Part 1

Chapter 2
17. p = 205 + 322 + 415
p = 942 cm
GEOMETRY
21.
2 2
c = 13.8 + 22.7 = 26.6 ft
2.1 Lines and Angles
1. ∠ ABE = 90
°
5. ∠EBD
and ∠ DBC are acute angles.
° °
9. The complement of ∠ CBD = 65 is 25 .
25. ∠ B = 90 − 23 = 67
29.
° ° °
A'
A/2
A
B
B/2
D
A/2
C'
C
13. ∠ AOB = 90 + 50 = 140
° ° °
° ° °
17. ∠ 1 = 180 − 145 = 35 =∠ 2 =∠ 4
° ° °
21. ∠ 3= 90 − 62 = 28
25. ∠ DEB = 44
29.
°
a 3.05 3.05
= ⇒ a = 4.75⋅ = 4.53 m
4.75 3.20 3.20
ΔADC ∼ ΔA' DC ' ⇒ DA' C ' = A /2
between bisectors = BA'
D
B
°
Δ BA' C ', + ( BA' D + A /2)
= 90
2
° ⎛ A B ⎞
from which BA' D = 90 − ⎜ + ⎟
⎝ 2 2 ⎠
° ⎛ A+
B⎞
or BA' D = 90 − ⎜ ⎟
⎝ 2 ⎠
33. ∠ BCD = 180 −47
= 133
° °
°
33.
D
B
37.
°
1+ 2+ 3=
180 ,
( 1, 2, and 3 form a straight line)
A
2 1
C
2.2 Triangles
1. ∠ 5= 45 ⇒∠ 3=
45
° °
∠ 2 = 180 −70 − 45 = 65
° ° ° °
A+ B
= 90
1+ B
= 90
⇒ A
= 1
redraw Δ BDC as
°
°
2
B
5. ∠ A = 180 −84 − 40 = 56
° ° ° °
C
1
D
1 1 7.6 2.2 8.4 ft
2 2
2
9. A= bh = ( )( ) =
1 1 3.46 2.55 4.41 ft
2 2
2
13. A= bh = ( )( ) =
1+ 2=
90
1+ B
= 90
⇒ 2
= B
and Δ ADC as
°
°
12

Section 2.3 Quadrilaterals 13
2
C
53. Redraw Δ BCP as
B
A
1
ΔBDC
and Δ ADC are similar.
37. Since MKL ∼ MNO; KN = KM −MN; 15 − 9 = 6
KM LM 6 LM
= KM ; = ; = ; 9LM = 72; LM = 8
MN MO 9 12
41.
45.
( )
2 76.6 + 30.6
s = = 91.9
2
A = 91.9 91.9 −76.6 91.6 −30.6
2
A = 1150 cm
Wall
2
( ) ( )
20
8
Ladder
Floor
2 2
distance up wall = 20 − 8 = 336 = 18.3 ft
D
P
Δ APD is
P
12.0 - PD
6.00
C
A
10.0
D
6.00 10.0
from which ΔBCP
∼ Δ ADP, so = 12.0 − PD PD
⇒ PD = 7.50 and PC = 12.0 − PD = 4.50
l = PB+ PA= 4.50 + 6.00 + 7.50 + 10.0
l = 20.0 ft
2.3 Quadrilaterals
2 2 2 2
49.
1.
5. p s ( )
= 4 = 4 65 = 260 m
4.5 5.4
=
z 1.2 + z
z = 6.0 m
x
y
= z + 4.5
2 2 2
x = 7.5 m
( )
2 2 2
= 1.2 + 6 + 5.4
y = 9.0 m
9. p l w ( ) ( )
13.
= 2 + 2 = 2 3.7 + 2 2.7 = 12.8 m
A= s = 2.7 = 7.3 mm
2 2 2
2
17. A= bh = 3.7( 2.5) = 9.3 m
21. p = 2b+
4a
2.5 H d
25. The parallelogram is a rectangle.
H
1.25 1.25

