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Answer Key
Geometry Common Core Regents
Course Workbook
2017-18 Edition Donny Brusca
© 2017 Donald Brusca. All rights reserved. ISBN 978-1545375358
www.CourseWorkbooks.com
3
Ch. Unit Page 2. PERIMETER AND AREA .................................................................................................... 4 3. LINES, ANGLES AND PROOFS ......................................................................................... 6 4. TRIANGLES ......................................................................................................................... 8 5. RIGHT TRIANGLES AND TRIGONOMETRY ............................................................... 17 6. OBLIQUE TRIANGLES ..................................................................................................... 25 7. QUADRILATERALS.......................................................................................................... 30 8. COORDINATE GEOMETRY ............................................................................................ 35 9. POLYGONS IN THE COORDINATE PLANE ................................................................. 40 10. RIGID MOTIONS ............................................................................................................... 48 11. DILATIONS ........................................................................................................................ 52 12. TRANSFORMATION PROOFS......................................................................................... 56 13. CIRCLES ............................................................................................................................. 61 14. SOLIDS................................................................................................................................ 71 15. CONSTRUCTIONS ............................................................................................................ 77 Notation A code next to each Regents Question number states from which exam the question came. For example, IA AUG ’09 [37] means the question appeared on the August 2009 Integrated Algebra Regents as question 37. Regents exam formats are abbreviated as follows:
GC = Geometry Common Core, GE = Geometry (pre-CC), CC = Algebra I Common Core, IA = Integrated Algebra, A2 = Algebra 2 / Trigonometry, MA = Math A, MB = Math B, S3 = Sequential III, 9Y = Ninth Year Math
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2. PERIMETER AND AREA 2.1 Perimeter and Circumference ~ Page 13
Practice Problems
1. ( 2) ( 3)x x x+ + + − Perimeter is 3 1x −
2.
4 ( 3) 3 1 348 2 34
8 324
x x xx
xx
+ + + − =+ =
==
4 16x = , 3 7x + = , 3 1 11x − = 3. Length of rectangle = diameter = 8 in. Perimeter = 1
2 2 2r l wπ + +
12 8 8 14 4 22 34.6π π= + + = + ≈ in.
4. Width of rectangle = radius = 2 cm. Perimeter = 1
2 2 2r l wπ + +
12 4 8 2 2 10 16.3π π= + + = + ≈ cm.
5. Length of arc SBT = 1 1
4 42 12 3rπ π π= = .
Let w RC= , so 8AR w= − . 6CT r RC w= − = − 6 (8 ) 2AS r AR w w= − = − − = − Since ABCR is a rectangle, its diagonals are congruent. Diagonal RB is a radius, so 6AC RB= = . Perimeter =
Regents Questions
1) IA AUG ’11 [28] Ans: 1 2) IA AUG ’12 [28] Ans: 1 3) IA AUG ’13 [26] Ans: 1 4) IA JUN ’14 [33]
Ans: 12 15 60 15 7.5 75 98.6π π⋅ + + = + ≈
5) IA AUG ’14 [32] Ans: 12 4 4 4 2 16 22.3π π⋅ + ⋅ = + ≈
6) IA JUN ’15 [32] Ans: 7 3
3 (6 ) ( 2) 63 10
SBT CT AS ACw wπ
π
+ + + =+ − + − + =+
5
2.2 Area ~ Page 18
Practice Problems
1. 2 25 (20)(10) 25 200r lwπ π π+ = + = + 2. 2 2 2 28 4 64 16s rπ π π− = − = − 3.
2 22 (20)(10) 2(5 ) 200 50lw rπ π π− = − = − 4.
2 21 1 1 14 2 4 2(4 ) (4)(4) 4 8r bhπ π π− = − = −
5. 1 12 2(12)(4) (9)(4) 30lw bh− = − = 6. 2 2 2 26 3 36 9s rπ π π− = − = −
Regents Questions
1) IA JAN ’11 [23] Ans: 2 2) IA JAN ’12 [17] Ans: 1 3) IA JAN ’13 [15] Ans: 3 4) IA JAN ’14 [7] Ans: 2 5) GC JAN ’17 [8] Ans: 1 6) IA JUN ’13 [39]
Ans:
2 21 12 2
5 6 5 16(16)(5) (3 )
80 4.5 65.86
lA lw rπ π
π
= + + == − = −= − ≈
7) IA JUN ’14 [31]
Ans: 2 2 2 2
2 2(2 )4
s r x xx x
π ππ
+ = += +
8) IA JAN ’15 [37] Ans:
2 2(36)(15) 4540 16 490
lw rπ ππ
− = −= − ≈
490 8.95 $4,385.50× =
6
3. LINES, ANGLES AND PROOFS 3.1 Parallel Lines and Transversals ~ Page 24
Practice Problems
1. (a) linear pair; supplementary (b) vertical angles; congruent (c) corresponding angles; congruent (d) alternate interior angles; congruent (e) alternate exterior angles; congruent 2. 180 (57 64) 59a = − + = ° b = 64° c = 57° d = 180 – 64 = 116°
3.
15 5 180 12515 5 55
15 604
xx
xx
− = −− =
==
7 27 125
7 9814
yyy
+ ===
Regents Questions
1) MA JAN '03 [20] Ans: 2 2) MA AUG '06 [13] Ans: 4 3) GE AUG '09 [1] Ans: 4 4) GE JUN '10 [7] Ans: 2 5) GE AUG '11 [9] Ans: 3
6) GE JUN '13 [20] Ans: 3 7) GC JUN '15 [17] Ans: 1 8) GC AUG ’16 [1] Ans: 2 9) MA JUN '01 [22] Ans: 57°
7
3.2 Deductive Reasoning ~ Page 30
Practice Problems
1. Statements Reasons ∠1 and ∠2 are complementary Given ∠2 and ∠3 are complementary Given ∠1 ≅ ∠3 Complements of the same angle are congruentm 1 m 3∠ = ∠ Definition of congruent angles
(Congruent angles are equal in measure)
2. Statements Reasons ∠1 and ∠3 are supplementary Given ∠1 and ∠2 are a linear pair Definition of linear pair ∠1 and ∠2 are supplementary Linear pairs are supplementary ∠2 ≅ ∠3 Supplements of the same angle are congruent ∠2 and ∠3 are alternate exterior angles Definition of alternate exterior angles
m n If two lines cut by a transversal form congruent alternate exterior angles, then the lines are parallel
3. Statements Reasons PCEG , PC GE≅ Given
CE CE≅ Reflexive Postulate PC = GE, CE = CE Definition of congruent segments PC + CE = GE + CE Addition Postulate PE = PC + CE, GC = GE + CE
Partition Postulate
PE = GC Substitution Postulate PE GC≅ Definition of congruent segments
Regents Questions
1) GE JAN '11 [8] Ans: 4 2) GE JUN '12 [6] Ans: 2 3) GE AUG '12 [7] Ans: 1 4) GE JUN '14 [27] Ans: 2 5) GC JAN '16 [6] Ans: 1
6) GC JAN '16 [33] Ans: (2) Parallel Postulate; (3) Alternate interior angles formed by parallel lines and a transversal are congruent; (4) Consecutive adjacent angles on a straight line add to 180°; (5) Substitution
8
4. TRIANGLES 4.1 Angles of Triangles ~ Page 39
Practice Problems
1. 180 (60 40) 80− + = °
2.
6 5 3 65
3 6020
x xxx
+ = +==
3. 60 60 120+ = °
4.
5 12 1807 12 1807 168
24
x x xxx
x
+ + + =+ ==
=
Angles are 24°, 36°, and 120°, so the triangle is obtuse.
5.
m 180 (90 65) 25ACB∠ = − + = ° m 180 (80 50) 50DEC∠ = − + = ° 180 (50 25) 105x = − + = °
6.
y = 56° (corresponding angles) z = 65° (linear pair) 180 (56 65) 59x = − + = °
Regents Questions
1) MA JUN '06 [6] Ans: 3 2) GE JUN '09 [1] Ans: 1 3) GE JUN '11 [7] Ans: 2 4) GE AUG '11 [11] Ans: 3 5) GE JAN '12 [6] Ans: 2 6) GE AUG '12 [6] Ans: 4 7) GE JAN' 13 [14] Ans: 1 8) GE JAN '14 [14] Ans: 2 9) GC AUG ’16 [4] Ans: 2
10) GE AUG '10 [35] Ans: Yes, m 180 (93 43) 44ABD∠ = − + =
19 2 6 3 5 1806 30 180
25
x x xx
x
+ + + + + =+ =
=
m 19 44BDC x∠ = + = Because alternate interior angles
ABD∠ and BDC∠ are congruent, AB DC .
11) GE JUN '12 [30] Ans: 180 – (90 + 63) = 27
9
12) GC FALL '14 [10] Ans: The sum of the measures of the angles of a triangle is 180°, som m m 180ABC BCA CAB∠ + ∠ + ∠ = °. Each interior angle of the triangle and its exterior angle form a linear pair. Linear pairs are supplementary, so m m 180ABC FBC∠ + ∠ = ° ,
m m 180BCA DCA∠ + ∠ = ° , and m m 180CAB EAB∠ + ∠ = ° . By addition, the sum of these linear pairs is 540°. When the angle measures of the triangle are subtracted from this sum, the result is 360°, the sum of the exterior angles of the triangle. 4.2 Medians, Altitudes and Bisectors ~ Page 47
Practice Problems
1. (a) BE (b) AD (c) CF
2. (a) circumcenter (b) centroid (c) orthocenter (d) incenter
3. (a) 48 (b) 16 (c) 24
4. (a)
2 3 7 4750 510
2(10) 3 232 46
x xx
xDHDG DH
+ = −=== + == ⋅ =
(b)
2
29 90
819
m 12 108
yy
yEFH y
+ ===
∠ = = °
Regents Questions
1) GE FALL '08 [25] Ans: 3 2) GE JUN '10 [25] Ans: 4 3) GE JUN '12 [14] Ans: 1 4) GE AUG '12 [24] Ans: 4
5) GE JUN '15 [34] Ans:
12m (84) 42REC∠ = =
m 180 (42 28) 110BRC∠ = − + = °
10
4.3 Isosceles and Equilateral Triangles ~ Page 54
Practice Problems
1. 180 120 302− = ° 2. (a) CE (b) AF (c) BD
3. (a) 60° (equilateral triangle)
(b) 180 50 652− = ° (base angle of isosceles triangle)
(c) 180 (60 65) 55− + = ° (parts of a straight angle) (d) 65 50 115+ = ° (exterior angle) (e) 60 60 120+ = ° (exterior angle) (f) 60° (vertical angles with BAC∠ ) 4.
180 54m 632
QRP −∠ = = °
m 180 63 117QRS∠ = − = °
180 117 31.52
x −= = °
5. m 90 60 30EFG∠ = − = ° 180 (90 30) 60x = − + = °
6. Statements Reasons
JKL△ , side JL extended through L to M, side KL extended through L to N, m 70K∠ = ° , m 55MLN∠ = °
Given
JLK∠ and MLN∠ are vertical angles Definition of vertical angles m mJLK MLN∠ = ∠ Vertical angles are equal in measure m 55JLK∠ = ° Substitution m m m 180J K JLK∠ + ∠ + ∠ = ° Sum of the angles of a triangle is 180° m 70 55 180m 55
JJ
∠ + °+ ° = °∠ = °
Substitution, and solving
m mJLK J∠ = ∠ Transitive property JKL△ is isosceles If two angles of a triangle are equal in
measure, then the triangle is isosceles
11
Regents Questions
1) GE JUN '12 [11] Ans: 1 2) GE JAN '15 [8] Ans: 4 3) GE JUN '15 [8] Ans: 3 4) GE AUG '15 [10] Ans: 2 5) GC JUN '16 [19] Ans: 2 6) GC AUG ’16 [11] Ans: 4 7) GE JAN '10 [29]
Ans: 180 46 672− = °
8) GE JAN '11 [29] Ans: 30
9) GE AUG '11 [35]
Ans: No, m mKGH GKH∠ ≠ ∠
10) GC FALL '14 [5]
Ans: In an isosceles triangle, the bisector of the vertex angle is also a median. Therefore, 1
2MO OP= , so MO = 8.
11) GC FALL '14 [24] Ans:
XYZ△ , XY ZY≅ and YW bisects XYZ∠ (Given). XYZ△ is isosceles (Definition of
isosceles triangle). YW is an altitude of XYZ△ (The angle bisector of the vertex of an isosceles triangle is also the altitude of that triangle). YW XZ⊥ (Definition of altitude).
YWZ∠ is a right angle (Definition of perpendicular lines).
12) GC JUN '15 [32] Ans: Since linear pairs are supplementary, m 65GIH∠ = ° . Since GH IH≅ , m 65IGH∠ = ° and m 180 (65 65) 50GHI∠ = − + = ° . Since EGB GHI∠ ≅ ∠ , the corresponding angles formed by the transversal and lines are congruent and AB CD .
13) GC JAN ’17 [30] Ans: m m 25DAC ECA∠ = ∠ = ° m 180 2(25) 130AXC∠ = − = °
12
4.4 Congruent Triangles ~ Page 62
Practice Problems
1. (2) by SSS 2. (1) by SAS 3. ASA
4. ASA AED∠ and CEB∠ are a pair of
vertical angles.
5. AAS
IJ LJ≅ by definition of bisector IJH LJK∠ ≅ ∠ by vertical angles
6. SAS CA AC≅ by Reflexive Property
7.
Statements Reasons
C D∠ ≅ ∠ (A) Given AC AD≅ (S) Given
A A∠ ≅ ∠ (A) Reflexive Property ACE ADB≅△ △ ASA
CE DB≅ CPCTC
13
Regents Questions
1) MB AUG '09 [7] Ans: 1 2) GE AUG '10 [1] Ans: 4 3) GE AUG '10 [7] Ans: 2 4) GE JAN '11 [22] Ans: 1 5) GE JUN' 11 [2] Ans: 3 6) GE AUG '11 [2] Ans: 2 7) GE AUG '11 [14] Ans: 4 8) GE JAN '12 [16] Ans: 4 9) GE JUN '14 [10] Ans: 4 10) GC AUG ’16 [22] Ans: 3 11) GC JAN ’17 [3] Ans: 1 12) MB JUN '02 [29]
Ans: 3. Perpendicular line segments form right angles; 6. If two parallel lines are cut by a transversal, the alternate interior angles are congruent; 8. AAS; 9. CPCTC
13) GE JUN '12 [35] Ans: B∠ and C∠ are right angles (perpendicular lines form right angles). B C∠ ≅ ∠ (all right angles are congruent). AEB DEC∠ ≅ ∠ (vertical angles are congruent).
ABE DCE≅△ △ (ASA). AB DC≅ (CPCTC).
