally,k � j � 48° � 180° (Triangle Sum Conjecture),so k � 84°.

24. Possible answers: (1, 0); (0, 1); (�1, 2); (�2, 3).

Find an equation for the line passing through the

points (2, �1) and (�3, 4), and then find points on

that line. First find the slope: m � �4��3(��12)

� � ��55� �

�1. Now use the slope and one of the points on the

line to find an equation for the line. Here, (2, �1)

is used.

�y �x �(�21)

� � �1

�xy �

�12� � �1

y � 1 � �1(x � 2)

y � 1 � �x � 2

y � �x � 1

To find points on this line in addition to the twothat are given, substitute any number for x otherthan 2 and �3 into the equation y � �x � 1 tofind the corresponding value for y. If x � 1, y � 0;if x � 0, y � 1; if x � �1, y � 2; if x � �2,y � 3. Therefore, four additional points on the lineare (1, 0), (0, 1), (�1, 2), and (�2, 3). Any three ofthese points would be sufficient; there are infinitelymany other points on the line.

25. y � �89�x � �

896�, or 8x � 9y � �86. The perpendicular

bisector of the segment with endpoints (�12, 15) and(4,�3) is the line through the midpoint of this segmentand perpendicular to it. First find the midpoint:

��122� 4�, �

15 �

2(�3)��

�

28�, �

122�� (�4, 6)

you have constructed. Label the intersection pointof the arc and the perpendicular line as E.Construct arcs of length PI and EI. Label theirintersection as I. Connect the points P, I, E, and Sto form the rectangle.

20. Converse: If the diagonals of a quadrilateral arecongruent and bisect each other, then the quadrilat-eral is a rectangle.

Given: Quadrilateral ABCD with diagonalsAC� � BD�. AC� and BD� bisect each other

Show: ABCD is a rectangle

Paragraph Proof: Because the diagonals arecongruent and bisect each other, AE� � BE� �DE� � EC�. Using the Vertical Angles Conjecture,�AEB � �CED and �BEC � �DEA. So �AEB ��CED and �AED � �CEB by SAS. Using theIsosceles Triangle Conjecture and CPCTC, �1 ��2 � �5 � �6, and �3 � �4 � �7 � �8. Eachangle of the quadrilateral is the sum of two angles,one from each set, so for example, m�DAB �m�1 � m�8. By the addition property of equality,m�1 � m�8 � m�2 � m�3 � m�5 � m�4 �m�6 � m�7. So m�DAB � m�ABC � m�BCD �m�CDA. So the quadrilateral is equiangular. Using�1 � �5 and the Converse of AIA, AB� � CD�. Using�3 � �7 and the Converse of AIA, BC� � AD��.Therefore ABCD is an equiangular parallelogram, soit is a rectangle.

21. If the diagonals are congruent and bisect each other,then the room is rectangular (Rectangle DiagonalsConjecture).

22. The platform stays parallel to the floor becauseopposite sides of a rectangle are parallel. (The Paral-lelogram Opposite Sides Conjecture applies becausea rectangle is a parallelogram.)

23. a � 54°, b � 36°, c � 72°, d � 108°, e � 36°,f � 144°, g � 18°, h � 48°, j � 48°, and k � 84°.First, a � 54° (Corresponding Angles (CA)Conjecture). Now look at the right triangle in whichthe measure of one of the acute angles is b. Theother acute angle in this triangle is the vertical angleof the angle of measure a, so its measure is 54°, andb � 90° � 54° � 36°. Next, 2c � b � 180°, so 2c �144°, and c � 72°. The angle with measure d formsa linear pair with the alternate interior angle of theangle with measure c, so d � 180° � 72° � 108°. Inthe triangle that contains the angle with measure e,both of the other angle measures have been foundto be 72°, so this is an isosceles triangle with vertexangle of measure e. Therefore, e � 2 � 72° � 180°,

A

E

B

CD

81 2

3

456

7



COORDINATE GEOMETRY 5

EXERCISES1. (�3, 1), (0, 3), (3, �1). Find the intersection point

for each pair of equations by solving that pairsimultaneously.

y � 3 � �23�x and y � ��

13�x: Substitute 3 � �

23�x for y

in the second equation and solve for x. Then substi-tute the value of x into the second equation to findthe corresponding value of y.

3 � �23�x � ��

13�x

x � �3

y � ��13�(�3) � 1

The first and second lines intersect at the point(�3, 1).

y � 3 � �23�x and y � ��

43�x � 3: Substitute 3 � �

23�x

for y in the second equation and solve for x; thenfind the corresponding value of y.

3 � �23�x � ��

43�x � 3

2x � 0

x � 0

y � 3 � �23�(0) � 3

The first and third lines intersect at thepoint(0, 3).

y � ��13�x and y � ��

43�x � 3: Substitute ��

13�x for y in

the third equation and solve for x; then find thecorresponding value of y.

��13�x � ��

43�x � 3

x � 3

y � ��13�(3) � �1

The second and third lines intersect at thepoint (3, �1).

Therefore, the vertices of the triangle are (�3, 1),(0, 3), and (3, �1).

2. a. � . The slope of the line containing

AC� is �xy22

�

�

yx1

1� � �1

90

��32� � �

68� � �

34�. The line passes

through the point A(2, 3), so �xy �

�32� � �

34�. Solve this

equation for y.

�xy �

�32� � �

34�

4(y � 3) � 3(x � 2)

4y � 12 � 3x � 6

4y � 3x � 6

y � �34�x � �

32�

y � �34�x � �

32�

y ��x � 12

The slope of this segment is

�4�

�

3(�

�

1152)

� � ��

1168

� � ��98

�

So, its perpendicular bisector will have slope �89�. Find

an equation of the line with m � �89� through (�4, 6).

�x �y �(�64)� � �

89�

�xy �

�64� � �

89�

9(y � 6) � 8(x � 4)

9y � 54 � 8x � 32

y � �89�x � �

896�, or 8x � 9y � �86

26. y � ��170�x � �

152�, or 7x � 10y � �24. The

median to AB� goes through C(8, �8) and themidpoint of AB� , which is (�2, �1). Find anequation of the line that contains (8, �8) and(�2, �1). First find the slope.

m � ��88��(�(�21))

� � ��170�

Now use the slope and either point to find the equa-tion of the line. Here, the point (�2, �1) is used.

