Download Geometry Exercises and Problem Solving and more Exercises Geometry in PDF only on Docsity!
Solutions Key
1 Foundations for Geometry
CHAPTER
ARE YOU READY? PAGE 3
1. C 2. E
3. A 4. D
5. 7 _^1 2
in. 6. 21 _ 2
cm
7. 100 yd 8. 10 ft 9. 30 in. 10. 15.6 cm 11. 8 y 12. - 2 x + 56 13. - x - 14 14. - 2 y + 31 15. x + 3 x + 7 x = 11 x = 11( - 5) = - 55 16. 5 p + 10 = 5(78) + 10 = 390 + 10 = 400 17. 2 a - 8 a = - 6 a = - 6(12) = - 72 18. 3 n - 3 = 3(16) - 3 = 48 - 3 = 45 19. (0, 7) 20. ( - 5, 4) 21. (6, 3) 22. ( - 8, - 2) 23. (3, - 5) 24. (6, - 4)
1-1 UNDERSTANDING POINTS, LINES,
AND PLANES, PAGES 6–
CHECK IT OUT!
1. Possible answer: plane and plane ABC 2. (^) M N 3. Possible answer: plane GHF 4.
THINK AND DISCUSS
1. By Post. 1-1-1, through any 2 pts. there is a line. Therefore any 2 pts. are collinear. 2. Post. 1-1- 3. Any 3 noncollinear pts. determine a plane. 4. AB ,
AB , AB , BA ; 0 planes 5. 1iviÊÊ/iÀÃÊÊ
iÊ Ű
Ű *ÌÊ *>iÊІÊ
A І
EXERCISES
GUIDED PRACTICE
1. Possible answer: the intersection of 2 floor tiles 2. S 3. A , B , C , D , E 4. Possible answer: AC , BD 5. Possible answer: ABC and 6. Possible answer: B , C , or D 7. (^) M N 8. (^) F G 9. Possible answer: AB 10. Possible answer: plane ABD 11. 12. (^) A B
C D
PRACTICE AND PROBLEM SOLVING
13. B , E , A 14. Possible answer: B , C , D , E 15. Possible answer: plane ABC 16.
X
Y
17. R
18. Possible answer: G , J , and 19. Possible answer: planes and 20.
R S
22a. Possible answer: tip of a stake b. Possible answer: string c. Possible answer: grid formed by string 23. M
R
S
T
25. U 26. U
27. U
28. If 2 pts. lie in a plane, then the line containing those pts. lies in the plane. 29. If 2 lines intersect, then they intersect in exactly 1 pt. 30. It is not possible. By Post. 1-1-2, any 3 noncollinear pts. are contained in a unique plane. If the 3 pts. are collinear, they are contained in infinitely many planes. In either case, the 3 pts. will be coplanar.
31. A;
32. N;
33. A; A B
34. S;
35. Post. 1-1- 36. There are 4 outcomes: ( A , B , C ), ( A , B , D ), ( A , C , D ), ( B , C , D ); only collinear outcome is ( A , B , C ). So probability is __^1 4
37. Post. 1-1- 38. Lines may not intersect: 0 pts. of intersection.
All 3 lines may intersect in 1 pt.
Two of the lines may not intersect, but they might each intersect a third line.
Each line may intersect each of the other lines.
TEST PREP
39. C; Other 3 sets are collinear. 40. F; Greatest number is when each pair of lines has separate intersection; there are 6 pairs of lines. 41. D; The 2 walls are planes, and they meet in a line. 42. 4; Greatest number is when each triple of pts. determines a separate plane; there are 4 triples of pts.
CHALLENGE AND EXTEND
43. 6 44. Among 10 pts, there are 45 pairs of pts. Therefore maximum is 45 segs. 45. Maximum = number of pairs of pts. 1st pt. can be chosen in n ways. 2nd pt. can be chosen in n - 1 ways. In this way, each pair is counted twice. Therefore maximum = _______ n ( n^ -^ 1) 2
46. Rescue teams can use the principles of Post. 1-1- 1 and Post. 1-1-4. A distress signal is received by 2 rescue teams. By Post. 1-1-1, 2 pts. determine a line. So 2 lines are created by the 3 pts., the locations of the rescue teams and the distress signal. By Post. 1-1-4, the unique intersection of the 2 lines will be the location of the distress signal.
SPIRAL REVIEW
47. Age of mother = a Age of each daughter = a - 25 a + 2( a - 25) = 58 3 a - 50 = 58 ______ +^^50 ____ +^^50 3 a = 108 ___ 3 a 3
=^108 ____ 3 a = 36 Mother is 36. Daughters are 36 - 25 = 11.
48. Yes, each element in the range is assigned to exactly one element in the range. 49. No, since the x -value 10 is assigned to the y -value 6 and the y -value - 6. 50. mean = Σ _ nx
= _^34 8 = 4. median = _^3 +^^5 2
mode: none
51. mean = Σ _ nx
= _2. 5 = 0. median = 0. mode = 0.
1-2 MEASURING AND CONSTRUCTING
SEGMENTS, PAGES 13–
CHECK IT OUT!
1a. XY = (^) ⎪ 5 - 1 _^1 2 ⎥^
b. XZ = ⎪- 3 - 1 _^1 2 ⎥
= (^) ⎪ 3 __^1 2 ⎥^
= (^) ⎪- 4 _^1 2 ⎥ = 3 _^1 2
= 4 _^1 2
2. Check students’ work. 3a. XZ = XY + YZ 3 = 1 __^1 3
+ YZ
____
- 11 __ 3 _________ - 1 __^1 3 12 __ 3
= YZ
b. DF = DE + EF 6 x = 3 x - 1 + 13 6 x = 3 x + 12 ____ -^^3 x^ ________ -^^3 x 3 x = 12 3 ___ x 3
=^12 ___ 3 x = 4 DF = 6 x = 6(4) = 24
4. Let current location be X , and location of new drinks station be T. XT = _^1 2
XY
= _^1 2
(1182.5) = 591.25 m
21. EF = _^1 2
( DF )
= _^1 2
( CD )
= _^1 2
22. GH = 2( DH )
4 x - 1 = 2(8) 4 x = 17 x = 4.
23. CF = 2( CD )
2 y - 2 = 2(3 y - 11) 2 y - 2 = 6 y - 22 20 = 4 y y = 5 CD = 3 y - 11 = 3(5) - 11 = 4
24. A;
25. S;
A M B A M B
26. A; X Y Z 27. AM MB is incorrect. The statement should be written as
AM
MB , not as two distances that are .