14) GE AUG '13 [35] Ans: ABC△ , BD bisects ABC∠ , BD AC⊥ (Given). CBD ABD∠ ≅ ∠ (definition of angle bisector). BD BD≅ (Reflexive property).
CDB∠ and ADB∠ are right angles (definition of perpendicular).
CDB ADB∠ ≅ ∠ (all right angles are congruent). CDB ADB≅△ △ (SAS). AB CB≅ (CPCTC).
15) GE JAN '15 [29] Ans: BE and AD intersect at point C, BC EC≅ , AC DC≅ , AB and DE are drawn (Given).
BCA ECD∠ ≅ ∠ (vertical Angles). ABC DEC≅△ △ (SAS).4.5 Similar Triangles ~ Page 74
Practice Problems
1.
2510 6
10 15015
AC
ACAC
=⋅ =
=
2.
50 4020
40 100025
XYXYXY
=⋅ =
=
3. m 25P∠ = ° and m 45R∠ = ° , so m 180 (25 45) 110Q∠ = − + = ° 4. QRS UTS△ △
Q U∠ ≅ ∠ and R T∠ ≅ ∠ by alternate interior angles formed by parallel lines. The triangles are similar by AA similarity. (Also, RSQ TSU∠ ≅ ∠ by vertical angles)
5. AEB CED△ △ A DCE∠ ≅ ∠ and B CDE∠ ≅ ∠ by
corresponding angles formed by parallel lines. The triangles are similar by AA similarity. (Also, E E∠ ≅ ∠ by Reflexive Property.)
14
Regents Questions
1) GE JAN '10 [19] Ans: 4 2) GE AUG '10 [6] Ans: 4 3) GE JAN '12 [9] Ans: 3 4) GE JAN '13 [7] Ans: 2 5) GE JUN '13 [24] Ans: 2 6) GE AUG '13 [17] Ans: 3 7) GE JAN '15 [28] Ans: 4 8) GC JUN '15 [15] Ans: 3 9) GC JUN '15 [21] Ans: 4 10) GC AUG '15 [19] Ans: 2 11) GC JAN '16 [13] Ans: 1 12) GC JAN '16 [20] Ans: 4 13) GC JAN '16 [24] Ans: 3 14) GC JUN '16 [5] Ans: 3 15) GC JUN '16 [17] Ans: 1 16) GC AUG ’16 [12] Ans: 3 17) MA JAN '09 [31]
Ans: 3021 7
ML= 10ML =
18) GE JAN '11 [36] Ans: B∠ and E∠ are right angles (definition of perpendicular lines).
B E∠ ≅ ∠ (all right angles are congruent). BFD∠ and DFE∠ are supplementary and ECA∠ and
ACB∠ are supplementary (linear pairs). DFE ACB∠ ≅ ∠ (angles supplementary to congruent angles are congruent). ABC DEF△ △ (AA Similarity).
19) GE AUG '11 [33] Ans: ACB AED∠ ≅ ∠ (Given).
A A∠ ≅ ∠ (Reflexive property). ABC ADE△ △ (AA Similarity).
20) GE JAN '16 [29]
Ans: 9 436 x
= 9 14416
xx
==
21) GC AUG '15 [29] Ans: 6 9
14 21= ; Yes (SAS) 22) GC JAN ’17 [29]
Ans: GI is parallel to NT , and IN intersects GT at A (Given);
I N∠ ≅ ∠ , G T∠ ≅ ∠ (parallel lines cut by a transversal form congruent alternate interior angles);
GIA TNA△ △ (AA Similarity).
15
4.6 Similar Triangle Theorems ~ Page 84
Practice Problems
1. (angle bisector theorem)
94 6
6 366
x
xx
=
==
2. (angle bisector theorem)
2 52 9
5( 2) 185 10 18
5 285.6
xxx
xx
=−
− =− =
==
3. (triangle midsegment theorem) 1
2 (24) 12x = =
4. (side-splitter theorem)
12 164
16 483
xxx
=
==
5. (side-splitter theorem)
138 42
MN MRNP RQ
xx
=
=−
13(42 ) 8546 13 8
546 2126
x xx x
xx
− =− =
==
MR = 26; RQ = 42 – 26 = 16
6. By the side splitter theorem,
96
ELBE
= , or 69
BEEL
= .
By the angle bisector theorem,
15
BE ELx
=
15x EL BE⋅ = ⋅
15 BExEL
= ⋅
By substitution, 6915 10x = ⋅ = ,
so BW = 10
16
Regents Questions
1) GE AUG '10 [3] Ans: 1 2) GE AUG '10 [27] Ans: 2 3) GE JUN '12 [11] Ans: 4 4) GE JUN '12 [16] Ans: 3 5) GE JAN '13 [11] Ans: 3 6) GE AUG '14 [23] Ans: 2 7) GE JAN '15 [21] Ans: 3 8) GE JUN '15 [20] Ans: 4 9) GE AUG '15 [7] Ans: 3 10) GC JUN '15 [11] Ans: 3 11) GC AUG '15 [17] Ans: 4 12) GC JUN '16 [21] Ans: 2 13) GC JAN ’17 [4] Ans: 4 14) MA AUG '00 [21]
Ans: 5 1210 x
= ; 24 miles
15) GE JAN '10 [33]
Ans: 39 15
5
x
x
=
=
16) GE JAN '11 [37] Ans:
16 320 5
16 80 20 60140 4
35
xx
x xx
x
−=+
+ = −==
3 32AC x= − =
17) GE JUN '11 [33]
Ans: 12
25 1816.7
EB
EB
=
≈
18) GE AUG '15 [31] Ans: 2 7 25
9x
x+ =
=
NT = 4.5 19) GC JUN '15 [31]
Ans:
1.654.15 16.6
6.6
x
x
=
≈
20) GC AUG '15 [27]
Ans: 120230 315
164
BC
BC
=
≈
21) GC JUN '16 [27]
Ans: 3.75 4.5
5 639.375 39.375
=
=
AB CD because AB divides the sides proportionately.
17
5. RIGHT TRIANGLES AND TRIGONOMETRY 5.1 Pythagorean Theorem ~ Page 97
Practice Problems
1.
2 2 2
25 7
7474 8.6
cc
c
+ === ≈
2.
2 2 2
2
2
24 26576 676
10010
bbb
b cm
+ =+ =
==
3.
2 2 2
2
2
8 1064 100
366
xxx
x
+ =+ =
==
2 2 2
2
2
4 1016 100
8484 9.2
yyy
y
+ =+ =
== ≈
9.2 6 3.2y x− ≈ − ≈ ft.
4.
2 2 2
2 2 2
2
(80 70) (40 15)10 25
725
ccc
− + − =+ =
=
725 26.93c = ≈ 26.93 6 162Area ≈ × ≈ sq. ft.
5.
2 2 2
26 8 ( )
100 ( )10
DFDF
DF
+ ===
Let x = EG. By angle bisector theorem,
68 10
6(8 ) 1048 6 10
48 163
xx
x xx x
xx EG
=−
− =− =
== =
2 2 2
23 6 ( )
45 ( )45 6.7
DGDG
DB
+ === ≈
18
Regents Questions
1) IA FALL ’07 [11] Ans: 1 2) IA JUN ’08 [25] Ans: 3 3) IA AUG ’08 [9] Ans: 4 4) IA JUN ’09 [9] Ans: 3 5) IA AUG ’09 [6] Ans: 1 6) GE JUN ’10 [24] Ans: 2 7) IA AUG ’10 [4] Ans: 2 8) IA JAN ’11 [7] Ans: 2 9) IA JUN ’11 [2] Ans: 3 10) GE AUG ’11 [27] Ans: 3 11) IA JAN ’12 [21] Ans: 1 12) IA JUN ’12 [23] Ans: 2 13) IA JUN ’13 [29] Ans: 3 14) IA AUG ’13 [23] Ans: 3 15) GC AUG '15 [11] Ans: 2 16) GC AUG ’16 [8] Ans: 3
17) MA JUN ’08 [32]
Ans:
2 2 2
29 18
24315.6
bb
b
+ ==≈
18) MA JAN ’09 [33]
Ans:
2 2 2
230 20
130036
ccc
+ ==≈
19) GC JAN '16 [32]
Ans: 169 20.6
36.6
x
x
=
≈
2 2 2
236.6 20.6
1763.9242
ccc
+ ==≈
5.2 Congruent Right Triangles ~ Page 105
Practice Problems
1. (1) 2. Right triangles MAT and HTM (Given) MT AH≅ (Given) AT AT≅ (Reflexive Property) MAT HTA≅△ △ (HL) M H∠ ≅ ∠ (CPCTC)
Regents Questions
1) MB JUN '03 [20] Ans: 1 1) GE JAN '14 [12] Ans: 1 2) GC JUN '16 [7] Ans: 3
19
5.3 Trigonometric Ratios ~ Page 109
Practice Problems
1. 35sin A = , 4
5cos A= , 34tan A= 2. 15
17sin B = , 817cos B = , 15
8tan B =
3. sin S and cos R 4. 725sin N =
5. 21
20tan P =
6.
2 2 2
230 40
250050
cc
c
+ ===
30 350 5sin B = =
Regents Questions
1) IA JAN ’10 [8] Ans: 3 2) IA JUN ’10 [9] Ans: 2 3) IA AUG ’10 [26] Ans: 2 4) IA JAN ’11 [9] Ans: 1 5) IA AUG ’11 [12] Ans: 2 6) IA JAN ’12 [26] Ans: 3
7) IA JUN ’12 [24] Ans: 4 8) IA JAN ’13 [7] Ans: 2 9) IA JUN ’13 [25] Ans: 4 10) IA AUG ’13 [29] Ans: 1 11) GC JUN '16 [15] Ans: 4 12) GC JAN ’17 [14] Ans: 3
20
5.4 Use Trigonometry to Find a Side ~ Page 116
Practice Problems
1. sin 3230x° =
30sin 32 15.9x = ° ≈ ft.
2. tan 6215x° =
15 tan 62 28.2x = ° ≈ ft.
3. 400tan11x
° =
400 2, 058tan11
x = š
ft.
4. tan 586x° =
6 tan 58 9.60x = ° ≈ ft. Area = 1 1
2 2 (6)(9.60) 28.8A bh= ≈ ≈ ft2
5. cos 7210x° =
10cos72 3.09 37x ft in= ° ≈ ≈ from base
sin 7210y° =
10sin 72 9.51 114x ft in= ° ≈ ≈ up wall
6. tan 28200
h° = (h = cliff + lighthouse)
200 tan 28 106.34h = ° ≈ ft.
tan18200
c° = (c = cliff height alone)
200 tan18 64.98c = ° ≈ ft. 106.34 64.98 41.4x h c= − = − ≈ ft.
Regents Questions
1) 9Y APR ’74 [36] Ans: 2 2) 9Y JUN ’75 [19B] Ans: 3 3) IA AUG ’09 [14] Ans: 2 4) IA JUN ’11 [8] Ans: 2 5) IA AUG ’11 [26] Ans: 2 6) GC JUN '15 [5] Ans: 3 7) GC JUN '16 [11] Ans: 4 8) GC JAN ’17 [7] Ans: 2 9) GC JAN ’17 [12] Ans: 3 10) IA JUN ’09 [37] Ans: 39, 63 11) IA AUG ’10 [39] Ans: 84, 71 12) IA JAN ’11 [33] Ans: 2.1 13) IA JAN ’13 [38] Ans: 8, 12
14) GC FALL '14 [13] Ans:
tan 3.47
6336384
M
M
=
≈
4960 + 384 = 5344
tan 0.64
20493229
A
A
=
≈
5344 – 229 = 5115
21
15) GC FALL '14 [22] Ans: x represents the distance between the lighthouse and the canoe at 5:00; y represents the distance between the lighthouse and the canoe at 5:05.
112 1.5tan 6
1051.3x
x
−=
≈
112 1.5tan 55
77.4y
y
−=
≈
1051.3 77.4 1955− ≈
16) GC AUG '15 [32]
Ans:
125tan 7
125 1018.0tan 7
AC
AC
=
= ≈
125tan16
125 435.9tan16
DC
DC
=
= ≈
1018.0 435.9 582AD AC AD= −≈ − ≈
17) GC JAN '16 [29]
Ans: 30sin 70
32x
x
=
≈
18) GC JAN '16 [36] Ans:
tan52.8
tan52.8 1.32
hx
h x x
=
= ≈
tan 34.98
( 8) tan 34.90.70( 8)
hx
h xx
=+
= +≈ +
1.32 0.7( 8)1.32 0.7 5.60.62 5.6
9.0
x xx xxx
= += +=≈
tan 52.89
9 tan 52.8 11.86
h
h
≈
≈ ≈
11.86 1.7 13.6+ ≈ 19) GC AUG ’16 [31]
Ans:
15sin 75
15sin 7515.5
xx
x
=
=
≈5.5 Use Trigonometry to Find an Angle ~ Page 127
Practice Problems
1. 3sin7
x = 25.4x ≈ ° 2. 420tan2000
x = 12x ≈ °
3. 2 2 212 16 r+ = 20r = 50 20 30s = − =
16sin30
x = 32x ≈ °
4. 6tan4
x = 56x ≈ °
sin 5615b° =
15sin 56 12b = ° ≈ ft.
22
Regents Questions
1) IA JUN ’08 [16] Ans: 2 2) IA AUG ’08 [29] Ans: 3 3) IA JUN ’11 [14] Ans: 1 4) GC FALL '14 [1] Ans: 1 5) GC JAN '16 [16] Ans: 3 6) IA JAN ’10 [32] Ans: 53 7) IA JUN ’10 [33] Ans: 37 8) IA AUG ’11 [35] Ans: 41.8 9) IA JUN ’12 [35] Ans: 78 10) IA AUG ’12 [38] Ans: 54, 23 11) IA JUN ’13 [35] Ans: 19 12) GC JUN '15 [28]
Ans: 4.5sin
11.7523
x
x
=
≈
13) GC JUN '16 [30]
Ans: 10tan4
68
x
x
=
≈
14) GC AUG ’16 [34] Ans:
12tan75
9.0972tan75
43.8334.7
x
x
y
yy x
=
≈
=
≈− ≈ °
5.6 Special Triangles (+) ~ Page 133
Practice Problems
1. 5 2 inches Each side is 5 inches. 45-45-90 triangle with factor of 5.
2. 4 units The altitude of an equilateral triangle
is also the angle bisector. 30-60-90 triangle with factor of 2.
Regents Questions
1) IA JAN ’09 [12] Ans: 3 2) GC JAN ’17 [9] Ans: 2
23
5.7 Cofunctions ~ Page 136
Practice Problems
1. m 90 25 65B∠ = − = ° 2.
