solutions key pages 1 5

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Properties and Attributes of Triangles

Solutions Key

ARE YOU READY?

1. E 2. C

3. A 4. D

5. B

6. acute 7. right

8. acute 9. obtuse

10. 8 2 = 64 11. (-12 ) 2 = 144

12. √  49 = 7 13. - √  36 =-6

14. √9  + 16 = √  25 = 5 15. √  100 - 36 = √  64 = 8

16.





_

81 25 =

√  81

_

√  25 =

_

9

5 17.

_

2 2 = 2

18. d + 5 < 1 d <-4

0 1 2 3

-1 -2 -3 -4 -5 -6

19. -4 ≤ w - 7 3 ≤ w w ≥ 3

0 1 2 3 4 5 6

-1 -2 -3

20. -3s ≥ 6 s ≤-2

0 1 2 3 -1

-2 -3 -4 -5 -6

21. -2 >

_

m 10 -20 > m m <-20

0 20 40 60

-60 -40 -20

22. Let p and q represent the following: p: Lines ℓ and m intersect.

q: Lines ℓ and m are not parallel.

Given: p → q, and p. So by the Law of Detachment,

q is true: Lines ℓ and m are not parallel.

23. Let p, q, and r represent the following: p : M is the midpoint of AB ̶̶

q : AM = MB r : AM =

_

1

2 AB and MB = 1

_

2 AB

Given: p → q and q → r. So by the Law of Syllogism,

p → r : If M is the midpoint of AB , then AM ̶̶ =

_

1 2 AB and MB =

_

1

2 AB.

PERPEnDicUlAR AnD AnglE

bisEcTORs

check it out!

1a. DG = EG

DG = 14.6

b. Since DG = GE and ℓ ⊥ DE , ̶̶ ℓ is the ⊥ bisector of DE by the Conv. of the ̶̶ ⊥ Bisector Thm. EF =

_

1

2 DE

EF =

_

1

2 (20.8) = 10.4

2a. WX = WZ

WX = 3.05

b. Since XW = ZW, XW ̶̶̶⊥ XY , and ̶̶ ZW ̶̶̶⊥ ZY , ̶̶ YW 

bisects ∠XYZ by the Conv. of the ∠ Bisector Thm.

m∠XYZ = 2m∠WYZ

m∠XYZ = 2(63°) = 126°

3. By the Conv. of the ∠ Bisector Thm., QS bisects 

∠PQR.

4.

x y

0 ₄ -4 -2

-4 -2 2 4

P

Q

(3, -1)

Step 1 Graph PQ .̶̶

The ⊥ bisector of PQ is ̶̶ ⊥ to PQ at its midpoint̶̶ Step 2 Find the midpoint of PQ .̶̶

midpoint of PQ ̶̶=

(

5

_

+ 1 2 ,

2

_

+ (-4)

2

)

= (3, -1)

Step 3 Find the slope of the perpendicularbisector.

slope of PQ ̶̶=

_

-4 - 2 1 - 5 =

-6

_

-4 =

3

_

2

Since the slopes of ⊥ lines are opposite reciprocals,

the slope of the ⊥ 1bisector is -

_

2

3 .

Step 4 Use point-slope form to write an equation.

The ⊥ bisector of PQ has slope ̶̶ -

_

2

3 and passes

through (3, -1).

y - y ₁ = m(x - x ₁ )

y - (-1) =-

_

2 3 (x - 3)

y + 1 =-

_

2

3 (x - 3)

think and diScuSS

1. Yes; no; since PY = QY = 3, Y is the midpoint of PQ , and thus by the def. of bisector, ̶̶ ℓ is a bisector of PQ . If ̶̶ ℓ is the ⊥ bisector of PQ , then PX ̶̶

would equal QX by the ⊥ Bisector Thm. However,

PX = 8.5 and QX = 8.4, so ℓ is not the ⊥ bisector

of PQ .̶̶

2. No; although MJ = ML, to apply the Conv. of the

∠ Bisector Thm., you must know that MJ ̶̶⊥ KJ ̶̶ and ML ̶̶⊥ KL .̶̶

5

CHAPTER

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3.

Conv.: If a pt. is equidistant from the endpoints of a

seg., then the pt. is on the ⊥ bisector of a seg.

Thm.: If a pt. is on the bisector of an ∠, then it is equidistant from

the sides of the ∠.