28. Let x be length of shorter piece. Let A and B be ends of dowel, and let C be cut point, nearest to A. AC + CB = 72 x + 5 x = 72 6 x = 72 6 ___ x 6
=^72 ___ 6 x = 12 AC = x = 12 CB = 5 x = 5(12) = 60 Dowel pieces are 12 cm long and 60 cm long.
29. MN = ⎪ x N - x M ⎥
4 = (^) ⎪ x (^) N - 2.5 ⎥ ± 4 = x (^) N - 2. x (^) N = 2.5 ± 4 = 6.5 or - 1.
30. D E F Possible answer:
DE +
EF =
DF
31. RS + ST = RT
7 y - 4 + y + 5 = 28 8 y + 1 = 28 8 y = 27 y = 3.
32. RS + ST = RT
3 x + 1 + __^1 2
x + 3 = 18
__ 7 2
x + 4 = 18 __ 7 2
x = 14
__ 2 7 (
7 __ 2
x ) =^2 __ 7
x = 4
33. RS + ST = RT
2 z + 6 + 4 z - 3 = 5 z + 12 6 z + 3 = 5 z + 12 z + 3 = 12 z = 9
34. B is not between A and C , because A , B , and C are not collinear. 35. Check students’ constructions.
TEST PREP
36. D Order of pts. P , Q , S , R , T. PQ + QS + SR + RT = PT 1 __ 2
QR + QR + RT = PT
1 __ 2
(8) + 8 + RT = 34
12 + RT = 34
RT = 22
37. J
AD = 2 AC
= 2(2 BC )
= 4 BC
38. B
Statement must refer to segments
XY and
YZ , and use symbol for congruence.
39. H Think: In AC = AB + BC , subst. BC for AB and CD for BC. AC = AB + BC = BC + CD = BD = 16 AC + CE = AE 16 + CE = 34 CE = 18
CHALLENGE AND EXTEND
40. JK = _^1 2
HJ
= _^1 2
(4 x ) = 2 x
HJ + JK = HK 4 x + 2 x = 78 6 x = 78 x = 13 JK = 2 x = 2(13) = 26
41. (^) X A D N 42. Race distance is: 13 + 9(8.5) + x = 100 89.5 + x = 100 x = 10.5 m 43. Race distance is: 13.72 + 9(9.14) + x = 110 95.98 + x = 110 x = 14.02 m 44. JK cannot be equal to JL because JK + KL = JL and KL ≠ 0.
SPIRAL REVIEW
45. ⎪ 20 - 8 ⎥^ = ⎪ 12 ⎥^ = 12 46. ⎪- 9 + 23 ⎥ = ⎪ 14 ⎥^ = 14
47. -⎪ 4 - 27 ⎥^ = -⎪- 23 ⎥^ = - 23
48. 8 a - 3(4 + a ) - 10 = 8 a - 12 - 3 a - 10 = 5 a - 22 49. x + 2(5 - 2 x ) - (4 + 5 x ) = x + 10 - 4 x - 4 - 5 x = - 8 x + 6 50. AB , CB 51.
AD ,
BD
52. A , B , D 53. CB
1-3 MEASURING AND CONSTRUCTING
ANGLES, PAGES 20–
CHECK IT OUT!
1. ∠ RTQ , ∠ T , ∠ STR , ∠ 1, ∠ 2
2a. m ∠ BOA = 40 ° ∠ BOA is acute.
b. m ∠ DOB = ⎪ 165 ° - 40 °⎥ = 125 ° ∠ DOB is obtuse. c. m ∠ EOC = 105 ° ∠ EOC is obtuse.
3. m ∠ XWZ = m ∠ XWY + m ∠ YWZ 121 ° = 59 ° + m ∠ YWZ 62 ° = m ∠ YWZ
4a. Step 1 Find y.
m ∠ PQS =^1 __ 2
m ∠ PQR
(5 y - 1) ° =^1 __ 2
(8 y + 12) ° 5 y - 1 = 4 y + 6 y - 1 = 6 y = 7
Step 2 Find m ∠ PQS. m ∠ PQS = 5 y - 1 = 5(7) - 1 = 34 °
b. Step 1 Find x. m ∠ LJK = m ∠ KJM ( - 10 x + 3) ° = ( - x + 21) ° 3 = 9 x + 21
- 18 = 9 x x = - 2 Step 2 Find m ∠ LJM. m ∠ LJM = 2m ∠ LJK = 2( - 10 x + 3) = (^2) (- 10( - 2) + (^3) ) = 46 °
THINK AND DISCUSS
1. Two with the same measure are. All rt. measure 90 ° , so any 2 rt. are. 2. m ∠ ABD = m ∠ DBC = _^1 2
m ∠ ABC
3. !NGLE !CUTE 'REATER THAN AND LESS
THAN
Ȝ A
Ȝ B
Ȝ C
Ȝ ABC
A
A B C
B
C
2IGHT
/BTUSE 'REATER THAN AND LESS
THAN
-EASURE $IAGRAM .AME
3TRAIGHT
EXERCISES
GUIDED PRACTICE
1. ∠ A , ∠ R , ∠ O 2. C ; CB , CD **
- ∠** AOB , ∠ BOA , or ∠ 1; ∠ BOC , ∠ COB , or ∠ 2; ∠ AOC or ∠ COA 4. m ∠ VXW = 15 ° ∠ VXW is acute. 5. m ∠ TXW = 105 ° ∠ TXW is obtuse. 6. m ∠ RXU = 110 ° ∠ RXU is obtuse. 7. m ∠ JKM = m ∠ JKL + m ∠ LKM = 42 ° + 28 ° = 70 ° 8. m ∠ JKM = m ∠ JKL + m ∠ LKM 82.5 ° = 56.4 ° + m ∠ LKM m ∠ LKM = 26.1 **°
- Step 1** Find x. m ∠ ABD = m ∠ DBC (6 x + 4) ° = (8 x - 4) ° 4 = 2 x - 4 8 = 2 x x = 4
Step 2 Find m ∠ ABD. m ∠ ABD = 6 x + 4 = 6(4) + 4 = 28°
10. Step 1 Find y. m ∠ ABD = m ∠ DBC (5 y - 3) ° = (3 y + 15) ° 2 y - 3 = 15 2 y = 18 y = 9
Step 2 Find m ∠ ABC. m ∠ ABC = 2m ∠ ABD = 2(5 y - 3) = 2 ( 5(9) - 3 ) = 84 °
PRACTICE AND PROBLEM SOLVING
11. ∠ 1 or ∠ JMK ; ∠ 2 or ∠ LMK ; ∠ M or ∠ JML 12. m ∠ CGE = ⎪ 110 - 20 ⎥^ = 90 ° ∠ CGE is a rt. ∠. 13. m ∠ BGD = ⎪ 113 - 20 ⎥^ = 93 ° ∠ BGD is obtuse. 14. m ∠ AGB = 20 ° ∠ AGB is acute. 15. m ∠ RSU = m ∠ RST + m ∠ TSU = 38 ° + 28.6 ° = 66.6 °
CHALLENGE AND EXTEND
46. Each hour interval measures 360 ÷^ 12 = 30 °. The angle formed at 7:00 is the (lesser) angle between the 12 and the 7, which measures 5(30) = 150 °. 47. m ∠ PQR = 2m ∠ PQS
x^2 = 2(2 x + 6) x^2 = 4 x + 12 x^2 - 4 x - 12 = 0 ( x - 6)( x + 2) = 0 x - 6 = 0 or x + 2 = 0 x = 6 or x = - 2 m ∠ PQR = x^2 = 36 ° or 4 °
48. 81 ° 24 ' 15 " - 42 ° 30 ' 10 " = (81 - 42) ° + (24 - 30) ' + (15 - 10) " = (80 - 42) ° + (60 + 24 - 30) ' + (15 - 10) " = 38 ° 54 ' 5 " 49. 2.25 ° = (60 · 60 · 2.25) " = 8100 50. ∠ ABC ∠ DBC m ∠ ABC = m ∠ DBC 3 ___ x 2
+ 4 = 2 x - 271 __ 4 31.25 =^1 __ 2
x
2(31.25) = (^2) (^1 __ 2
x )
x = 62. No; x = 62.5, and substituting this value into the expressions for the ∠ measures gives a sum of 195.5.