�xy �

�((��12))� � �

�107�

�xy �

�12� � �

�107�

10(y � 1) � �7(x � 2)

10y � 10 � �7x � 14

7x � 10y � �24, or y � ��170�x � �

152�

DEVELOPING MATHEMATICAL REASONINGThink about which pairs of items can be together. Onesolution: Take the rabbit across and leave it there. Goback. Take the carrots across and return with the rabbit.Leave the rabbit on the original side. Take the dogacross, leave it there, and go back. Finally, bring therabbit across again.

EXPLORATION • TURNING TWO QUADRILATERALSINTO A PARALLELOGRAM

DEVELOPING MATHEMATICAL REASONINGLook at each net and see if adjacent faces are related asthey are in the cube. A good problem-solving techniquehere is to eliminate choices. For example, the net in 1Acan be eliminated because the base of the green face istouching the red face, unlike in the cube.

1. D 2. C




�y �x �(�23)

� � �23�

�xy �

�32� � �

23�

3(y � 3) � 2(x � 2)

3y � 9 � 2x � 4

3y � 2x � 13

y � �23�x � �

133�

4. (4, �7). Find the perpendicular bisectors of any twosides of the triangle, and then find the point wherethey intersect. Any two of the perpendicular bisec-tors can be chosen; in this solution, the perpendicu-lar bisectors of TR� and RM�� will be used to find thecircumcenter.

First find the perpendicular bisector of TR�.

Midpoint of TR� � ��22� 4�, �

1 �23

�� (1, 2)

Slope of TR�� 4 �3 �(�12)� � �

26� � �

13�

The slope of the perpendicular bisector of TR� is theopposite reciprocal of �

13�, which is �3. Find the

equation in the form y � mx � b for the line withslope �3 that passes through (1, 2).

�xy �

�21� � �3

y � 2 � �3(x � 1)

y � 2 � �3x � 3

y � �3x � 5

Now find the perpendicular bisector of RM��.

Midpoint of RM�� 4 �2(�4)�, �

3 �2(�1)�� (0, 1)

Slope of RM�� 14

��

34� � �

��48� � �

12�

The perpendicular bisector of RM�� has slope �2 andy-intercept 1, so its equation is y � �2x � 1.

Now solve the system formed by the equations ofthe perpendicular bisectors.

�To solve this system by the substitution method,substitute �2x � 1 for y in the first equation, andsolve for x.

�2x � 1 � �3x � 5

x � 4

Now substitute 4 for x in the second equation, andsolve for y.

y � �2(4) � 1 � �7

y � �3x � 5y � �2x � 1

The slope of the line containing BD� is �xy22

�

�

yx1

1� �

�48

�� 4

8� � ��44� � �1. The line passes through the

point B(8, 4), so �xy �

�48� � �1. Solve this equation

for y.

�xy �

�48� � �1

y � 4 � �1(x � 8)

y � �x � 12

b. (6, 6). The system from part a may be solved byeither the substitution or elimination method. Tosolve by substitution, substitute �

34�x � �

32� for y in the

second equation. Multiply both sides by the leastcommon denominator, 4, to eliminate fractions.

�34�x � �

32� � �x � 12

4��34�x � �

32�� 4(�x � 12)

3x � 6 � �4x � 48

7x � 6 � 48

7x � 42

x � 6

Now substitute 6 for x in the second equation ofthe system from 11a and solve for y. y � �6 �12 � 6.

c. (6, 6); (6, 6). Use the Coordinate Midpoint Prop-erty to find the midpoint of each diagonal.

Midpoint of AC�: � ��x1 �

2x2�, �

y1 �

2y2��

� ��2 �210

�, �3 +29

�� 122�, �

122��

� (6, 6)

Midpoint of BD�: � ��x1 �

2x2�, �

y1 �

2y2��

� ��8 +24

�, �4 +28

�� 122�, �

122��

� (6, 6)

d. The diagonals intersect at their midpoints, whichsupports the conjecture that the diagonals of aparallelogram bisect each other.

3. y � �23�x � �

133�. The perpendicular bisector of RE�

is the line perpendicular to RE� and passing

through its midpoint. The midpoint of RE� is

��0 �24

�, �0 �

2(�6)�� (2, �3), and the slope of RE� is

��46��00

� � ��46� � ��

32�. The slope of the perpendicular

bisector is the opposite reciprocal of ��32�, which is �

23�.

Find the equation in the form y � mx � b of the

line with slope �23� passing through (2, �3).



�16x � 72 � �2x � 3

�14x � �69

x � �6194�

Now substitute �6194� for x in the second equation, and

solve for y.

y � �2��6194�� 9 � �

�769� � �

673� � ��

67�

Therefore, the circumcenter of �FGH is ��6194�, ��

67��.

6. �3, ��32��. Find the perpendicular bisectors of MO��

and NO��, and then find their point of intersection.

Midpoint of MO�� 4 �2

10�, �

0 �2(�3)��

� �3, ��32��

Slope of MO�� 10�

�3 �(�04)� � �

�143�

Slope of perpendicular bisector of MO�� 134�

Solve the equation � �134� for y to rewrite

this equation in the form y � mx � b :y � �

134�x � �

321�.

Midpoint of NO�� 0 �210

�, �5 �

2(�3)�� (5, 1)

Slope of NO�� 103

��05

� � ��108� � ��

45�

Slope of perpendicular bisector of NO�� 54�

Solve the equation �xy �

�15� � �

54� for y to obtain

y � �54�x � �

241�.

Now solve the system formed by the equations ofthe two perpendicular bisectors.

�Substitute �

54�x � �

241� for y in the first equation.

�54�x � �

241� � �

134�x � �

321�

12��54�x � �

241�� 12��

134�x � �

321��

15x � 63 � 56x � 186

�41x � �123

x � 3

To find the value of y, substitute 3 for x in the firstequation of the system.

y � �134�(3) � �

321� � 14 � �

321� � �

228� � �

321� � ��

32�

The circumcenter is �3, ��32��.

y � �134�x � �

321�

y � �54�x � �

241�

y � ��32��x � 3

The solution of the system gives the coordinates ofthe point where the two perpendicular bisectorsintersect, which is the circumcenter of the triangle.Therefore the circumcenter of �TRM is (4, �7).

5. ��6194�, ��

67��. Follow the method given in the solution

for Exercise 13. In this solution, the perpendicularbisectors of FG� and FH� will be found, but youcould choose any pair of perpendicular bisectors.

Midpoint of FG�� 0 �23

�, ��6

2� 6��

32�, 0�

Slope of FG�� 6 �3 �(�06)

� � �132� � 4

The slope of the perpendicular bisector of FG�is ��

14�.