15 5 902 10 90
2 8040
x xx
xx
+ + − =+ =
==
Regents Questions
1) S3 AUG ’96 [33] Ans: 2 2) S3 JAN ’97 [29] Ans: 1 3) S3 JUN ’98 [25] Ans: 3 4) GC JUN '15 [12] Ans: 4 5) GC AUG '15 [4] Ans: 1 6) GC JAN '16 [9] Ans: 4 7) GC AUG ’16 [6] Ans: 1 8) S3 AUG ’97 [4]
Ans: 90 72 18x = − = 9) S3 JUN ’99 [12]
Ans: 2 25 55 9030
xx
− + ==
10) S3 JUN ’03 [10]
Ans: 2 20 40 9015
xx
+ + ==
11) S3 JAN ’04 [4]
Ans: 3 60 9010
xx
+ ==
12) GC FALL '14 [7]
Ans: 2 0.1 4 0.7
0.8 20.4
x xx
x
+ = −==
A and B are complementary angles, and cofunctions of complementary angles are equal.
13) GC FALL '14 [20] Ans: The acute angles in a right triangle are always complementary. The sine of any acute angle is equal to the cosine of its complement.
14) GC JUN '16 [28] Ans: 90 73 17R = − = ° Cofunctions of complementary angles are equal.
15) GC JAN ’17 [27] Ans: Yes, because 28º and 62º angles are complementary. The sine of an angle equals the cosine of its complement. 5.8 Mean Proportional ~ Page 139
Practice Problems
1.
2 3 9 27
27 3 3xx
= ⋅ == =
2.
2 12 3 36
36 6xx
= ⋅ == =
24
3.
2
2
2
10 ( 21)100 21
21 100 0( 25)( 4) 0
25
x xx x
x xx xx
= += +
+ − =+ − =
= − 4x =
4.
2 4 5 20
20 2 5xx
= ⋅ == =
2 9 4 36
36 6yy
= ⋅ == =
2 9 5 45
45 3 5zz
= ⋅ == =
5.
2 24 6 144
144 12xx
= ⋅ == =
2 30 6 180
180 6 5yy
= ⋅ == =
2 24 30 720
720 12 5zz
= ⋅ == =
6.
(a) 2
12
1 21 2
xx
x
===
(b)
312 2
2 3 312 2 4
34
2y
h xy
h
= − =
= = ⋅ =
=
Regents Questions
1) GE JAN '11 [23] Ans: 4 2) GE JUN '11 [28] Ans: 1 3) GE JAN '13 [8] Ans: 1 4) GE JUN '13 [27] Ans: 3 5) GE AUG '13 [26] Ans: 3 6) GE AUG '14 [10] Ans: 3 7) GE JUN '15 [6] Ans: 3 8) GE AUG '15 [28] Ans: 3 9) GC JAN '16 [22] Ans: 2 10) GC JUN '16 [13] Ans: 2 11) GC AUG ’16 [10] Ans: 2 12) MB JUN '08 [28]
Ans:
2
10 1212
(10 ) 14410 144 0
( 18)( 8) 0
xx
x xx x
x x
+ =
+ =+ − =+ − =
x = 8 miles
13) GE FALL '08 [29]
Ans: 2 (3)(4)
12 2 3xx
== =
14) MB AUG '09 [32] Ans: Let x = AD.
2
2( 6) 46 16 0
( 8)( 2) 02
x xx xx x
x
+ =+ − =
+ − ==
Since DC = 8, AC = 10. 15) GE JUN '14 [31]
Ans:
2
28(10 8)14412
xx
x
= +==
16) GC JUN '15 [34]
Ans: 2 20.55 0.25 0.49− ≈
No, 20.49 0.25
0.9604y
y==
0.9604 + 0.25 < 1.5
25
6. OBLIQUE TRIANGLES 6.1 Trigonometric Ratios of Obtuse Angles (+) ~ Page 148
Practice Problems
1. 180° – 100° = 80° 2. 140° 3. (1) 4. (4)
Regents Questions
1) S3 JUN ’02 [15] Ans: 4 2) MB AUG ’05 [11] Ans: 4 6.2 SAS Sine Formula for Area of a Triangle (+) ~ Page 151
Practice Problems
1. 1 12 46 8 6Area = ⋅ ⋅ ⋅ = 2. 1
2 12 15 sin150 45Area = ⋅ ⋅ ⋅ =
3. 12 7 10 sin 25 14.8Area = ⋅ ⋅ ⋅ ≈ 4. m 180 2 75 30C∠ = − ⋅ =
12 10 10 sin30 25Area = ⋅ ⋅ ⋅ =
Regents Questions
1) S3 JAN ’01 [27] Ans: 4 2) S3 AUG ’01 [20] Ans: 1 3) S3 JAN ’02 [19] Ans: 2 4) MB JUN ’07 [4] Ans: 4 5) A2 FALL ’09 [7] Ans: 2 6) A2 JAN ’13 [16] Ans: 3 7) A2 JUN ’14 [3] Ans: 2 8) A2 JUN ’16 [2] Ans: 2 9) S3 JUN ’01 [8]
Ans:
1 12 242 24
42 67
bb
b
= ⋅ ⋅ ⋅==
10) MB JAN ’02 [25] Ans:
12 12 31 sin 62 164.2Area = ⋅ ⋅ ⋅ ≈
11) S3 AUG ’02 [16] Ans:
12 12 8 sin 40 30.9Area = ⋅ ⋅ ⋅ ≈
12) MB AUG ’02 [26] Ans:
12 16 21 sin58 142.5Area = ⋅ ⋅ ⋅ ≈
13) S3 JUN ’03 [6]
Ans:
1 12 212 6
12 1.58
cc
c
= ⋅ ⋅ ⋅==
14) MB AUG ’03 [24] Ans:
12 14 16 sin30 56Area = ⋅ ⋅ ⋅ =
15) S3 JAN ’04 [3] Ans:
12 8 4 sin 30 8Area = ⋅ ⋅ ⋅ =
16) MB JUN ’05 [25] Ans:
12 11 13 sin 70 67Area = ⋅ ⋅ ⋅ ≈
26
17) MB JUN ’08 [25] Ans:
12 (20.4)(20.4)sin80 204.9Area = ≈
18) MB JUN ’10 [22] Ans:
12 (12)(20.5)sin 73 117.6Area = ≈
19) A2 JAN ’16 [33] Ans:
12 (15)(31.6)sin125 194Area = ≈
6.3 Law of Sines (+) ~ Page 156
Practice Problems
1. 8n ≈ 2. 40a =
3. m 45 30 15ADB∠ = − =
30sin 30 sin15
a = 58.0a ≈ sin 4558
58sin 45 41
h
h
=
= ≈
4. 10 12sin 42 sin B
=
1sin 0.803m sin 0.803 53
BB −
≈∠ ≈ ≈
5.
4sin 26 sin 67
n= 1 14 8
20a =
3 3 2sin 45 sin C
=
13sin 3 22
3sin 3sin 1
C
CC
= ⋅
==
1m sin 1 90C −∠ = = °
27
Regents Questions
1) MB JAN ’02 [12] Ans: 2 2) S3 JUN ’02 [23] Ans: 3 3) MB AUG ’02 [14] Ans: 2 4) MB AUG ’01 [31]
Ans: 7
sin 65 sin 70a = 6.75a ≈
5) MB AUG ’04 [21]
Ans: angles measure 20°, 30°, and 130°.
0.75sin130 sin 30
a = 1.15a ≈ miles
6) MB JUN ’05 [27] Ans: angles measure 48°, 105°, and 27°.
110sin105 sin 27
a = 234a ≈
7) MB JAN ’06 [31] Ans: base angle = 27.4°.
246.75sin125.2 sin 27.4
a = 438a ≈
8) MB JUN ’06 [22] Ans:
3b ≈
9) MB JUN ’07 [28] Ans:
6 10sin 30 sin P
=
1sin 0.833m sin 0.833 56.4
PP −
≈∠ ≈ ≈
m 180 (30 56.4) 93.6R∠ ≈ − + ≈ 6
sin 30 sin 93.6PQ= 12PQ ≈
10) MB JUN ’09 [22] Ans:
11) MB JAN ’10 [27] Ans:
30sin 40 sin105
r =
19.96r ≈
12) A2 JAN ’11 [37]
Ans: 12 10
sin 32 sin B=
1sin 0.442m sin 0.442 26.2
BB −
≈∠ ≈ ≈
m 180 (32 26.2) 121.8C∠ ≈ − + ≈ 12
sin 32 sin121.8c= 19.2c ≈
13) A2 JAN ’12 [36] Ans: angles are 32°, 115°, and 33°.
100sin 32 sin 33
w = 97.3w ≈ ft.
sin 6597.3
t= 88t ≈
14) A2 JAN ’13 [38] Ans: angles are 43°, 105°, and 32°.
100sin 32 sin105
b= 182.3b ≈
100sin 32 sin 43
a= 128.7a ≈
15) A2 JUN ’13 [37] Ans:
15sin 42 sin103
a = 10.3a ≈
12 (15)(10.3)sin35 44Area ≈ ≈
12 (7)(6.75)sin 45 16.71Area = ≈
10sin 53 sin14
b=
10 70.6 sin B
=
10sin 4.2sin 0.42
BB
==
12 (30)(19.96)sin 35 172Area = ≈
28
6.4 Law of Cosines (+) ~ Page 164
Practice Problems
1.
2 2 2 18
2
4 5 2(4)(5)
39.756.3
t
tt
= + −
=≈
2.
2 2 2
25 8 2(5)(8)cos60497
cc
c
= + −==
3.
2 2 24 3 3 2(3)(3)cos16 18 18cos2 18cos
0.111 cosm 84
AA
AA
A
= + −= −
− = −≈
∠ ≈ °
4. The largest angle is always across from
the largest side (9).
2 2 29 6 7 2(6)(7)cos81 85 84cos
4 84cos0.048 cosm 87
AA
AA
A
= + −= −
− = −≈
∠ ≈ °
5.
2 2 2250 312 490 2(312)(490)cos62,500 337,444 305,760cos
274,944 305,760cos0.899215 cos
m 25.945
AA
AA
A
= + −= −
− = −≈
∠ ≈
12 (312)(490)sin 25.94533, 443
Area =≈
29
Regents Questions
1) S3 JAN ’02 [33] Ans: 1 2) MB JUN ’08 [17] Ans: 3 3) A2 AUG ’10 [17] Ans: 4 4) A2 JUN ’11 [10] Ans: 1 5) A2 JAN ’15 [1] Ans: 2 6) MA JUN '00 [27]
Ans: various answers, such as
7) S3 JAN ’02 [40]
Ans: 2 2 220 14.5 2(20)(14.5)cos 45
14.1a
a= + −≈
8) MB JAN ’02 [27] Ans:
2 2 23.2 5.6 2(3.2)(5.6)cos403.8
aa
= + −≈
9) S3 JUN ’03 [16] Ans:
2 2 212 14 10 2(14)(10) cos0.543 cosm 57
FF
F
= + −≈
∠ ≈
10) MB AUG ’03 [29] Ans:
2 2 270 130 2(70)(130)cos117.4174
aa
= + −≈
11) S3 JAN ’04 [42] Ans:
2 2 275 120 2(75)(120)cos135181
aa
= + −≈
181 75sin135 sin B
=
sin 0.293m 17
BB
≈∠ ≈
12) MB JUN ’04 [34] Ans:
2 2 2400 200 2(400)(200)cos50312
aa
= + −≈
12 (400)(200)sin 5030,642
Area =≈
13) MB JAN ’05 [28] Ans:
2 2 23800 2900 2(3800)(2900)cos110
5,513a
a= + −≈
14) MB JAN ’09 [29] Ans:
2 2 224 36 30 2(36)(30) cos0.75 cos
m 41.4
AA
A
= + −=
∠ ≈
15) A2 JUN ’10 [39] Ans:
2 2 210 6 2(10)(6)cos8010.7
aa
= + −≈
6 10.7sin sin 80x
= 33x ≈
16) A2 JUN ’15 [36] Ans:
2 2 228 47 34 2(47)(34) cos0.808 cosm 36
AA
A
= + −≈
∠ ≈
17) A2 AUG ’15 [36] Ans:
2 2 28 11 2(8)(11)cos8212.67
aa
= + −≈
8 12.67sin sin 82x
= 38.7x ≈
18) A2 JUN ’16 [36] Ans:
2 2 211 6 7 2(6)(7)cos0.429 cosm 115.4
AA
A
= + −− ≈
∠ ≈
30
7. QUADRILATERALS 7.1 Properties of Quadrilaterals ~ Page 171
Practice Problems
1. (2) An isosceles trapezoid has a pair of parallel bases and a pair of congruent legs, but may not be a parallelogram.
2. In a parallelogram, diagonals bisect each other, so AM = 5.
3. In a parallelogram, opposite angles are congruent, so
6 30 4 10
2 4020
x xxx
− = +==
m 6(20) 30 90A∠ = − = m m 90C A∠ = ∠ = Consecutive angles in a parallelogram
are supplementary, so m 180 m 90m m 90
B AD B
∠ = − ∠ =∠ = ∠ =
ABCD is a rectangle because it is a parallelogram with four right angles.
Note: It is sufficient to show that one angle of a parallelogram is a right angle. As long as one angle in a parallelogram is a right angle, then all four angles must be right angles.
4. In a parallelogram, diagonals bisect each other, so AE = 5.
In a rhombus, diagonals are perpendicular, so AEB△ is a right triangle with a right angle at E.
2 2 2
2 2( ) ( ) ( )
8 5 3939 6.24
BE AB AE
BE
= −= − == ≈
2 12.5BD BE= ⋅ ≈
5. Since ABCD is a rhombus, all sides are congruent, so AB = AD = 24.
Diagonals of a rhombus are perpendicular, so AEB△ is a right triangle with a right angle at E.
sin 3024
24sin 30 12
BE
BE
=
= =
Diagonals bisect each other, so DE = BE = 12
6. Since ABCD is a rhombus, the diagonals are angle bisectors, so
12m m 30ACB DCB∠ = ∠ = .
Diagonals of a rhombus are perpendicular, so AEB△ is a right triangle with a right angle at E.
18sin 30
18 36sin 30
BCBC
=
= =
All sides of a rhombus are congruent, so DC = BC = 36.