Conv.: If a pt. in the int. of an ∠ is equidistant from the sides of the ∠,

then the pt. is on the bisector of the ∠. Thm.: If a pt. is on

the ⊥ bisector of a seg., then it is equidistant from the endpoints of

the seg.

⊥Bisector ∠Bisector

exerciSeS

guided practice

1. perpendicular bisector

2. Since PS = QS and m ⊥ PQ , m is the ̶̶ ⊥ bisector of PQ by the Conv. of the ̶̶ ⊥ Bisector Thm. PQ = 2QT

PQ = 2(47.7) = 95.4

3. SP = SQ SP = 25.9

4. PS = QS 4a = 2a + 26 2a = 26 a = 13

So QS = 2(13) + 26 = 52.

5. AD = CD

AD = 21.9

6. Since AD = CD, AC ̶̶⊥ AB , and ̶̶ CD ̶̶⊥ BC , ̶̶BD 

bisects ∠ABC by the Conv. of the ∠ Bisector Thm.

m∠CBD =

_

1

2 m∠ABC m∠CBD =

_

1

2 (48°) = 24°

7. Since DA = DC, AD ̶̶⊥ AB , and ̶̶ CD ̶̶⊥ BC , ̶̶BD 

bisects ∠ABC by the Conv. of the ∠ Bisector Thm.

m∠DBC = m∠DBA

10y + 3 = 8y + 10 2y + 3 = 10 2y = 7 y =

_

7

2 So m∠DBC = [10

(

_

7

2

)

+ 3]°= 38°

8. The braces can be installed so that PK ̶̶⊥ JL ,̶̶

PM ̶̶⊥ NL , and PK ̶̶ = PM. Then by the Conv. of the

∠ Bisector Thm., P will be on the bisector of ∠JLN.

9.

x y

4 2 -2

-4 -2 4 M

N (-2, 1)

Step 1 Graph MN .̶̶̶

The ⊥ bisector of MN is ̶̶̶ ⊥ to MN at its midpoint.̶̶̶ Step 2 Find the midpoint of MN .̶̶̶

midpoint of MN ̶̶̶=

(

_

-5 + 1

2 ,

4 + (-2)

_

2

)

= (-2, 1)

Step 3 Find the slope of the perpendicular bisector.

slope of MN ̶̶̶=

_

-2 - 4 1 - (-5) =

-6

_

6 =-1

the slope of the ⊥ bisector is 1.

Step 4 Use point-slope form to write an equation.

The ⊥ bisector of MN has slope 1 and passes ̶̶̶ through (-2, 1).

y - y ₁ = m(x - x ₁ ) y - 1 = 1[x - (-2)] y - 1 = x + 2

10.

x y

0 2 -2

-4 2

V

U (3, -3)

Step 1 Graph UV .̶̶

The ⊥ bisector of UV is ̶̶ ⊥ to UV at its midpoint̶̶ Step 2 Find the midpoint of UV .̶̶

midpoint of UV ̶̶=

(

_

2 + 4 2 ,

-6 + 0

_

2

)

= (3, -3)

slope of UV ̶̶=

_

0 - (-6) 4 - 2 =

6

_

2 = 3

the slope of the ⊥ bisector is -

_

1

3 .

The ⊥ bisector of UV has slope ̶̶ -

_

1

3 and passes

through (3, -3).

y - y ₁ = m(x - x ₁ )

y - (-3) =-

_

1 3 (x - 3)

y + 3 =-

_

1

3 (x - 3)

11.

x y

0 ₂ -2 -4 -6

-4 -2 2 J

K (-3, 2)

Step 1 Graph JK .̶̶

The ⊥ bisector of JK is ̶̶ ⊥ to JK at its midpoint̶̶ Step 2 Find the midpoint of JK .̶̶

midpoint of JK ̶̶=

(

_

-7 + 1

2 ,

5 + (-1)

_

2

)

= (-3, 2)

slope of JK ̶̶=

_

-1 - 5 1 - (-7) =

-6

_

8 =- 3

_

4

the slope of the ⊥ bisector is

_

4 3 .

The ⊥ bisector of JK has slope ̶̶

_

4

3 and passes

through (-3, 2).

y - y ₁ = m(x - x ₁ )

y - 2 =

_

4

3 [x - (-3)] y - 2 =

_

4

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practice and problem Solving

12. GJ = GK

GJ = 8.25

13. JG = KG x + 12 = 3x - 17

12 = 2x - 17 29 = 2x

14.5 = x

So KG = 3(14.5) - 17 = 26.5.