Find the equation in the form y � mx � b for theline with slope ��

14� that passes through the point

��32�, 0�.

� ��14�

� ��41�

4y � �1�x � �32��

4y � �x � �32�

y � ��14�x � �

38�

Midpoint of FH�� 0 �212

�, ��6

2� 0�� (6, �3)

Slope of FH�� 01�2 �(�06)

� � �162� � �

12�

The slope of the perpendicular bisector of FH� is�2. Find the equation in the form y � mx � b forthe line with slope �2 that passes through (6, �3).

�y �x �(�63)

� � �2

y � 3 � �2(x � 6)

y � 3 � �2x � 12

y � �2x � 9

Now solve the system formed by the equations ofthe perpendicular bisectors.

�To solve this system by substitution, substitute�2x � 9 for y in the first equation, and solve for x.

�2x � 9 � ��14�x � �

38�

8(�2x � 9) � 8��14�x � �

38��

y � ��14�x � �

38�

y � �2x � 9

y�x � �

32�

y � 0�x � �

32�




LESSON 5.6

EXERCISES1. Start dealing from the bottom of the deck in the

opposite direction, starting with yourself, until allcards are dealt.

2. Completion of flowchart proof: 1. WA�� RT�,Given; 2. WR�� AT�, Given; 3.WT�� WT��, Samesegment; 4. �WRT � �TAW by SSS; 5. �1 � �2by CPCTC; 6. RT� � WA�� by Converse of theParallel Lines Conjecture; 7. �4 � �3 by CPCTC;8. RW�� TA� by Converse of the Parallel LinesConjecture; 9. Definition of parallelogram

Paragraph Proof: Because WA�� RT� andWR�� AT� (both given) and also WT�� WT��,�WRT � �TAW by SSS. Then �1 � �2 byCPCTC, so RT� � WA�� by the Converse of the ParallelLines Conjecture. Similarly, �3 � �4, so RW�� TA�.Because both pairs of opposite sides are parallel,WATR is a parallelogram.

3. parallelogram; opposite sides are congruent

4. Flowchart Proof

5. parallelogram

6. Flowchart Proof

Paragraph Proof: Look at two overlappingtriangles: �GOY and �IYO. IY� � GO�� becauseopposite sides of a rectangle are congruent.(Opposite sides of a parallelogram are congruent,and a rectangle is a parallelogram.) �GOY � �IYOby the definition of a rectangle. (All angles of arectangle are congruent.) Therefore, �GOY ��IYO by SAS, and YG� � OI� by CPCTC.

Opposite sides of rectangle are congruent.

Definition of rectangle

SAS

Same segment

4

CPCTC

5

31 2 /GOY > /IYO ### ### YO > YO ### ###IY > GO

### ###YG > IO

nGOY > nIYO

Given

1 ### ###SP i OA

Converse of �AIA Conjecture

7 ### ### PA i SO

Definition of parallelogram

8 SOAP is a parallelogram �

Given

2 ### ###SP > OA

SAS

5nSOP > nAPO

Same segment

4 ### ###OP > OP

AIA Conjecture

3/1 > /2

CPCTC

6/3 > /4

7. The circumcenter is the midpoint of the sideopposite the right angle (the hypotenuse). For aright triangle, the perpendicular bisectors of the twolegs adjacent to the right angle are two of themidsegments of the triangle. As with any pair oftriangle midsegments, they intersect at the midpointof the third side.

8. Find the slope of sides:

OA� � � 0;

AB� � ;

AB� � .

Find the slope of ⊥ bisectors:

HE�� is undefined;

GF�� ;

KD�� .

Find the midpoints:

OA� � (a, 0);

AB� � � � � (b � a, c);

OB� � � � � (b, c).

Find the equations of ⊥ bisectors:

HE��: x � a;

KD��: ;

GF��: .

Point of Intersection of GF�� and

HE��: � �Point of Intersection of GF�� and

KD��: � �The three perpendicular bisectors of the sides areconcurrent.

DEVELOPING MATHEMATICAL REASONING

r p o l y

s e l l a

d e n s p

o s i t e

s t r u c

g o n s

t i o n

i r a l

s i d e

t i o n

s t a

t e s

g o l

o p p

c o n

a,c2 � ab � c2

c

a,c2 � ab � c2

c

y �a � bc

x �c2 � b2 � a2

c

y � �bcx �

c2 � b2

c

2b � 0

2,2c � 0

2

2b � 2a

2,2c � 0

2

bc

b � ac

�a � bc

2c � 0

2b � 0�c

b

2c � 0

2b � 2a�

c

b � a

0 � 0

2a � 0



9. Flowchart Proof

Paragraph Proof: Look at the overlapping trianglesin the figure, �RGT and �HTG. It is given thatGTHR is an isosceles trapezoid with nonparallelsides GR� � TH��. Also, GT� � GT� (same segment),and �RGT � �HTG by the Isosceles TriangleConjecture. Then �RGT � �HTG by SAS, soGH�� TR� by CPCTC.

10. 30° angles in 4-pointed star, 30° angles in 6-pointedstar; yes

4-pointed star: At each vertex of the tiling, twohexagons, one square, and one vertex of the starmeet. Each interior angle of the hexagon measures

� 120°, and each interior angle of thesquare measures 90°. The sum of the angles aroundany point is 360°, so the measure of each angle ofthe 4-pointed star is 360° � 2 � 120° � 90° � 30°.

6-pointed star: At each vertex of the tiling, threesquares and three stars meet. Again, the sum ofthe angles around any point is 360°, so the measureof each acute angle of the 6-pointed star is�13�(360° � 3 � 90°) � �

13�(90°) � 30°.

The acute angles of both stars are the same, 30°.

11. He should measure the alternate interior angles tosee if they’re congruent. If they are, the edges areparallel.

12. ES��: y � –2x – 3, QI��:y � �

12�x � 2. The two

diagonals of anyquadrilateral connectnonconsecutive vertices,so the diagonals ofrhombus EQSI are ES�and QI�. The graphillustrates this.

The slope of ES�� is �x

y2

2

�

�

y

x1

1� � �

��1 �1 �(�33)� � �

�24� � �2.

The line passes through the point E(�3, 3), so

�x �

y �

(�33)� � �2. Solve this equation for y.