SPIRAL REVIEW
51. 35 · 64% = 35 · 0.64 = 22. 52. _33. 280
53. A B C
M N
55.
Q
56. JK + KL = JL x + 3 x = 2 x + 4 2 x = 4 x = 2 JK = x = 2 57. KL = 3 x = 3(2) = 6 58. JL = 2 x + 4 = 2(2) + 4 = 8
1-4 PAIRS OF ANGLES, PAGES 28–
CHECK IT OUT!
1a. ∠ 5 and ∠ 6 are adjacent angles. Their noncommon sides, PQ and PT , are opposite rays, so ∠ 5 and ∠ 6 also form a linear pair. b. ∠ 7 and ∠ SPU share
SP but are on the same side of it, so ∠ 7 and ∠ SPU are not adjacent angles. c. ∠ 7 and ∠ 8 share vertex P but do not have a common side, so ∠ 7 and ∠ 8 are not adjacent angles. 2a. (90 - y ) ° 90 ° - (7 x - 12) ° = 90 ° - 7 x + 12 ° = (102 - 7 x ) ° b. (180 - x ) ° 180 ° - 116.5 ° = 63.5 °
3. Step 1 Let m ∠ A = x °. Then ∠ B , its supplement, measures (180 - x ) °. Step 2 Write and solve an equation. x = __^1 2
(180 - x ) + 12
x = 90 - __ x 2
x = 102 - __ x 2 3 __ 2
x = 102 __ 2 3 (
3 __ 2
x ) =^2 __ 3
x = 68 m ∠ A = x ° = 68 °
4. 1 Understand the Problem Answers are measures of ∠ 1, ∠ 2, and ∠ 4. List important information: · ∠ 1 ∠ 2 · ∠ 1 and ∠ 3 are comp., and ∠ 2 and ∠ 4 are comp. · m ∠ 3 = 27.6 ° 2 Make a Plan If ∠ 1 ∠ 2, then m ∠ 1 = m ∠ 2. If ∠ 1 and ∠ 3 are comp., then m ∠ 1 = (90 - 27.6) °. If ∠ 2 and ∠ 4 are comp., then m ∠ 4 = (90 - 27.6) °. 3 Solve m ∠ 1 = m ∠ 2 = (90 - 27.6) ° = 62.4 ° m ∠ 3 = m ∠ 4 = (90 - 62.4) ° = 27.6 ° 4 Look Back Answer makes sense because 27.6 ° + 62.4 ° = 90 ° , so ∠ 1 and ∠ 3 are comp., and ∠ 2 and ∠ 4 are comp. Thus m ∠ 1 = 62.4 ° , m ∠ 2 = 62.4 ° , and m ∠ 4 = 27.6 °. 5. Possible answer: ∠ EDG and ∠ FDH are vert. angles and appear to have the same measure. ∠ EDG ≈ ∠ FDH ≈ 45 °.
THINK AND DISCUSS
1. All rt. measure 90 ° , so the sum of the measures of any 2 rt. is 180 °. Therefore any 2 rt. are supp. 2. Vert. cannot be adj. because the def. of vert. states that they are nonadj. formed by intersecting lines. 3.
,IN PAIR A PAIR OF ADJ ѓ WHOSE NONCOMMON SIDES ARE OPP RAYS
6ERT ѓ NONADJ ѓ FORMED BY INTERSECTING LINES
3UPP ѓ ѓ WHOSE MEASURES HAVE A SUM OF
*>ÀÃÊvÊ}iÃ
!DJ ѓ ѓ IN THE SAME PLANE WITH A COMMON VERTEX AND A COMMON SIDE BUT NO COMMON INTERIOR PTS
#OMP ѓ ѓ WHOSE MEASURES HAVE A SUM OF
EXERCISES
GUIDED PRACTICE
1. (90 - x ) ° ; (180 - x ) ° 2. BC 3. ∠ 1 and ∠ 2 are adj. angles. Their noncommon sides, EG^ and EJ , are opposite rays, so ∠ 1 and ∠ 2 also form a lin. pair. 4. ∠ 1 and ∠ 3 share vertex E but do not have a common side, so ∠ 1 and ∠ 3 are not adj. angles. 5. ∠ 2 and ∠ 4 share vertex E but do not have a common side, so ∠ 2 and ∠ 4 are not adj. angles. 6. ∠ 2 and ∠ 3 are adj. angles. Their noncommon sides, EF^ and EH , are not opposite rays, so ∠ 1 and ∠ 2 are only adj. angles. 7. (180 - x ) ° 180 ° - 81.2 ° = 98.8 ° 8. (90 - x ) ° 90 ° - 81.2 ° = 8.8 ° 9. (180 - y ) ° 180 ° - (6 x - 5) ° = 180 ° - 6 x + 5 = (185 - 6 x) ° 10. (90 - y ) ° 90 ° - (6 x - 5) ° = 90 ° - 6 x + 5 = (95 - 6 x ) **°
- Step 1** Let m ∠ A = x °. Then ∠ B , its comp., measures (90 - x ) °. Step 2 Write and solve an equation. x = 3(90 - x ) + 6 x = 270 - 3 x + 6 x = 276 - 3 x 4 x = 276 x = 69 m ∠ A = x = 69 **°
- 1 Understand the Problem Answers** are measures of ∠ 2, ∠ 3, and ∠ 4. List Important information: · ∠ 1 ∠ 2 · ∠ 1 and ∠ 3 are comp., and ∠ 2 and ∠ 4 are comp. · m ∠ 1 = 47.5 ° 2 Make a Plan If ∠ 1 ∠ 2, then m ∠ 1 = m ∠ 2. If ∠ 1 and ∠ 3 are comp., then m ∠ 3 = (90 - 47.5) °. If ∠ 2 and ∠ 4 are comp., then m ∠ 4 = (90 - 47.5) °. 3 Solve m ∠ 1 = m ∠ 2 = 47.5 ° m ∠ 3 = m ∠ 4 = (90 - 47.5) ° = 42.5 ° 4 Look Back Answer makes sense because 18.5 ° + 71.5 ° = 90 ° , so ∠ 1 and ∠ 3 are comp., and ∠ 2 and ∠ 4 are comp. Thus m ∠ 2 = 18.5 ° , m ∠ 3 = 71.5 ° , and m ∠ 4 = 71.5 °. 13. ∠ ABE and ∠ CBD are vert. ; ∠ ABC and ∠ EBD are vert. .
