geometry rbc a answers - mr....

Answers

Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers

A61

93. 94

24y x= − 94. 8 2y x= −

95. a. 2224 ft b. 12 ft by 14 ft c. 2168 ft

96. a. 10 ft b. 24 ft c. 224 ft

Chapter 6 6.1 Start Thinking

The roof lines become steeper; The two top chords will get longer as the king post gets longer, but the two top chords will always be congruent to each other. The angles formed by the top chords and the king post are congruent. The angles formed by the top chords and the bottom chord are congruent.

6.1 Warm Up

1. 5 2. 8 3. 2 4. 1−

6.1 Cumulative Review Warm Up

1.

2.

6.1 Practice A

1. P lies on the perpendicular bisector of ;RS The

markings show that TP

satisfies the definition of a perpendicular bisector.

2. P lies on the perpendicular bisector of ;RS Because

T is equidistant from the endpoints R and S, T lies

on the perpendicular bisector of RS by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). Because only one line can be

perpendicular to RS at U, TU

must be the

perpendicular bisector of ,RS and P is on .TU

3. P lies on the angle bisector of ;DEF∠ P is

equidistant from sides ED

and EF

of angle ,DEF∠ so P is on the angle bisector of DEF∠ by

the Converse of the Angle Bisector Theorem (Thm. 6.4).

4. 20; Because D is on the perpendicular bisector of

,AC D is equidistant from A and C.

5. 17; By the Perpendicular Bisector Theorem (Thm. 6.1), .GJ GH= Solving 4 5 2 11x x+ = + gives 3,x = so 2 11 17.x + =

6. 14; Q is on the angle bisector of ,PSR∠ so Q is

equidistant from and .SP SR

7. 76 ;° By the Converse of the Angle Bisector

Theorem (Thm. 6.4), GE

bisects .DGF∠

Solving 3 8 5 12x x+ = − gives 10.x = So,

( ) ( )3 10 8 5 10 12 76 .m DGF∠ = + + − ° = °

8. 4 7y x= − +

9. Because any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment, you can draw an isosceles triangle by drawing segments from each endpoint of the segment to the same point on the perpendicular bisector.

10. yes; In a right triangle, the bisector of the right angle is also the perpendicular bisector of the hypotenuse when the right angle is isosceles. The figure shows that when right of C ABC∠

is bisected by , CD ACD BCD≅ by either

the ASA Congruence Theorem (Thm. 5.10) or the SAS Congruence Theorem (Thm. 5.5). Then the corresponding sides

andAC BC must be

congruent, so ABC is isosceles.

STATEMENTS REASONS

1. P is the midpoint

of MN and .TQ

1. Given

2. MP NP≅ 2. Definition of segment midpoint

3. PT PQ≅ 3. Definition of segment midpoint

4. MPQ NPT∠ ≅ ∠ 4. Vertical Angles Congruence Theorem (Thm. 2.6)

5. MQP NTP≅ 5. SAS Congruence Theorem (Thm. 5.5)

STATEMENTS REASONS

1. ,AB DC≅

AC DB≅

1. Given

2. BC BC≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)

3. ABC DCB≅ 3. SSS Congruence Theorem (Thm. 5.8)

A

C B

D

Answers

Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A62

6.1 Practice B

1. neither; Because PQ

is not marked perpendicular

to ,RS you cannot be certain that point P is on the

perpendicular bisector of .RS

2. neither; Because there is no indication that ,PD PF= you cannot be certain that point P is

on the angle bisector of .DEF∠

3. P lies on the bisector of ;DEF∠ Using the

HL Congruence Theorem (Thm. 5.9),

,DEQ FEQ≅ so DQ FQ= and EQ

bisects

DEF∠ by the Converse of the Angle Bisector

Theorem (Thm. 6.4). The figure shows that point P

is on .EQ

4. 36; Point D lies on the perpendicular bisector of

AC by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). 90m DBC∠ = ° by the

Triangle Sum Theorem (Thm. 5.1), so BD

is the

perpendicular bisector of .AC Solving

( )25 3 2 8 ,x x− + = − gives 7,x = so

36.AC =

5. 45 ;° Because L is 5

2 units from each side NK

and ,NM NL

bisects .KNM∠ Because

( )190 , 90 45 .

