section 10.1 points, lines, planes, and...
TRANSCRIPT
2/13/2012 Section 10.1 1
Section 10.1
Points, Lines, Planes, and Angles
Objectives
1. Understand points, lines, and planes as the basis of
geometry.
2. Solve problems involving angle measures.
3. Solve problems involving angles formed by parallel
lines and transversals.
2/13/2012 Section 10.1 2
Defining Points, Lines and Planes
• Point
– Represented as a small dot
– Has no length, width or thickness
• Line
– Connecting two points along the shortest path
– Has no thickness and infinite length
• Plane
– Flat surface
– Has no thickness and no boundaries
2/13/2012 Section 10.1 3
Lines
• A line may be named using any two of its points.
• A half line is formed when a point divides a line
• A ray is a half-line with its endpoint included
• A line segment is a portion of a line joining two points.
• In the diagrams below, a closed circle indicates that the
point is included. An open circle indicates that the point
is not included.
2/13/2012 Section 10.1 4
Angles
• Angle
– Formed by the union of two rays.
– One ray is called the initial side.
– The other ray is called the terminal side.
2/13/2012 Section 10.1 5
Measuring Angles Using Degrees
• Angles are measured by the amount of rotation from the
initial side to the terminal side.
• Angles are measured in degrees, symbolized by º.
• There are 360º in a full rotation (circle).
• 1 degree is 1 of a complete rotation. 360
2/13/2012 Section 10.1 6
Example 1
Using Degree Measure
• The hour hand of a clock moves
from 12 to 2 o’clock. Through
how many degrees does it move?
• Solution: Moving from 12 to 2 o’clock is 2 or 1 12 6
of a complete revolution. Thus,
the hour hand moves,
606
360360
6
1
2/13/2012 Section 10.1 7
Classifying Angles by their Degree
Measurement
2/13/2012 Section 10.1 8
Protractor
• Protractors are used for finding the degree measure of an angle.
• We measure the angle by placing the center point of the protractor on the vertex of the angle and the straight side of the protractor along one side of the angle. We read the degree by reading where the other side of the angle intercepts the edge of the protractor.
• Choose the number based on whether the angle is obtuse or acute.
2/13/2012 Section 10.1 9
Special Pairs of Angles
• Complementary Angles are two angles whose
sum is 90º. To find the complement of an angle,
subtract it from 90º.
– The complement of 70º is:
90º - 70º = 20º.
• Supplementary Angles are two angles whose sum
is 180º.
– The supplement of 110º is:
180º - 110º = 70º.
2/13/2012 Section 10.1 10
Example 2
Angle Measures and Complements
• Find m DBC
• Solution:
• m DBC = 90º - 62º = 28º
2/13/2012 Section 10.1 11
Example 3
Angle Measures and Supplements
m ABD is 66º greater than
m DBC and they are
supplementary angles.
Find the measure of each angle.
Solution:
m DBC + m ABD = 180º
x + (x + 66º ) = 180º
2x + 66º = 180º
2x =114º
x = 57º
m DBC = 57º
m ABD=57º + 66º = 123º
2/13/2012 Section 10.1 12
Example 4
Vertical Angles
• When two lines intersect, the opposite angles formed
are called Vertical Angles.
• Vertical Angles are equal.
• The angle on the left measures 68º.
• Find the other angles.
• Solution:
From the figure we see that:
1 = 68º ( Vertical angles are equal )
1 + 2 = 180º ( Supplementary angles)
2 = 180º - 68º = 112º
3 = 2 = 112º (Vertical angles are equal)
2/13/2012 Section 10.1 13
Special Line Relationships
• Parallel Lines
– Lines that lie in the same
plane and have no points
in common.
• Intersecting Lines
– Two lines that are not
parallel and have a single
point in common.
• Transversal
– A line that intersects two
parallel lines.
2/13/2012 Section 10.1 14
Names of Angle Pairs formed by a
Transversal Intersecting Parallel Lines
Name Description Sketch Angle Pairs
Described
Property
Alternate interior
angles
Interior angles that
do not have a
common vertex on
alternate sides of
the transversal
3 and 6
4 and 5
Alternate interior
angles have the
same measure
. 3 = 6
4 = 5
Alternate exterior
angles
Exterior angles
that do not have a
common vertex on
alternate sides of
the transversal
1 and 8
2 and 7
Alternate exterior
angles have the
same measure.