�xy �

�

33

� � �2

y

x

S (21, 21)I (26, 21)

E (23, 3) Q (2, 3)

4 �180°�6

Given Same segment Isosceles TrapezoidConjecture

SAS

5

CPCTC

42

Given

/RGT > /HTG3 #### #### GT > GT

6 #### #### GH > TR

1 Isosceles Trapezoid GTHR

### ####GR > TH

nRGT > nHTG

7. Flowchart Proof

Paragraph Proof: RE� � AB� is given, andBR� � EA� and AR� � EB� because opposite sides ofa parallelogram are congruent (ParallelogramOpposite Sides Conjecture). Therefore,�EBR � �ARB � �RAE � �BEA by SSS, so�EBR � �ARB � �RAE � �BEA by CPCTC.Thus, all four angles of parallelogram BEAR arecongruent, and therefore, BEAR is a rectangle bythe definition of a rectangle.

8. 1 PA i TR

Given

5 ZART is a parallelogram

Definition of parallelogram

3 TZ i RA

Given

4 /2 > /3

CA Conjecture

9 /1 > /2

Isosceles TriangleConjecture

10 /1 > /3

Substitution

2 ZA i TR

Z is on PA

6 ZT > AR

ParallelogramOpposite Sides Conjecture

8 ZT > PT

Substitution

7 PT > AR

Given

Given Opposite sides of parallelogram arecongruent.

Opposite sides of parallelogram arecongruent.

SSS

5

CPCTC

6

42

Given

### ###AR > EB3 ### ### BR > EA

1 BEAR is a parallelogram.

### ###RE > AB

Definition of rectangle

7 BEAR is a rectangle

nEBR > nARB > nRAE > nBEA

/EBR > /ARB > /RAE > /BEA




LESSON 5.7

EXERCISES1. A triangle has three midsegments, one parallel to

each of its sides, while a trapezoid has just one,which is parallel to the two bases.

2. 28. By the Triangle Midsegment Conjecture,PO � �

12�(RA) � 10. From the figure, PT � PR � 8,

and OT � OA � 10, so the perimeter of �TOP is8 � 10 � 10 � 28.

3. x � 60°, y � 140°. By the Triangle MidsegmentConjecture, the midsegment shown is parallel to thethird side, so x � 60° (AIA Conjecture). Also, theangle that forms a linear pair with the angle ofmeasure y measures 40°, so y � 180° � 40° � 140°.

4. 65°. By the Three Midsegments Conjecture, themidsegments divide the large triangle into foursmaller congruent triangles. The three angles of thelarge triangle measure 65°, 42°, and 180° � (65° �42°) � 73°. Because all four triangles are congruent,each must have angle measures of 65°, 42°, and 73°.By the Triangle Midsegment Conjecture, each of thethree midsegments is parallel to the third side. Byusing the CA Conjecture, you can find the anglemeasures in the three outer triangles. (Notice thatall three of these triangles are oriented the sameway, while the inner triangle is oriented differently.)From these angle measures, you can see that 42° �z � 73° � 180°, so z � 65°.

5. 23. By the Triangle Midsegment Conjecture, eachside of �TEN is half the length of the parallel sideof �UPA: TN � �

12�(PA) � 9, TE � �

12�(UA) � 8, and

NE � �12�(UP) � 6. Therefore, the perimeter of �TEN

is 9 � 8 � 6 � 23.

6. m � 129°, n � 73°, p � 42 cm. By the TrapezoidMidsegment Conjecture, the midsegment isparallel to the two bases, so n � 73° (CA Conjec-ture). By the Interior Supplements Conjecture,m � 180° � 51° � 129°. By the TrapezoidMidsegment Conjecture, p � �

12�(36 cm � 48 cm) �

�12�(84 cm) � 42 cm.

7. 35. By the Trapezoid Midsegment Conjecture, 24 ��12�(13 � q), so 48 � 13 � q and q � 35. Another wayto find q is to notice that the length of the shorterbase is 11 less than the length of the midsegment,so the length of the longer base must be 11 morethan the midsegment: q � 24 � 11 � 35.

8. 3. Triangle Midsegment Conjecture; 4. OA� � RD�;5. LN� � RD�

9. Coordinates: E(2, 3.5), Z(6, 5); both slopes are �38�.

E is the midpoint of RY�, so its coordinates are

��4 �20

�, �7 �20

�� (2, 3.5), and Z is the midpoint of

y – 3 � �2(x � 3)

y � �2x � 3

Use the same method to find the equation ofQI��. The slope of QI�� is �

x

y2

2

�

�

y

x1

1� � �

32

��

((��16))� � �

48� � �

12�.

The line passes through the point Q(2, 3), so

�xy �

�

32� � �

12�. Solve this equation for y.

�xy �

�

32

� � �12�

y � 3 � �12

�(x � 2)

y � �12

�x � 2

13. The container is �182� � �

23� full, so it will be �

23� full no

matter which of the faces it rests on, which tells youthat the height of the liquid in the new position willbe �

23�(9 in.) � 6 in.

14. a and b.

c. The quadrilaterals are congruent.

d. quadrilateral ABCD � quadrilateral FEHG

e. (x, y) → (�x � 1, y � 8)

f. The quadrilaterals are congruent.

15. They didn’t use the three points to get two sides ofthe parallelogram. They constructed three sides oftheir parallelogram. Bisecting the angle at (�1, 4)on the map and marking off distances congruent tothe lengths between the points would put the fourthvertex of the parallelogram at the point (5, 2), agood place to dig.

y

x

A

F

C

F'and A'

D

GE

B

C'and H'

E'andB'

H

D'and G'

9

12 3

6



15. Parallelogram; It is also a parallelogram when thequadrilateral is concave. Because the midsegment ofa triangle is by definition parallel to the third sideyou will always get a parallelogram.

16. Use the coordinates of the midpoints and themidpoint formula to write a system of equations.Solve for x and y respectively. Repeat for each sideof the triangle. In other words, the three equationsin x1, x2, and x3 give us a system of equations thatwe can solve to find the values of x1, x2, and x3; andlikewise the three equations in y1, y2, and y3 give usa system of equations that we can solve to find thevalues of y1, y2, and y3.

Another way would be to construct the threemidsegments of the triangle to get an image of theoriginal triangle that can then be dilated by a factorof 2.

17. Possible explanation: Use theoriginal picture to find thevolume of water,V � 60 � 80 � 30 �144,000 cm3. Let x be theheight of the water when thecontainer is resting on its smallestface. The volume must be the same, so 144,000 �40 � 60 � x. Therefore, x � �

14404•,06000

� � 60 cm.