31
Regents Questions
1) MA JUN ’01 [6] Ans: 1 2) MA AUG ’05 [17] Ans: 4 3) MA JAN ’09 [19] Ans: 4 4) GE AUG ’09 [7] Ans: 1 5) GE JUN ’10 [8] Ans: 4 6) GE JAN ’11 [4] Ans: 3 7) GE JAN ’11 [12] Ans: 1 8) GE JUN ’11 [11] Ans: 3 9) GE JUN ’11 [16] Ans: 3 10) GE JUN ’11 [25] Ans: 1 11) GE AUG ’11 [21] Ans: 1 12) GE AUG ’11 [28] Ans: 3 13) GE JAN ’12 [4] Ans: 2 14) GE JUN ’13 [26] Ans: 2 15) GE JAN ’14 [25] Ans: 3 16) GE JUN ’14 [14] Ans: 2 17) GE AUG ’14 [2] Ans: 3 18) GE AUG ’14 [19] Ans: 3 19) GE JAN ’15 [5] Ans: 1 20) GE JAN ’15 [22] Ans: 2 21) GE JUN ’15 [11] Ans: 3
22) GE AUG ’15 [17] Ans: 1 23) GC AUG '15 [1] Ans: 2 24) GC AUG '15 [8] Ans: 3 25) GC JAN '16 [3] Ans: 3 26) GC AUG ’16 [24] Ans: 1 27) GC JAN ’17 [16] Ans: 1 28) MA JAN ’08 [35]
Ans: 1∠ is a base angle of an isosceles triangle.
m 2 96∠ =
29) GC JUN '15 [26] Ans: Opposite angles in a parallelogram are congruent, so m 118O∠ = ° . The interior angles of a triangle equal 180°. 180 – (118 + 22) = 40
7.2 Trapezoids ~ Page 180
Practice Problems
1. (a) Diagonals are congruent, so BD = 25. (b) m m 180 75 105D C∠ = ∠ = − = .
2.
36 20 8
2AE −= =
2 2 2( ) 17 8 225
225 15BE
BE= − =
= =
32
Regents Questions
1) GE JAN ’12 [19] Ans: 4 2) GE JUN ’12 [4] Ans: 1 3) GE JUN ’13 [1] Ans: 2
4) MB AUG ’07 [26] Ans: Draw altitude AT . RAT△ is an isosceles right triangle with legs of 6.
2 2 26 672 6 2
cc
= += =
7.3 Parallelogram Proofs ~ Page 184
Practice Problems
1. Yes, one pair of opposite sides are both parallel and congruent.
2. Yes, both pairs of opposite sides are congruent.
3.
ABCD is a parallelogram (Given) AB CD≅ and AD CB≅ (opposite
sides of parallelograms are congruent) BD BD≅ (Reflexive Property) ABD CDB≅△ △ (SSS)
4. AOB COD≅△ △ (Given) AB CD≅ (CPCTC) OAB OCD∠ ≅ ∠ (CPCTC) AB CD (if two lines cut by a
transversal form congruent alternate interior angles, then they are parallel)
ABCD is a parallelogram (if a quadrilateral has a pair of opposite sides that are both parallel and congruent, then it is a parallelogram)
5. ABCD is a parallelogram, DF = EB (Given)
AEB DFC (opposite sides of a parallelogram are parallel)
AB DC≅ (opposite sides of a parallelogram are congruent)
AB = DC (definition of congruence) AB – EB = DC – DF
(Subtraction Postulate) AE = AB – EB, FC = DC – DF
(Partition Postulate) AE = FC (substitution) AE FC≅ (definition of congruence) AECF is a parallelogram (if a
quadrilateral has a pair of opposite sides that are both parallel and congruent, then it is a parallelogram)
6. ABCD is a parallelogram, CEBF is a rhombus (Given)
BE CE≅ (all sides of a rhombus are congruent)
BE = CE (definition of congruence) 2 2BE CE⋅ = ⋅ (Multiplication
Property) 2BD BE= ⋅ , 2AC CE= ⋅ (diagonals
of a parallelogram bisect each other) BD = AC (Substitution) BD AC≅ (definition of congruence) ABCD is a rectangle (if the diagonals
of a parallelogram are congruent, then it is a rectangle)
33
Regents Questions
1) GE JUN ’12 [28] Ans: 3 2) GC JUN '15 [13] Ans: 4 3) GC JUN '16 [9] Ans: 1 4) GC AUG ’16 [7] Ans: 3 5) GC JAN ’17 [5] Ans: 4 6) GC AUG '15 [28]
Ans: Parallelogram ABCD, diagonals AC and BD intersect at E (Given). DC AB ; DA CB (opposite sides of a parallelogram are parallel).
ACD CAB∠ ≅ ∠ (alternate interior angles formed by parallel lines and a transversal are congruent).
7) GC AUG '15 [35] Ans: Parallelogram ABCD,
, , (Given). (perpendicular lines form right angles, which are congruent).
(reflexive property). (ASA).
(CPCTC). ABCD is a rhombus (a parallelogram with consecutive congruent sides is a rhombus).
8) GC JAN '16 [35] Ans: Parallelogram ANDR with and bisecting and at points W and E (Given).
, (Opposite sides of a parallelogram are congruent). (Opposite angles of a parallelogram are congruent).
, , so
(Definition of bisect and division property).
(SAS).
, , so
(Definition of bisect and division property). AER NWD
(Opposite sides of a parallelogram are parallel). AWDE is a parallelogram (A quadrilateral with a pair of opposite sides that are congruent and parallel is a parallelogram).
9) GC JUN '16 [33] Ans: Parallelogram ABCD, , and diagonal (Given);
(vertical angles); (opposite sides of a
parallelogram are parallel); (alternate interior
angles are congruent); (AA)
34
10) GC JUN '16 [35] Ans: Quadrilateral ABCD with diagonals and that bisect each other, and (Given); quadrilateral ABCD is a parallelogram (if the diagonals of a quadrilateral bisect each other, it is a parallelogram); quadrilateral ABCD is a rhombus (if a diagonal of a parallelogram bisects its angle, it is a rhombus); AD DC≅ (the sides of a rhombus are congruent); is an isosceles triangle (definition of isosceles triangle); (the diagonals of a rhombus are perpendicular); is a right angle (perpendicular lines form a right angle); is a right triangle (definition of right triangle).
11) GC JAN ’17 [35] Ans: Quadrilateral ABCD, ,
, and and are perpendicular to diagonal at points F and E (given). and
are right angles (perpendicular lines form right angles).
(All right angles are congruent). ABCD is a parallelogram (A quadrilateral with one pair of sides congruent and parallel is a parallelogram). (Opposite sides of a parallelogram are parallel). (Parallel lines cut by a transversal form congruent alternate interior angles). (Opposite sides of a parallelogram are congruent).
(AAS). (CPCTC).
35
8. COORDINATE GEOMETRY 8.1 Forms of Linear Equations ~ Page 194
Practice Problems
1. No.
(3) 3( 2) 15?3 9
= − +≠
2. y mx b= + 4 5y x= − +
3. 2 3( 1)y x+ = − −
4.
2 1 272 1
5 35 2
y ymx x
− − += = = −− +
273 ( 2)y x+ = − +
5. 2 5
2 5x yx y
− + = −− =
6.
( )3 1
4 24 44 3 2
3 4 23 4 2
y xy x
x yx y
= += +
− + =− = −
Regents Questions
1) IA AUG '10 [29] Ans: 2 2) IA AUG '08 [36]
Ans:
254 ( 5)y x− = −
8.2 Parallel and Perpendicular Lines ~ Page 197
Practice Problems
1. equation (1)
2 2 6
2 2 63
y xy xy x
+ == − += − +
2. equation (1)
3 2 52 3 5
2( 2 3 5)4 6 10
y xx yx yx y
− = +− − =
− − − =+ = −
3. equation (2) 4. equation (1)
15
5 255 25
5
x yy xy x
− =− = − +
= −
5. 2 2y x= − + 6. 2y x= −
1 1( )y y m x x− = −
1 1( )y y m x x− = −
36
Regents Questions
1) GE JAN '14 [10] Ans: 2 2) GE JUN '14 [17] Ans: 2 3) GE JUN '14 [20] Ans: 1 4) IA AUG '14 [27] Ans: 3 5) GE AUG '14 [4] Ans: 2 6) GE AUG '14 [14] Ans: 2 7) GE AUG '14 [26] Ans: 1 8) GE JAN '15 [20] Ans: 4 9) GE JUN '15 [7] Ans: 1 10) GE JUN '15 [18] Ans: 4 11) GE JUN '15 [28] Ans: 4 12) GE JAN '16 [13] Ans: 4 13) GC JUN '15 [9] Ans: 1 14) GC AUG '15 [10] Ans: 1 15) GC JAN '16 [2] Ans: 4 16) GC JUN '16 [14] Ans: 4 17) GC JAN ’17 [1] Ans: 3
18) GE JAN '12 [31] Ans: The slope of 2 3y x= + is 2.
The slope of 2 6y x+ = is 12− .
Since the slopes are opposite reciprocals, the lines are perpendicular.
19) GE JUN '12 [31] Ans: The slope of 2 4x y+ = is 1
2− .
The slope of 4 2 12y x− = is 12 .
Since the slopes are neither equal nor opposite reciprocals, the lines are neither parallel nor perpendicular.
20) GE JAN '15 [32] Ans: 1
3 5y x= + 21) GE JAN '15 [34]
Ans: 1 34 8
x − −= , so 12x = −
8.3 Distance Formula ~ Page 205
Practice Problems
1.
2 2(6 2) ( 3 3)
64 36 100 10d = + + − −= + = =
2.
2 2(8 3) (10 5)
25 25 50 5 2d = − + −= + = =
3.
2 2(4 3) (25 1)
49 576 625 25d = + + −= + = =
4.
2 2(146 4) (52 2)22,500 2,500 25,000
50 10
d = + + −= + ==
5. 5m⊥ = − Equation of ⊥ line: Solve system for x: Solve for y: Distance for (6,–2) to (5,3):
2 5( 6)5 28
y xy x
+ = − −= − +
155 28 25
x xx
− + = +=
15 (5) 23
yy
= +=
2 2(5 6) (3 2)
1 25 26d = − + +
= + =
37
Regents Questions
1) GE JUN '11 [9] Ans: 2 2) GE AUG '11 [7] Ans: 3 3) GE JAN '12 [5] Ans: 1 4) GE JUN '12 [17] Ans: 3 5) GE JAN '13 [13] Ans: 2 6) GE JUN '14 [11] Ans: 3 7) GE AUG '14 [15] Ans: 2 8) GE JAN '15 [7] Ans: 1 9) GC JUN '15 [3] Ans: 3 10) GC JAN '16 [15] Ans: 2 11) MA AUG '00 [30]
Ans: 2 2(5 3) 7 113 10.6+ + = ≈ 12) MA JUN '03 [30]
Ans: 2 23 6 45 6.7+ = ≈
13) MA JUN '06 [33]
Ans: 2 23 4 25 5+ = = 14) GE AUG '12 [32]
Ans: 2 2( 4 2) (3 5)
36 4 40 2 10− − + −
= + = =
15) GE AUG '13 [31]
Ans: 2 2( 1 3) (4 2)
16 36 52 2 13− − + +
= + = =
16) GE JAN '14 [31]
Ans: 2 2(3 7) ( 4 2)
16 36 52 2 13− + − −
= + = =
17) GE JUN '15 [33]
Ans: 2 2(6 3) ( 1 8)
9 81 90 3 10− + − −
= + = =8.4 Midpoint Formula ~ Page 211
Practice Problems
1.
1 2 1 2,2 22 6 3 3, (2,0)2 2
x x y y+ +
− + − = =
2.
( )
1 2 1 2
1 12 2
,2 2
3 8 5 10, 5 ,72 2
x x y y+ +
+ + = =
3. The third terms of the arithmetic sequences x: 1, 3, … and y: –5, 5, … are 5 and 15, so B(5,15).
4. M is the midpoint of BC. Sequences x: 4, 0, … and y: –3, –2, … give us third terms of –4 and –1, so C(–4,–1).
Regents Questions
1) GE JUN '09 [19] Ans: 4 2) GE AUG '09 [10] Ans: 2 3) GE JAN '11 [6] Ans: 2 4) GE AUG '11 [15] Ans: 1 5) GE AUG '12 [3] Ans: 4 6) GE JAN '13 [5] Ans: 3 7) GE JUN '13 [12] Ans: 2 8) GE JAN '14 [1] Ans: 2 9) GE JUN '14 [2] Ans: 1 10) GE AUG '14 [7] Ans: 3 11) GE JAN '15 [2] Ans: 4
12) GE JUN '15 [5] Ans: 2 13) MA JAN '06 [33]
Ans: x: –1, 2, 5 and y: 5, 3, 1, so (5,1)
14) MA AUG '08 [34] Ans: x: 6, 2, –2 and y: 2, 0, –2, so (–2,–2)
15) GE JAN '10 [31] Ans: x: 1, 3.5, 6 and y: 8, 2, –4, so R(6,–4)
38
8.5 Perpendicular Bisectors ~ Page 216
Practice Problems
1. Midpoint is 3 9 5 17, (6,11)
2 2+ + =
17 5 12 29 3 6
m −= = =−
12m⊥ = −
Equation of perpendicular bisector is 1211 ( 6)y x− = − −
2. Midpoint is 2 6 3 3, (2,0)2 2
− + − =
34
3 3 66 2 8
m − − −= = = −+
43m⊥ =
Equation of perpendicular bisector is 43 ( 2)y x= −
3. Midopoint is 2 8 6 12, (5,9)
2 2+ + =
12 6 6 18 2 6
m −= = =−
1m⊥ = −
Equation of perpendicular bisector is
9 ( 5)9 5
14
y xy x
y x
− = − −− = − +
= − +
y-intercept is 14
4. Midopoint is 4 2 5 5, ( 1,5)2 2
− + + = −
The segment is horizontal with a 0 slope. Its perpendicular bisector is vertical, with
an equation of 1x = − .
Regents Questions
1) GE AUG '11 [26] Ans: 1 1) GE JAN '12 [25] Ans: 4 2) GE AUG '13 [27] Ans: 3 3) GC JUN '16 [12] Ans: 1 4) GE AUG '09 [35]
Ans: 1 7 1 ( 5), (3, 2)2 2
M − + + − = = −
34
1 ( 5)1 7
m − −= = −− −
43m⊥ =
432 ( 3)y x+ = −
5) GE JAN '13 [34]
Ans: 3 3 1 5, (3,2)2 2
M + − + = =
Line segment is vertical, so its perpendicular bisector is horizontal, with equation y = 2.
