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.