14. Since GJ = GK and t ⊥ JK , t is the ̶̶ ⊥ bisector of JK ̶̶

by the Conv. of the ⊥ Bisector Thm.

JK = 2JH

JK = 2(26.5) = 53

15. RQ = TQ RQ = 1.3

16. Since RQ = TQ, RQ ̶̶⊥ RS , and ̶̶ TQ ̶̶⊥ TS , ̶̶ SQ  bisects ∠RST by the Conv. of the Bisector Thm.

m∠RST = 2m∠RSQ

m∠RST = 2(58°) = 116°

17. m∠QSR = m∠QST 9a + 48 = 6a + 50 3a + 48 = 50

3a = 2 m∠QST = 6

(

_

2

3

)

+ 50 = 54°

18. They can position Main St. so that the ∠ formed by

Elm St. and Main St. is ≅ to the ∠ formed by Grove

St. and Main St. Then by the ∠ Bisector Thm., every

point on Main St. will be equidistant from Elm St. and Grove St.

19. Step 1 Graph EF .̶̶

The ⊥ bisector of EF is ̶̶ ⊥ to EF at its midpoint̶̶ Step 2 Find the midpoint of EF .̶̶

(

_

x 1 + x 2 2 ,

y ₁ + y ₂

_

2

)

midpoint of EF ̶̶=

(

_

-4 + 0 2 ,

-7 + 1

_

2

)

= (-2, -3)

slope =

_

y 2 - y 1 x ₂ - x ₁

slope of EF ̶̶=

_

1 - (-7) 0 - (-4) =

8

_

4 = 2

the slope of the ⊥ bisector is -

_

1 2 .

The ⊥ bisector of EF has slope ̶̶ -

_

1

2 and passes through (–2, –3).

y - y ₁ = m(x - x ₁ )

y - (-3) =-

_

1

2 [x - (-2)]

y + 3 =-

_

1

2 (x + 2)

20. Step 1 Graph XY .̶̶

The ⊥ bisector of XY is ̶̶ ⊥ to XY at its midpoint̶̶

Step 2 Find the midpoint of XY .̶̶

(

x

_

1 + x 2

2 ,

y ₁ + y ₂

_

2

)

midpoint of XY ̶̶=

(

-

_

7 + (-1)

2 ,

5 + (-1)

_

2

)

= (–4, 2)

slope = y

_

2 - y 1 x ₂ - x ₁

slope of XY ̶̶=

_

-1 - 5

-1 - (-7) =

-6

_

6 = –1

the slope of the ⊥ bisector is 1.

The ⊥ bisector of XY has slope ̶̶ - 1

_

2 and passes

through (-2, -3). y - y ₁ = m(x - x ₁ ) y - 2 = 1[x - (-4)] y - 2 = x + 4

21. Step 1 Graph MN .̶̶̶

The ⊥ bisector of MN is ̶̶̶ ⊥ to MN at its midpoint̶̶̶ Step 2 Find the midpoint of MN .̶̶̶

(

x

_

1 + x 2

2 ,

y ₁ + y ₂

_

2

)

midpoint of MN ̶̶̶=

(

-

_

3 + 7

2 ,

1 + (-5)

_

2

)

= (2, -3)

Step 3 Find the slope of the ⊥ bisector.

slope = y

_

2 - y 1 x ₂ - x ₁

slope of MN ̶̶̶= -

_

5 - (-1) 7 - (-3) =

-4

_

10 = - 2

_

5

the slope of the bisector is 5

_

2 .