14 Chapter 2 GEOMETRY
29. The diagonal always divides the rhombus into two
congruent triangles. All outer sides are always
equal.
33.
13. A πr
π ( )
= = 0.0952 = 0.0285 yd
2 2 2
17. ∠ CBT = 90 −∠ ABC = 90 − 65 = 25
21. ( )
° ° ° °
ARC BC = 2 60 = 120
° °
37.
w+ 2.5 = 4w−4.7
w = 2.4 ft
4w
= 9.6 ft
1.74
1.46
d
1.86
2.27
d = 2.27 + 1.86
2 2
° ⎛ π ⎞
25. 022.5 ⎜ 0.393 rad
° ⎟ =
⎝180
⎠
1 r
P = 2 r + 2 r = π + 2 r
4 2
29. ( π )
33. All are on the same diameter.
37. C πr
π ( )
= 2 = 2 3960 = 24,900 mi
41. c = 112; c = π d; d = c/ π = 112 / π = 35.7 in.
3
45. A of room = A of rectangle + A of circle
4
3
A = 24( 35) + π ( 9.0) 2
4
2
A = 1000 ft
1
For right triangle, A = ( 2.27)( 1.86)
2
1.46 + 1.74 + d
For obtuse triangle, s =
2
and A= s( s−1.46)( s−d)( s−1.74)
A of quadrilateral = Sum of areas of two triangles,
1
A= ( 2.27 )( 1.86 ) + s( s− 1.46 )( s−d)( s−
1.74 )
2
2
A = 3.04 km
_____________________________________
2.4 Circles
1. ∠ OAB + OBA +∠ AOB = 180
∠ OAB + 90 + 72 = 180
°
∠ OAB = 18
° ° °
5. (a) AD is a secant line.
(b) AF is a tangent line.
°
_____________________________________
2.5 Measurement of Irregular Areas
1. The use of smaller intervals improves the approximation
since the total omitted area or the total
extra area is smaller.
5. A = ⎡ + ( ) + ( ) + ( ) + ( )
A
trap
trap
2.0 0.0 2 6.4 2 7.4 2 7.0 2 6.1
2
⎣
⎤⎦
⎡⎣+ 25.2 ( ) + 25.0 ( ) + 25.1 ( ) + 0.0⎤⎦
2
= 84.4 = 84 m to two significant digits
9. A = ⎡ + ( ) + ( ) + ( ) + ( )
A
trap
trap
0.5 0.6 2 2.2 2 4.7 2 3.1 2 3.6
2
⎣
⎤⎦
⎡⎣+ 2( 1.6) + 2( 2.2) + 2( 1.5)
+ 0.8⎤⎦
2
= 9.8 m
9. c πr
π( )
= 2 = 2 275 = 1730 ft