1 = 8
2 = 7
Corresponding
angles
One interior and
one exterior angle
on the same side
of the transversal
1 and 5
2 and 6
3 and 7
4 and 8
Corresponding
angles have the
same measure.
1 = 5
2 = 6
3 = 7
4 = 8
2/13/2012 Section 10.1 15
Parallel Lines and Angle Pairs
• If parallel lines are cut by a
transversal,
– Alternate interior angles
have the same measure
– Alternate exterior angles
have the same measure
– Corresponding angles
have the same measure
2/13/2012 Section 10.1 16
Example 4
Finding angle measure when parallel lines are
intersected by a transversal
• Find the measure of all the angles:
• Solution:
m 1 = 35º
m 6 = 180º - 35º = 145º
m 7 = 145
m 2 = 35º
m 3 = 145º
m 5 = 35º
m 4 = 180º - 35º = 145º
2/13/2012 Section 10.2 1
Section 10.2
Triangles
Objectives
1. Solve problems involving angle relationships
in triangles.
2. Solve problems involving similar triangles.
3. Solve problems using the Pythagorean
Theorem.
2/13/2012 Section 10.2 2
Triangle
• A closed geometric figure that has three sides,
all of which lie on a flat surface or plane.
• Closed geometric figures
– If you start at any point and trace along the
sides, you end up at the starting point.
2/13/2012 Section 10.2 3
Euclid’s Theorem
•Theorem: A conclusions that is proved to be true through
deductive reasoning.
•Euclid’s assumption: Given a line and a point not on the
line, one and only one line can be drawn through the given
point parallel to the given line.
2/13/2012 Section 10.2 4
Euclid’s Theorem (cont.)
Euclid’s Theorem: The sum of the measures of the three
angles of any triangle is 180º.
Proof:
m 1 = m 2 and m 3 = m 4 (alternate interior angles)
Angles 2,5,4 form a straight angle (180º)
Therefore, m 1+ m 5 + m 3 =180º ( by substitution)
2/13/2012 Section 10.2 5
Example 1
Using Angle Relationships in Triangles
•Find the measure of
angle A for the triangle
ABC.
•Solution:
m A + m B+ m C =180º
m A + 120º + 17º = 180º
m A + 137º = 180º
m A = 180º - 137º = 43º
2/13/2012 Section 10.2 6
Example 2
Using Angle Relationships in Triangles
•Find the measures of angles 1
through 5.
•Solution:
m 1 = 90º (supplementary to right angle)
m 1+ m 2 + 43º =180º (sum = 180º)
m 1 + m 2 + 43º = 180º
90º + m 2 + 43º = 180º
m 2 + 133º = 180º
m 2 = 47º
m 3 = m 2 = 47º (vertical angles)
m 3 + m 4+ 60º =180º
47º + m 4 + 60º = 180º
m 4 = 180º-107º = 73º
m 4 + m 5 = 180º
73º + m 5 = 180
m 5 = 107º
2/13/2012 Section 10.2 7
Triangles and Their Characteristics
2/13/2012 Section 10.2 8
Similar Triangles
• Similar figures have the same shape, but not
necessarily the same size.
• ln similar triangles, the angles are equal but the sides
may or may not be the same length.
• Corresponding angles are angles that have the same
measure in the two triangles.
• Corresponding sides are the sides opposite the
corresponding angles.
2/13/2012 Section 10.2 9
Similar Triangles (continued)
FE and CB
Triangles ABC and DEF are similar:
Corresponding Angles Corresponding Sides
Angles A and D Sides
Angles C and F Sides
Angles B and E Sides
DE and AB
DF and AC
2/13/2012 Section 10.2 10
Example 3
Using Similar Triangles
13.5x
13.516
216
16
16x
21624916x
24
16
x
9
•Find the missing length x.
•Solution: Because the triangles are similar, their corresponding sides are proportional:
Apply the cross-products principle:
2/13/2012 Section 10.2 11
Pythagorean Theorem
•The sum of the squares
of the lengths of the legs
of a right triangle equals
the square of the length of
the hypotenuse.
•If the legs have lengths a
and b and the hypotenuse
has length c, then
a² + b² = c²
2/13/2012 Section 10.2 12
Example 4
Using the Pythagorean Theorem
Find the length of the
hypotenuse c in this right
triangle:
Solution:
Let a = 9 and b = 12
15225c
225
14481
129
ba
c
c
c
c
2
2
222
222
C
2/13/2012 Section 10.3 1
Section 10.3
Polygons, Perimeter, and Tessellations
Objectives
1. Name certain polygons according to the number of sides.
2. Recognize the characteristics of certain quadrilaterals.
3. Solve problems involving a polygon’s perimeter.
4. Find the sum of the measures of a polygon’s angles.
5. Understand tessellations and their angle requirements.
2/13/2012 Section 10.3 2
Polygons and Perimeter
• Polygon: Any closed shape in the plane formed by three
or more line segments.