18. If a quadrilateral is a kite, then exactly one diagonalbisects a pair of opposite angles. Yes, both the originaland converse statements are true. (By definition, a kiteis not a rhombus.)

19. a � 54°, b � 72°, c � 108°, d � 72°, e � 162°,f � 18°, g � 81°, h � 49.5°, i � 130.5°, k � 49.5°,m � 162°, and n � 99°. a � 54° (Alternate InteriorAngles Conjecture). 2a � b � 180°, so b � 180° �2 � 54° � 72°. The angle with measure c is a remoteexterior angle of base angles in the isosceles trianglewhose base angles measure 54°, so c � 54° � 54° �108° (Triangle Exterior Angle Conjecture). Theangles with measures b and d are correspondingangles formed when the parallel lines are cut by atransversal, so d � b, and d � 72° (CorrespondingAngles Conjecture). (Also, d � 180° � b � 180° �108° � 72° by the Linear Pair Conjecture.) Nextlook at the right triangle in which the angle with

80 cm

40 cm 60 cm

60 cm

M

C

AD Q

P

B

N15

4 3 2

RT�, so its coordinates are ��4 �28

�, �7 �23

�� (6, 5).

The slope of EZ�� 56��32.5

� � �14.5� � �

38�, and the slope

of YT�� 38

��

00� � �

38�.

10. The coordinates of the midpoints are D(2, 6) andZ(�6, 2); all three slopes are �

12�. D is the midpoint

of AR�, so its coordinates are ��0 �24

�, �7 �25

��, or (2, 6).Z is the midpoint of PT�, so its coordinates are

, or (�6, 2).

The slope of AP� � .

The slope of TR� � .

The slope of ZD� � .

11. Parallelogram. Draw a diagonal of the originalquadrilateral. The diagonal forms two triangles.Each of the two midsegments is parallel to the diag-onal, and thus the midsegments are parallel to eachother. Now draw the other diagonal of the originalquadrilateral. By the same reasoning, the secondpair of midsegments is parallel. Therefore, thequadrilateral formed by joining the midpoints is aparallelogram.

12. The length of the edge of the top base is 30 mby the Trapezoid Midsegment Conjecture.If x represents the length of the edge of the top,�52

2� x� � 41, so 52 � x � 82, and x � 30. You can

also figure this out by noticing that the bottom baseis 11 meters longer than the midsegment, so the topedge must be 11 meters shorter than the midseg-ment: 41 � 11 � 30.

13. Ladie drives a stake into the ground to create atriangle for which the trees are the other twovertices. She finds the midpoint from the stake toeach tree. The distance between these midpoints ishalf the distance between the trees.

14. Construct auxiliary line PQ parallel to NB�. PQBN isa parallelogram because opposite sides are parallel.Opposite sides of a parallelogram are congruent, soPQ� � BN� � CN� � BN� by definition of a midseg-ment. So, by substitution, CN� � PQ�. Because theyare corresponding angles �1 � �2 and �3 � �2,so �3 � �1 by substitution. �4 � �5 byCorresponding Angles. Therefore �CPN � �PDQby AAS. CP� � PD�� by CPCTC.

Cabin

2 � 6�6 � 2 � �4

�8 � 12

5 � 04 � (�6) � 5

10 � 12

7 � 4

0 � (�6)�

1

2

a�6 � (�6)

2,4 � 0

2b




22. Possible construction: Copy FN�,which will form one of thediagonals of the kite, and thenconstruct its perpendicularbisector. Because the diagonalsof a kite are perpendicular, theother diagonal, RK�, will liealong this perpendicular bisector,but you must locate the points Rand K. To do this, draw an arccentered at N with radius NK.Label the point where this arcintersects the perpendicular bisector as K. Now useyour compass to mark off a distance equal to RKalong the perpendicular bisector to locate K.Connect the vertices F, R, N, and K to form the kite.

There is only one kite, but more than one way toconstruct it.

EXTENSIONTriangle: Place a triangle on a coordinate system withone vertex at the origin and one side along the x-axis.Draw the midsegment that connects the endpoints ofthe two sides of the triangle that don’t lie on the x-axis.Assign coordinates to the three vertices of the triangle,and use the midpoint formula to find the coordinates ofthe endpoints of the midsegment.

Trapezoid: Place a trapezoid on a coordinate system withone vertex at the origin and one side along the x-axis.Draw the midsegment of the trapezoid, which connectsthe midsegments of the two nonparallel sides. Assigncoordinates to the four coordinates of the trapezoid.The two vertices that do not lie on the x-axis will lieon the same horizontal line and therefore have the samey-coordinate because the two bases of a trapezoid areparallel. Use the midpoint formula to find the coordi-nates of the endpoints of the midsegment.

In both the triangle and the trapezoid, the twoendpoints of the midsegment have the samey-coordinates, so each of the midsegments lies on

x

y

(0, 0)

(b, c) (d, c)

(a, 0)

a 1 d_____ 2, 22 c_

2, 22 c_

2b_2

x

y

(0, 0)

(b, c)

(a, 0)

a 1 b_____ 2, 22 c_

2, 22 c_2

b_2

K

R

NF

measure d is one of the acute angles. The otheracute angle of this triangle measures 90° � 72° �18°. Then e � 180° � 18° � 162° (Linear PairConjecture), and f � 18° (Vertical Angles Conjec-ture). The triangle containing angles with measuresf and g is isosceles with f the measure of the vertexangle, so 2g � f � 180°, and g � 81°. Now look atthe isosceles triangle that contains the angle withmeasure h as one of its base angles. In this triangle,the vertex angle measures 81° (Vertical AnglesConjecture), so 2h � 81° � 180°, and therefore,2h � 99°, so h � 49.5°. The angle with measure iforms a linear pair with one of the base angles ofthat isosceles triangle, so i � 180° � 49.5° � 130.5°.The angle with measure k and the angle of thesmall isosceles triangle that forms a linear pair withthe angle of measure i are corresponding anglesformed when the parallel lines are cut by a trans-versal, so k � 49.5°. Next look at the pentagon thatcontains angles of measures i, k, and m. The twounmarked angles each measure 99° because each ofthem forms a linear pair with one of the base anglesof the isosceles triangle that contains the angles ofmeasures f and g; and in that triangle, it was foundearlier that the base angles each measure 81°. By thePentagon Sum Conjecture, the sum of the measuresof the interior angles of a pentagon is 540°, so k �i � 2 � 99° � m � 540°, or 49.5° � 130.5° � 198° �m � 360°, so m � 162°. Finally, look at thepentagon that contains angles of measures b, c, e,and n. The unmarked angle in this pentagon formsa linear pair with the angle of measure g, so itsmeasure is 180° � 81° � 99°. Then, by the PentagonSum Conjecture, b � c � e � 99° � n � 540°, or72° � 108° � 162° � 99° � n � 540°. Therefore,441° � n � 540°, and n � 99°.

