journal chapter 5

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Journal Chapter 5 TRIANGLES……. TRIANGLES…. AND… MORE… TRIANGLES. Menu options ahead. Full screen to listen to music

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Journal Chapter 5 . Full screen to listen to music. TRIANGLES……. … TRIANGLES…. … AND… … MORE… TRIANGLES. Menu options ahead. Option #1. Perpendicular Bisector:. What is it? A line that bisects a segment and is perpendicular to that same segment. . THEOREM: . - PowerPoint PPT Presentation

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Page 1: Journal Chapter 5

JournalChapter 5

TRIANGLES…….…TRIANGLES….…AND……MORE…TRIANGLES.

Menu options ahead.

Full screen to listen to music

Page 2: Journal Chapter 5

Option #1

Page 3: Journal Chapter 5

What is it? A line that bisects a segment and is perpendicular to that same segment.

* Any point that is on the perpendicular bisector is equidistant to both of the endpoints of the segment which is perpendicular to.

Converse:*If the point is equidistant from both of the endpoints of the segment, then it is a perpendicular bisector.

THEOREM:

Extra:A perpendicular bisector is the locus of all points in a plane that are equidistant from the endpoints of a segment.

Perpendicular Bisector:

Page 4: Journal Chapter 5

SegmentPerpendicular bisectorAre both congruent.

7cm

7cm

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EXAMPLESa

b

cm d

1. Given that AC= 25.5, BD=20, and BC= 25.5. Find AB.

AB= 51

2. Given that m is the perpendicular bisector of AB and BC= 30 . Find AC.

AC= 30

3. Given that m is the perpendicular bisector of AB , AC= 4a, and BC= 2a+26. Find BC.

2a+26=4a-2a=-26a=13 plug in both equationsBC= 52

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Option #2

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Angle BisectorsWhat is it?A ray or any line that cuts an angle into two congruent angles.

Theorem : * Any point that lies on the angle bisector is equidistant to both of the sides of the angle.

Converse: * If it is equidistant to both sides of the angle, then it lies on the angle bisector.

MUST: the point always has to lie on the interior of the angle.

Page 8: Journal Chapter 5

Angle point that Angle bisector lies on the bisector is equidistant to both sides of the angle

1.5 cm

1.5 cm

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EXAMPLES

A

D

C

B

1. Given that BD bisects ∠ABC and CD = 21, find AD. AD= 212. Given that AD =65 , CD= 65 and m∠ABC = 50˚, find

m∠CBD.m DBC∠ = 25˚

3. Given that DA=DC , m∠DBC= (10y+3)˚ , and m ∠DBA= (8y+10) ˚, find m∠DBC.10y+3=8y+102y+3=102y=7 y=3.5 plug in both equations. m∠DBC=38˚

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Option #3

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example. G is the point of concurrency.

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What is a circumcenter?It is the point where all the 3 perpendicular bisectors of a triangle meet.

What makes it special?It is the point of concurrency and it is equidistant to all 3 vertices.

EXAMPLE

F is the circumcenter in this equilateral triangle. The circumcenter is on the inside of the triangle.

1. Knowing what circumcenter is, it would help you if you are a business person to know a strategic place in between 3 cities to place your business. That way it would be located in the same distance from all 3 cities for customers to come.

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f

d

Examples:2.In an acute the circumcenter is on the inside of the triangle

C is the circumcenter

3.In a right the circumcenter is on the midpoint of the hypotenuse

D is the circumcenter

4.In an obtuse the circumcenter is on the outside of the triangle.

F is the circumcenter

c

Page 16: Journal Chapter 5

The angle bisectors of a triangle meet at a point that is equidistant from the three sides of the triangle.

Concurrency of Angle bisectorsOf a triangle theorem

1. What is the distance from H to G, if F to H is 10 cm?

10cm from H to G

2. H to E is 15cm , then what would be H to G doubled?

Real: H to G would be 15cm and doubled would be 30cm

3. H in this triangle is what?The incenter of triangle ABC

Page 17: Journal Chapter 5

•The point of concurrency where the three angle bisectors of a triangle meet. It is equidistant (same distance) from each three sides of the triangle

EXAMPLES:G.

H

K

LJ

M

P

1. P is the incenter

2. If you want to open a new restaurant in town you have to place it at the incenter of three major highways so people can go to your restaurant.

TRIANGLEIncenter of a

3. If the distance from K to P is 2cm, what is the distance from P to L?From P to L = 2cm

Page 18: Journal Chapter 5

Option #4

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Median of a triangle

Median

Is the segment that goes from the vertex of a triangle to the opposite midpoint

Ex.1 :Median

Ex.3 : Ex. 2: all medians are congruent. Shown in yellow

Page 20: Journal Chapter 5

CentroidThe point where the medians of a triangle intersect. Is the center of balance. The distance from the vertex to the centroid is the double the distance from the centroid to the opposite side.