2m KNM m LNM∠ = ° ∠ = ° = °

6. 42 ;° TV

bisects UTW∠ because .VU VW= Solving 2 3 5 24x x+ = − gives 9,x = so

( )2 2 9 3 42 .m UTW∠ = + ° = °

7. 1 43 3

y x= − +

8. no; yes; Point Q does not lie on line m because Q is not the same distance from P and R. Point S lies on

line m, the perpendicular bisector of ,PR because S is the same distance from P and R.

9. Install the fountain at the point where the bisectors of the angles of the triangle intersect. That point is the same distance from each side of each angle, or equivalently, each side of the triangle.

6.1 Enrichment and Extension

1. 73

2. ( ) ( ) ( ) ( )2 2 2 21 5 5 2 ;x y x y+ + − = − + −

12

2y x= −

3. 12, 8x y= = 4. 3, 2x y= =

5.

STATEMENTS REASONS

1. PQRST is a regular polygon.

SV RV≅

1. Given

2. TP QP

TS QR

S R

≅

≅∠ ≅ ∠

2. Definition of regular polygon

3. DrawTV

and .QV

3. Through any points there exists exactly one line.

4. TSV QRV≅ 4. SAS Congruence Theorem (Thm. 5.5)

5. VT VQ≅ 5. Corresponding parts of congruent triangles are congruent.

6. VT VQ

PT PQ

==

6. Definition of congruent segments

7. V lies on the perpendicular

bisector of .TQ

P lies on the perpendicular

bisector of .TQ

7. Converse of the Perpendicular Bisector Theorem (Thm. 6.2)

8. PV lies on the perpendicular

bisector of .TQ

8. Perpendicular Postulate (Post. 3.2)

Answers


A63

6.

6.1 Puzzle Time

TOOK HIS MEDICINE AND FORGOT TO SHAKE IT

6.2 Start Thinking

They intersect at one point that appears to be the center of the circle. They also appear to be both angle bisectors and perpendicular bisectors.

6.2 Warm Up

1. 2.

3.

4.


1.

2.

6.2 Practice A

1. 9; Circumcenter Theorem (Thm. 6.5)


3. 2; Incenter Theorem (Thm. 6.6)

4. ( )3, 2 5. ( )2, 4−

BAY

X

64°A

126° B

STATEMENTS REASONS

1. LQ NQ≅ 1. Given

2. 2 3∠ ≅ ∠ 2. Base Angles Theorem (Thm. 5.6)

3. 1 4∠ ≅ ∠ 3. Given

4. 1 2

3 4

MLR

MNR

∠ + ∠ = ∠∠ + ∠ = ∠

4. Angles Addition Postulate (Post. 1.4)

5. 4 3 MLR∠ + ∠ = ∠ 5. Substitution

6. MLR MNR∠ ≅ ∠ 6. Substitution

7. MN MN≅ 7. Converse of the Base Angles Theorem (Thm. 5.7)

8. MQ

is the

perpendicular

bisector of .LN

8. Converse of Perpendicular Bisector Theorem (Thm. 6.2)

9. LP NP≅ 9. Converse of Perpendicular Bisector Theorem (Thm. 6.2)

x

y

42

2

4

−4

−4

A

A′

B′

C′

D′

B

D

C

y = x

CA

B

A′

B′

C′

x

y

62

6

−2

−2

y = 2

Answers


6. 12 7. 50

8. Sample answer:

9. Construct the circumcenter of the triangle formed by the locations of the three buildings. The tower is located at the circumcenter.

10. no; The circumcenter of an obtuse triangle lies outside the triangle, and the circumcenter of a right triangle lies on the hypotenuse.

6.2 Practice B



3. 11; Incenter Theorem (Thm. 6.6)

4. ( )9 32 2, 5. 10, 10MG NG= =

6. 18, 17GJ GE= =

7. a. Find the incenter by finding the intersection of two angle bisectors.

b. Find the circumcenter by finding the intersection of the perpendicular bisectors of two sides.

c. the circumcenter; The circumcenter is centered between the three vertices, whereas the incenter is toward the bottom of the triangular lawn in the figure.

8. The circumcenter lies outside the triangle when the triangle is obtuse.

9. 8


1. ( )23115 5

,Q 2. 13 units

3. ( )15.4, 21.2 4. ( )3.2,11.0

5.