20. (3, 8). Find the y-coordinate first. Because TR� � CA�,

TR� is a horizontal segment, so the y-coordinate of

T must be 8. Because of the symmetry of an

isosceles trapezoid, the slopes of TC� and RA�

must be opposite signs. Because the slope of RA� is

�182

��

015� � �

�83� � ��

83�, the slope of TC� is �

83�. Let x

represent the x-coordinate of T. Then �8x �

�00

� � �83�,

so x � 3, and the coordinates of T are (3, 8).

21. (0, �8). By the Kite Diagonals Conjecture, HS� is theperpendicular bisector of ER�. The diagonals of thiskite intersect at the origin, so (0, 0) is the midpointof ER�. Therefore, because E is 8 units directly abovethe origin, Rmust be 8 units directly below the origin.This means that the coordinates of R are (0, �8).



CHAPTER 5 REVIEW

EXERCISES

1. a. True

b. True

c. False

d. False

e. Paradox

2. Divide 360° by the number of sides.

3. Sample answers: Using the measure of an interiorangle, set the expression in the interior anglemeasure formula equal to the angle measure andsolve for n. Using the measure of an exterior angle,divide the angle measure into 360° to find n.

4. Trace both sides of the ruler as shown below.

5. Make a rhombus using the double-edged straightedge,and draw a diagonal connecting the vertex of the orig-inal angle to the opposite vertex of the rhombus.

6. Sample answer: Measure the diagonals with string tosee if they are congruent and bisect each other.

7. Draw a third point and connect it with each of thetwo points to form two sides of a triangle. Find themidpoints of the two sides and connect them toconstruct the midsegment. The distance between thetwo points is twice the length of the midsegment.

8. x � 10°, y � 40°. This figure is a kite, so thediagonals are perpendicular and form two pairsof congruent right triangles. In the larger righttriangles (on the right), one of the acute angles ofeach triangle measures 80°, so the other acute anglemeasures 10°, that is, x � 10°. In the smaller righttriangles (on the left), one of the acute angles ofeach triangle measures 50°, so the other acute anglesmeasure 40°. Thus, y � 40°.

9. x � 60 cm. 2x � 52 � 64 � 266, so 2x � 120, andx � 60 cm.

10. a � 116°, c � 64°. This figure is an isosceles trape-zoid, so a � 116° (Isosceles Trapezoid Conjecture)and c � 180° � 116° � 64° (Trapezoid ConsecutiveAngles Conjecture).

a horizontal line (or has slope 0) and is thereforeparallel to the x-axis. Therefore, the midsegment of thetriangle is parallel to the third side, and the midsegmentof the trapezoid is parallel to the two bases.

PERFORMANCE TASK

Using the property that the diagonals of a rhombusbisect the angles of the rhombus and a DESE toconstruct the angle bisector of an angle, lay the DESEalong one side of the angle and draw a line on the otheredge. Do the same on the other side of the angle. Thetwo lines and the sides of the angle form a rhombuswhose diagonal bisects the angle.

Students might also align one edge of the straightedgewith one side of the angle and draw a line along theother edge of the straightedge. Repeat this process withthe other side of the angle. The point where the linesintersect is equidistant from the sides of the angle. Theray from the angle vertex through this intersection pointis the angle bisector.

Using the property that the diagonals of a rhombus areperpendicular bisectors of each other and a DESE toconstruct the perpendicular bisector of the segment,draw a segment that is longer than the width of theDESE. Lay the DESE so that the edges of the DESE areon each endpoint of the segment. Draw lines on eachedge of the DESE. Turn the DESE the opposite way andrepeat. The four lines form a rhombus with the segmentas one diagonal. Draw the other diagonal, the perpendi-cular bisector of the angle.

Students might also put the straightedge on the linesegment so that one end of the line segment touchesone side of the straightedge and the other end touchesthe other side. Draw lines. Repeat with the sidestouching opposite ends.




16. a � 120°, b � 60°, c � 60°, d � 120°, e � 60°,f � 30°, g � 108°, m � 24°, and p � 84°. First, lookat the equiangular hexagon on the left. Because a isthe measure of an interior angle of this hexagon,

a � � � 120°

The angle with measure b forms a linear pair withone of the interior angles of the hexagon, so b �180° � 120° � 60°. The third angle in the trianglecontaining angles with measures b and c also formsa linear pair with an interior angle of the hexagon,so this angle also measures 60°. Thus b � c � 60°� 180°, so c � 60°. The angle that forms a linearpair with the angle of measure c is the alternateinterior angle of the angle with measure d. There-fore, d � 180° � 60°, so d � 120°. The measure ofthe angle that forms a linear pair with the angle ofmeasure d is 60°, so 2e � 60° � 180°, and e � 60°.Now look at the triangle that contains the angle ofmeasure f. One of the angles of this triangle forms alinear pair with an angle that is marked as a rightangle, so this angle of the triangle is also a rightangle and this is a right triangle. The measure ofone of the acute angles of this triangle is f and themeasure of the other acute angle is e � 60°, so e �f � 90°, and f � 30°. Now look at the equiangularpentagon on the right. Here, g is the measure of oneof the interior angles, so

g � � � 108°

Next, look at the small triangle with threeunmarked angles, one of which forms a linear pairwith the angle of measure d. That angle measures180° � d � 60°, and another angle of the triangleis an exterior angle of the equiangular pentagon,so its measure is �

3650°� � 72° (or 180° � 108° �

72°). Thus, the measure of the third angle of thistriangle is 180° � 60° � 72° � 48°. Then m � 108°� 48° � 180°, so m � 24°. Finally, look at the smalltriangle that contains an angle of measure m.Another angle of this triangle measures 72° (exterior

(5 � 2) � 180°��5

3 � 180°�5

(6 � 2) � 180°��6

4 � 180°�6

11. Perimeter � 100. By the Triangle MidsegmentConjecture, MS � �

12�(OI), so OI � 36. Also, because

S is the midpoint of IT� and M is the midpoint ofOT� (definition of midsegment), IS � 20 andOM � 26. Therefore, the perimeter of MOIS (whichis a trapezoid) is 36 � 20 � 18 � 26 � 100.