PRACTICE AND PROBLEM SOLVING, PAGES 32–
14. ∠ 1 and ∠ 4 are adj. angles. Their noncommon sides are opposite rays, so ∠ 1 and ∠ 4 also form a lin. pair. 15. ∠ 2 and ∠ 3 are adj. angles. Their noncommon sides are opposite rays, so ∠ 2 and ∠ 3 also form a lin. pair. 16. ∠ 3 and ∠ 4 are adj. angles. Their noncommon sides are not opposite rays, so ∠ 3 and ∠ 4 are only adj. angles. 17. ∠ 3 and ∠ 1 share a vertex but do not have a common side, so ∠ 3 and ∠ 1 are not adj. angles. 18. (180 - x ) ° 180 ° - 56.4 ° = 123.6 ° 19. (90 - x ) ° 90 ° - 56.4 ° = 33.6 ° 20. (180 - y ) ° 180 ° - (2 x - 4) ° = 180 ° - 2 x + 4 = (184 - 2 x ) ° 21. (90 - y ) ° 90 ° - (2 x - 4) ° = 90 ° - 2 x + 4 = (94 - 2 x ) **°
- Step 1** Let m ∠ A = x ° , so m ∠ B = (90 - x ) °. Step 2 Write and solve an equation. x = 3(90 - x ) x = 270 - 3 x 4 x = 270 x = 67. m ∠ A = x = 67.5 ° m ∠ B = 90 - 67.5 ° = 22.5 **°
- ∠** 1 ∠ 2 m ∠ 1 = m ∠ 2 = 22.3 ° m ∠ 4 = m ∠ 3 = 90 ° - 22.3 ° = 67.7 **°
- ∠** PTU , ∠ VTR ; ∠ UTQ , ∠ STV ; ∠ QTR , ∠ PTS ; ∠ PTQ , ∠ STR ; ∠ UTR , ∠ PTV ; ∠ QTV , ∠ UTS
41. C
m ∠ A + m ∠ B = 180 ° 3 y + 2(3 y ) = 180 9 y = 180 (^9) ___ y 9
= ____^180 9 y = 20
42. H
7 x + 5 x = 180 12 x = 180 12 ____ x 12
= ____^180 12 x = 15 m ∠ 2 = 5(15) = 75
CHALLENGE AND EXTEND
43. 4 pairs of 45 ° + 4 pairs of 90 ° + 4 pairs of 135 ° = 12 pairs of vert. 44. Let ∠ measure be x °. Then 180 - x = 2(90 - x ) + 4 180 - x = 180 - 2 x + 4 180 - x = 184 - 2 x x = 4 m ∠ = x = 4 ° 45. m ∠ 1 + m ∠ 2 = 90 ° 2m ∠ 2 + m ∠ 2 = 90 ° 3m ∠ 2 = 90 ° m ∠ 2 = 30 ° m ∠ 1 - m ∠ 2 = 2m ∠ 2 - m ∠ 2 = m ∠ 2 = 30 ° 46. Let ∠ measure be x °. Then 180 - x = (^2) ( 180 - (90 - x ) ) - 36 180 - x = 2(180 - 90 + x ) - 36 180 - x = 144 + 2 x 36 = 3 x x = 12 (180 - x ) ° 180 ° - 12 ° = 168 °
SPIRAL REVIEW
47. 4 x + 10 = 42 4 x = 32 x = 8 48. 5 m - 9 = m + 4 4 m - 9 = 4 4 m = 13 m = 3. 49. 2( y + 3) = 12 2 y + 6 = 12 2 y = 6 y = 3 50. - ( d + 4) = 18 - d - 4 = 18 - d = 22 d = - 22 51. XY + YZ = XZ 3 x + 1 + 2 x - 2 = 84 5 x - 1 = 84 5 x = 85 x = 17 52. XY = 3 x + 1 = 3(17) + 1 = 52 53. YZ = 2 x - 2 = 2(17) - 2 = 32 54. m ∠ XYZ = m ∠ WYX = 26 ° 55. m ∠ WYZ = m ∠ WYX + m ∠ XYZ = 26 ° + 26 ° = 52 °
READY TO GO ON? PAGE 35
X Y
M P
3. 4. 5. Possible answer: T , V , W 6. XZ and WY 7. plane TVX 8. 9.
SV = (^) ⎪ 5 - ( - 1.5) ⎥ = ⎪ 6.5 ⎥^ = 6. 10.
TR = (^) ⎪ 2 - ( - 4) ⎥ = ⎪ 6 ⎥^ = 6 11.
ST = (^) ⎪ 2 - ( - 1.5) ⎥ = ⎪ 3.5 ⎥^ = 3.
