Download Section 8 – 1 Lines and Angles Point is a specific location in space

Basic College Mathematics (ALEKS)
Section 8 – 1
Chapter Eight GEOMETRY
Lines and Angles
Point is a specific location in space.
.
Line is a straight path (infinite number of points).
…A……B…..
Line Segment is part of a line between TWO points.
A…………B
Ray is part of the line that includes an ENDPOINT.
A………B…..>
∙
Point A
Line AB (no Endpoints)
Ray with Endpoint A
Line Segments with Endpoints A and B
Angle is formed by TWO Rays with a common Endpoint. (
Rays: and Vertex is the common Endpoint. Vertex = B
Angle ABC, Angle CBA, Angle B,
ABC,
Circle = 360°
1
Chapter 8
CBA
)
Basic College Mathematics (ALEKS)
Right Angle = 90°
Box at vertex shows 90°
Straight Angle = 180°
Acute Angle is > 0° and < 90° (small)
Obtuse Angle is > 90° and < 180° (Fat)
Congruent Angles has the same measure.
Complementary Angles measures add to 90°.
Supplementary Angles measures add to 180°.
What is the supplement of 35°?
180° − 35° = 145°
What is the complement of 52°?
90° − 52° = 38°
Parallel Lines never intersect.
2
Chapter Eight GEOMETRY
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
Two intersecting Lines form four angles.
A and
C are vertical angles.
m
A=m
C
B and
D are vertical angles.
m
B=m
D
Adjacent Angles share side (Ray).
A is adjacent to
B and
D
Perpendicular Lines intersect at 90°.
When a Line intersects two Parallel Lines, Eight Angles are formed.
1=
4=
Section 8 – 2
5=
8 and
2=
3=
7
Triangles and the Pythagorean Theorem
Triangle is a three sided figure.
The sum of the angles of a triangle = 180°.
3
6=
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
What is measure of angle a?
90 + 43 + X = 180
133 + X = 180
X = 180 – 133 = 47°
What are the measures of angles a and b?
100 + a = 180
a = 80
39 + 80 + b = 180
119 + b = 180
b = 180 – 119 = 61
Names of Triangles:
Acute – all angles acute.
Right – has a right angle.
Obtuse – has an obtuse angle.
Equilateral – all three sides equal.
Isosceles – two sides equal.
Scalene – NO sides equal.
SQUARE ROOTS
Square of a number = that number times itself.
N2 = N x N
Reverse is the Square Root.
4
Chapter 8
Basic College Mathematics (ALEKS)
√ = 02 = 0 √0 = 0
12 = 1 √1 = 1
22 = 4 √4 = 2
32 = 9 √9 = 3
42 = 16 √16 = 4
52 = 25 √25 = 5
62 = 36 √36 = 6
72 = 49 √49 = 7
Chapter Eight GEOMETRY
√ x √ = N
82 = 64 √64 = 8
92 = 81 √81 = 9
102 = 100 √100 = 10
112 = 121 √121 = 11
122 = 144 √144 = 12
132 = 169 √169 = 13
142 = 196 √196 = 14
152 = 225 √225 = 15
PYTHAGOREAN THEOREM AND APPLICATIONS
Right Triangles have a special property.
Side a and side b are called legs and are on either side of the 90° angle.
Side c is the Hypotenuse and is opposite the 90° angle.
(Leg)2 + (Leg)2 = (Hypotenuse)2
a2 + b 2 = c2
5
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
4 2 + 3 2 = c2
16 + 9 = c2
25 = c2
√25 = √ 5=c
? = a or b 26 = c
a2 + 242 = 262
a2 + 576 = 676
a2 = 100
= √100
a = 10
The bottom of a 17 foot ladder is
placed 8 feet from the building.
How far up is the ladder on the
building?
c = 17, a = 8, b = building
82 + b2 = 172
64 + b2 = 289
b2 = 225
√ = √225
b = 15
Section 8 – 3
Quadrilaterals, Perimeter, and Area
Quadrilaterals:
6
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
Rectangle = Box with four right angles.
Opposite sides are equal, adjacent sides are not.
Square = Rectangle with all side equal.
Parallelogram = four sided figure with two sets of parallel sides.
Rhombus = parallelogram with all sides equal.
Trapezoid = four sided with ONE set of parallel sides.
Perimeter:
Square = S + S + S + S = 4S
7
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
Rectangle = L + W + L + W = 2L + 2W
What is the perimeter of a square
with a side of 4.3 inches?
P = 4(4.3 inches) = 17.2 inches
Find the perimeter of a rectangle
with a length of 3 yards and a
width of 2 feet?
P = 2(3 yards)+2(2 feet)
P = 6 yards + 4 feet
P = 18 feet + 4 feet = 22 feet
Area:
Rectangle = L x W
Square = S x S = S2
Parallelogram = Base x Height = B x H
Triangle =
!"
#"$%ℎ' = #
Note: Height can drawn and measured
outside the figure (parallelogram too).
Trapezoid =
8
(!)* +, -.//"/ !$0"!1 #"$%ℎ' =
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
( + 1 #
Find the area of a parallelogram
with 10.4 in and height of 3.1 in?
A = BxH = (10.4 in)x(3.1 in)
A = 32.24 in2
A side of a house must be painted.
The length = 40 feet and the width
= 12 feet. There are two 4 by 4
foot windows (not painted). What
is the area to be painted?