6. Begin by drawing a line segment from point B to

point F, as shown. You are given FC is the

perpendicular bisector of AB and FE is the

perpendicular bisector of .BD By the Perpendicular Bisector Theorem (Thm. 6.1) you know that AF FB= and .FD FB= Using the Transitive

Property of Equality, you can conclude that .AF FD= By the definition of congruent

segments, you know that .AF FD≅

6.2 Puzzle Time

HAD A BYTE

C

A

C

B E

F

D

STATEMENTS REASONS

1. GJ is the ⊥

bisector of .HK

1. Given

2. HJ JK≅ 2. Definition of segment bisector

3. MH MK≅ 3. Perpendicular Bisector Theorem (Thm. 6.1)

4. GH GK≅ 4. Perpendicular Bisector Theorem (Thm. 6.1)

5. GM GM≅ 5. Reflexive Property of Segment Congruence (Thm. 2.1)

6. GHM

GKM

≅

6. SSS Congruence Theorem (Thm. 5.8)

7. GHM

GKM

∠ ≅∠

7. Corresponding parts of congruent triangles are congruent.

Answers


A65

6.3 Start Thinking

Two of the altitudes coincide with the two legs of the right triangle, and the three altitudes intersect at the vertex of the right angle.

6.3 Warm Up

1.

2.


1. ( )32

4 2y x− = + 2. 5x =

3. ( )34

9 1y x+ = − − 4. ( )12

7 1y x+ = − −

5. ( )53

2 3y x+ = − − 6. 32

y = −

6.3 Practice A

1. 8, 4BP PL= = 2. 8, 24PL CL= =

3. 18, 9AP PL= = 4. 51, 153PL BL= =

5. ( )4, 4 6. ( )2, 1

7. on; ( )0, 0 8. outside; ( )1, 5−

9. orthocenter; When the strings are pulled tight, right angles are formed on opposite sides of each vertex. Three altitudes are formed, which are concurrent at the orthocenter.

10. no; The orthocenter and the centroid are the same point in an isosceles triangles.

6.3 Practice B

1. 42, 63QL AL= = 2. 48, 24JQ QA= =

3. 5, 15QA KA= = 4. ( )3, 4−

5. inside; ( )1, 3 6. outside; ( )92

8,−

7. 1; Sample answer: The two vertices give one side of the triangle. The centroid and each given vertex can be used to find the exact location of the midpoint of another side. The two midpoints and the two given vertices can be used to find the third vertex of the triangle.

8. 1; Sample answer: The two vertices give one side of the triangle. Each vertex and the orthocenter can be used to draw a line containing the altitude from that vertex. The line perpendicular to that line and passing through the other vertex contains another side of the triangle. The third vertex is the intersection of the new sides.

9. ( )10, 2− 10. ( )3, 6−

11. no; Sample answer: A congruent pair of adjacent sides in a triangle have a perpendicular bisector, angle bisector, median, and altitude from the shared vertex that are all the same segment. An equilateral triangle has three pairs of congruent adjacent sides, so the special segments will all be the same, as will their points of concurrency.

12. yes; Sample answer: The median is the same as the altitude only when the sides that share the vertex are congruent, so the triangles formed by this median will be congruent by the HL Congruence Theorem (Thm. 5.9).

13. no; The circumcenter of an obtuse triangle is outside the triangle and the incenter of any triangle is inside the triangle.


1. 1 2 3 1 2 3 1 2 3, ,3 3 3

x x x y y y z z zN

+ + + + + +

2. , ,3 3 3

a b cN

m

P

Q

Answers


3. , , 0 ;2 2

a bT

The distance of

2 2 2

2 2 2

2

3 3 3

4

3

a b cFN

a b c

= + +

+ +=

and the distance of

2 2 2

2 2 2

2

6 6 6

4,

6

a b cNT

a b c

= + +

+ +=

so 2 .FN NT= •

4. 2 5. 1 or 7 6. 1− or 5.5

7. 8 1794

3 square units

8. about 2.237 square units

9. 2 6

,2 2

,

6;

2 2

c acAM y x

b a b a

cBN y x

a b

c acCP y x

b a b a

= = − − −

= = +

= = − − −

( )2 2 , 2a b c+

6.3 Puzzle Time

A BOOKWORM

6.4 Start Thinking

The measure of MN is one-half the measure of .BC

6.4 Warm Up

1. ( )3, 0− 2. ( )32

1, 3. 73 4. 65


1. 79° 2. 48 , 42° ° 3. 88 , 50 , 42° ° °

6.4 Practice A

1. The slope of 1 4

3,2 1

ED−= = −−

and the slope of

2 43.

5 3BC

− −= = −−

So, || .ED BC

( ) ( )2 22 1 1 4 10,ED = − + − =

( ) ( )2 25 3 2 4 2 10,BC = − + − − = so

1.