12. HJ + JK = HK 4 x + 6 + 9 = 39 4 x + 15 = 39 4 x = 24 x = 6 HJ = 4 x + 6 = 4(6) + 6 = 30 13. Check students’ work.
14. PR = 2 PQ
8 z - 12 = 2(2 z ) 8 z - 12 = 4 z
- 12 = - 4 z z = 3 PQ = 2 z = 2(3) = 6 PR = 8 z - 12 = 8(3) - 12 = 12 15. ∠ LMN , ∠ NML , or ∠ 1; ∠ NMP , ∠ PMN , or ∠ 2; ∠ LMP or ∠ PML 16. acute 17. obtuse 18. obtuse 19. m ∠ QRS = m ∠ SRT 3 x + 8 = 9 x - 4 12 = 6 x x = 2 m ∠ SRT = 9 x - 4 = 9(2) - 4 = 14 ° 20. Check students’ work. 21. ∠ 1 and ∠ 2 are adj. angles. Their noncommon sides are opposite rays, so ∠ 1 and ∠ 2 also form a lin. pair. 22. ∠ 4 and ∠ 5 are adj. angles. Their noncommon sides are not opposite rays, so ∠ 4 and ∠ 5 are only adj. . 23. ∠ 3 and ∠ 4 share a vertex but do not have a common side, so ∠ 3 and ∠ 4 are not adj. angles. 24. (180 - y ) ° 180 ° - (5 x - 10) ° = 180 ° - 5 x + 10 = (190 - 5 x ) ° 25. (90 - y ) ° 90 ° - (5 x - 10) ° = 90 ° - 5 x + 10 = (100 - 5 x ) °
1-5 USING FORMULAS IN GEOMETRY,
PAGES 36–
CHECK IT OUT!
1. P = 4 s = 4(3.5) = 14 in.
A = s^2 = (3.5) 2 = 12.25 in 2
2. The area of one rectangle is A = w = (6.5)(2.5) = 16.25 in 2. The total area of the 4 rectangles is 4(16.25) = 65 in 2.
3. C = 2 π r = 2 π (14) = 28 π ≈ 88.0 m
A = π r^2 = π (14) 2 = 196 π ≈ 615.8 m 2
THINK AND DISCUSS
1. Possible answer: A rect. with length 8 in. and width 2 in.; a square with sides 4 in. long; a with base 4 in. and height 8 in. 2.
A ?? bh
2ECTANGLE 3QUARE (^) #IRCLE 4RIANGLE
&ORMULAS
P Ű X A ŰX
P s A s
C ûr A ûr
EXERCISES
GUIDED PRACTICE
1. Both terms refer to the dist. around a figure. 2. base and height 3. P = 2 + 2 w = 2(11) + 2(4) = 22 + 8 = 30 mm
A = w = (11)(4) = 44 mm 2
4. P = 4 s = 4( y - 3) = 4 y - 12
A = s^2 = ( y - 3) 2 = y^2 - 6 y + 9
5. P = a + b + c = 5 + ( x + 3) + 13 = ( x + 21) m
A = _^1 2
bh
= _^1 2
( x + 3)(4) = (2 x + 6) m 2
6. Area of one is A =^1 _ 2
bh =^1 _ 2
(3)(4) = 6 in 2.
There are 80(20) = 1600 in total. The total area of the 1600 is 1600(6) = 9600 i n 2.
7. C = 2 π r = 2 π (2.1) = 4.2 π ≈ 13.2 m
A = π r^2 = π (2.1) 2 = 4.41 π ≈ 13.9 m 2
8. C = 2 π r = 2 π (7) = 14 π ≈ 44.0 in.
A = π r^2 = π (7) 2 = 49 π ≈ 153.9 in 2
9. r = _ d 2
=^16 ___ 2
= 8 cm C = 2 π r = 2 π (8) = 16 π ≈ 50.3 cm
A = π r^2 = π (8) 2 = 64 π ≈ 201.1 cm 2
PRACTICE AND PROBLEM SOLVING
10. P = 4 s = 4(7.4) = 29.6 m
A = s^2 = (7.4) 2 = 54.76 m 2
11. P = 2 + 2 w
= 2( x + 6) + 2 x = 4 x + 12
A = w = ( x + 6) x = x^2 + 6 x
12. P = a + b + c = 5 x + 8 + 4 x = 9 x + 8
A =^1 __ 2
bh
=^1 __ 2
(8)(3 x ) = 12 x
13. Area of one is^1 __ 2
bh = __^1 2
(3)(1.5) = 2.25 in 2. Area of 32 is 32(2.25) = 72 in 2.
14. C = 2 π r = 2 π (12) = 24 π ≈ 75.4 m
A = π r^2 = π (12) 2 = 144 π ≈ 452.4 m 2
15. r = _ d 2
= 6.25 ft C = 2 π r = 2 π (6.25) = 12.5 π ≈ 39.3 ft A = π r^2 = π (6.25) 2 = 39.0625 π ≈ 122.7 ft 2
16. r = _ d 2
= _^1 4
mi C = 2 π r = 2 π ( _^1 4 )^
=^1 _ 4
π
≈ 1.6 mi
A = π r^2 = π ( _^1 4 )
2 = _^1 16
π
≈ 0.2 mi 2
17. A = s^2 = (9.1) 2 = 82.81 yd 2 18. A = s^2 = ( x + 1) 2 = x^2 + 2 x + 1 19. A =^1 __ 2
bh
=^1 __ 2
(5.5)(2.25) = 6.1875 in 2
20. A =^1 __ 2
bh
6.75 =^1 __ 2
(3) h
6.75 =^3 __ 2
h
__ 2 3
(6.75) =^2 __ 3 (
3 __ 2
h )
h = 4.5 m
21. A = w 347.13 = 20.3 w w = 17.1 cm 22. Possible answer: A = π r^2 64 π = π r^2 64 = r^2 r = √ 64 = 8 23. A = π (8 ) 2 is incorrect. The radius is 4 cm, not 8 cm. A = π r^2 = π (4) 2 = 16 π cm 2 24. r = _ d 2
= 14 m A = π r^2 = π (14) 2 = 196 π m 2
25. A = π r^2 = π (3 y ) 2 = 9 y^2 π 26. Dist. around Earth at equator is the circumference. C = 2 π r = 2 π (3964) = 7928 π ≈ 24,907 mi 27. For a square, the length and width are both s , so P = 2 + 2 w = 2 s + 2 s = 4 s and A = w = s ( s ) = s^2. 28. P = 2 + 2 w = 2( x + 1) + 2( x - 3) = 4 x - 4
A = w = ( x + 1)( x - 3) = x^2 - 2 x - 3
52a. P = 2 + 2 w P - 2 = 2 w
______ P - 2
= w
b. w = P ______^ -^^2
9 _______ - 2(3)
= 1.5 ft
53. Think: assume that ≥ w. Values for ( , w ) satisfy 12 = 2 + 2 w or + w = 6: (5, 1), (4, 2), or (3, 3). So possible areas are A = w = (5)(1) = 5; or (4)(2) = 8; or (3)(3) = 9 54. A actual = π r^2 = π (4.5) 2 = 20.25 π A estimate = s^2 = 8 2 = 64
Percent error =
A ______________estimate - A actual
A actual
=^64 ___________^ -^ 20.25 π
20.25 π
≈ 0.6% (overestimate)
55. Step 1 Write an equation relating and w.
w = _^4
Step 2 Substitute in formula for area of a rectangle and solve to find . A = w
320 = (^4 __
320 =^4 __
^2
__ 5
(320) =^5 __
__ 4
^2 )
400 = ^2
= 20 in. Step 3 Find w.
w = _^4
= __^4
(20) = 16 in.