6) GE AUG '15 [36]
Ans: 4 8 2 6, (6,4)2 2
M + + = =
6 2 18 4
m −= =−
1m⊥ = −
4 ( 6)y x− = − −
39
8.6 Directed Line Segments ~ Page 219
Practice Problems
1. 56k =
56
56
( )
3 (9 3) 8x x x xP A B A= + −
= + − =
56
56
( )
5 (17 5) 15y y y yP A B A= + −
= + − =
P(8,15)
2. 35k =
3535
( )
2 (3 2) 1x x x xS R T R= + −
= − + + =
35
35
( )
2 ( 8 2) 4y y y yS R T R= + −
= + − − = −
S(1,–4)
3. 38k =
3838
( )
2 (6 2) 1x x x xP L M L= + −
= − + + =
38
3 38 4
( )
3 ( 3 3)y y y yP L M L= + −
= + − − =
( )341,P
4. 25k =
25
25
( )
1 (6 1) 3x x x xG F H F= + −
= + − =
252 15 5
( )
3 (5 3)y y y yG F H F= + −
= − + + =
( )153,G
Regents Questions
1) GC FALL '14 [14] Ans: 4 2) GC AUG ’16 [18] Ans: 4 3) GC JAN ’17 [20] Ans: 1 4) GC JUN '15 [27]
Ans: 256 (4 6) 2xP = − + + = − 255 (0 5) 3yP = − + + = −
P(–2,–3) 5) GC AUG '15 [31]
Ans: 251 (16 1) 7xE = + − =
254 (14 4) 8yE = + − =
E(7,8)
6) GC JAN '16 [27] Ans:
232 (4 2) 2xJ = − + + =
231 (7 1) 5yJ = + − =
J(2,5) 7) GC JUN '16 [26]
Ans: 494 (22 4) 12xP = + − =
492 (2 2) 2yP = + − =
P(12,2)
40
9. POLYGONS IN THE COORDINATE PLANE 9.1 Triangles in the Coordinate Plane ~ Page 224
Practice Problems
1.
43
11 75 2ABm −= =
−
34
8 119 5BCm −= = −
−
AB BC⊥ , so B∠ is a right angle. Therefore, ABC is a right triangle.
2. 2 2(5 2) (11 7) 25 5AB = − + − = = 2 2(9 5) (8 11) 25 5BC = − + − = = 2 2(9 2) (8 7) 50 5 2AC = − + − = =
Two sides are equal in length, so ABC△ is an isosceles triangle.
3. 2 2(5 4) (5 2) 50 5 2DE = − + + = = 2 2( 1 5) (3 5) 40 2 10EF = − − + − = = 2 2( 1 4) (3 2) 50 5 2DF = − − + + = =
Two sides are equal in length, so DEF△ is an isosceles triangle.
4. JK KL⊥
14
6 46 2JKm −= =
+, so 4KLm = −
2 6 468 4( 6)8 4 24
32 48
xx
xx
x
− − = −−
− = − −− = − +
− = −=
5. Recognize that PR is horizontal
(since 2y = − for both endpoints). 8 4 4PR = − = An altitude may be drawn from the
opposite vertex, Q, to point S(–6,–2). QS = 4 ( 2) 6− − =
Therefore, the area is 12 (4)(6) 12=
square units.
6. 2 2
2 2
(5 2) (1 2) 10
(4 1) ( 5 4) 10
AB
DE
= − + − =
= − + − + =
2 2
2 2
(4 5) (5 1) 17
(3 4) ( 1 5) 17
BC
EF
= − + − =
= − + − + =
2 2
2 2
(4 2) (5 2) 13
(3 1) ( 1 4) 13
AC
DF
= − + − =
= − + − + =
AB DE≅ , BC EF≅ , and AC DF≅ , so ABC DEF≅△ △ by SSS.
41
Regents Questions
1) GE JUN '11 [15] Ans: 2 2) GE AUG '12 [26] Ans: 2 3) GE JAN '13 [28] Ans: 3 4) GC JAN '16 [18] Ans: 1 5) GC AUG ’16 [15] Ans: 3 6) MB JAN '10 [29]
Ans: To prove that JEN△ is a right triangle, prove that its legs are perpendicular by showing their slopes are opposite reciprocals:
3 1 22 4JEm − −= = −
− +
12
1 32 2ENm − += =
+
To prove that JEN△ is an isosceles triangle, prove that it legs are congruent by using the distance formula:
2 2( 2 4) ( 3 1) 20JE = − + + − − = 2 2(2 2) ( 1 3) 20EN = + + − + =
7) GE JUN '13 [31] Ans:
2 2(7 3) ( 8 0) 80 4 5− + − − = =
8) GE JAN ’16 [37] Ans:
84
1 7 23 1RSm − −= = − = −
+
3 16 2
2 19 3STm += = =
−
The slopes are opposite reciprocals, so they are perpendicular lines forming a right angle at S. Since S∠ is a right angle, RST△ is a right triangle.
9) GC AUG '15 [33] Ans:
32BCm = − 2
3m⊥ = The right angle may be at B or C. Right angle at B means:
231 ( 3)1
bb
− = − +=
233 13
xx
= +=
Right angle at C means: 23
103
4 ( 1) b
b
− = − +
= −
1023 3192
3
9.5
x
x
= −
= =
Either answer, 3x = or 9.5x = , is correct.
42
9.2 Quadrilaterals in the Coordinate Plane ~ Page 229
Practice Problems
1. 8 0 21 5ABm − −= = −
− + 4 4 2
3 7CDm += = −−
12
4 87 1BCm − += =
+ 1
24 03 5ADm −= =
+
AB BC⊥ , BC CD⊥ , CD AD⊥ , andAD AB⊥ , so all 4 angles are right angles. Therefore, ABCD is a rectangle.
2. 2 2
2 2
2 2
2 2
(1 6) (0 3) 58
(4 1) (7 0) 58
( 3 4) (4 7) 58
( 3 6) (4 3) 58
AB
BC
CD
AD
= + + + =
= − + − =
= − − + − =
= − + + + =
37
0 31 6ABm += =
+ 7
37 04 1BCm −= =
−
4 congruent sides but not 4 right angles = Rhombus
3. 2 2
2 2
2 2
2 2
(2 5) (0 6) 85
(11 2) (9 0) 162
(4 11) (3 9) 85
(4 5) (3 6) 162
AB
BC
CD
AD
= + + + =
= − + − =
= − + − =
= + + + =
67
0 62 5ABm += =
+ 9 0 1
11 2BCm −= =−
2 pairs of opposite sides congruent but not 4 right angles = Parallelogram
4. 2 2
2 2
2 2
2 2
(5 1) (2 1) 17
(6 5) ( 2 2) 17
(2 6) ( 3 2) 17
(2 1) ( 3 1) 17
AB
BC
CD
AD
= − + − =
= − + − − =
= − + − + =
= − + − − =
14
2 15 1ABm −= =
− 2 2 4
6 5BCm − −= = −−
AB BC⊥ , so B∠ is a right angle. 4 congruent sides and a right angle = Square
5. 2 2
2 2
(7 4) ( 5 0) 146
( 1 4) ( 8 3) 146
PR
QS
= + + − − =
= − − + − − = 5
115 0
7 4PRm − −= = −+
115
8 31 4QSm − −= =
− −
Diagonals are congruent. Diagonals are perpendicular.
Midpoint of PR is ( )3 52 2
4 7 0 5, ,2 2
− + − = −
and of QS is ( )3 52 2
4 1 3 8, ,2 2− − = −
Diagonals bisect each other.
43
Regents Questions
1) GE JUN '10 [28] Ans: 2 2) GE JUN '12 [9] Ans: 1 3) GC AUG '15 [22] Ans: 4 4) GC AUG ’16 [14] Ans: 1 5) GC JAN ’17 [19] Ans: 3 6) MA AUG '00 [32]
Ans:
Both pairs of opposite sides of a parallelogram are parallel. Parallel lines have the same slope. The slope of side BC is 3. For side AD to have a slope of 3, the coordinates of point D must be (1,3).
25
2 05 0ABm −= =
− 2
53 51 6CDm −= =
−
3 0 31 0ADm −= =
− 5 2 3
6 5BCm −= =−
7) MB JUN '03 [27] Ans: To prove that ABCD is a rhombus, show that all sides are congruent using the distance formula:
2 2(8 1) (2 5) 130AB = + + + = 2 2(11 8) (13 2) 130BC = − + − = 2 2(2 11) (6 13) 130CD = − + − =
2 2(2 1) (6 5) 130AD = + + + =
8) MB JUN '06 [33] Ans: To prove that ABCD is a parallelogram, show that both pairs of opposite sides of the parallelogram are parallel by showing the opposite sides have the same slope:
38
5 26 2ABm −= =
+ 3
83 04 4CDm − −= =
− −
52
3 24 2ADm − −= =
− + 5
20 54 6BCm −= =
−
Because 38 and 5
2 are not opposite reciprocals, the consecutive sides of ABCD are not perpendicular, and ABCD is not a rectangle.
9) GE AUG '10 [38] Ans:
6 6 06 5ABm −= =
+
3 3 03 8CDm − += =
− −
92
3 63 5ADm − −= = −
− +
92
3 68 6BCm − −= = −
−
AB CD and AD BC because they have equal slopes. ABCD is a parallelogram because opposite side are parallel. AB = 11 and 85BC = . ABCD is not a rhombus because AB BC≠ . AB and BC are not perpendicular because their slopes are not opposite reciprocals. Therefore, ABCD is not a rectangle because B∠ is not a right angle.
44
10) GE JAN '11 [38] Ans: The length of each side of quadrilateral is 5. Since each side is congruent, quadrilateral MATH is a rhombus. The slope of MH is 0 and the slope of HT is 4
3− . Since the slopes are not opposite reciprocals, the sides are not perpendicular and do not form rights angles. Since adjacent sides are not perpendicular, quadrilateral MATH is not a square.
11) GE AUG '11 [38] Ans:
6 2 2 8, ( 2,3)2 2
D D− + − + = −
2 6 8 2, (4,3)2 2
E E+ − =
6 6 2 2, (0, 2)2 2
F F− + − − = −
To prove that ADEF is a parallelogram, show that both pairs of opposite sides of the parallelogram are parallel by showing the opposite sides have the same slope:
54
3 22 6ADm += =
− +
54
2 30 4EFm − −= =
−
AF DE because horizontal lines have the same slope. AF = 6 and
2 2( 2 6) (3 2) 41AD = − + + + = ADEF is not a rhombus because not all sides are congruent.
12) GE JUN '13 [34]
Ans: 34
4 14 2ABm −= =
−
Since AB BC⊥ , 23BCm = − .
13) GE AUG '13 [38] Ans: Use the midpoint formula to find M(–5,5), N(0,3), P(2,–4), Q(–3,–2). Use the slope formula to find
25MNm = − 2
5PQm = −
75MQm = − 7
5NPm = −
Since both opposite sides have equal slopes and are parallel, MNPQ is a parallelogram. Use the distance formula to find
29MN = and 53NP = . MN NP≠ , so MNPQ is not a rhombus since not all sides are congruent.
14) GE JUN '14 [38] Ans: Use the slope formula to find
32JKm = − 3
2LMm = − 12JMm = 1
2KLm = Since both opposite sides have equal slopes and are parallel, JKLM is a parallelogram. Use the distance formula to find
45JM = and 52LM = . JM LM≠ , so JKLM is not a rhombus since not all sides are congruent.
15) GE AUG '15 [34] Ans: Midpoint of KM is
( )12
0 1 4 4, ,02 2+ − =
45
16) GC FALL '14 [11] Ans: Midpoint of MT is
( )52
0 4 1 6, 2,2 2+ − + =
.
74
6 14 0MTm += =
−. In a rhombus,
diagonals are perpendicular, so 47AHm = − .
5 42 7 ( 2)y x− = − −
The diagonals, MT and AH , of rhombus MATH are perpendicular bisectors of each other.
17) GC JUN '15 [36] Ans:
53TSm = − 3
5SRm =
Since the slopes of TS and SR are opposite reciprocals, they are perpendicular and form a right angle.
RST△ is a right triangle because S∠ is a right angle.
P(0,9) 53RPm = − 3
5PTm = Since the slopes of all four adjacent sides are opposite reciprocals, they are perpendicular and form right angles. Quadrilateral RSTP is a rectangle because it has four right angles.
18) GC JAN ’17 [31] Ans: (1,3) and (–3,–3)
46
9.3 Perimeter and Area using Coordinates ~ Page 239
Practice Problems
1.
2 2
2 2
2 2
(4 1) (1 4) 18 3 2
( 3 4) (0 1) 50 5 2
( 3 1) (0 4) 32 4 2
AB
BC
AC
= − + − = =
= − − + − = =
= − − + − = =
Perimeter = 3 2 5 2 4 2 12 2+ + = 2. Method 1: enclose in rectangle Method 2: area of right triangle
44 1CAm = = 3
3 1ABm = − = − Slopes are opposite reciprocals, so CA AB⊥ and A∠ is a right angle. In problem 1, 4 2AC = and 3 2AB = . Area of right triangle ABC = 1
2 ( )( )AC AB
12 (4 2)(3 2) 12= = square units.
Area of rectangle = 7×4 = 28 Area of I = 1
2 (4)(4) 8=
Area of II = 12 (3)(3) 4.5=
Area of III = 12 (7)(1) 3.5=
Area of ABC△ = 28 – (8 + 4.5 + 3.5) = 12 square units 3.
2 2
2 2
(6 3) (10 6) 25 5
(18 6) (5 10) 169 13
EF
FG
= − + − = =
= − + − = =Since EFGH is a parallelogram, GH = EF and EH = FG. Perimeter of EFGH = 2×5 + 2×13 = 36 Area of rectangle = 15×9 = 135 Area of I = Area of III = 1
2 (3)(4) 6=
Area of II = Area of IV = 12 (5)(12) 30=
Area of EFGH = 135 – (2×6 + 2×30) = 63 square units
47
4. vertex x y upper lower difference
K –7 –7 –14 35 –49 L –5 2 –30 6 –36 M 3 6 –9 6 –15 N 1 –3 –7 21 –28 K –7 –7
Area = ( 49) ( 36) ( 15) ( 28) 128
642 2
− + − + − + −= = square units.
Regents Questions
1) GC JUN '16 [22] Ans: 3 2) GE JUN '09 [36]
Ans: 2 23 4 25 5AB = + = = 2 28 ( 6) 100 10BC = + − = =
Perimeter = 5 10 5 5 15 5 5+ + = +
2 211 ( 2) 125 5 5AC = + − = =
48
10. RIGID MOTIONS 10.1 Translations ~ Page 245
Practice Problems
1. (2 6, 4 1) ( 4,5)− + = − 2. 2,2T of Q is (6, 4)Q′ − 3. ( 5, 5) (3,0)R R′− − → is 8,5T ( 5 8,1 5) (3,6)U ′− + + = 4.