2ED BC=

2. ( ) ( )1, 4 , 4,1E F

3. The slope of 1 4

1,4 1

EF−= = −−

and the slope of

( )2 4

1.5 1

AC− −= = −

− − So, || .EF AC

( ) ( )2 24 1 1 4 3 2,EF = − + − =

( ) ( )2 25 1 2 4 6 2,AC = − − + − − = so

1.

2EF AC=

4. ( ) ( )2,1 , 4,1D F

5. The slope of 1 1

0,4 2

DF−= =−

and the slope of

( )4 4

0.3 1

AB−= =

− − So, || .DF AB

4 2 2,DF = − = and ( )3 1 4,AB = − − = so

1.

2DF AB=

6. 32 7. 34 8. 23

9. 17 10. 12 11. 21

12. yes; Because each midsegment is half as long as the corresponding side, the sum of the lengths of the midsegments (the perimeter of the midsegment triangle) will be half the sum of the lengths of the corresponding sides (the perimeter of the original triangle).

13. 116 ft

A

B

M N

C

Answers


A67

6.4 Practice B

1. ( ) ( ) ( )2, 2 , 1, 2 , 4, 1D E F− − − − −

2.

3. The slope of ( )( )

2 1 3,

2 4 2FD

− −= =

− − − and the

slope of ( )( )

1 5 3.

1 3 2CB

− −= =

− − So, || .FD CB

The slope of ( )( )

2 1 1,

1 4 3FE

− − −= = −

− − − and the

slope of ( )

1 3 1.

1 5 3AB

−= = −− −

So, || .FE AB

The slope of ( )

2 24,

1 2DE

− −= = −− − −

and the

slope of ( )

5 34.

3 5AC

− −= = −− − −

So, || .DE AC

4. [ ]( ) [ ]( )2 22 4 2 1 13,FD = − − − + − − =

[ ]( ) [ ]( )2 21 3 1 5 2 13,CB = − − + − − = so

1.

2FD CB=

[ ]( ) [ ]( )2 21 4 2 1 10,FE = − − − + − − − =

[ ]( ) ( )2 21 5 1 3 2 10,AB = − − + − = so

1.

2FE AB=

[ ]( ) ( )2 21 2 2 2 17,DE = − − − + − − =

[ ]( ) ( )2 23 5 5 3 2 17,AC = − − − + − − = so

1.

2DE AC=

5. 9 6. 4 7. 64 8. 8

9. Sample answer: The two pairs of equal measures determine the midpoints of two sides of a triangle. The midsegment of the triangle is represented by the surface of the bottom step. So, the bottom step is parallel to the floor, which represents the bottom side of the triangle.

10. no; Sample answer: The midsegments of a triangle with side lengths of 2a, 2b, and 2c divide the triangle into 4 triangles with side lengths of a, b, and c. So, the smaller triangle has only one-fourth the area of the larger triangle.


1. a.

b.

c.

1

242

n

y = •

d. 1

2

n

y w = •

2. ( ) ( ) ( )4, 4 , 0, 2 , 8, 0G H J

3. perimeter 17.1 units,= 2area 12.5 units=

Stage n 0 1 2 3 4 5

Midsegment length

24 12 6 3 3

2

3

4

2

y

−4

1 x−2

C

A

BD

EF

C

A

D E

F GH J

24 B

y

4

8

12

16

18

20

24

0 3 4 5 621 x

Stage

Mid

seg

men

t le

ng

th

Midsegments of Triangles

Answers


4. You are given .ABC DEF≅ Because

corresponding parts of congruent triangles are

congruent, you can conclude ,AB DE≅

,BC EF≅ and .CA FD≅ By the definition of congruent segments, you can also conclude

,AB DE= ,BC EF= and .CA FD= You are also given that T, U, and V are the midpoints of

ABC and X, Y, and Z are midpoints of .DEF

So, 12

TV BC= and 12

.XZ EF= By the

Substitution Property of Equality, you have 12

TV EF= and .TV XZ= You know

12

UV AB= and 12

.YZ DE= So, by the


UV DE= and .UV YZ= Finally, you know

12

TU CA= and 12

.XY FD= So, by the


TU FD= and .TU XY= By the definition of

congruent segments, you can conclude ,TV XZ≅

UV YZ≅ and .TU XY≅ By the SSS Congruence Theorem (Thm. 5.8), you can conclude

.TUV XYZ≅

6.4 Puzzle Time

STICK WITH ME AND WE WILL GO PLACES

6.5 Start Thinking

1. BC is the longest side and A∠ is the largest

angle.