SPIRAL REVIEW
56. D: {2, - 5, - 3} R: {4, 8}
57. D: {4, - 2, 16}
R: { - 2, 8, 0}
58. plane 59. line or segment 60. Let a and b be the lengths. Write 2 equations. a = 4 b 10 = a + b 10 = 4 b + b 10 = 5 b
___ 10
=^5 ___ b
b = 2 yd a = 4 b = 4(2) = 8 yd
61.
AB
BC
AB = BC
⎪- 2.5^ -^ ( - 8) ⎥ =^ ⎪ C^ -^ ( - 2.5) ⎥ 5.5 = (^) ⎪ C - ( - 2.5) ⎥ Think: B > A , so C > B. Therefore C - ( - 2.5) = 5. C = - 2.5 + 5.5 = 3
62. m ∠ = 2(180 - m ∠ ) + 9 m ∠ = 360 - 2m ∠ + 9 3m ∠ = 369 m ∠ = 123 °
1-6 MIDPOINT AND DISTANCE IN THE
COORDINATE PLANE, PAGES 43–
CHECK IT OUT!
1. M (
_ x 1 +^ x 2
_ y 1 +^ y 2
_ -^2 +^^5
_^3 +^ ( - 3)
2 )^
= ( _^3
, _^0
= ( _^3
2. Step 1 Let coords. of T equal ( x , y ). Step 2 Use Mdpt. Formula.
( - 1, 1) = ( _ -^6 +^ x
_ -^1 +^ y
Step 3 Find x -coord.
- 1 = _ -^6 +^ x
2( - 1) = 2 ( _ -^6 +^ x
- 2 = - 6 + x x = 4
Find y -coord. 1 =
_ -^1 +^ y
- _ 1 + y
2 = - 1 + y y = 3 The coordinates of T are (4, 3).
3. Step 1 Find coords. of each point. E ( - 2, 1), F ( - 5, 5), G ( - 1, - 2), and H (3, 1) Step 2 Use Dist. Formula.
d = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
EF = √ ( - 5 - ( - 2)) 2 + (5 - 1) 2
GH = √ ( 3 - ( - 1)) 2 + (1 - ( - 2)) 2
Since EF = GH ,
EF
GH.
4a. Method 1 Use Dist. Formula. Subst. values for coords. of R and S into Dist. Formula. RS = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
= √ ( - 3 - 3 )^2 + ( - 1 - 2 )^2
= √ ( - 6 )^2 + ( - 3 )^2
Method 2 Use Pyth. Thm. Count the units for the legs of the rt. formed by R and S. a = 6 and b = 3 c^2 = a^2 + b^2 = 62 + 3 2 = 36 + 9 = 45 c = √ 45 ≈ 6.
b. Method 1 Use Dist. Formula. Subst. values for coords. of R and S into Dist. Formula. RS = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
= √ (2 - ( - 4)) 2 + ( - 1 - 5 )^2
= √ ( - 6 )^2 + ( - 6 )^2
Method 2 Use Pyth. Thm. Count the units for the legs of the rt. formed by R and S. a = 6 and b = 6 c^2 = a^2 + b^2 = 6 2 + 3 2 = 36 + 36 = 72 c = √ 72 ≈ 8. 5.
H (0,0) F (90,0)
T (0,90) S (90,90)
Let the pitching mound be M (42.8, 42.8). The dist. MH from center of mound to home plate is the length of hyp. of a rt. . MH = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
= √ (42.8 - 0 ) 2 + (42.8 - 0 ) 2
= √ 42.8 2 + 42. 8 2
= √ 3663.68 ≈ 60.5 ft
THINK AND DISCUSS
1. yes;
_ x 1 +^ x 2
_ x 2 +^ x 1
and
_ y 1 +^ y 2
_ y 2 +^ y 1
2. s and t represent lengths of legs. r represents length of hyp. 3. Yes; you can use either method to find dist. between 2 pts. 4. Possible answer: to make locating addresses easier 5.
`«Ì°ÊÀÕ>
ÀÕ>Ã
*ÞÌ °Ê/ iÀi
ÃÌ°ÊÀÕ>
M ?????? x^ ^ x^ ^ ????? y^ ^ y^ A x (^) y (^)
A x (^) y (^)
B x (^) y (^)
M
a b c
a
b c
d
d Ȗеееееееее x (^) x (^) y (^) y (^) ^
B x (^) y (^)
EXERCISES
GUIDED PRACTICE
1. hypotenuse
2. M (
_ x^1 +^ x^2
_ y^1 +^ y^2
_^4 +^ ( - 4)
, _ -^6 +^^2
2 )^
= ( _^0
, _ -^4
3. M (
_ x^1 +^ x^2
_ y^1 +^ y^2
_^0 +^^3
, _ -^8 +^^0
2 )^
= ( _^3
, _ -^8
= ( 11 _
4. Step 1 Let coords. of N equal ( x , y ). Step 2 Use Mdpt. Formula:
(0, 1) = ( _ -^3 +^ x
_ -^1 +^ y
Step 3 Find x -coord.
0 = - _^3 +^ x
2(0) = 2 ( _ -^3 +^ x
0 = - 3 + x x = 3
Find y -coord. 1 =
- _ 1 + y
- _ 1 + y
2 = - 1 + y y = 3 The coordinates of N are (3, 3).
5. Step 1 Let coords. of C equal ( x , y ). Step 2 Use Mdpt. Formula.
(-^1
_ 1
, 1 ) = (- ______^3 +^ x
4 _____ + y
Step 3 Find x -coord.
- 1 _^1
= _ -^3 +^ x
2 ( _^3
2 )^
= 2 (- _^3 +^ x
- 3 = - 3 + x x = 0
Find y -coord. 1 =
_^4 +^ y
4 _ + y
2 = 4 + y y = - 2 The coordinates of C are (0, - 2).
6. Step 1 Find coords. of each point. F (5, 4), G (3, - 1), J ( - 4, 0), and K (1, - 2) Step 2 Use Dist. Formula. d = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
JK = √ (1 - ( - 4)) 2 + ( - 2 - 0) 2
FG = √ (3 - 5) 2 + ( - 1 - 4) 2
Since JK = FG ,
JK
FG.