• Regular Polygon: Has sides which are all the same
length and angles of all the same measure.
• Perimeter of a Polygon: The sum of the lengths of its
sides.
2/13/2012 Section 10.3 3
Regular Polygons
Name Picture
Triangle
3 sides
Quadrilateral
4 sides
Pentagon
5 sides
Name Picture
Hexagon
6 sides
Heptagon
7 sides
Octagon
8 sides
2/13/2012 Section 10.3 4
Types of Quadrilaterals
Name Characteristics Representation
Parallelogram Quadrilateral in which both pairs of opposite
sides are parallel and have the same
measure. Opposite angles have the same
measure
Rhombus Parallelogram with all sides having equal
lengths.
Rectangle Parallelogram with four right angles.
Because a rectangle is a parallelogram,
opposite sides are parallel and have the
same measure.
Square A rectangle with all sides having equal
length. Each angle measures 90 , and the
square is a regular quadrilateral.
Trapezoid A quadrilateral with exactly one pair of
parallel sides.
2/13/2012 Section 10.3 5
Example 1
An Application of Perimeter
• Fencing costs $5.25 per foot.
Find the cost to enclose the field
with fencing.
• Solution:
a. Find the perimeter of the rectangle in yards.
b. Convert to feet.
c. Multiply by the cost per foot.
$2205420($5.25)
foot
$5.25
1
feet 420Cost
420ft3ft1401yd
3ft
1
yd 140140yd
yd 140yd 282yd 422P
2w2lP
2/13/2012 Section 10.3 6
Example 2
The Sum of the Measures of a Polygon’s Angles
• The sum of the measures of the angles of a polygon with
n sides is (n-2)180º.
• Find the sum of the measures of the angles of an
octagon.
• Solution:
Using the formula (n-2)180º with n = 8, the sum of the
measures of an octagon’s angles is:
(8-2)180º = 6∙ 180º = 1080 º
2/13/2012 Section 10.3 7
Tessellations
(Tiling)
• A pattern consisting of the repeated use of the same
geometric figures to completely cover a plane, leaving no
gaps and no overlaps.
• The sum of the measures of the angles that come
together at each vertex is always 360º.
• Most restrictive condition in creating tessellations is that
just one type of regular polygon may be used.
2/13/2012 Section 10.3 8
Examples of Tessellations
2/13/2012 Section 10.3 9
Example 3
Angle Requirements of Tessellations
• Explain why a tessellation cannot be
created using only regular pentagons.
• Solution: Applying (n-2)180º to find
the measure of each angle of a regular pentagon.
Each angle measures (5-2)180º = 3∙180º =108º 5 5
A requirement for the formation of a tessellation is that
the measures of the angles that come together at each
vertex is 360º. Note that the three regular pentagons fill
in 3∙108º = 324º leaving a gap of 36º.
2/13/2012 Section 10.4 1
Section 10.4
Area and Circumference
1. Use area formulas to compute the areas
of plane regions and solve applied
problems.
2. Use formulas for a circle’s circumference
and area.
2/13/2012 Section 10.4 2
Area of a Rectangle and a Square
• The area, A, of a
rectangle with length l
and width w is given by
the formula A = lw.
• The area, A, of a square
with one side measuring
s linear units is given by
the formula A = s2.
2/13/2012 Section 10.4 3
Example 1
Solving an Area Problem
• You decide to cover the path shown
in bricks. Find the area of the path.
• Solution: We begin by drawing a
dashed line to divide the path into 2
rectangles. Then use the length
and width of each rectangle to find
its area. The area is found by
adding the areas of the two
rectangles together.
Area of path = 39 ft² + 27 ft²
= 66 ft²
2/13/2012 Section 10.4 4
Area of a Parallelogram
• The area, A, of a parallelogram with height h and base b
is given by the formula A = bh.
• The height of a parallelogram is the perpendicular
distance between two of the parallel sides. It is not the
length of a side.
2/13/2012 Section 10.4 5
Example 2
Using the Formula for a Parallelogram’s Area
• Find the area of the parallelogram.