Chapter 2 Review Exercises 15
13. Atrap
= [ + ( ) + ( ) + ( ) + ( )
2
+ 2( 350) + 2( 330) + 2( 290) + 230]
17.
A
trap
45 170 2 360 2 420 2 410 2 390
= 120,000 ft
]
2
Atrap
0.500
= 0.0 2 1.732
2
⎡⎣ + + 2 2.000 + 2 1.732
+ 0.0
2
= 2.73 in.
( ) ( ) ( )
2
This value is less than 3.14 in. because all of the
trapezoids are inscribed.
_____________________________________
2.6 Solid Geometric Figures
1. ( )( )( )
V1 = lwh1, V2 = 2l w 2h = 4lwh = 4V1
The volume is four times as much.
3
33. V = πr = π ( d /2)
37.
4 4
3
3 3
4
3
= π ( 165 / 2)
3
6 3
= 2.35×
10 ft
29.8
c = 2π
r = 29.8⇒ r =
2π
3
4 3 4 ⎛29.8⎞
= πr
= π⎜ ⎟
V
3 3 ⎝ 2π
⎠
3
V = 447 in.
Chapter 2 Review Exercises
1. ∠ CGE = 180 − 148 = 32
5.
9.
2 2
c = 9 + 40 = 41
° ° °
2 2
c = 6.30 + 3.80 = 7.36
5.
V = e = 7.15 = 366 ft
3 3 3
13. P s ( )
= 3 = 3 8.5 = 25.5 mm
4 4
= = 0.877 = 2.83 yd
3 3
9. V πr
π ( )
3 3 3
= 1 = 1 76 130 = 250,000 in.
3 3
13. V Bh ( )( )
2 3
3
3 3
1 4 2 0.83
17. V = ⎛ ⎜ πr
⎞ ⎟= π
⎛ ⎜ ⎞
⎟ = 0.15 yd
2⎝3 ⎠ 3 ⎝ 2 ⎠
3 3
4 3 4 ⎛d
⎞ 4 d
21. V = πr
= π⎜
⎟ = π
3 3 ⎝ 2⎠
3 8
1 3
V = π d
6
25.
( r) 2
final surface area 4π
2 4
= =
2
original surface area 4π
r 1
2
29. V = πr h= π ( d / 2) 2 h = π ( 4.0 / 2) 2
( 3,960,000)
= 5.0×
10 ft or 0.00034 mi
7 3 3
17. C πd
π( )
= = 98.4 = 309 mm
1 26.0 34.0 14.0 6190 cm
2
3
21. V = Bh = ( )( )( ) =
25. A e ( )
29.
= 6 = 6 0.520 = 1.62 m
2 2
°
50
∠ BTA = = 25
2
33. ∠ ABE = 90 − 37 = 53
° ° °
2
37. ( ) π ( )
°
1
= + + + = + + + π
2
2 2 2 2
P b b 2a 2a b b 4a a
41. A square is a rectangle with four equal sides and a
rectangle is a parallelogram with perpendicular
intersecting sides so a square is a parallelogram.
A rhombus is a parallelogram with four equal sides
and since a square is a parallelogram, a square is
a rhombus.
15

16 Chapter 2 GEOMETRY
45.
49.
53.
57.
B
a
E
A
d
C
b
c
BEC = AED, vertical ' s.
BCA = ADB, both are inscribed in AB
CBE = CAD, both are inscribed in CD
a b
which shows ΔAED
∼ ΔBEC
⇒ =
d c
2 2
L = 1.2 + 7.8 = 7.9 m
AB 42
=
38 54
38( 42)
AB = = 30 m
54
The longest distance in inches between points on
the photograph is,
2 2
8.00 10.0 12.8 in. from which
+ =
x 18,450
=
12.8 1
⎛ 1 ft ⎞⎛ mi ⎞
x = ( 12.8)( 18,450 ) in. ⎜ ⎟⎜ ⎟
⎝12 in. ⎠⎝5280 ft ⎠
x = 3.73 mi
D
2 1 4 3
69. V = πr h+ ⋅ πr
2 3
2 3
⎡ ⎛2.50 ⎞ ⎛ 2.50 ⎞ 1 4 ⎛2.50
⎞ ⎤
= ⎢π⎜ ⎟ ⎜4.75
− ⎟+ ⋅ ⋅π⎜ ⎟ ⎥
⎢⎣
⎝ 2 ⎠ ⎝ 2 ⎠ 2 3 ⎝ 2 ⎠ ⎥⎦
⎛7.48 gal ⎞
⎜ 3 ⎟
⎝ ft ⎠
= 159 gal
73. Label the vertices of the pentagon ABCDE. The
area is the sum of the areas of three triangles, one
with sides 921, 1490, and 1490 and two with sides
921, 921, and 1490. The semi-perimeters are given
by;
921+ 921+
1490
s1
= = 1666 and
2
921+ 1490 + 1490
s2
= = 1950.5.
2
( )( )( )
A = 2 1666 1666 −921 1666 −921 1666 −1490
( )( )
+ 1950.5 1950.5 −1490 1950.5 −1490
= 1, 460, 000 ft
( 1950.5 921)
+ −
2
2
1.0
= 4.0 8.0 −2π
⋅ = 30 ft
4
61. A ( )( )
2
2 4.3
3
= π = π ⎛ ⎜
⎞
⎟ 13 = 190 m
⎝ 2 ⎠
65. V r h ( )
2