17. Step 1 Find coords. of each point. D ( - 4, 0), E (0, - 2), R ( - 3, - 4), and S (2, - 2). Step 2 Use Dist. Formula.
DE = √ (0 - ( - 4)) 2 + ( - 2 - 0 ) 2
RS = √ ( 2 - ( - 3)) 2 + ( - 2 - ( - 4)) 2
Since DE ≠ RS ,
DE
RS.
18. Method 1 Dist. Formula.
UV = √ ( - 3 - 0 ) 2 + ( - 9 - 1 ) 2
= √ ( - 3 ) 2 + ( - 10 ) 2
Method 2 Pyth. Thm. Count the units for the legs of the rt. formed by U and V. a = 3 and b = 10 c^2 = a^2 + b^2 = 32 + 10 2 = 100 + 9 = 109 c = √ 109 ≈ 10.
19. Method 1 Dist. Formula.
MN = √ (2 - 10 ) 2 + ( - 5 - ( - 1) ) 2
Method 2 Pyth. Thm. Count the units for the legs of the rt. formed by M and N. a = 8 and b = 4 c^2 = a^2 + b^2 = 82 + 4 2 = 64 + 16 = 80 c = √ 80 ≈ 8.
20. Method 1 Dist. Formula.
PQ = √ ( 5 - ( - 10)) 2 + (5 - 1 ) 2
Method 2 Pyth. Thm. Count the units for the legs of the rt. formed by P and Q. a = 15 and b = 4 c^2 = a^2 + b^2 = 152 + 4 2 = 225 + 16 = 241 c = √ 241 ≈ 15.
21. Use Pyth. Thm. a = 11 and b = 14 c^2 = a^2 + b^2 = 112 + 14 2 = 121 + 196 = 317 c = √ 317 ≈ 18 in. 22. Step 1 Find coords. of each point. A ( - 4, 2), B (1, 4), C (2, 5), D (4, 1), E ( - 2, - 2), and F (3, - 1). Step 2 Use Dist. Formula.
AB = √ ( 1 - ( - 4)) 2 + (4 - 2 ) 2
CD = √ (4 - 2 ) 2 + (1 - 5 ) 2
2 + ( - 4 ) 2
EF =
2
2
2 + 1
2
CD ,
EF ,
AB
23. a = 2 and b = 4 c^2 = 2 2 + 4 2 = 20 c = √ 20 ≈ 4.
_ a^ +^ ( -^5 a )
, _^3 a^ +^^0
2 )^
= ( _ -^4 a
, _^3 a
= (- 2 a , _^3
a )
25. Divide each coord. by 2. 26. Coords. of Cedar City are (2, 3). Coords. of Milltown are (3, - 3). Dist. = √ (3 - 2) 2 + ( - 3 - 3) 2 = √ 1 2 + ( - 6) 2 = √ 37 ≈ 6.1 mi 27. Coords. of Jefferson are ( - 2, - 2).
Dist. =^1 __
(dist. from Jefferson to Milltown)
=^1 __
=^1 __
=^1 __
√ 26 ≈ 2.5 mi
28. Use Pyth. Thm. a = 960 and b = 750 c^2 = a^2 + b^2 = 9602 + 750 2 = 1,484, c = √ 1,484,100 ≈ 1218 m 29. Possible answer: seg. with endpts. (1, 2) and ( - 1, - 2)
30. Step 1 Find AB , BC , and AC.
AB = √ ( - 2 - 1 ) 2 + ( - 1 - 4 ) 2
= √ ( - 3 ) 2 + ( - 5 ) 2
BC = √ ( - 3 - ( - 2)) 2 + ( - 2 - ( - 1)) 2
AC = √ ( - 3 - 1 ) 2 + ( - 2 - 4 ) 2
Step 2 Find perimeter. P = AB + BC + AC = √ 34 + √ 2 + √ 52 ≈ 14.
31. A = _^1
bh
= _^1
( BC ) (√ 2 )
= _^1
= _^1
(2) = 1 square unit
32. When 2 pts. lie on a horiz. or vert. line, they share a common y -coord. or x -coord. To find the dist. between the pts., find the difference of the other coords. 33. Let M be the mdpt. of
AC.
AM = MC = _^1
(10) = 5.0 ft
MB = MD = √ 5 2 + 4 2 = √ 41 ≈ 6.4 ft
TEST PREP
34. B GH = √ 3 2 + 3 2 = √ 18 ≈ 4. 35. G Coords. of mdpt. of
LM are (2.5, 1); coords. of mdpt. of
JK are (1.5, - 2.5).
Dist. = √ (2.5 - 1.5 ) 2 + ( 1 - ( - 2.5)) 2
36. D
_^7 +^ ( - 5)
, _ -^3 +^^6
2 )^
= ( _^2
, _^3
= ( 1, 1_^1
37. J
Dist. = √ ( 3 - ( - 5) ) 2 + (5 - 1 ) 2
CHALLENGE AND EXTEND
38a. x -coord. of Q = x -coord. of P = _^1 +^^4
y -coord. of Q = y -coord. of R = _^1 +^^3
coords. of Q are (2.5, 2). b. Area of PBRQ = w = (1.5)(1) = 1.5 square units c. DB = √ 3 2 + 2 2 = √ 13 ≈ 3.
39. XY^2 = ( a + 1 - ( a - 5)) 2 + (2 a - ( - 2 a ) ) 2
10 2 = 6 2 + (4 a ) 2 100 - 36 = (4 a ) 2 4 a = ±√ 64 = ± 8 a = ± 2
40. Let coords. of pt. on y -axis be (0, y ). 5 2 = (4 - 0) 2 + (2 - y ) 2 25 = 16 + 4 - 4 y + y^2 0 = y^2 - 4 y - 5 0 = ( y - 5)( y + 1) y = 5 or - 1 Coords. of 2 pts. are (0, 5) and (0, - 1). 41. By Pyth. Thm., AB^2 = AC^2 + BC^2 = x^2 + y^2 AB = √ x^2 + y^2
SPIRAL REVIEW
42. y 3 x - 1 4 3( - 1) - 1 4 - 4 no 43. f ( x ) 5 - x^2 4 5 - ( - 1) 2 4 4 yes 44. g ( x ) x^2 - x + 2 4 ( - 1 ) 2 - ( - 1) + 2 4 4 yes
45. m ∠ ABD = _^1
(180) = 90 ° ; rt.