2. YZ is the longest side and X∠ is the largest angle.

3. NP is the longest side and M∠ is the largest

angle.

The largest angle is always opposite the longest side.

6.5 Warm Up

1. If there is no right angle in a triangle, then it is not a right triangle.

2. If two lines do not have the same slope, then they are not parallel.

3. Sample answer: If a quadrilateral does not have four right angles, then the quadrilateral is not a rectangle.

4. Sample answer: If no two angles of a triangle are congruent, then the triangle is a scalene triangle.

5. If the sum of the measures of the interior angles of a polygon is not 180°, then the polygon is not a triangle.

6. Sample answer: If a triangle does not contain three congruent angles, then it is not equiangular.


1. 110 2. 28 3. 2−

6.5 Practice A

1. , ,N L M∠ ∠ ∠ 2. , ,F D E∠ ∠ ∠

3. , ,AB CA BC 4. , ,QP PR RQ

5. no; The sum of the first two sides is 15 37 52,+ = which is not greater than 53.

6. yes; The sum of the lengths of any two sides of the triangle is greater than the length of the third side.

7. Assume that a triangle has more than one obtuse angle. An obtuse angle is an angle that is greater than 90 .° This makes the sum of the angles in the triangle greater than 180 .° However, the sum of the angles of a triangle must be equal to, not greater than 180 .° This is a contradiction, so the assumption that a triangle has more than one obtuse angle must be false, which proves that a triangle has, at most, one obtuse angle.

8. 14x >

9. , , ;ACD D A∠ ∠ ∠ Using the properties of exterior

angles, you can solve for 14,x = and then obtain all of the angle measures.

10. Sample answer: In terms of the Triangle Inequality Theorem (Thm. 6.11), this can be thought of as a direct route between two points by traveling along one side, or an indirect route between two points by traveling along the other two sides. Because the sum of the lengths of any two sides of a triangle is greater than the length of the third side, the direct route along a single side is the shortest distance between two points.

6.5 Practice B

1. , ,L M N∠ ∠ ∠ 2. , ,V U W∠ ∠ ∠

3. , ,QS RQ SR 4. , ,BC DB CD

Answers


A69

5. Assume temporarily that a right triangle has three acute angles. So the measure of each angle is less than 90 ,° but by the definition of a right triangle, one angle must have a measure of 90 .° So, the assumption that a right triangle has three acute angles must be false. Next assume temporarily that a right triangle has exactly one acute angle, which means the triangle has one right angle and one obtuse angle. The measure of an obtuse angle plus 90° is greater than 180 .° By the Triangle Sum Theorem (Thm. 5.1), 180 .m A m B m C∠ + ∠ + ∠ = ° So,

the assumption that a triangle has exactly one acute angle must be false, which proves that a right triangle has exactly two acute angles.

6. yes; Substituting the given value of x into the expressions for the measures of the sides gives 60, 107, and 122. Because the sum of any two of these lengths is greater than the third, it is possible to construct such a triangle.

7. , ;DEF FGC Use the Triangle Inequality

Theorem (Thm. 6.11). For , ,DEF DE DF EF+ > ,DE EF DF+ > and

.DF EF DE+ > For , ,FGC FC CG FG+ >

,CG FG FC+ > and .FG FC CG+ >

8. yes; If you know all three angles measures, you can use a protractor and a straightedge to construct a triangle with the given angles that obeys the Triangle Inequality Theorem (Thm. 6.11).


1. , , , ,CD BC BD AB AD

2. , , , , , ,DE AE AD AB BD BC CD

3. Sample answer:

STATEMENTS REASONS

1. ABC and median

‘ AM

1. Given

2. Extend AM to point D

such that .AM DM≅ Draw .CDB

2. Construction

3. MB MC≅ 3. Definition of median

4. AMB DMC∠ ≅ ∠ 4. Vertical Angles Congruence Theorem (Thm. 5.5)

5. AMB DMC≅ 5. SAS Congruence Theorem (Thm. 5.5)

6. AB DC≅ 6. Corresponding parts of congruent triangles are congruent.

7. , AB DC

AM DM

==

7. Definition of congruent segments

8. AM MD AD+ = 8. Segment Addition Postulate (Post. 1.2)

9. AM AM AD+ = 9. Substitution Property of Equality

10. 2AM AD= 10. Simplify.