12. x � 38 cm. Use the Trapezoid Midsegment Conjecture.

32 � �12�[x � (x � 12)]

32 � �12�(2x � 12)

32 � x � 638 � x

Thus, x � 38 cm.

13. y � 34 cm, z � 51 cm. The key to finding y and zis to look for smaller polygons contained within thelarge triangle. First look at the triangle in which yis one of the lengths of the sides. In this triangle,17 cm is the length of the midsegment and y is thelength of the third side, so 17 � �

12�y, or y � 2 � 17 �

34 cm by the Triangle Midsegment Conjecture. Nowlook at the trapezoid in which one of the bases haslength 17 cm and the other has length z. In thistrapezoid, x is the length of the midsegment, so bythe Trapezoid Midsegment Conjecture, y � �

172� z�.

Because y � 34 cm, this equation becomes34 � �

172� z�, or 68 � 17 � z. Therefore, z � 51 cm.


15. a � 72°, b � 108°. The measure of each inte-rior angle of a regular decagon is

� � 144°

By looking at a common vertex of two isoscelestrapezoids and the decagon along the inner edge ofthe frame and by using the fact that the sum of allangles around a point is 360°, you can see that2b � 144° � 360°, so 2b � 216° and b � 108°.Then, by the Trapezoid Consecutive AnglesConjecture, a � b � 180°, so a � 72°.

(10 � 2) �180°��10

8 �180°�10


Chapter 5 Review, Exercise 14

Kite Isosceles trapezoid Parallelogram Rhombus Rectangle Square

Opposite sides are parallel No No Yes Yes Yes Yes

Opposite sides are congruent No No Yes Yes Yes Yes

Opposite angles are congruent No No Yes Yes Yes Yes

Diagonals bisect each other No No Yes Yes Yes Yes

Diagonals are perpendicular Yes No No Yes No Yes

Diagonals are congruent No Yes No No Yes Yes

Exactly one line of symmetry Yes Yes No No No No

Exactly two lines of symmetry No No No Yes Yes No


Another way to find b is to notice that three angleswith this measure meet at a common vertex, andthe sum of the measures of all the angles around apoint is 360°, so 3b � 360°, and thus b � 120°.

21. Recall that the diagonals of a rhombus are perpen-dicular and bisect each other (Rhombus DiagonalsConjecture); use this conjecture as the basis of yourconstruction.

First, copy the segment of length y and label thecopied segment as SR�. Construct the perpendicularbisector of SR�. Also, bisect the given segment oflength x. Using the midpoint of SR� as the centerand �

12�x as the radius, mark off arcs on both sides of

SR� along its perpendicular bisector. Label the pointswhere these arcs intersect the perpendicular bisectoras E and Q. To form the rhombus, connect thevertices S, Q, R, and E.

22. Possible construction: Duplicate �F and draw anarc centered at F with radius x, intersecting bothsides of the angle. Label the points where this arcintersects the angle as R and L. Bisect �F. Duplicate�L with vertex at L and one side along FL�. Labelthe point where this angle intersects the anglebisector of �F as Y. Connect the vertices F, L, Y,and R to form the kite.

23. Construct LP� of length z and duplicate �L withone side along LP�. Placing your compass point at L,mark off a distance of x along the other side of�L and label the point where the arc intersects theangle as N. Construct the line parallel to LP��through N. Placing your compass point at N, markoff a distance of y along this parallel line to formNE�. Connect the vertices E and P to form thefourth side of the trapezoid.

PL z

yN E

x

FL

Y

R

FLx

Y

R

x12

E

RS

Q

x12

angle of pentagon or supplement of g), and thethird angle measures p (Vertical Angles Conjecture),so p � m � 72° � 180°, and p � 84°.

17. 15 stones. Let x represent the measure of an interiorangle in the regular polygon. Two obtuse anglesmeasuring 96° (from two adjacent isosceles trape-zoids) and one interior angle of the regular polygonmeet at a common vertex, so 2 � 96° � x � 360°,which gives x � 168°. Use algebra to find thenumber of sides in a regular polygon in which eachinterior angle measures 96°.

� 168°

180°n � 360° � 168°n

12°n � 360°

n � 30

Thus, the regular polygon is a 30-gon, but becausethe arch is semicircular, the original arch contained15 stones.

18. (1, 0). Sketch the kite and observe that �A and �Care the vertex angles.

By the Kite Diagonal Bisector Conjecture, AC� is theperpendicular bisector of BD�, so the diagonals ofthe kite intersect at the midpoint of BD�, which hascoordinates ��0 �

22

�, �2 �(

2�2)�� (1, 0).

19. When the swing is motionless, the seat, the bar atthe top, and the chains form a rectangle. When youswing left to right, the rectangle changes to a paral-lelogram. The opposite sides stay equal in length, sothey stay parallel. The seat and the bar at the topare also parallel to the ground.

20. a � 60°, b � 120°. Corresponding (and thereforecongruent) angles of six of the pentagons meet ata common vertex in the center of the figure, so6a � 360°, and a � 60°. By looking at the tiling,you can see that four of the interior angles of eachpentagon are congruent, with measure b, while thefifth angle has measure a. The sum of the measuresof all the interior angles of any pentagon is 540°(Pentagon Sum Conjecture), so 4b � 60° � 540°,and thus b � 120°.

y

x

B(2, –2)A(–3, –2)

C(3, 1)

D(0, 2)

(n � 2) � 180°��n




28. Possible answer:

Given: Parallelogram ABCD

Show: AB� � CD�� and AD�� CB�

Flowchart Proof

Paragraph Proof: By the definition of parallelo-gram, AD�� BC� and AB� � CD��. Because parallel linesAD�� and BC�� are cut by transversal BD��, �1 � �3 bythe AIA Conjecture. Also, because parallel lines AB��and CD�� are cut by transversal BD��, �4 � �2 by theAIA Conjecture. We also have BD� � BD� becausethey are the same segment. Therefore, �ABD ��CDB by ASA, and thus AB� � CD�� and AD�� CB�by CPCTC.

29. Construct auxiliary NH�� so that NH ⊥ AB�. �NHBis a right angle by definition of perpendicular.CE�� ⊥ AB� and CE�� ⊥ MN�� by definition of altitudeand corresponding angles of parallel lines. �CPNand �CEB are right angles by definition of perpen-dicular, so �CPN � �CEB because all right anglesare congruent. CD�� AB� by definition of a trape-zoid. MN�� AB� by Trapezoid Midsegment Conjec-ture. Therefore, �CNP � �NBH by correspondingangles. CN�� NB� by definition of midpoint.�CPN � �NHB by AAS. CP� � NH�� by CPCTC.PE�� NH�� by definition of parallel lines. Bysubstitution, CP�� PE��.