Regents Questions
1) MA JUN '03 [9] Ans: 1 2) MA JUN ’04 [2] Ans: 1 3) MA JAN ’05 [9] Ans: 4 4) MB AUG '05 [8] Ans: 4 5) MA JAN '06 [14] Ans: 4 6) MA AUG ’06 [9] Ans: 4 7) GE FALL '08 [3] Ans: 1 8) GE FALL ’08 [18] Ans: 4 9) GE JAN '11 [7] Ans: 3 10) GE JUN ’15 [4] Ans: 1 11) GE AUG ’15 [4] Ans: 2 12) GE JAN ’16 [17] Ans: 2 13) MA SPR '98 [23]
Ans: B(–6,4) and D'(–5,–4)
14) GE JUN ’12 [29] : ( 6,3)T ′ − , ( 3,3)A′ − , ( 3, 1)P′ − −
15) GE AUG '14 [31]
Ans: A(0,1)
49
10.2 Reflections ~ Page 251
Practice Problems
1. (2,3) 2. P’(–4,1) 3. M’(2,8) 4. (–2,5) 5. 6.
Regents Questions
1) MA SPR '98 [3] Ans: 2 2) MA JAN '00 [7] Ans: 3 3) MB JUN '03 [6] Ans: 4 4) MA AUG '07 [13] Ans: 1 5) MA JAN '09 [18] Ans: 2 6) GE JUN '09 [5] Ans: 3 7) GE AUG '11 [8] Ans: 2 8) GE AUG '11 [13] Ans: 1 9) MA JAN '01 [24]
Ans:
A'(0,–2) and B'(4,–6)
10) MA JUN '01 [35] Ans:
S'(0,6), U'(–3,5), N'(–3,0)
50
11) MA JUN '05 [37] Ans:
A'(7,–2), B'(2,–1), C'(3,–2), D'(2,–4)
12) MA AUG '06 [37] Ans:
A'(–2,0), B'(–1,7), C'(–5,1)
13) MA JUN '07 [39] Ans:
2 2(7 2) (8 6) 29AB = − + − =
Area = ( )229 29=
14) GE JUN '10 [32] Ans: X'(5,1), Y'(4,4), Z'(7,4)
15) GE JAN '11 [30] Ans: A'(–2,–2), B'(–1,3), C'(4,0)
16) GE JUN '15 [30] Ans: A'(1,–1), B'(3,1), C'(1,4)
17) GE AUG '15 [29] Ans: R'(–2,3) and S'(–5,–1)
18) GC JAN '16 [25] Ans:
51
10.3 Rotations ~ Page 260
Practice Problems
1. 90 : ( , ) ( , )R x y y x° → − (2,4) ( 4,2)→ −
2. A clockwise rotation of 90° is equivalent to a counterclockwise rotation of 270°.
270 : ( , ) ( , )R x y y x° → − ( 2,5) (5,2)− →
3. A clockwise rotation of 180° is equivalent to a counterclockwise rotation of 180°.
180 : ( , ) ( , )R x y x y° → − − ( 2,1) (2, 1)− → −
4. 180 : ( , ) ( , )R x y x y° → − − (0,4)A′ , ( 4,2)B′ − , ( 5,4)C′ − , ( 1,6)D′ −
5. (2) 6. (4) Regents Questions
1) MA AUG '07 [21] Ans: 3 2) MB JUN '08 [9] Ans: 3 3) GE JUN '13 [4] Ans: 4 4) GE JAN '14 [21] Ans: 4 5) GC FALL '14 [2] Ans: 4 6) GC JAN '16 [11] Ans: 4 7) GC AUG ’16 [5] Ans: 1 8) GE JAN '12 [32]
Ans: R'(–3,–2), S'(–4,4), T'(2,2) 9) GE AUG '12 [30]
Ans: A'(–2,1), B'(–3,–4), C'(5,–3)
10) GE JUN '14 [29] Ans: B'(–1,5) and C'(2,–3)
11) GC AUG ’16 [29] Ans: m m 47P L∠ = ∠ = ° because rotations preserve angles. m 180 (47 57) 76M∠ = − + = because the angles of a triangle add to 180°.
10.4 Map a Polygon onto Itself ~ Page 270
Practice Problems
1. (3) 2. 180°
3. 360 725
= ° 4. 360 22.516
= °
Regents Questions
1) GC FALL '14 [15] Ans: 2 2) GC JUN '15 [10] Ans: 1 3) GC AUG '15 [5] Ans: 1 4) GC JAN ’17 [17] Ans: 4
5) GC AUG ’16 [27]
Ans: 360 606
=
52
11. DILATIONS 11.1 Dilations of Line Segments ~ Page 275
Practice Problems
1. (15,–10) 2. ( )32 , 1−
3. (1) 4.
Regents Questions
1) MA JAN ’07 [25] Ans: 1 2) MB JAN ’08 [3] Ans: 2 3) GC JUN '15 [18] Ans: 1 4) GC JAN '16 [10] Ans: 2 5) GC JUN '16 [2] Ans: 4
6) GC AUG ’16 [21] Ans: 4 7) GC JAN ’17 [13] Ans: 1 8) S3 AUG ’02 [2]
Ans: (4,12)
53
11.2 Dilations of Polygons ~ Page 280
Practice Problems
1.
2.
3.
54
Regents Questions
1) GE JAN ’15 [24] Ans: 3 2) GC JUN '15 [16] Ans: 2 3) GC AUG '15 [6] Ans: 4 4) GC AUG '15 [20] Ans: 1 5) GC AUG '15 [23] Ans: 1 6) MA AUG ’01 [28]
Ans:
7) MB JUN ’07 [33]
Ans: ( 1,1)A′ − , (4, 2)B′ − , (3, 5)C′ − , ( 2, 2)D′ − −
35
2 14 1A Bm ′ ′
− −= = −+
35
2 52 3C Dm ′ ′
− += = −− −
2 1 32 1A Dm ′ ′
− −= =− +
5 2 33 4B Cm ′ ′
− += =−
' 'A B C D′ ′ and A D B C′ ′ ′ ′ A B C D′ ′ ′ ′ is a parallelogram because both pairs of opposite sides are parallel.
8) MA JAN ’09 [37] Ans:
( 2,6)A′ − , (4,0)B′ , (4, 2)C′ − , ( 6, 2)D′ − −
9) GE AUG ’13 [29] Ans: (2, 2)A′ , (3,0)B′ , (1, 1)C′ −
10) GE AUG ’14 [29] Ans:
11) GC JAN ’17 [32]
Ans:
A dilation preserves slope, so the slopes of QR and Q R′ ′ are equal. Therefore, Q R QR′ ′ .
55
11.3 Dilations of Lines ~ Page 288
Practice Problems
1. 3 20y x= − 2. 2 43 32 3 4x y y x+ = → = − +
Equation of its image is 23 4y x= − +
3. C(1,–1) is a point on the line: ( 1) 3(1) 4− = − Therefore, the equation of the image is
the same as the pre-image, 3 4y x= − .
4. C(1,0) is not on the line.
Find the point on the line where x = 1: 2(1) 2 4y = + = , so A(1,4). CA = 4, so 2(4) 8CA′ = = and (1,8)A′ 8 2(1) 6b b= + → = Equation of image is 2 6y x= + .
Regents Questions
1) GC FALL '14 [3] Ans: 2 2) GC FALL '14 [16] Ans: 2 3) GC JUN '15 [22] Ans: 1 4) GC AUG '15 [24] Ans: 4
5) GC JAN '16 [31]
Ans: : 3 4: 3 8y x
m y x= −= −
ℓ
56
12. TRANSFORMATION PROOFS 12.1 Properties of Transformations ~ Page 291
Practice Problems
1. (a) reflection (b) dilation (c) translation (d) rotation (e) dilation
Regents Questions
1) MA JAN '05 [5] Ans: 1 2) GE JAN ’10 [3] Ans: 2 3) GE JUN ’10 [18] Ans: 4 4) GE JUN ’12 [1] Ans: 2 5) GE JAN ’13 [4] Ans: 3 6) GE AUG ’14 [5] Ans: 3 7) GE JAN ’15 [3] Ans: 3
8) GE JUN ’15 [1] Ans: 3 9) GE AUG ’15 [6] Ans: 4 10) GC AUG '15 [13] Ans: 2 11) GC JAN ’16 [5] Ans: 3 12) GC JUN '16 [4] Ans: 1 13) GC AUG ’16 [2] Ans: 2
57
12.2 Sequences of Transformations ~ Page 299
Practice Problems
1. Rotation 90R ° with center at origin, followed by reflection x axisr −
2. Rotation 270R ° around point A, followed by translation 5, 2T −
3. Reflection 12yr = followed by rotation
270R ° around the point (1,–3)
4.
5.
6.
58
Regents Questions
1) GE JAN '11 [26] Ans: 3 2) GE JUN ’11 [3] Ans: 4 3) GC JUN '15 [4] Ans: 4 4) GC AUG '15 [7] Ans: 1 5) GC JAN '16 [8] Ans: 1 6) GC JUN '16 [8] Ans: 4 7) GC JAN ’17 [10] Ans: 3 8) MA AUG '00 [28]
Ans: If AB is reflected first, the coordinates are (2, 6)A′ − and
(4, 2)B′ − . When the reflection is dilated, the coordinates are (1, 3)A′′ − and (2, 1)B′′ − . If AB is dilated first, the coordinates are (1,3)A′ and (2,1)B′ . When the dilation is reflected, the coordinates are (1, 3)A′′ − and (2, 1)B′′ − . The images are the same.
9) S3 AUG ’01 [41] Ans:
(3, 4), ( 2, 4), (3, 2)A B C′ ′ ′− ( 1,7), ( 6,7), ( 1,5)A B C′′ ′′ ′′− − −
10) GE JAN ’16 [35] Ans:
180R °
11) GC JUN '16 [25] Ans: translation 6 units right and reflection over x-axis
12) GC AUG ’16 [26] Ans:
13) GC JAN ’17 [26]
Ans: 0, 2T − and y axisr −
12.3 Transformations and Congruence
Practice Problems
1. Translation so that A maps onto D. 2. Reflection over AC . 3. Rotation of 180° around point J.
4. Reflection over the bisector of HLK∠ . (Note that HLK△ is isosceles.)
5. Reflection over the vertical line passing through P, followed by the translation 2, 4T − (or vice versa). No, it is not possible because the triangles have opposite orientations and both translations and rotations preserve orientations.
59
Regents Questions
1) GC JUN '15 [24] Ans: 3 2) GC JUN '16 [16] Ans: 3 3) GC JAN ’17 [6] Ans: 4 4) GC FALL '14 [4]
Ans: Translate ABC△ such that point C maps onto point F, then reflect over DF . or Reflect ABC△ over the perpendicular bisector of EB .
5) GC FALL '14 [21] Ans: a) , , and
(Given). LCA∠ and DCN∠ are right angles
(Definition of perpendicular lines). LAC△ and DNC△ are right
triangles (Definition of a right triangle). LAC DNC≅△ △ (HL). b) Rotate LAC△ counterclockwise 90º about point C such that point L maps onto point D.
6) GC JUN '15 [30] Ans: Reflections are rigid motions that preserve congruency.
7) GC JUN '15 [33] Ans: Quadrilateral ABCD is a parallelogram with diagonals and intersecting at E (Given). AD BC≅ (Opposite sides of a parallelogram are congruent).
AED CEB∠ ≅ ∠ (Vertical angles are congruent). BC DA (Definition of parallelogram).
DBC BDA∠ ≅ ∠ (Alternate interior angles are congruent).
AED CEB≅△ △ (AAS). 180° rotation of AED△ around point E.
8) GC AUG '15 [30] Ans: The transformation is a rotation of 180° around the origin. Rotations are rigid motions that preserve congruency.
9) GC AUG '15 [34] Ans: Translations preserve distance, so if point D is mapped onto point A, point F would map onto point C.
DEF ABC≅△ △ since DEF△ can be mapped onto ABC△ by a sequence of rigid motions.
10) GC JAN '16 [28] Ans: Yes. The sequence of transformations consists of a reflection y axisr − and a translation
0, 3T − , which are rigid motions which preserve congruency.
11) GC AUG ’16 [33] Ans:
( , ) rotation point
( , ) rotation point(2, 9) (2, 3) (0, 6)
(6,0) (2, 3) (8, 3)(6, 8) (2, 3) (4, 5)
(5, 4) (2, 3) (7,1)
x yy x
CC
BB
− →− +− − − = − →
′+ − = −− − − = − →
′− + − =
DEF A B C′ ′ ′≅△ △ because DEF△ is a reflection of A B C′ ′ ′△ over
1x = − and reflections preserve congruency.
LA DN≅ CA CN≅DAC LCN⊥
ACBD
60
12.4 Transformations and Similarity ~ Page 314
Practice Problems
1. 12
D with A as the center of dilation
followed by 4,0T mapping A A′→ .
2. 12
D with B as the center of dilation, then
a translation 0, 2T − to map B B′→ , and
finally a reflection B Cr ′ ′ .
Regents Questions
1) GC JUN '15 [2] Ans: 4 2) GC AUG '15 [2] Ans: 3 3) GC AUG '15 [14] Ans: 4 4) GC AUG ’16 [9] Ans: 4 5) GC JAN ’17 [2] Ans: 2 6) GC FALL '14 [17]
Ans: Circle A can be mapped onto circle B by first translating circle A such that A maps onto B, and then dilating circle A, centered at A, by a scale factor of 5
3 . Since there exists a sequence of transformations that maps circle A onto circle B, circle A is similar to circle B.
7) GC FALL '14 [19] Ans: Let X Y Z′ ′ ′△ be the image of
XYZ△ after a rotation about point Z such that ZX ′ coincides with ZU . Since rotations preserve angle measure, ZY ′ coincides with ZV . Then, dilate X Y Z′ ′ ′△ by a scale
factor of ZUZX ′
with its center at
point Z. Since dilations preserve angles, X Y′ ′ maps onto UV . Therefore, XYZ UVZ△ △ .
8) GC JUN '16 [34] Ans: A dilation of 5
2 about the origin. Dilations preserve similarity.
61
13. CIRCLES 13.1 Circumference and Rotation ~ Page 322
Practice Problems
1. D = 100 ft. = 1200 in. 8C π= in.
1200 47.78
D inRC inπ
= = ≈
47 clocks can be framed.
2. D = 2 mi. = 10,560 ft. 394C π= ft.
10,560 . 8.5394 .
ftRftπ
= ≈
Regents Questions
1) MB JAN ’02 [15] Ans: 2 2) GC JAN '16 [23] Ans: 1 3) MA AUG ’00 [27]
Ans: 2C π= 2 1100.5 6,914.65D C R ftπ= ⋅ = ⋅ ≈
16,914.65 1.35, 280
mileft milesft
× ≈
4) MA JAN ’04 [37] Ans: 10C π=
1000 31.810
DRC π
= = ≈
32 revolutions needed
13.2 Arcs and Chords ~ Page 325
Practice Problems
1. 2 112 6= 1
6 360 60⋅ ° = ° 2. m 30ABC∠ = °
3. m 90A∠ = ° , so m 35C∠ = ° 4. 132 82m 107
2ADC +∠ = = °
5.