11. AD AC CD< + 11. Triangle Inequality Theorem (Thm. 6.11)

12. 2AM AC AB< + 12. Substitution

13. ( )12

AM AC AB< + 13. Division Property of Equality

14. ( )( )

12

12

AB AC

AB AC BC

+ <

+ +

14. Properties of real numbers

15.

( )12

AMAB AC BC<

+ + 15. Transitive

Property of Inequality

Answers


4. Sample answer: In ,ABC let , , andAX BY CZ

be medians. By the Centroid Theorem (Thm. 6.7) and the Triangle Inequality Theorem (Thm. 6.11), 2 2 2 23 3 3 3

, ,AX BY AB AX CZ AC+ > + > and

2 23 3

.BY CZ BC+ > Adding the left sides and

right sides of the three inequalities gives

( )43

.AX BY CZ AB AC BC+ + > + +

Multiplying each side by 34

gives

( )34

.AX BY CZ AB AC BC+ + > + + Finally,

3 14 2

> and the Transitive Property of Inequality

gives ( )12

,AX BY CZ AB AC BC+ + > + +

as desired.

5. 60 180; 0 60x x° < < ° < < °

6. a. 34

8x< <

b. 2x >

7. If a line segment is perpendicular to a plane, then it is perpendicular to every line segment in the plane. So, ,PC DC⊥ and PCD is a right triangle. The largest angle in a right triangle is the right angle, so .m PCD m PDC∠ > ∠ Finally, you can conclude that PD PC> because if one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.

6.5 Puzzle Time

A JUMP ROPE

6.6 Start Thinking

The length increases; The angle must be less than 180 degrees; 8

6.6 Warm Up

1. yes; andABE DCE are congruent by the

AAS Congruence Theorem (Thm. 5.11).

2. no 3. no

4. yes; andCDB CEA are congruent by the

SAS Congruence Theorem (Thm. 5.5).


1. rotation 2. dilation 3. translation

6.6 Practice A

1. ;AC DF> By the Hinge Theorem (Thm. 6.12),

because AC is the third side of the triangle with

the larger included angle, it is longer than .DF

2. ;m HGI m IGJ∠ = ∠ The triangles are congruent

by the SSS Congruence Theorem (Thm. 5.8). So, because corresponding parts of congruent triangles are congruent, .m HGI m IGJ∠ = ∠

3. 1 2;m m∠ < ∠ By the Converse of the Hinge Theorem (Thm. 6.13), because 1∠ is the included angle in the triangle with the shorter third side, its measure is less than that of 2.∠

4. ;KL MN< By the Hinge Theorem (Thm. 6.12),

because KL is the third side of the triangle with

the smaller included angle, it is shorter than .MN

5. 7 2 3, 10x x x+ < − >

6. ( )6 1 14 10, 2x x x+ > − <

7.

STATEMENTS REASONS

1. TV UW≅ 1. Given

2. UV UV≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)

3. TU VW> 3. Given

4. m TVU m WUV∠ > ∠ 4. Converse of the Hinge Theorem (Thm. 6.13)

Answers


A71

8.

9. no; You cannot apply either the Hinge Theorem (Thm. 6.12) or the Converse of the Hinge Theorem (Thm. 6.13) in this situation; These theorems require that two sides of one triangle are congruent to two sides of the other triangle. In this case, the ladders are different heights, so you only have one pair of congruent sides.

6.6 Practice B

1. ;BC DE> By the Hinge Theorem (Thm. 6.12),

because BC is the third side of the triangle with

the larger included angle, it is longer than .DE

2. ;JI GH> By the Hinge Theorem (Thm. 6.12),

because JI is the third side of the triangle with the

larger included angle, it is longer than .GH

3. 1 2;m m∠ > ∠ By the Converse of the Hinge Theorem (Thm. 6.13), because 1∠ is the included angle in the triangle with the longer third side, its measure is greater than that of 2.∠

4. ;m U m R∠ < ∠ By the Converse of the Hinge

Theorem (Thm. 6.13), because U∠ is the included angle in the triangle with the shorter third side, its measure is less than that of .R∠

5. ( )2 3 8 14, 6x x x− > + >

6. ( )2 22 3 18, 62x x x+ > − <

7.

8. Sailboat A; Because 151 129 ,° > ° the distance Sailboat A traveled is a greater distance than the distance Sailboat B traveled by the Hinge Theorem (Thm. 6.12)

9. Each theorem refers to the included angles of two triangles when two sides of one triangle are congruent to two sides of the other triangle, The Hinge Theorem (Thm. 6.12) refers to the case when the included angle of the first is larger than the included angle of the second. The SAS Congruence Theorem (Thm. 5.5) refers to the case when the included angles are congruent.