The bisector of ⊥ has slope - 1

_

2 and passes through

(-2, -3).

y - y ₁ = m(x - x ₁ )

y - (-3) = 5

_

2 (x - 2)

y + 3 = 5

_

2 (x - 2)

22. PS = PT 3m + 9 = 5m - 13

9 = 2m - 13

22 = 2m

11 = m

QS = QT

6n - 3 = 4n + 14 2n - 3 = 14

2n = 17 n = 8.5

23. JK = LK JK = 38

24. GN = 2GZ

GN = 2(36) = 72

25. MK = HK ML + LK = HJ + JK

ML = HJ ML = 38

26. HY = MY HY = 24

27. JL = 2LX JL = 2(12) = 24

28. NK = GK NM + ML + LK = 114 NM + 38 + 38 = 114 NM = 38

29. Possible answer: C(3, 2); AC = √  26 ; BC =   26 ;

so AC = BC, and by the Conv. of the ⊥ Bisector

Thm., C is on the ⊥ bisector of AB .̶̶

30. Draw line ℓ ⊥ to AB through X. So m̶̶ ∠AYX = 90°

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31. Statements reasons

1. PS bisects  ∠QPR, SQ ̶̶⊥ PQ ,  SR ̶̶⊥  PR

1. Given

2. ∠QPS ≅∠RPS 2. Def. of ∠

bisector 3. ∠SQP and ∠SRP are rt. . 3. Def. of ⊥

4. ∠SQP ≅∠SRP 4. Rt. ∠≅ Thm.

5. PS ̶̶≅ PS ̶̶ 5. Reflex. Prop. of ≅

6. △PQS ≅△PRS 6. AAS

7. SQ ̶̶≅ SR ̶̶ 7. CPCTC

8. SQ = SR 8. Def. of ≅ segs.

32. Possible answer: By stating that the point must be

in the int. of the ∠, the thm. implies that it must be

in the same plane as the ∠. It is possible for a point

to be equidistant from the sides of an ∠ but to lie in

a different plane. In the diagram, ∠ABC is in plane

Z, and P is equidistant from the sides of ∠ABC, but P does not lie in plane Z. Thus P cannot be on the

bisector of the ∠, because the bisector must lie in

the same plane as the ∠.

ZN

P

A C B

33a. Step 1 Graph AC .̶̶

The ⊥ bisector of AC is ̶̶ ⊥ to AC at its midpoint̶̶ Step 2 Find the midpoint of AC .̶̶

(

x

_

1 + x 2

2 ,

y ₁ + y ₂

_

2

)

midpoint of AC = ̶̶

(

-

_

3 + 3

2 ,

-2 + 6

_

2

)

= (0, 2)

Step 3 Find the slope of the perpendicular

bisector.

slope = y

_

2 - y 1 x ₂ - x ₁

slope of AC ̶̶= 6

_

- (-2) 3 - (-3) =

8

_

6 = 4

_

3

Since the slopes of ⊥ lines are opposite

reciprocals, the slope of the ⊥ bisector is - 3

_

4 .

The ⊥ bisector of AC has slope ̶̶ - 3

_

4 and passes

through (0, 2). y - y ₁ = m(x - x ₁ )

y - 2 =- 3

_

4 (x – 0)

y =- 3

_

4 x + 2

b. There are 2 points on the ⊥ bisector that are 4 mi dist. from the midpoint of AC . ̶̶

c. Distance of warehouse from midpoint of AC ̶̶= 4 mi.

Distance of midpoint of AC from A ̶̶

=  3  2 + 4 2 = 5 mi.

Distance of warehouse from A =  4  2 + 5 2 ≈ 6.4 mi.

34. In the construction of the perpendicular bisector

of AB , the same compass setting is used to draw an ̶̶

arc from each end point of the segment. So in the

diagram, AX = BX and AY = BY. By the Converse

of the Perpendicular Bisector Theorem, both X and Y lie on the perpendicular bisector of AB . So ̶̶ ℓ is the perpendicular bisector of AB . ̶̶

A B

Y X

P

teSt prep

35. D;

J is on the perpendicular bisector of XY , so by the ̶̶

Perpendicular Bisector Theorem, JX = JY.

36. G;

37. Possible answer: All locations that are equidistant

from Park St. and Washington Ave. lie on the

bisector of the ∠ formed by the 2 streets. All

locations that are equidistant from the museum and the library lie on the perpendicular bisector of a segment formed by the museum and the library. So the visitor center should be built at point V, where the angle bisector and the perpendicular bisector intersect.

challenge and extend

38a. The dist. from P to BA and from P to  BC are both  2 √  5. So P is equidistant from BA and  BC , and  therefore by the Converse of the Angle Bisector

Theorem, P is on the bisector of ∠ABC.

b. Possible answer: y = 3x - 6.

39. The distance of a point (x, y) from x-axis is _⎜y_⎟ , and its distance from y-axis is ⎜x⎟ . So locus is _⎜y_⎟ = ⎜x⎟ , or the lines y = x and y =-x.