15 36024
xx
= °= °
m 48FE = ° and m 168GD = °
12m m m 24D G FE∠ = ∠ = = ° or 1
2m m m 84E F GD∠ = ∠ = = °
62
Regents Questions
1) GE JAN ’11 [24] Ans: 4 2) GE AUG ’13 [2] Ans: 1 3) GE AUG ’13 [8] Ans: 4 4) GE JUN ’14 [9] Ans: 3 5) GE AUG ’14 [3] Ans: 4 6) GE AUG ’14 [13] Ans: 1 7) GE JAN ’15 [23] Ans: 3 8) GE AUG ’15 [18] Ans: 1 9) GE JAN ’16 [2] Ans: 2 10) GE JUN ’15 [16] Ans: 2 11) GC JUN '15 [8] Ans: 1 12) GC JUN '16 [10] Ans: 2 13) GC AUG '15 [15] Ans: 3 14) GC AUG ’16 [23] Ans: 1 15) GE AUG ’11 [29]
Ans: 180 80 502− = °
16) GE AUG ’14 [32] Ans: m 86 2 172GFE = × = °
m 180 86 94F∠ = − = ° 17) GC JAN '16 [26]
Ans: 120° B C∠ ≅ ∠ (alternate interior angles) A B∠ ≅ ∠ (base angles of isosceles
triangle formed by two radii)
13.3 Tangents ~ Page 338
Practice Problems
1. Perimeter = 78.
2. ABC△ is a right triangle.
2 2 2
2( ) 8 15( ) 289
17
ACAC
AC
= +==
AD is a radius. 17 8 9CD AC AD= − = − =
3. m 180 54 126RS = − = ° . 4. The measure of minor arc DC is 360 246 114− = ° . m 180 114 66DEC∠ = − = ° .
63
Regents Questions
1) GE JAN ’12 [8] Ans: 4 2) GE AUG ’12 [14] Ans: 2 3) GE JAN ’14 [19] Ans: 3 4) GE JAN ’14 [28] Ans: 4 5) GE JUN ’15 [13] Ans: 2 6) GC JUN '15 [20] Ans: 1 7) GC AUG '15 [12] Ans: 3 8) GC JAN '16 [21] Ans: 3
9) GE JAN ’13 [30] Ans:
10) GC AUG ’16 [25]
Ans: 38 56 21⋅ =
13.4 Secants ~ Page 347
Practice Problems
1. (a) tangent (b) secant (c) diameter (d) chord
2.
2
2
2
( ) ( )( )( ) (4)(16)( ) 64
8
PA PB PCPAPA
PA
====
3. 86 44 212
x −= = ° 4. 140 m30
2CD−= , so m 80CD = ° .
m 360 (140 80) 140DE = − + = ° . 5.
( )( ) ( )( )
(3)(12) ( )(9)4
BC AC EC DCEC
EC
===
6.
2
23( 18)3 54
x xx x
= += +
2 3 54 0
( 9)( 6) 09
x xx x
x
− − =− + =
=
64
Regents Questions
1) GE JAN ’13 [25] Ans: 1 2) GE JUN ’13 [28] Ans: 1 3) GE AUG ’15 [25] Ans: 3 4) GC JAN ’17 [15] Ans: 2 5) GE JUN ’12 [37]
Ans: Let x = TM.
2( 2) (12)(2)
2 24 0( 6)( 4) 0
4
x xx xx x
x
+ =+ − =
+ − ==
RT = 6 + 4 = 10 2
2( ) (8)(18)( ) 144
12
PSPS
PS
===
6) GE AUG ’12 [38] Ans: 52, 40, 80. 360 – (56 + 112) = 192 14 192 48× = 56 48 52
2+ =
192 112 402− =
112 48 802+ =
7) GE JAN ’15 [33] Ans: 8 (6)(24)8 144
18
xxx
===
8) GC JAN ’17 [28]
Ans: 152 56 482− = °
65
13.5 Circle Proofs ~ Page 354
Practice Problems
1. Statements Reasons
Circle O, AB AC≅ Given
AB AC≅ (S) If two arcs are congruent, their chords are congruent
AO AO≅ (S) Reflexive Property OC OB≅ (S) All radii in a circle are congruent
AOC AOB≅△ △ SSS Alternately, AB AC≅ could be replaced with AOC AOB∠ ≅ ∠ (central angles
of congruent arcs are congruent), and then AOC AOB≅△ △ by SAS. 2.
Statements Reasons Circle Q, PQR ST⊥ Given
PQR PQR≅ (S) Reflexive Property PRS∠ , PRT∠ are right angles ⊥ segments form right angles PRS PRT∠ ≅ ∠ (A) All right angles are congruent
QR bisects ST If a radius is perpendicular to a chord, it bisects the chord
RS RT≅ (S) Definition of bisector PRS PRT≅△ △ SAS
PS PT≅ CPCTC
Regents Questions
1) MB JAN ’07 [32] Ans: (Givens omitted) m m 35RDB YDB∠ = ∠ = (The measure of an inscribed angle is half its intercepted arc).
BRD∠ and DYB∠ are right angles (an inscribed angle of a semicircle is a right angle).
BRD DYB∠ ≅ ∠ (all right angles are congruent). BD DB≅ (Reflexive property).
RBD YDB≅△ △ (AAS). Alternately, BR YD≅ (if two arcs are congruent, their chords are congruent), which leads to the triangles congruent by HL.
2) GE FALL ’08 [38] Ans: (Givens omitted) AD BC≅ (parallel chords intersect congruent arcs).
BDC ACD∠ ≅ ∠ (inscribed angles that intercept congruent arcs are congruent). AD BC≅ (congruent chords intersect congruent arcs).
DAC DBC∠ ≅ ∠ (inscribed angles that intercept the same arc are congruent).
ACD BDC≅△ △ (AAS).
66
3) MB JUN ’09 [34] Ans: (Givens omitted) AB DC≅ (parallel chords intersect congruent arcs). AB DC≅ (congruent chords intersect congruent arcs).
BEA CED∠ ≅ ∠ (vertical angles are congruent).
BAC CDB∠ ≅ ∠ (inscribed angles that intercept the same arc are congruent).
BAE CDE≅△ △ (AAS). BE CE≅ (CPCTC).
4) GE JUN ’11 [38] Ans: (Givens omitted) OA OB≅ (all radii in a circle are congruent). OP OP≅ (Reflexive property). OA PA⊥ and OB PB⊥ (a tangent is perpendicular to a radius at the point of tangency).
PAO∠ and PBO∠ are right angles (perpendicular lines form right angles).
PAO PBO∠ ≅ ∠ (all right angles are congruent).
AOP BOP≅△ △ (HL). AOP BOP∠ ≅ ∠ (CPCTC).
5) GE JAN ’14 [38] Ans: 2. The diameter of a circle is perpendicular to a tangent at the point of tangency. 4. An angle inscribed in a semicircle is a right angle. 5. All right angles are congruent. 7. AA Similarity. 8. Corresponding sides of similar triangles are in proportion. 9. The product of the means equals the product of the extremes.
6) GC FALL '14 [26] Ans: (Givens omitted) Chords BC and BD are drawn (auxiliary lines).
A A∠ ≅ ∠ , BC BC≅ (Reflexive property).
12m mBDC BC∠ = (an inscribed
angle is half the measure of its intercepted arc).
12m mCBA BC∠ = (an angle formed
by a tangent and chord measures half the intercepted arc). m mBDC CBA∠ = ∠ (substitution).
BDC CBA∠ ≅ ∠ (definition of congruence).
ABC ADB△ △ (AA similarity). AB ADAC AB
= (corresponding sides of
similar triangles are in proportion). 2AC AD AB⋅ = (product of the
means equals the product of the extremes).
67
7) GC AUG ’16 [35] Ans: (Givens omitted) Chords BC and BD are drawn (auxiliary lines).
A A∠ ≅ ∠ , BC BC≅ (Reflexive property).
12m mBDC BC∠ = (an inscribed
angle is half the measure of its intercepted arc).
12m mCBA BC∠ = (an angle formed
by a tangent and chord measures half
the intercepted arc). m mBDC CBA∠ = ∠ (substitution).
BDC CBA∠ ≅ ∠ (definition of congruence).
ABC ADB△ △ (AA similarity). AB ADAC AB
= (corresponding sides of
similar triangles are in proportion). 2AC AD AB⋅ = (product of the
means equals the product of the extremes).
13.6 Arc Lengths and Sectors ~ Page 360
Practice Problems
1.
90360 30
Lπ
= 7.5L π= feet
2.
40360 12
Lπ
= 43L π= m
3. Central angle = 180 – 150 = 30°
30360 34
Lπ
= 9L ≈ cm
4.
14360 36θ π
π= 140θ = °
5.
40360 36
Sπ
= 4S π= sq. units
6. Central angle = 360 – 45 = 315°
315360 16
Sπ
= 14S π= sq. cm
7. Area of sector:
90360 100
Sπ
= 25S π=
Area of triangle: 1
2 (10)(10) 50= Area of segment = 25 50π − sq. units
8. Area of sector:
120360 144
Sπ
= 48S π= sq. units
Area of triangle: 1
2 (12)(12)sin120 36 3° =
Area of segment = 48 36 3π − sq. units
68
Regents Questions
1) MB JUN ’07 [16] Ans: 3 2) A2 JUN ’12 [12] Ans: 3 3) GC JUN '15 [23] Ans: 2 4) GC AUG '15 [18] Ans: 3 5) GC JAN '16 [12] Ans: 3 6) GC JUN '16 [24] Ans: 3 7) GC AUG ’16 [19] Ans: 2 8) GC JAN ’17 [21] Ans: 4 9) MB AUG ’04 [26]
Ans:
360 2L
rθ
π=
°
9360 2 3954
Lπ
=⋅
621.1L ≈ miles 10) MB AUG ’05 [24]
Ans:
360 2L
rθ
π=
°
165360 2 2.4
Lπ
=⋅
6.9L ≈ meters
11) MB JUN ’05 [31] Ans:
360 2L
rθ
π=
°
247360 2 150θ
π=
⋅
94θ ≈ ° 12) GC FALL '14 [23]
Ans: 180 20m 80
2BOD −∠ = = °
2360Sr
θπ
=°
80360 36
Sπ
=
8S π= 13) GC JUN '15 [29]
Ans:
2360Sr
θπ
=°
12360 36θ π
π=
120θ = ° 13.7 Radians (–) ~ Page 368
Practice Problems
1.
a) 45180 4π π⋅ = rad
b) 3270180 2π π⋅ = rad
c) 5150180 6π π⋅ = rad
2.
a) 180 306π
π⋅ = °
b) 5 180 2254π
π⋅ = °
c) 3 180 1085π
π⋅ = °
3.
12 34
L π π= ⋅ = inches
4.
8 10π θ= ⋅ 45πθ = rad
5. 65 5r= r = 13 feet
6.
22Sr
θπ π
= 0.972 81
Sπ π
=
S = 39.285 cm2
69
Regents Questions
1) MB JUN ’08 [18] Ans: 3 2) MB JAN ’09 [10] Ans: 1 3) MB JUN ’09 [1] Ans: 3 4) A2 FALL ’09 [22] Ans: 4 5) MB JAN ’10 [6] Ans: 1 6) A2 JUN ’10 [2] Ans: 2 7) A2 JUN ’11 [25] Ans: 3 8) A2 JAN ’12 [20] Ans: 1 9) A2 JUN ’13 [2] Ans: 2 10) A2 JAN ’14 [27] Ans: 1 11) A2 JAN ’15 [26] Ans: 2 12) A2 JAN ’16 [11] Ans: 3 13) GC FALL '14 [4] Ans: 3 14) MB JUN ’10 [26]
Ans: 3418 r=
r = 24 cm
15) A2 JAN ’11 [29]
Ans: 1802.5 143.2π
≈ °
16) A2 JUN ’12 [32]
Ans: 216 3.8180π ≈
rad
17) A2 JUN ’16 [32]
Ans: 243
rππ =
r = 6 ft. 18) GC JUN '16 [29]
Ans:
4
4
A
A
ππ
= ⋅
=
13 6.58
4
B
B
π
π
= ⋅
=
Yes, both angles are equal. 13.8 Circles in the Coordinate Plane ~ Page 373
Practice Problems
1. (a) (0,0) and 10 (b) 2 23 ( 1) 10+ − = Yes.
2. (2,–3) and 6
3. (0,7) and 4 2 4. r = 3 2 2( 1) ( 2) 9x y− + + = 5.
2 2
2 2
2 2
2 2
2 2
6 4 12( 6 9) 4 12 9( 3) 4 21( 3) ( 4 4) 21 4( 3) ( 2) 25
x x y yx x y yx y yx y yx y
+ + − =+ + + − = +
+ + − =+ + − + = ++ + − =
Center is (–3,2), radius is 5
6.
2 2
2 2
2 2
2 2
2 2
2
2 10 12 10 1
( 2 1) 10 1 1( 1) 10 2( 1) ( 10 25) 2 25( 1) ( 5) 27
x x y yx x y yx x y yx y yx y yx y
− = − + +− + − =− + + − = +
− + − =− + − + = +− + − =
Center is (1,5), radius is 3 3
70
Regents Questions
1) MA JAN ’06 [25] Ans: 4 2) MB JUN ’09 [11] Ans: 3 3) A2 JUN ’10 [16] Ans: 2 4) A2 JAN ’14 [15] Ans: 3 5) GE JUN ’14 [28] Ans: 1 6) GE JAN ’15 [11] Ans: 2 7) GE JAN ’15 [14] Ans: 3 8) GE JUN ’15 [10] Ans: 1 9) GE JUN ’15 [15] Ans: 3 10) GE JUN ’15 [24] Ans: 2 11) GE AUG ’15 [2] Ans: 3 12) GE AUG ’15 [20] Ans: 2 13) A2 AUG ’16 [26] Ans: 4 14) GC JUN '15 [14] Ans: 2 15) GC AUG '15 [9] Ans: 3 16) GC JAN '16 [17] Ans: 4
17) GC JUN '16 [3] Ans: 2 18) GC JUN '16 [23] Ans: 1 19) GC AUG ’16 [16] Ans: 1 20) GC JAN ’17 [18] Ans: 1 21) GC JAN ’17 [22] Ans: 3 22) MB JUN ’08 [31]
Ans: ( 4,2) ( 8,4)− → −
81 9= ; new radius is 18 23) GC AUG ’16 [30]
Ans: Yes. 2 2 2
2 2 2( 1) ( 2) 4
(3.4 1) (1.2 2) 45.76 10.24 16
x y− + + =− + + =
+ =
71
14. SOLIDS 14.1 Volume ~ Page 382
Practice Problems
1. 6 8 4 192V Bh= = ⋅ ⋅ = ft3 2. 3 34 43 3 (9) 972V rπ π= = = m3
3. 2 2(4) (10) 502.65V r hπ π= = ≈ in3 4. 1 13 3 (10)(8)(6) 160V Bh= = = in3
5. 3 8 24V = ⋅ = cm3 6. 2 21 13 3 (3) (8) 24V r hπ π π= = = in3
7.