1. a. never b. never c. always

d. never e. never f. sometimes

STATEMENTS REASONS

1. PQ SR≅ 1. Given

2. PS SR PR+ > 2. Triangle Inequality Theorem (Thm. 6.11)

3. PR PQ

QR

= + 3. Segment Addition Postulate (Post. 1.2)

4. PS SR

PQ QR

+ >+

4. Substitution

5. PQ SR= 5. Definition of congruent segments

6. PS SR

SR QR

+ >+

6. Substitution

7. PS QR> 7. Subtraction Property of Inequality

8. m PQS

m RSQ

∠ >∠

8. Converse of the Hinge Theorem (Thm. 6.13)

STATEMENTS REASONS

1. BD BE≅ 1. Given

2. AB BC≅ 2. Definition of segment midpoint

3. 1 2m m∠ > ∠ 3. Given

4. AD CE> 4. Hinge Theorem (Thm. 6.12)

5. DF EF≅ 5. Given

6. DF EF= 6. Definition of congruent segments

7. AD DF CE DF+ > + 7. Addition Property of Inequality

8. AD DF CE EF+ > + 8. Substitution

9. ,AD DF AF

CE EF CF

+ =+ =

9. Segment Addition Postulate (Post 1.2)

10. AF CF> 10. Substitution

Answers


2. AB AD≅ and AC AE≅ because they are radii of the circles and all radii are congruent. By the Hinge Theorem (Thm. 6.12), because m BAC∠ > , then .m DAE BC DE∠ >

3. 16.5 37.5y< <

4.

6.6 Puzzle Time

DUCK

Cumulative Review

1. 9c > or 3c < − 2. 2 11s< <

3. 10 7m− < ≤ − 4. 1 15p≤ <

5. 6j ≤ 6. 8 13v< <

7. 5 12r− ≤ < 8. 17b ≥ or 3b <

9. 3

2m = − 10. 1m =

11. m is undefined. 12. 19

5m = −

13. 2

13m = − 14.

3

2m = −

15. 6

5m = − 16.

15

4m = −

17. m is undefined. 18. 9

7m = −

19. 3

5m = − 20. 7

6m =

21. 2

3m = − 22. 1

15m = −

23. 1

24m = − 24. 2 10y x= − −

25. 2y x= − 26. 1 5

2 2y x= −

27. 1 1

3 3y x= − − 28. 6x = −

29. 3x = − 30. 1

32

y x= −

31. 51

3y x= − −

32. a. 1.25 12.50C t= +

b. $16.25

c. $18.75

33. 1 37

8 8y x= − + 34.

3 39

2 2y x= − −

35. 6y = 36. 0y =

37. 47

7y x= − 38. 4y x= − −

STATEMENTS REASONS

1. and ADB CDA∠ ∠ are supplementary.

1. Linear Pair Postulate (Post. 2.8)

2.

180

m ADB

m CDA

∠ +∠ = °

2. Definition of supplementary angles

3.

180

m CDA

m ADB

∠ =° − ∠

3. Subtraction Property of Equality

4. 100m ADB∠ = ° 4. Given

5.

180 100

m CDA∠ =° − °

5. Substitution

6. 80m CDA∠ = ° 6. Subtraction Property of Equality

7. m ADB

m CDB

∠ >∠

7. Definition of congruent angles

8. D is the midpoint

of .BC

8. Given

9. BD CD≅ 9. Definition of midpoint

10. AD AD≅ 10. Reflexive Property of Segment Congruence (Thm. 2.1)

11. AB CA> 11. Hinge Theorem (Thm. 6.12)

12. m C m A∠ > ∠ 12. Triangle Longer Side Theorem (Thm. 6.9)

Answers


A73

39. 1 5

4 4y x= − 40.

3 5

2 2y x= − +

41. 152, 28x y= = 42. 77, 103x y= =

43. 26, 77x y= = 44. 4, 82x y= =

45. 115, 13x y= = 46. 33, 103x y= =

47. a. 67° b. 113°

48. a. 46° b. 92°

49. ( )3, 3A′ − 50. ( )4, 6B′ −

51. ( )5, 6C − − 52. ( )10, 8D −

53. ( )5, 8A′ − 54. ( )3, 9B′

55. ( )5, 3C − − 56. ( )4, 2D −

57.