• Solution:
The base is 8 centimeters and the
height is 4 centimeters.
Thus,
b = 8 and h = 4.
A = bh
A = 8 cm ∙ 4 cm = 32 cm²
2/13/2012 Section 10.4 6
Area of a Triangle
• The area, A, of a triangle with height h and base b is
given by the formula
A = ½bh.
2/13/2012 Section 10.4 7
Example 3
Using the Formula for a Triangle’s Area
• Find the area of the triangle..
• Solution:
The base is 16 meters and the
height is 10 meters.
Thus,
b = 16 and h = 10.
A = ½ bh
A = ½ ∙ 16 m ∙ 10 m
= 80 m²
2/13/2012 Section 10.4 8
Area of a Trapezoid
• The area, A, of a trapezoid with parallel bases a and b
and with height h is given by the formula
A = ½h(a +b).
2/13/2012 Section 10.4 9
Example 4
Finding the Area of a Trapezoid
• Find the area of the trapezoid.
• Solution:
The height is 13 ft. The
lower base, a, is 46 ft and
the upper base, b, is 32 ft.
Thus,
A = A = ½h(a +b).
A = ½ ∙ 13 ft ∙ (46 ft + 32 ft)
= 507 ft²
2/13/2012 Section 10.4 10
Circle
• A Circle is a set of points in the plane equally distant from a given point, its center.
• The radius, r, is a line segment from the center to any point on the circle. All radii in a given circle have the same length.
• The diameter, d, is a line segment through the center whose endpoints both lie on the circle. It is twice the radius. All diameters in a given circle have the same length.
2/13/2012 Section 10.4 11
Example 5
Finding the Distance Around a Circle
• The Circumference, C, of a circle with diameter d
and radius r is
C = πd or C = 2πr
Where π is an irrational number ≈ 3.14 derived by
dividing the circumference by the diameter.
• Find the circumference of the circle
with diameter = 40 yards.
• Solution:
C = πd
C = π(40 yds) = 40π yd ≈ 125.7 yd
2/13/2012 Section 10.4 12
Example 6
Finding the Area of a Circle • The Area, A, of a circle with radius r is A = πr²
• Which is a better buy? A large pizza with a 16-inch
diameter for $15.00 or a medium pizza with an 8-inch
diameter for $7.50?
• Solution: The better buy is the pizza with the lower price
per square inch. The radius of the large pizza is 8
inches and the radius of the medium pizza is 4 inches.
Large pizza:
A = πr² = π(8 in)² = 64 π in² ≈ 201 in²
Small pizza:
A = πr² = π(4 in)² = 16 π in² ≈ 50 in²
2/13/2012 Section 10.4 13
Example 6 continued
For each pizza, the price per square inch is found by
dividing the price by the area:
Price per square inch for large pizza =
$15.00 ≈ $15.00 ≈$0.07
64π in² 201 in² in²
Price per square inch for medium pizza =
$7.50 ≈ $7.50≈ $0.15
16π in² 50 in² in²
The large pizza is the better buy!
2/13/2012 Section 10.5 1
Section 10.5
Volume
Objectives
1. Use volume formulas to compute the
volumes of three-dimensional figures and
solve applied problems.
2. Compute the surface area of a three-
dimensional figure.
2/13/2012 Section 10.5 2
Formulas for Volume
• Volume of a rectangular solid, V,
is the product of its length, l, its width, w,
and its height, h:
V = lwh
• Volumes of Boxlike Shapes
2/13/2012 Section 10.5 3
Example 1
Solving a Volume Problem
• You are about to begin work on a swimming pool in
your yard. The first step is to have a hole dug that is 90
feet long, 60 feet wide, and 6 feet deep.
• You will use a truck that can carry 10 cubic yards of dirt
and charges $12 per load. How much will it cost you to
have all the dirt hauled away?
• Solution: Begin by converting feet to yards:
Similarly, 60 ft = 20 yd and 6 ft = 2 yd.
yd 30yd3
90
ft 3
1yd
1
ft 90ft 90
2/13/2012 Section 10.5 4
Example 1 continued
Next, we find the volume of dirt that needs to be dug out and hauled off.
V = lwh = 30 yd∙20 yd∙2 yd = 1200 yd³
Now, find the number of truckloads by dividing the number of cubic yards of dirt by 10 yards.
