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Introduction Geometry is a part of mathematics concerned with questions of size, shape, and relative position of figures and with properties of space

# Introduction Geometry is a part of mathematics concerned with questions of size, shape, and relative position of figures and with properties of space

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Introduction

Geometry is a part of mathematics concerned with questions of size,

shape, and relative position of figures and with properties of space

Initially a body of practical knowledge concerning lengths, areas, and volumes, in the third century BC geometry was put

into an axiomatic form by Euclid, whose treatment—Euclidean geometry—set a

standard for many centuries to follow.

Geometry being used in :

Computer graphics is based on geometry - how images are transformed

when viewed in various ways

Structural engineering. What shapes are rigid or flexible, how they respond to forces and stresses. Statics (resolution of forces) is essentially geometry. This goes over into all levels of design, form, and function of many

things

This a pictures with some basic geometric structures. This is a modern reconstruction of the English Wigwam. As you can there the door way is a rectangle, and the wooden panels on the side of the house are made up of planes and lines. Except

for really planes can go on forever. The panels are also shaped in the shape of squares. The house itself is half a

cylinder.

Construct a segment congruent to a given segment.Given:

A B

Procedure:

1. Use a straightedge to draw a line. Call it l.

2. Choose any point on l and label it X.

3. Set your compass for radius AB and make a mark on the line where B lies. Then, move your compass to line l and set your pointer on X. Make a mark on the line and label it Y.

Construct: XY = AB

X Yl

Construction #1Construction #1

This is our compass.

Construct an angle congruent to a given angle

Given: Procedure:1) Draw a ray. Label it RY.

Construct:

E

D

2) Using B as center and any radius, draw an arc intersecting BA and BC. Label the points of intersection D and E.

4) Measure the arc from D to E.E2

Construction #2

RY

3) Using R as center and the SAME RADIUS as in Step 2, draw an arc intersecting RY. Label point E2 the point where the arc intersects RY

5) Move the pointer to E2 and make an arc that that intersects the blue arc to get point D2

D2

A

CB

6) Draw a ray from R through D2

How do I construct a Bisector of a given angle?

Given:

C

A

B

Procedure:1. Using B as center and any radius, draw and arc that intersects BA at X and BC at point Y.

X

Y

2. Using X as center and a suitable radius, draw and arc. Using Y as center and the same radius, draw an arc that intersects the arc with center X at point Z.

Z

3. Draw BZ.

Construction #3

How do I construct a perpendicular bisector to a given segment?

Given:

Procedure:1. Using any radius greater than 1/2 AB, draw four arcs of equal radii, two with center A and two with center B. Label the points of intersection X and Y.

A B

X

Y

2. Draw XY

Construction #4

How do I construct a perpendicular bisector to a given segment at a given point?

Given: C k

Procedure: 1. Using C as center and any radius, draw arcs intersecting k at X and Y.

X Y

2. Using X as center and any radius greater than CX,draw an arc. Using Y as center and the same radius,

draw and arc intersecting the arc with center X at Z.

Z

3. Draw CZ.

Construction #5

How do I construct a perpendicular bisector to a given segment at a given point outside the line?

Given: k

P

Procedure: 1. Using P as center, draw two arcs of equal radii that intersect k at points X and Y.

X Y

2. Using X and Y as centers and a suitable radius, draw arcs that intersect at a point Z.

Z

3. Draw PZ.

Construction #6

How do I construct a line parallel to a given line through a given point?

Given: k

P

Procedure: 1. Let A and B be two points on line k. Draw PA.

A B

2. At P, construct <1 so that <1 and <PAB are congruent corresponding angles. Let l be the line containing the ray you just constructed.

1 l

Construction #7

Parallel Lines – Two lines in the same plane which never intersect.

Symbol: “ // ”

Transversal – A line that intersects two // lines.

8 Special Angles are formed.

1 23 4

5 687

Interior Portion of the // Lines

t

m

n

Corresponding Angles Most Important Angle Relationship

Always Congruent

Cut the Transversal & lay the top part onto the bottom part. Overlapping Angles are Corresponding.

1 23 4

5 67 8

Corresponding Angles

1 & 5

2 & 6

3 & 7

4 & 8

If a Transversal Intersects two // lines, then the corresponding angles are Congruent.

1 23 4

6587

1 23 4

6587

4 Pairs of Vertical Angles

Are Congruent

1 4 2 3

5 8 6 7

3 & 6

4 & 5Alternate Interior Angles

3 & 5

4 & 6Same-Sided Interior Angles

Special Interior Angles

Are congruent

Are Supplementary

(= 180)

The (3-1) Alternate Interior Angle Theorem

If a Transversal intersects two // lines, then the alternate interior angles are congruent. 1 2

3 4

5 67 8

1. l // m

2. 3 7

3. 7 6

4. 3 6

Given: l // m

Prove: 3 6

1. Given

2. Corrsp. Angles are Congruent

3. Def. of Vertical Angles

4. Subs

l

m

The (3-2) Same-Sided Interior Angle TheoremIf a Transversal intersects two // lines, then the same-sided interior angles are supplementary.

Given: l // m Prove: 4 & 6 are Supplementary

1 23 4

5 67 8

Statements

1. l // m

2. m4 + m2 = 180

3. m2 = m6

4. m4 + m6 = 180

5. 4 & 6 are Supplementary

Reasons

1. Given

3. Corrsp. s are

4. Subs

5. Def of Supplementary

l

m

Th (3-3) Alternate Exterior Angle Theorem

If a Transversal intersects two // lines, then the alternate exterior angles are congruent.

Given: l // mProve: 1 & 8 are Congruent.

1 23 4

5 67 8

Statements

1. l // m

2. m1 = m5

3. m5 = m8

4. m1 = m 8

Reasons

1. Given

2. Corrsp. s are 3. Vert. ’s are

4. Transitive Property

l

m

Examples 1 & 2

Solve for the missing s

• Solve for x, then for each .

5x - 20

3x

l

m

14x - 5

13x

l

m

5x – 20 +3x = 180

8x = 200

x = 25

14x – 5 = 13x

-5 = -x

5 = x

.

42

1

2

m1 = 42 (Corrsp. )

m1 + m2 = 180

42 + m2 = 180

m2 = 138

2. Solve for angles a, b, c if l//m

l

m

m

l

65 40

a bc

ma = 65 (Alt. Inter. )

mc = 40 (Alt. Inter. )

ma + mb + mc = 180

65 + mb + 40 = 180

m = 75

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