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§4.6 Area & Volume

The student will learn about:

area postulates, Cavalieri’s Principle,

1

and the areas of basic shapes.

Postulates

2

1. Existence of area and volume.

2. Relative size.

3. Additively.

4. Congruency

5. Basic unit of measure is 1.

6. Cavalieri’s Principle.

Cavalieri’s Principle for Area

3

L x L’x

If, in two shapes of equal altitude, the sections made by lines at the same distance from their respective bases are always equal in length, then the areas of the shapes are equal.

Cavalieri’s Principle for Volmue

4

If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal

5

Area P1 = 22.82 cm2

Area ABC = 22.82 cm2

I'

H'

G'

K

J

A

B

C

E

D

F

G

H

I

Moment for Discovery p. 293Assigned for homework!

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§4.8 Modern Triangles

The student will learn about:

linearity of numbers, Ceva’s Theorem,

6

and the Fermat Point of a triangle.

Menelaus’ Theorem,

Introduction

7

In ABC with points D, E and F on sides BC AC and AB respectively, define the linearity number of the points D, E, and F with respect to ABC as

ABC AF BD CE

DEF FB DC EA

C

A

BD

EF

Ceva’s Theorem

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The Cevians AD, BE, and CF of ABC are concurrent iff

C

A

BD

EF

ABC1

DEF

AF BD CE1

FB DC EA

Theorem 2 – Menelaus’ Theorem

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If points D, E, and F lie on sides BC, AC, and AB of ΔABC respectively, then D, E, and F are collinear iff ABC

1DEF

C

A

FE

DB

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§5.1 Euclid’s Superposition Proof and Plane Transformations.

The student will learn:

the basic concepts of transformations.

1010

TransformationDefinition. If a mapping f: Π → Π from a plane Π to itself is both one-to-one and onto, then f is called a plane transformation.

Definition. If a transformation maps lines onto lines, it is called a linear transformation.

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The inverse mapping of a transformation f, denoted f -1, is the mapping which associates Q with P for each pair of points (P, Q) specified by f. That is, f -1 (Q) = P iff f (P) = Q.

Inverse Transformation

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The Identity

Definition. A transformation of the plane is called the identity mapping iff every point of the plane is a fixed point. This transformation is denoted e.

13

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§5.1 Euclid’s Superposition Proof and Plane Transformations.

The student will learn the basic concepts of reflections.

14

15

Introduction to Line Reflections

Definition. Let l be a fixed line in the plane. The reflection R (l) in a line l is the transformation which carries each point P of the plane into the point P’ of the plane such that l is the perpendicular bisector of PP’. The line l is called the axis (or mirror or axis of symmetry) of the reflection.

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Introduction to Point Reflections

Definition. Let C be a fixed point in the plane. The reflection R (C) in a point C is the transformation which carries each point P of the plane into the point P’ of the plane such that C is the midpoint of PP’. The point C is called the center of the reflection.

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Isometry

Definition - A transformation of the plane that preserves distance is called an isometry. If P and Q are points in the plane and a transformation maps them to P’ and Q’ respectively so that m (PQ) = m (P’Q’) then that transformation is an isometry.

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Lemma 1. An isometry preserves collinearity.

If A, B, and C are pints then A, B, and C, are collinear iff m(AB) + m(BC) = m(AC). This also means that B is between A and C which is written A-B-C.

Isometry Facts

Lemma 2. An isometry preserves betweenness.

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Theorem 1Reflections are

A. Angle-measure preserving.

B. Betweeness preserving.

C. Collinearity preserving.

D. Distance preserving.

We need only show distance preservation to get the other three.

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The product of two line reflections R (l) and R (m), where l and m are parallel is distance and slope preserving and maps a given line n into one that is parallel to it.

Theorem 3

A line reflection is an isometry by lemma and hence distance preserving. We will prove slope preserving but first let’s look at a figure of what this product of two line reflections looks like.

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5.3 Translations and Rotations

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A translation, is the product of two reflections R (l) and R (m) where l and m are parallel lines.

DEFINITION

23

an isometry, and

is a direct transformation, and

has no fixed points.

A translation is

Theorems

Proven in section 5.2. (Homework.)

24

Given two parallel lines you should be able to construct the translation of any set of points and describe that translation.

25

Given a translation you should be able to construct the two lines whose reflections produce the necessary transformation. They are not unique.

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Rotations

θ

O

27

A rotation is the product of two line reflections R (l) • R (m), where l and m are not parallel. The center of the rotation is O = l m . The direction of the rotation is about O from l toward m, and the angular distance of rotation is twice the angle from l to m.