58.

59.

2 4 6 8

−2

−4

−6

4

6

2

−2−8 x

y

C

A

B

A′

B′

C′

2 4

−2

−4

4

6

2

−2−4 x

y

C

A

B

A′

B′C′

4 6 8

−2

−4

−6

−8

4

6

8

2

−2 x

y

C

A

BA′

B′

C′

Answers


60.

61.

62.

63.

8 10

−2

−6

−8

−10

4

6

8

10

−2−4−6−8−10 x

y

C

A

B

A′

B′

C′

124 16

−12

−16

12

16

8

4

−4−8−12−16 x

y

D

F

E

F′

E′

D′

2 4 6 8

−6

−8

6

8

2

−2−4−6−8 x

y

A

C

B B′

A′

C′

2 4

−2

−4

−6

−8

6

8

−6−8 x

yC

A

B

A′

B′

C′

Answers


A75

64.

65.

66.

67.

68.

69.

−10

−30

1010

20

−20 20 x

y

S

U

U′

T

V

V′

S′

T′

−20

20

−20 2010 x

y

J

J′

L′

M′

L

M

128

−12

−8

12

8

−8−12 x

y

WY

X

W′

X′

Y′

Z′

Z

128 16

−12

−8

−4

−16

12

16

8

4

−8−16 x

y

A

B B′C′

A′

C

12

−12

−4

12

8

−12 x

y

E H

H′

G

G′

FF′E′

128 16

−8

−4

−16

12

16

−4−8−12−16 x

y

NM

P

N′

M′

P′

Answers


70. equilateral triangle

71. acute isosceles triangle

72. right scalene triangle

73. obtuse scalene triangle

74. 55 ;° obtuse scalene triangle

75. 76 ;° acute isosceles triangle

76. 24 ;° right scalene triangle

77. 95 ;° obtuse scalene triangle

78. 33° 79. 52° 80. 76°

Chapter 7 7.1 Start Thinking

1. 540° 2. 720° 3. 900°

7.1 Warm Up

1. 120 2. 70 3. 119


1. 1x = 2. 1y = −

3. ( )21 5

3y x+ = − 4. ( )11 14

12 5

y x− = +

7.1 Practice A

1. 900° 2. 19-gon

3. interior: 168 ,° exterior: 12°

4. 84 5. 125

6. 75m X m Y∠ = ∠ = °

7. 135m X m Y∠ = ∠ = °

8. 76 9. 88

10. 120° 11. 144 people

7.1 Practice B

1. 103 2. 68

3. 116m X m Y∠ = ∠ = °

4. 130m X m Y∠ = ∠ = °

5. 56 6. 55

7. interior: 165 ,° exterior: 15°

8. 20

9. 14; The sum of the interior angle measures of the polygon is ( ) ( )2 90 3 180 1440 2160 .° + ° + ° = °

So, the polygon has 2160

2 14 sides.180

° + =°

10. a. 720° b. 135


1. 61° 2. 130° 3. 58° 4. 50°

5. 84° 6. 85° 7. 146° 8. 145°

9. regular decagon 10. 12 sides

11. You know that 180 ,a b c+ + = °

180 ,d e f+ + = ° and 180g h i+ + = °

because the sum of the interior angles of a triangle equals 180 .° You can add those three equations to obtain 540 .a b c d e f g h i+ + + + + + + + = °

,m YZV f i∠ = + ,m ZVW h∠ =

,m VWX g d a∠ = + + ,m WXY b∠ = and

,m XYZ c e∠ = + so

m YZV m ZVW m VWX m WXY∠ + ∠ + ∠ + ∠ +540 .m XYZ∠ = °

7.1 Puzzle Time

THEY DIDN’T WANT TO WAIT FORTY YEARS FOR A TRAIN

7.2 Start Thinking

yes; Sample answer: The scout could use the Pythagorean Theorem to determine the distance that should be between opposite corner posts, the length of the hypotenuse. It

should be approximately 15 feet 716

7 inches. Or, the

scout could make sure that the distances between the two pairs of opposite corners are the same and not be concerned about the exact measure. This method uses the SSS Congruence Theorem (Thm. 5.8).

7.2 Warm Up

1.

STATEMENTS REASONS

1. ,MN PO

NO MP

≅

≅

1. Given

2. NP NP≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)

3. PMN NOP≅ 3. SSS Congruence Theorem (Thm. 5.8)

geometry rbc a answers - mr....

Documents