2
3 2
( 2)( 1)(2 )( 2)(2 )2 2 4
x x xx x xx x x
− += − −= − −
8. 3 135 (25)(6) 175V = + = cm3
9. (20)(15)(10) 3,000containerV = = in3 2(2) (5) 20cupV π π= = in3 3,000 20 47.7π÷ ≈ 47 cups
Regents Questions
1) IA FALL ’07 [12] Ans: 4 2) GE FALL '08 [8] Ans: 3 3) IA JUN ’08 [9] Ans: 2 4) GE JUN '09 [4] Ans: 4 5) IA AUG ’11 [5] Ans: 3 6) IA AUG ’12 [24] Ans: 4 7) IA JAN ’13 [12] Ans: 3 8) GE AUG '14 [28] Ans: 1 9) GC AUG '15 [21] Ans: 4 10) GC JAN '16 [4] Ans: 2 11) GC JAN '16 [7] Ans: 2 12) GC JAN '16 [14] Ans: 3 13) GC JUN '16 [6] Ans: 4 14) GC AUG ’16 [20] Ans: 4 15) GC JAN ’17 [11] Ans: 2 16) GC JAN ’17 [24] Ans: 1 17) IA JAN ’09 [36]
Ans: 2V r hπ=
342 9 hπ= 38hπ
=
38 3636 2.9738
ππ
÷ = ≈
2 cans
18) IA AUG ’09 [32] Ans: (30)(12)(16) (12)(6)(9)5,760 648 5,112
− =− =
19) GE AUG '09 [30] Ans: 21
3 (12) (42) 2,016V = = 20) GE JUN '10 [34]
Ans: 13288 (8)(6)h=
288 16h= h = 18 cm
21) IA JUN ’12 [37] Ans:
Mike: (5)(3.5)(7) 122.5
prismV BhV
== =
Carol: 2
2(2.5) (7) 137.4cylinderV r h
V
π
π
=
= ≈
Carol’s, by 14.9 22) IA JUN ’13 [32]
Ans: 2(6.5) (24) 1,014V π π= =
72
23) GC JUN '16 [32]
Ans: 2(11.25) (33.5) 57.7
231π ≈
24) GC JUN '16 [36] Ans: Similar triangles are required to model and solve a proportion. Let x = height of the rest of the cone height of cone height below glassradius of top radius of bottom
=
51.5 1x x+ = x = 10
h = 10 + 5 = 15 glass cone below glassV V V= −
2 21 13 3(1.5) (15) (1) (10)24.9
V π π= −≈
25) GC JAN ’17 [34] Ans:
21 13 3
231.416 2
31.416 52
(25)(13) 340
C rr
r
V r h
ππ
ππ π
==
= ≈
= = ≈
14.2 Density ~ Page 392
Practice Problems
1.
33
11.34 g100 cm1 cm
1,134 g
W VD=
= ⋅
=
2.
33
89.6 g 8.96 g/cm10 cm
WDV
= = =
3.
33
94.44 g 7.87 g/cm12 cm
WDV
= = =
4.
9.8 14 mL0.7
WVD
= = =
5.
31 13 3(24)(10) 80 inV Bh= = =
380 inW =
2 g⋅
31 cm
3 3(2.54) cm⋅31 in
1 kg⋅1000 g
2.2 lbs1 kg
⋅
5.8 lbs≈
73
Regents Questions
1) GC JUN '15 [7] Ans: 3 2) GC AUG '15 [16] Ans: 1 3) GC JAN '16 [19] Ans: 2 4) GC JUN '16 [18] Ans: 2 5) GC JUN '16 [20] Ans: 1 6) GC AUG ’16 [17] Ans: 2 7) GC FALL '14 [6]
Ans: 5.1 10.2 20.3 1,056.006× × = cm3 500 1,056.006 528,003× = cm3
33 3
31m528,003cm 0.528003m
100cm× =
33
1920 kg 0.528003m 1013kgm
× ≈
No, the weight of the bricks is greater than 900 kg
8) GC FALL '14 [25] Ans:
1m25cm 0.25m100cm
r = × =
2(0.25) (10) 0.625V π π= = m3 3
3380 kg0.625 m 746.1kg
mW π= × ≈
$4.75 746.1 $3,544× ≈ per tree $50,000 14.1$3,544
≈
Need to sell 15 trees 9) GC JUN '15 [35]
Ans: tan 478.5x° = 9.115x ≈
213 (8.5) (9.115) 689.6coneV π= ≈
2(8.5) (25) 5674.5cylinderV π= ≈
( )31 42 3 (8.5) 1286.3hemisphereV π= ≈
3689.6 5674.5 1286.37650ft
towerV ≈ + +≈
33
62.4 lb85% 7650ft 405,756lbft
× × =
No, the weight exceeds 400,000 lbs
10) GC AUG '15 [25]
Ans: 33
137.8 0.638g/cm6
≈ ; Ash
11) GC AUG '15 [36] Ans:
( )2 3313 2 (8) 18.85inconeV π= ≈
Total 3100 18.85 1885in≈ × ≈ (0.52 1885) $0.10 $98.02× × = (100 $1.95) ($98.02 $37.83)
$59.15× − +
=
12) GC JAN '16 [30] Ans: Dish A
240,000 19.6(25.5)Ad
π= ≈
272,000 16.3(37.5)Bd
π= ≈
13) GC AUG ’16 [36] Ans:
( ) ( )2 38.3 8.31 1 43 2 2 3 2
3
(10.2)
183.961 149.693 333.65 cm
V π π= + ⋅
≈ + ≈
3333.65 50 16,682.7cm× = 16,682.7 0.697 11,627.8g× = 11,627.8 3.83 $44.53× =
14) GC JAN ’17 [36] Ans:
2 2(26.7) (750) (24.2) (750)95,437.5
cylinderV
π ππ
= −=
33
2.7 1 $0.3895,437.51000 11
$307.62
g kgcmg kgcm
× × ×
=
2 2(40 )(750) (35 )(750) 281,250prismV
= − =3
32.7 1 $0.38281,250
1000 11$288.56
g kgcmg kgcm
× × ×
=Prisms cost less. 307.62 288.56 $19.06− =
74
14.3 Lateral Area and Surface Area ~ Page 400
Practice Problems
1.
2.
3.
2
22 5 11110 ft
LA rhππ
π
== ⋅ ⋅=
2
2
250 110160 502.7 ft
SA r LAππ ππ
= += += ≈
4. 3 64 4s = = 26 4 96SA = ⋅ = in2
5. 2V r hπ= , so 2Vhrπ
=
Radius r Height h Surface Area SA
2 2144 36
4Vhr
πππ
= = = 22 2
8 144 152SA r rhπ π
π π π= += + =
4 2144 916
Vhr
πππ
= = = 22 2
32 72 104SA r rhπ π
π π π= += + =
6 2144 436
Vhr
πππ
= = = 22 2
72 48 120SA r rhπ π
π π π= += + =
The cylinder with the radius of 4 has the least surface area.
Regents Questions
1) IA JUN ’08 [27] Ans: 4 2) IA JAN ’10 [29] Ans: 4 3) IA AUG ’12 [16] Ans: 4 4) IA AUG ’13 [25] Ans: 2 5) GC JUN '15 [19] Ans: 2 6) IA AUG ’10 [35]
Ans: 310 2 4 80cmV = ⋅ ⋅ =
22 10 2 2 10 4 2 2 4136cm
SA = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅=
7) IA JUN ’11 [36] Ans: 2( 3)( 4)
2( 3)(5)2( 4)(5)
x xxx
+ −+ ++ −
2
2
2 2 2410 3010 40
2 18 34
x xxx
x x
− −+
+ −
+ −
8) IA JAN ’12 [31] Ans:
22(5.5)(3) 2(5.5)(6.75) 2(3)(6.75)
33 74.25 40.5 147.75cm+ +
= + + =
2
2 2 22 8 6 2 4 8 2 4 696 64 48 208 ft
SA lw hl hw= + += ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅= + + = 2
22 6 9108 ft
LA rhππ
π
== ⋅ ⋅=
2
2
272 108180 ft
SA r LAππ ππ
= += +=
75
14.4 Rotations of Two-Dimensional Objects ~ Page 406
Practice Problems
1. sphere
2. cylinder diameter = 10 in and height = 5 in 2 2(5) (5) 125V r hπ π π= = = in3
3. cylinder with a hemisphere on top
( )2 31 42 3
27 18 45
cylinder hemisphereV V
r h rπ ππ π π
+
= +
= + =
4. 12 inch radius sphere with a 6 inch radius spherical cutout in its center
3 34 43 3(12) (6)2,304 288 2,016
outer sphere cutoutV V
π ππ π π
−
= −= − =
5. cylinder with a cone on top The cone has a height of 24 – 12 = 12. Use the Pythagorean Theorem to find the radius of the cone: 2 2 212 13r + = 5r = The cylinder has a 2.5 radius and height of 12.
213 (5) (12) 100coneV π π= =
2(2.5) (12) 75cylinderV π π= = 175object cone cylinderV V V π= + = .
Regents Questions
1) MA JAN '04 [17] Ans: 4 2) GC JUN '15 [1] Ans: 4 3) GC AUG '15 [3] Ans: 4
4) GC JUN '16 [1] Ans: 3 5) GC AUG ’16 [3] Ans: 1
14.5 Cross Sections ~ Page 411
Practice Problems
1. (4) 2. (2) 3. (1) 4. (4) 5. (1) 6. (a) pentagon (b) rectangle
Regents Questions
1) GC JUN '15 [6] Ans: 2 2) GC JAN '16 [1] Ans: 1
3) GC AUG ’16 [13] Ans: 3 4) GC JAN ’17 [23] Ans: 4
76
14.6 Cavalieri’s Principle (–) ~ Page 416
Practice Problems
1. 3(4)(3)(6) 72 inV Bh= = = 2. 2 2 3(2) (5) 20 inV r hπ π π= = = Regents Questions
1) GC FALL '14 [18] Ans: Each quarter in both stacks has the same base area. Therefore, each corresponding cross-section of the stacks will have the same area. Since the two stacks of quarters have the same height of 23 quarters, the two volumes must be the same.
77
15. CONSTRUCTIONS 15.1 Copy Segments, Angles, and Triangles ~ Page 419
Practice Problems
1.
2.
3. SAS 4. AAS
Regents Questions
1) GE AUG '15 [32] Ans:
2) GC JAN '16 [34]
Ans: solutions vary, such as
SAS
3) GC AUG ’16 [32] Ans:
A C′ ′ is twice AC
78
15.2 Construct an Equilateral Triangle ~ Page 426
Practice Problems
1.
2.
Regents Questions
1) GE JUN ’10 [12] Ans: 1 2) GE JAN '12 [7] Ans: 1 3) GE JAN '13 [9] Ans: 3 4) GE AUG ’10 [32]
Ans:
5) GE JUN '13 [32] Ans:
79
15.3 Construct an Angle Bisector ~ Page 431
Practice Problems
1.
2.
3.
80
Regents Questions
1) GE JUN '09 [25] Ans: 3 2) GE AUG '12 [5] Ans: 2 3) GE JAN '15 [9] Ans: 2 4) GE JAN '11 [33]
Ans:
5) GE JAN '12 [33]
Ans:
6) GE JUN '12 [32]
Ans:
7) GE AUG '13 [30] Ans:
8) GE JUN '14 [37]
Ans:
9) GE JAN ’16 [34]
Ans:
81
15.4 Construct a Perpendicular Bisector ~ Page 436
Practice Problems
1.
2. Construct the perpendicular bisector of XY . Then, construct the bisector of each half.
Regents Questions
1) GE JAN '11 [20] Ans: 1 2) GE JUN '13 [5] Ans: 2 3) GE JAN '14 [2] Ans: 3 4) GE AUG '11 [30]
Ans:
5) GE JAN '14 [30] Ans:
82
6) GE AUG '14 [37] Ans:
7) GE JAN '15 [30]
Ans:
8) GE JUN '15 [32]
Ans:
9) GC AUG ’16 [28] Ans:
10) GC JAN ’17 [25]
Ans:
83
15.5 Construct Lines Through a Point ~ Page 443
Practice Problems
1.
2.
3. (a) 5 (b) 2 (c) 1 (d) 4 (e) 3 (f) 6 Regents Questions
1) GE JUN '12 [8] Ans: 2 2) GE JUN '15 [12] Ans: 2 3) GE AUG '15 [12] Ans: 3 4) GE AUG '12 [33]
Ans:
5) GE JAN '13 [33]
Ans:
6) GC FALL '14 [9] Ans: The altitude is perpendicular to the extended side.
7) GC JUN '16 [31]
Ans: The tangent is OA⊥ .
84
15.6 Construct Tangents to a Circle (+) ~ Page 451
Practice Problems
1. a) Draw AB
.
b) Construct CD
, the perpendicular bisector of AB , and label the midpoint of AB point E.
c) Place the tip of the compass on E, set the compass to the length of AE , and mark points F and G on the circle.
d) Draw tangents BF
and BG
.
15.7 Construct Inscribed Regular Polygons ~ Page 453
Practice Problems
1.
2.
3.
a) Construct two perpendicular diameters (as if constructing an inscribed square).
b) Construct angle bisectors of the 90° central angles, forming eight 45° central angles.
c) The endpoints of the four diameters are the vertices of the regular octagon. Draw the chords between them.
85
Regents Questions
1) GC FALL '14 [12] Ans:
The sides of the square are four congruent chords in the circle, so they intercept four congruent arcs. Each arc therefore measures one-fourth of 360°, or 90°. Therefore, an arc intercepted by two adjacent sides measures 2 90 180× ° = ° .
2) GC JUN '15 [25] Ans:
3) GC AUG '15 [26] Ans:
4) GC JAN ’17 [33]
Ans:
COF is a diameter, so FBC∠ is an inscribed angle of a semicircle, and is therefore a right angle. This means FBC△ is a right triangle.
86
15.8 Construct Points of Concurrency ~ Page 458
Practice Problems
1.
2.
15.9 Construct Circles of Triangles ~ Page 460
Practice Problems
1. (a) 2 (b) 1 (c) 5 (d) 4 (e) 3 Regents Questions
1) GE AUG '09 [25] Ans: 4 2) GE AUG '10 [28] Ans: 1