Definition

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Given two lines that are not parallel you should be able to reflect a set of points in the first line and then again in the second line.

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Transformations in General

Theorem. Given any two congruent triangles, ΔABC and ΔPQR, there exist a unique isometry that maps one triangle onto the other.

A B

CP Q

R

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Prove that given three points A, B, and C and their images P, Q, and R, there exist a unique isometry that maps these points onto their images.

A B

C P

Q

R

B’

C’

A’

Theorem 1

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The previous proof shows that every isometry on the plane is a product of at most three line reflections; exactly two if the isometry is direct and not the identity.

Theorem 2: Fundamental Theorem of Isometries.

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A nontrivial direct isometry is either a translation or a rotation.

Corollary

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A nontrivial indirect isometry is either a reflection or a glide reflection. (Glide reflections are covered in the next section.)

Corollary

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DefinitionA glide reflection, G (l, AB) is the product of a line reflection R (l) and a translation T (AB) in a direction parallel to the axis of reflection. That is, ABl.

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Theorems

A glide reflection is an isometry,

and is an opposite transformation, and

there are no invariant points under a glide reflection.

Proof: Use the components of a glide reflection.

Defect

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Since the sum of the angles of a triangle is less than 180 we can define the defect of the triangle as that difference, or

D(ABC) = 180 - ∡A - ∡ B - ∡ C

The defect of a convex polygon P1 P2 . . . Pn is the number

D(P1 P2 . . . Pn ) = 180(n -2) – P1 – P2 . . . - Pn

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Definition of Area

37

The area of a convex polygon P1 P2 . . . Pn is defined y the number:

K = k [180 (n – 2) – P1 – P2 . . . – Pn ]Where k is some predetermined constant for the entire plane (not depending on each given polygon). The value for k is frequently taken as π/180, which converts degree measure to radian measure.

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§6.4 Hyperbolic Models

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Intro to Poincare's Disk Model.

• A

Point – Any interior point of circle C (the ordinary points of H or h-points)

Line – Any diameter of C or any arc of a circle orthogonal to C in H are h-lines.

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§6.4 Hyperbolic Half-Plane Model

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The half-plane model can be thought of as taking the disk model at a point on the circle and separating the disk and stretching the boundary circle into a straight line.

In essence it is all points P(x, y) for which y > 0 or the upper half-plane. All such points are the h-points in the model. The x-axis is not part of the geometry.

4040

§6.4 Hyperbolic Half-Plane ModelFinding h-lines.

Find the h-line through the h-points A (2, 3) and B(9, 4).

B (9, 4)

M (11, 0)

A (2, 3)

N (1, 0)

••

A semicircle with center on the x-axis has equation

x 2 + y 2 + ax = b

2 2 + 3 2 + a2 = b and 9 2 + 4 2 + a9 = b and solving yields

a = - 12 and b = - 11

x 2 + y 2 - 12x = - 11

Check your work!

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§6.4 Hyperbolic Half-Plane Model

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DistanceVertical ray

AMAB* ln

BM

Semicircle

AM BN

AB* lnAN BM

A B

B

A

M MN

• •

•

•

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§6.4 Hyperbolic Half-Plane Model

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Angle MeasureAngle Measure – Let

m ABC * = m A’BC’ where BA’ and BC’ are the Euclidean rays tangent to the sides of ABC as in the disk model.

A

B

C

A’

•

C’

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Parallel PostulateHyperbolic Geometry is all about the parallel postulate so let’s look at an example. Given the h-line x2 + y2 = 25, and the point P (1, 7), find the two h-lines through P parallel to the h-line.

1. Find M and N from the equation of the h-line.

Let y = 0 and then x = 5 and M (-5, 0), N (5, 0)

2. Find the equation of the line PM. 1 + 49 + a = b

25 + 0 - 5a = bHence a = - 4.1666 and b = 45.833

NM

P [x]•

One parallel is x2 + y2 – 4. 17 x = 45.83

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Parallel PostulateHyperbolic Geometry is all about the parallel postulate so let’s look at an example. Given the h-line x2 + y2 = 25, and the point P (1, 7), find the two h-lines through P parallel to the h-line.

3. Find the equation of the line PN.

1 + 49 + a = b

25 + 0 + 5a = b

Hence a = 6.25 and b = 56.25

NM

P [x]•

The other parallel is x2 + y2 + 6.25 x = 56.25