EXAMPLES:1. Balance a triangular piece of glass on a triangular coffee table. 2. G is the centroid. 3.

by

px

az

c

AP= 2/3 AY BP=2/3BZCP=2/3CX

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Concurrency of Medians of a Triangle Theorem

The medians of a triangle intersect at a point that is 2/3 of the distance from each vertex to the midpoint of the opposite side

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Option #5

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ALTI

TUDE

A segment that goes from the vertex perpendicular to the line containing the opposite side.

Where the altitudes of a triangle intersect.

altitu

de

hypotenuse

ORTHO

CENTER

TRIANGLES !!!!!!!

!!!!!!!!

!!!!!!!!

!!!!!!!!

!!!!!

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Concurrency of Altitudes of triangles Concurrency

of Altitudes of a Triangle Theorem The lines containing the altitudes of a triangle are concurrent

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Option #6

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Midsegments: A segment that joins the midpoints of two sides of the triangle.

Every triangle has 3midsegments that form the midsegment triangle.

Midsegment Theorem The segment joining the midpoints of two sides of a triangle is parallel to the third side, and is half the length of the third side.

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Option #7

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In any triangle the longest side is always opposite from the largest angle. The small side is opposite from a small angle.

Side – Angle Relationship EXAM

PLES:

Page 29: Journal Chapter 5

Option #8

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Exterior angle inequality

The exterior angle is bigger (grater) than the non-adjacent interior angles of the triangle .

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Triangle Inequality Theorem The sum of the lengths of two sides of a triangle is greater than the length of the third side.

TRIANGLE INEQUALITY

TELL WHEATHER A TRIANGLE CAN HAVE THE SIDES WITH THE GIVEN LENGTHS. EXPLAIN

1. 3,5,7 2. 4, 6.5,11 3. 7, 4, 53 + 5 > 7 4 +6.5> 11 5+4 > 7 8>7 10.5>11 9>7

YES, the sum of each NO. by the 3rd triangle Yes , 5+4 is grater than 7 soPair of the lengths is inequality thm. A triangle there can be a triangle. Grater than the 3rd <. Cannot have these side lengths.

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Option #9

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How to write an indirect proof

1. Assume that what you are proving is FALSE

2. Use that as your given, and start proving it.

3. When you come to a contradiction that’s when you stop and have proven that it is true.

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PROVE: A TRIANGLE CANNOT HAVE 2 RIGHT ANGLES

1. assume a triangle has 2 right angles <1+<2….

GIVEN2. M<1 =M<2 = 90..

DEFF OF RT. <3. M<1+M<2+M<3=180

…..TRIANGLE + THM4. M<3= 0

CONTRADICTION So a triangle cannot have

2 right angles.

Examples: PROVE: A SCALENE TRIANGLE CANNOT HAVE TWO CONGRUENT ANGLES

1. Assume a scalene triangle can have two congruent sides

2. a triangle with 2 congruent angles… deff. of an isosceles triangle

3. Deff. Of isosceles and scalene triangles… therefore a scalene triangle cannot have two congruent angles.

PROVE: IF A>0 THAN 1/A >0

1. Assume a>0 than 1/a<0

2. (a) 1/a <0a a 1 <0CONTRADICTION3. Therefore if a>0

than 1/a>0

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Option #19

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HINGE THEOREM If 2 triangles have 2 sides that are congruent, but the third side is not congruent, then the triangle with the large included angle has the longer side.

(What changes is the incidence and the side….) Theorem:If two sides of one triangle are congruent to the two sides of another triangle and the third sides are not congruent, then the larger included angle is across from the

longer third side. Converse :

Page 37: Journal Chapter 5

Option #11

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45

45909030

6045˚- 45 ˚- 90˚ 90˚- 30˚- 60˚Both legs of the triangle are congruent and the length of the hypotenuse is the length of the two legs √2

The length of the hypotenuse is 2x the length of the shorter leg, and the length of the longer leg is the length of the shorter leg time √3

Examples: 18√360

x

y

1. 18√3 =2x 9 √3=x y=x √3 y=2760

30

x

y

2 . 15= √3 15/ √3=x 5 √3=x y= 2xY=2(5 √3)Y= 10 √3

30y

5X=5 √3Y=2(5)Y=10

Page 39: Journal Chapter 5

"The only angle from which to approach a problem is the TRY-Angle"