B

N

C

P

H

D

M

AE

Definition ofparallelogram

AIA Conjecture

1 ABCD is a parallelogram

Given

5

AIA Conjecture

/2 > /4

ASA

7nABD > nCDB

CPCTC

8

4/1 > /3

Definition ofparallelogram

3 ### ####AB i CD2 ### #### AD i CB

Same segment

6 #### #### BD > BD

### #### ### #### AD > CB and AB > CD

12

43

A B

D C

24. Compare this situation with Exercise 23. Notice thatExercise 23 gives the lengths of both bases of thetrapezoid, whereas Exercise 24 gives only the lengthof one possible base. Use the same method asoutlined in the solution for Exercise 23, but noticethat now you can make RY� any length you wish.The figures below show two possible trapezoidsobtained with different choices for RY.

25. 20 sides. Notice that a square, a regular pentagon,and the third regular polygon with an unknownnumber of sides meet at the common vertex, B. Themeasure of each interior angle of a regular pentagonis � 108°, and the measure of each interiorangle of a square is 90°. Because the sum of themeasures of the angles around any point is 360°, themeasure of an interior angle of the third polygon is360° � 108° � 90° � 162°. Now find the number ofsides that a regular polygon has if the measure of itsinterior angles is 162°.

� 162°

180°n � 360° � 162°n

18°n � 360°

n � 20

Therefore, the third polygon has 20 sides.

26. 12 cm. First look at the two triangles with solidsides and ignore the dashed segments. The lowertriangle is isosceles, so the shared side of the twotriangles has length 48 cm (Converse of theIsosceles Triangle Conjecture). The middle of thethree dashed segments is the midsegment of theupper solid triangle, so its length is �

12�(48) � 24 cm

(Triangle Midsegment Conjecture). Now look at thesmaller triangle in which the dashed segment oflength x is the midsegment and the dashed segmentof the larger triangle (of length 24 cm) is the thirdside. Apply the Triangle Midsegment Conjecture tothis smaller triangle to find x: x � �

12�(24) � 12 cm.

27. 1. Given; 2. DE� � DI� by definition of rhombus;3. NE� � NI� by definition of rhombus; 4. DN�� DN��, Same segment; 5. �DEN � �DIN by SSS;6. �1 � �2, �3 � �4 by CPCTC; 7. definition ofangle bisector

(n � 2) � 180°��n

3 �180°�5

Fx

z

R

D

Y

Fx

z

R

D

Y



times the sum of the interior angles of each trianglewould equal the sum of the interior angles of ann-gon regardless of the number of dents.

4.

As shown in TAL 2 and 3, the sum of the interiorangles of a concave polygon is (n – 2) 180°. Eachvertex of the n-gon equals 360°. The sum of theexterior angles would be n(360°) – the sum of theinterior angles � n(360°) – (n – 2) 180°. Thenumber of dents does not affect the number oftriangles drawn inside the polygon, therefore thesum is the same.

5. a. The sum of the interior angles of a concavepolygon is (n – 2) 180°. Each vertex of the n-gonequals 360°. The sum of the exterior angleswould be n(360°) – the sum of the interiorangles � n(360°) – (n – 2) 180°. The number ofdents does not affect the number of trianglesdrawn inside the polygon, therefore the sum isthe same.

b.

Using hexagon ABDFHJ as an example:

Using the result from TAL 2, the sum of the interiorangles of the hexagon � (6 – 2) 180° � 720°.

From the diagram, the sum of the interior anglesof the hexagon � a � b � d � f � h � k �180° � 720°.

By substitution, a � b � d � f � h � k �180° � 720°.

Simplify to get a � b � d � f � h � 540° � k.

The sum of the exterior angles � (180° � a) �(180° � b) � (180° � d) � (180° � f) � (180° �h) � k.

Simplify to get the sum � 5(180°) – (a � b � d �f � h) – k.

By substitution the sum � 5(180°) – (540° � k) –k � 360°.

d

f

h

a k

b

d

a

c

b

TAKE ANOTHER LOOK

1.

a. By the Quadrilateral Sum Conjecture, the sum ofthe interior angles is 360°. Each reflex angle isequal to 360° � the measure of the interiorangle. Therefore, the sum of the reflex angles of aquadrilateral is equal to 360° � a � 360° � b �360° � c � 360° � d. Simplifying the equation,we get 1440° � (a � b � c � d). By substitu-tion, the sum of the reflex angles of a quadrilat-eral is 1440° � 360° � 1080°.

b. By similar reasoning, the sum of the reflex anglesof a pentagon would equal 5(360°) – 540° �1260°

c. For any polygon, the sum of the reflex anglesequals n(360°) – (n – 2) (180°) � 360°n –180n � 360° � 180° n � 360° or (n � 2) 180°

2.

Using the quadrilateral, draw an auxiliary line tocreate 2 triangles. Since the measures of the interiorangles of the two triangles add up to the sum of theinterior angles of the quadrilateral, use the TriangleSum Conjecture to find that a � b � c � d � e �f � 180° � 180° � 360°.

For any polygon, auxiliary lines can be drawn tocreate triangles. The number of triangles is n – 2.Therefore following the reasoning for a quadrilat-eral, the sum of the interior angles of a concaven-gon would be (n – 2) 180°.

3.

(n – 2) 180°. Following similar reasoning to TAL 2,draw auxiliary lines to create triangles. Each n-gonwould have n – 2 triangles. The number of triangles

d e

a

c

b f

d

a c

b




7. BD� � AC� because diagonals of a rectangle arecongruent. DM�� BM�� and AM�� CM�� becausediagonals of rectangle bisect each other.DM � MB � BD and AM � CM � AC bysegment addition. By substitution, DM � MB �AM � CM and DM � DM � AM � AM;2 DM � 2 AM; DM � AM. Therefore,DM � AM � BM.

With similar reasoning for each polygon, the sum ofthe exterior angles measured in this way is alwaysequal to 360°.

6. C-34: Nonvertex angles are congruent. True. Split-ting the dart into two triangles, these triangles arecongruent by SSS. Therefore the nonvertex anglesare congruent.

C-35: No. The diagonals do not intersect.

C-36: No. Diagonals do not intersect.C-37: No. Diagonals do not intersect.



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