A = (40)(12) – 2(4)(4)
A = 480 – 32 = 448 ft2
Figure is triangle.
B = 7 m and H = 4 m
A=
#
A=
7*
A =14 m2
What is the area of this figure?
9
Chapter 8
4*
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
Figure is a trapezoid.
A = 11 cm, B = 15 cm, H = 6 cm
Area = ( + 1 #
What is the area of this
Figure?
A = (11 + 1516 = (261(61
2
A = 78 cm
Find the area of the shaded
Region?
Looks like a rectangle with a triangle cut
out of the end. So, area of the rectangle
minus the area of the triangle.
AR = 15x10 = 150 ft2
AT = (101(61 = 30 ft2
A = 150 – 30 = 120 ft2
Paint the deck. One gallon covers
160 ft2 and cost $ 11.95. What is
the
cost to paint the deck?
Section 8 – 4
Looks like a rectangle with a
trapezoid on the end.
AR = 31ft x 8 ft = 248 ft2
AT = (8 + 161(61 = 72 ft2
A = 248+72 = 320 ft2
DEFFGH
320 ft2 x
M = 2 gallons
IJ KL
$.OP
2 gal x
DEF
= $23.90
Circles, Circumference, and Area
Definitions:
Circle – all points located the same distance from a fixed point.
10
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
Center – that fixed point.
Radius (r) – a line from the center to the circle.
Diameter (d) – a line from the circle through the center on to the
circle.
.=
2r = d
Q
Circumference ( C ): - distance around a circle (perimeter).
Π = (Pi – Greek Letter) – if left as the symbol it is exact.
Π = 3.1415926535… (not exact – rounded)
Π = 3.14 or
R
will normally be used.
C=ST
C = 2πR
The line is the diameter.
d =10.4 m
J.U V
r=
= 5.2 *
What is the radius?
11
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
The line is the radius.
P
r = X0
W
P
P
W
U
d = 2( X01 =
X0=1 yd
U
What is the diameter?
C = πD
a)
C = π10 = 10π ft (exact answer)
b)
C = 3.14(10) = 31.4 ft (approximate)
Find the circumference?
a) exact answer in S
b) approximate answer using 3.14
C = 2πR
R = 4.7 cm
a) C = 2(π)(4.7 cm) = 9.4π cm
b) C = 2(3.14)(4.7 cm) = 29.516 cm
Find the circumference?
a) exact answer in S
b) approximate answer using 3.14
AREA:
A = πR2 = πRxR
Y
Y[Y
U
A = π( )2 = π(
12
)
(rarely used)
Chapter 8
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
21 in = radius
A = πR2 = π(R)(R)
a)
A = π(21 in)(21 in) = 441π in2
b)
Find the area of the circle?
[ [ A=
= 1386 in2
a) exact
b) S =
R
R
(approximate)
9.4 in = diameter
Y
O.U \H
R= =
= 4.7 $]
a)
A = π(4.7 in)(4.7 in)
A =22.09π in2
Find the area of the circle?
b)
a) exact
A=(3.14)(4.7)(4.7)=69.3 in2
b) S = 3.14 (approximate)
A = 69 in2
round to nearest whole unit
A rectangle with two half
circles
Removed. Find perimeter and
Area, use π = 3.14
Section 8 – 5
Perimeter is two staright lines and two
half circles. Two half circle = one
circle. C = πD A=πRR
P = 2(12 cm) + πD
P = 24 cm + (3.14)(6) cm
P = 24 cm + 18.84 cm = 42.84 cm
Area of rectangle – Area Circle
A=(12)(6) – (3.14)(3)(3)
A= 72 – 28.26 = 43.74 cm2
Volume
1 cm3 = 1 cc = 1ml 1 in3 (volume has units X3) “Cubic”
13
Chapter 8
Basic College Mathematics (ALEKS)
Rectangular Solid
V = LxWxH
Cube
V = S3 = SxSxS
Right Circular Cylinder
V = πR2H
Right Circular Cone
V = S_ #
^
Sphere
U
V = S_ ^
^
14
Chapter 8
Chapter Eight GEOMETRY
Basic College Mathematics (ALEKS)
Chapter Eight GEOMETRY
What is the volume of a rectangular V = L x W x H
Solid 8 in, by 11 in by 2 in?
V = (8 in) x (11 in) x (2 in)
V = 176 in3
Find the volume of a right cylinder V = πR2H
with π = 3.14, Radius = 2 in, and
V = (3.14)(2 in)(2 in) (4.5 in)
Height = 4.5 in?
V = 56.52 in3
Find the volume of a sphere. Use
π = 3.14 and Radius = 3 cm.
4
` = S_ ^
3
4
` = (3.141(3 1(31(31*^
3
V = 113.04 cm3
Find the volume of a cone.
Use π = 3.14, diameter = 8 cm and
18 cm height.
=
= 4 *
1
` = S _ #
3
V = (3.141(4cm)(4cm)(18cm)
^
V = 301.44 cm3
15
R=
Chapter 8
Y
W aV
Basic College Mathematics (ALEKS)
.U VV
Chapter Eight GEOMETRY
= 1.2 **
a)
R=
b)
R=
c)
Vouter – VInner = (3.14)(1.2)(1.2)(2) – (3.14)(0.8)(0.8)(2)
.I VV
= 0.8 **
V = 9.0432 – 4.0192 = 5.0240 mm3 ≈ 5 mm3
16
Chapter 8