40. Statements reasons

1. VX ̶̶⊥ YX ,  VZ ̶̶⊥ YZ ,  VX = VZ

1. Given

2. ∠VXY and ∠VZY are rt. .

2. Def. of ⊥

3. YV ̶̶≅ YV ̶̶ 3. Reflex. Prop. of ≅

4. △YXV ≅△YZV 4. HL

5. ∠XYV ≅∠ZYV 5. CPCTC

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41. It is given that KN is the perpendicular bisector of ̶̶ JL ̶̶ and LN is the perpendicular bisector of ̶̶

KM . By the Perpendicular Bisector Theorem, JK ̶̶ = KL and KL = ML. Thus JK = ML by the Trans. Prop. of =. By the definition of ≅ segs., JK ̶̶≅ ML . By the ̶̶

Seg. Add. Post., JR + RL = JL and .MT + TK =

MK. By the definition of the perpendicular bisector, R is the midpoint of JL and T is the midpoint of ̶̶ MK . Thus ̶̶ JR ̶̶≅ RL and ̶̶ MT ̶̶≅ TK . By the definition ̶̶

of cong segs., JR = RL and MT = TK. By Subst.,

JR + JR = JL and MT + MT = MK. It is given

that JR ̶̶≅ MT . So JR ̶̶ = MT by definition of ≅ segs.

By Subst., JR + JR = MK. By the Trans. Prop. of =,

JL = MK, so JL ̶̶≅ MK by the definition of ̶̶ ≅ segs. By the Reflex. Prop. of ≅, JM ̶̶≅ JM . Therefore ̶̶

△JKM ≅△MLJ by SSS, and ∠JKM ≅∠MLJ by CPCTC.

bisEcTORs Of TRiAnglEs

check it out!

1a. GM = MJ = 14.5

b. GK = KH = 18.6

c. Z is circumcenter of △GHJ, By the Circumcenter Theorem, Z is equidistant from the vertices of

△GHJ.

JZ = GZ = 19.9

2.

x y

8 -8 -4

-8 4 8

H

G O

(4, -4.5) y= -4.5

x= 4

0

Step 1 Graph the △.

Step 2 Find equations for two perpendicular

bisectors. Since two sides of △ lie along the axes,

use the graph to find the perpendicular bisectors of these two sides. the perpendicular bisector of GO

is y =-4.5, and the perpendicular bisector of OH

is x = 4.

Step 3 Find the intersection of the two equations.

The lines y =-4.5 and x = 4 intersect at (4, -4.5),

the circumcenter of △GOH.

3a. X is the incenter of △PQR. By the Incenter

Theorem, X is euqidistant from the sides of △PQR.

The distance from X to PR is 19.2, so ̶̶ the distance from X to PQ is also 19.2.̶̶

b. m∠PRQ = 2m∠PRX

m∠PRQ = 2(12°) = 24°

m∠RQP + m∠PRQ + m∠QPR = 180°

52 + 24 + m∠RQP = 180° m∠RQP = 104°

m∠PQX =

_

1

2 m∠RQP

m∠PQX =

_

1

2 (104°) = 52°

4. By the Incenter Theorem, the incenter of a △ is

equidistant from the sides of the △. Draw the △

formed by the streets and draw the ∠ bisectors

to find the incenter, point M. The city should place the monument at point M.

M

think and diScuSS

1. Possible answer:

2. Q; P. Possible answer: the incenter is always inside

the △, so Q cannot be the incenter. Therefore

P must be the incenter, and Q must be the circumcenter.

3.

Definition The pt. of concurrency of the ⊥ bisectors Equidistant from the vertices of the Can be inside, outside, or on the

Circumcenter

The pt. of concurrency of the ∠ bisectors Equidistant from the sides of the Inside the

Incenter

Distance

Location (Inside, Outside, or On)

exerciSeS

guided practice

1. They do not intersect at a single point.

2. circumscribed about

3. N is the circumcenter of △PQR. By the Circumcenter Theorem, N is equidistant from

vertices of △PQR. NR = NP = 5.64

4. RV = PV = 5.47

5. TR = QT = 3.95

6. N is the circumcenter of △PQR. By the Circumcenter Theorem, N is equidistant from

vertices of △PQR.

QN = NP = 5.64