The truck charges $12 per trip, the cost to have all the dirt hauled away is:
120 trips ∙$12 = 120($12) = $1440
trips120trips10
1200
10yd
trip
1
1200yd
trip
10yd
1200yds truckloadofNumber
3
3
3
3
2/13/2012 Section 10.5 5
Example 2
Volume of a Pyramid • The Transamerica Tower in
San Francisco is a pyramid
with a square base.
It is 256 feet tall and each side
of the square base is 52 meters long.
Find its volume.
• Solution: The area of the square base is:
B = 52m ∙52 m = 2704 m²
The volume of the pyramid is:
V = ⅓Bh = ⅓∙2704 m² ∙ 256m ≈ 230,742 m³
2/13/2012 Section 10.5 6
Example 3
Volume of a Right Circular Cylinder
• Find the volume of this cylinder
with diameter = 20 yards and
height = 9 yards.
• Solution.
The radius is ½ the diameter
= 10 yards
V = πr²h = π(10 yd)²∙ 9 yd
= 900π yd³ ≈2827 yd³
2/13/2012 Section 10.5 7
Volumes of a Cone and a Sphere
• The Volume, V of a right
circular cone that has
height h and radius r is
given by the formula:
• The Volume, V of a
sphere of radius r is given
by the formula:
2/13/2012 Section 10.5 8
Example 4
Volumes of a Sphere and Cone
• An ice cream cone is 5 inches
deep and has a radius of 1 inch.
A spherical scoop of ice cream
also has a radius of 1 inch. If the
ice cream melts into the cone, will
it overflow?
• Solution: The ice cream will
overflow if the volume of the ice
cream, a sphere, is greater than
the volume of the cone. Find the
volume of each.
2/13/2012 Section 10.5 9
Example 4 continued
Vcone= ⅓ πr²h
= ⅓π(1 in.)² ∙5 in.
= 5π in.³ ≈ 5 in.³ 3 Vsphere = 4πr³ 3 = 4π(1 in.)³ 3 = 4π in.³ ≈ 4 in.³ 3 The volume of the spherical scoop of ice cream is less
than the volume of the cone so there will be no overflow.
2/13/2012 Section 10.5 10
Surface Area
• The area of the outer surface of a three-dimensional
object.
• Measured in square units
2/13/2012 Section 10.5 11
Example 5
Finding the Surface Area of a Solid
• Find the surface area of this rectangular solid.
• Solution:
The length is 8 yards, the width is 5 yards,
and the height is 3 yards.
Thus, l = 8, w = 5, h = 3.
SA = 2lw + 2lh + 2wh
= 2 ∙ 8 yd ∙ 5 yd + 2 ∙ 8 yd ∙3 yd + 2∙5 yd∙3 yd
= 80 yd² + 48 yd²+ 30 yd² = 158 yd²
2/13/2012 Section 10.6 1
Section 10.6
Right Triangle Trigonometry
Objectives
1. Use the lengths of the sides of a right triangle to find trigonometric ratios.
2. Use trigonometric ratios to find missing parts of right triangles.
3. Use trigonometric ratios to solve applied problems.
2/13/2012 Section 10.6 2
Ratios in Right Triangles
• Trigonometry means measurement of
triangles.
• Trigonometric Ratios: Let A represent an
acute angle of a right triangle, with right
angle, C, shown here.
For angle A, the trigonometric ratios are
defined as follows:
2/13/2012 Section 10.6 3
Example 1
Becoming Familiar with the Trigonometric Ratios
• Find the sine, cosine, and tangent of A.
• Solution: Using the Pythagorean
Theorem, find the measure of the
hypotenuse c.
• Now apply the trigonometric ratios:
12
5
Aangle to adjacent side
Aangle opposite sidetanA
13
12
hypotenuse
Aangle to adjacent sidecosA
13
5
hypotenuse
Aangle opposite sidesinA
13 169 c
169 144 25 12² 5² b² a² c²
2/13/2012 Section 10.6 4
Example 2
Finding a Missing Leg of a Right Triangle
• Find a in the right triangle
• Solution: Because we have
a known angle, 40º, and an
unknown opposite side, a,
and a known adjacent side,
150 cm, we can use the
tangent ratio.
tan 40º = a 150
a = 150 tan 40º ≈ 126
2/13/2012 Section 10.6 5
Applications of the Trigonometric Rations
• Angle of elevation: Angle formed by a horizontal line and the line of sight to an object that is above the horizontal line.
• Angle of depression: Angle formed by a horizontal line and the line of sight to an object that is below the horizontal line.