46. m ∠ CBE = _^1
m ∠ CBD = _^1
(180 - 90) = 45 ° ; acute
47. m ∠ ABE = 180 - m ∠ CBE = 135 ° ; obtuse 48. P = 4 s 20 = 4 s s = 5 in. A = s^2 = 5 2 = 25 in 2 49. b = 2 h = 2(2) = 4 ft
A = _^1
bh
= _^1
(4)(2) = 4 ft 2
50. A = w = ( x )(4 x + 5) = 4 x^2 + 5 x
PRACTICE AND PROBLEM SOLVING
8. rotation: DEFG → D ′ E ′ F ′ G ′
9. reflection: WXYZ → W ′ X ′ Y ′ Z ′
x
y
J K
M L
J Ī K Ī
M Ī L Ī
translation; each pt. moves the same dist. right and the same dist. down.
11. Step 1 Apply the rule to find the vertices of the image.
A ′ ( - 4 + 3, 1 - 2) = A ′ ( - 1, - 1)
B ′ (1 + 3, 1 - 2) = B ′ (4, - 1)
C ′ (1 + 3, - 2 - 2) = C ′ (4, - 4)
D ′ ( - 4 + 3, - 2 - 2) = D ′ ( - 1, - 4)
Step 2 x
y
D Ī
A B
B Ī
C Ī
D C
A Ī
12. Step 1 Choose 2 pts.
Choose a pt. A on the preimage and a corr. pt. A ′ on
the image. A has coords. ( - 5, 1) and A ′ has coords.
Step 2 Translate.
To translate A to A ′ , 11 units are added to the x -
coord. and 4 units are subtracted from the y -coord. Therefore, the translation rule is ( x , y ) → ( x + 11, y - 4).
13. reflection 14. translation 15. reflection
16. Vertices of image are F ′ (3, - 5), G ′ ( - 1, - 4), and
H ′ (5, 0).
(^)
G^ F
H
G Ī F Ī
H Ī
x
y
17. Vertices of image are F ( - 3, 5), G (1, 4), and
H ( - 5, 0).
x
y
^ ^
F Ī^ F
H Ī H
G G Ī
18. Vertices of 1 are (1, 1), (3, 1), and (2, 3). 1 to 2: ( x , y ) → ( x , - y ) 2 to 3: ( x , y ) → ( - x , y ) 3 to 4: ( x , y ) → ( x , - y ) 19. B 20. A 21. D 22. C
23. R ′ (1 - 2, - 4 - 8) = R ′ ( - 1, - 12)
S ′ ( - 1 - 2, - 1 - 8) = S ′ ( - 3, - 9)
T ′ ( - 5 - 2, 1 - 8) = T ′ ( - 7, - 7)
24. Both translations move the preimage right and up. The second translation moves the preimage farther in each direction.
25. M ′ (2 + 2, 8 - 5) = M ′ (4, 3)
N ′ ( - 3 + 2, 4 - 5) = N ′ ( - 1, - 1)
y
x
N
M
N Ī
M Ī
26. K ′ ( - 1 - 4, 1 + 3) = K ′ ( - 5, 4)
L ′ (3 - 4, - 4 + 3) = L ′ ( - 1, - 1)
y
x
K
K
L
L
27. Find the coords. of the preimage. Then, find the coords. of the image after translation. Plot the vertices of image pts. and use a straightedge to draw the image . 28. Possible answer: 2 reflections (across the y -axis and across EC )
TEST PREP
29. A 30. H
E ′ ( - 3 - 2, - 3 + 1) = E ′ ( - 5, - 2)
31. A
32. H
CHALLENGE AND EXTEND
33a. R (( - 2 - 1) + 4, ( - 2 + 3) - 1 ) = R (1, 0)
S (( - 3 - 1) + 4, (1 + 3) - 1 ) = S (0, 3)
T ((1 - 1) + 4, (1 + 3) - 1 ) = T (4, 3)
b. ( x , y ) → (( x - 1) + 4, ( y + 3) - 1 ) = ( x + 3, y + 2)
34. m ∠ =^12 ___ 60
35. Transformation is ( x , y ) → ( - y , x ). So vertices of
image are A ′ (0, 1), B ′ (0, 5), and C ′ ( - 2, 2).
x
y
^ ^
C Ī
B Ī
A Ī
36. ( x , y ) → ( x , - y ) 37. ( x , y ) → ( - x , y )
SPIRAL REVIEW
38. 0 = x^2 + 12 x + 35 0 = ( x + 7)( x + 5) x = - 7 or - 5 39. 0 = x^2 + 3 x - 18 0 = ( x + 6)( x - 3) x = - 6 or 3 40. 0 = x^2 - 18 x + 81 0 = ( x - 9) 2 x = 9 41. 0 = x^2 - 3 x + 2 0 = ( x - 2)( x - 1) x = 2 or 1 42. m(supp. of ∠ A ) = 180 - m ∠ A = 180 - 76.1 = 103.9 ° 43. m(comp. of ∠ A ) = 90 - m ∠ A = 90 - 76.1 = 13.9 **°
- Method 1** Use Dist. Formula. Subst. values for coords. of 2 pts. into Dist. Formula. AB = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
= √ (4 - 2 ) 2 + (6 - 3 ) 2
= √ 2 2 + 3 2
Method 2 Use Pyth. Thm. Count the units for the legs of the rt. formed by 2 pts. a = 3 and b = 2 c^2 = a^2 + b^2 = 32 + 2 2 = 9 + 4 = 13 c = √ 13 ≈ 3.
45. Method 1 Use Dist. Formula. Subst. values for coords. of 2 pts. into Dist. Formula. AB = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
Method 2 Use Pyth. Thm. Count the units for the legs of the rt. formed by 2 pts. a = 1 and b = 4 c^2 = a^2 + b^2 = 12 + 4 2 = 1 + 16 = 17 c = √ 17 ≈ 4.
46. Method 1 Use Dist. Formula. Subst. values for coords. of 2 pts. into Dist. Formula. AB = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
Method 2 Use Pyth. Thm. Count the units for the legs of the rt. formed by 2 pts. a = 3 and b = 9 c^2 = a^2 + b^2 = 32 + 9 2 = 9 + 81 = 90 c = √ 90 ≈ 9.
47. Method 1 Use Dist. Formula. Subst. values for coords. of 2 pts. into Dist. Formula. AB = √ ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2
= √ ( - 1 - 5 ) 2 + (3 - 1 ) 2
= √ ( - 6 ) 2 + 2 2
Method 2 Use Pyth. Thm. Count the units for the legs of the rt. formed by 2 pts. a = 6 and b = 2 c^2 = a^2 + b^2 = 62 + 2 2 = 36 + 4 = 40 c = √ 40 ≈ 6.