2/13/2012 Section 10.6 6
Example 3
Problem Solving using an Angle of Elevation
• Find the approximate height of
this tower.
• Solution: We have a right
triangle with a known angle,
57.2º, an unknown opposite
side, and a known adjacent
side, 125 ft.
Using the tangent ratio:
tan 57.2º = a
125
a = 125 tan 57.2º ≈ 194
2/13/2012 Section 10.6 7
Example 4
Determining the Angle of Elevation
• A building that is 21 meters tall
casts a shadow 25 meters long.
Find the angle of elevation of the
sun.
• Solution: We are asked to
find m A.
Use the inverse tangent key
(TAN -1)21 ≈ 40º 25
2/13/2012 Section 10.7 1
Section 10.7
Objective
1. Gain an understanding of some of the general ideas of other kinds of geometries.
2/13/2012 Section 10.7 2
The Geometry of Graphs
(Graph Theory)
• The Swiss mathematician Leonhard Euler (1707 – 1783)
proved that it was not possible to stroll through the city of
Kőnigsberg, Germany by crossing each of 7 bridges
exactly once.
• His solution opened up a new kind of geometry called
graph theory.
2/13/2012 Section 10.7 3
Graph Definitions
• Vertex is a point.
• Edge is a line segment or curve that starts or ends at a
vertex.
• Graph consists of vertices and edges
• Odd vertex has an odd number of attached edges.
• Even vertex has an even number of attached edges.
2/13/2012 Section 10.7 4
Rules of Traversability
1. A graph with all even vertices is traversable. One can
start at any vertex and end where one began.
2. A graph with two odd vertices is traversable. One must
start at either of the odd vertices and finish at the other.
3. A graph with more than two odd
vertices is not traversable.
A graph is traversable if it can be traced
without lifting the pencil from the paper
and without tracing an edge more than once.
2/13/2012 Section 10.7 5
Example 1
To Traverse or Not to Traverse?
• Is this graph traversable?
• Solution
Begin by determining if each vertex is even or odd.
This graph has two odd vertices, by Euler’s second rule, it is traversable.
• Describe the path to traverse it.
• Solution
By Euler’s second rule, start at one of the odd vertices and finish at the other.
2/13/2012 Section 10.7 6
Topology
The Study of Shapes
• Objects are classified according to the number of holes
in them, called their genus.
• Genus is the number of cuts that can be made in the
object without cutting the object in two pieces.
• Topologically Equivalent objects have the same
genus.
• The topology of knots is used to identify viruses and how
they invade our cells.
2/13/2012 Section 10.7 7
Examples of Topologically Equivalency
• The three shapes below have the same genus: 0. No
complete cuts can be made without cutting these
objects into two pieces.
• A doughnut and a coffee cup have a genus of 1.
2/13/2012 Section 10.7 8
Klein Bottle
• The figures below show the transformation into the figure
called the Klein bottle. Because the inside surface loops
back on itself to merge with the outside, it has neither an
outside nor an inside. It passes through itself without the
existence of a hole, which is impossible in three-
dimensional space. It only exists when generated on a
computer.
2/13/2012 Section 10.7 9
Comparing the Three Systems of Geometry
(Euclidean and non-Euclidean)
Euclidean Geometry
Euclid (300 B.C.)
Hyperbolic Geometry
Lobachevsky, Bloyai (1830)
Elliptic Geometry
Riemann(1850)
Given a point not on a
line, there is one and
only one line through
the point parallel to the
given line.
Given a point not on a line, there
are an infinite number of lines
through the point that do not
intersect the given line.
There are no parallel
lines
Geometry is on a
plane:
Geometry is on a pseudosphere: Geometry is on a
sphere:
2/13/2012 Section 10.7 10
Comparing the Three Systems of Geometry
(Euclidean and non-Euclidean)
Euclidean Geometry
Euclid (300 B.C.)
Hyperbolic Geometry
Lobachevsky, Bloyai (1830)
Elliptic Geometry
Riemann(1850)
The sum of the
measures of the angles
of a triangle is 180 .
The sum of the measures of
the angles of a triangle is
less than 180 .
The sum of the
measures of the
angles of a triangle is
greater than 180 .
2/13/2012 Section 10.7 11
Fractal Geometry
• Developed by Benoit Mandelbrot
using computer programming.
• Geometry of natural shapes.
• Self-similarity is the quality of
smaller versions of an object
appearing in the object itself.
• Iteration is the process of
repeating a rule again and again to
create a self-similar fractal.