1 problem 1a problem 2a problem 3a problem 4a problem 1b problem 4b problem 2b problem 3b triangles...

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PROBLEM 1A

PROBLEM 2A

PROBLEM 3A

PROBLEM 4A

PROBLEM 1B

PROBLEM 4B

PROBLEM 2B

PROBLEM 3B

TRIANGLES AS POLYGONS: CLASSIFICATION

Standards 4 and 5

REFLEXIVE, SYMMETRIC AND TRANSITIVE PROPERTIES

EXTERIOR ANGLE THEOREM

CPCTC

ANGLE SUM THEOREM

PROBLEM 5A PROBLEM 5BEND SHOW

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

2

STANDARD 4:Students prove basic theorems involving congruence and similarity.

ESTÁNDAR 4: Los estudiantes prueban teoremas que involucran congruencia y semejanza.

STANDARD 5:Students prove that triangles are congruent or similar, and they are able to use the concept of corresponding parts of congruent triangles.

ESTÁNDAR 5: Los estudiantes prueban que son triángulos congruentes o semejantes y son capaces de usar el concepto de partes correspondientes de triángulos congruentes.



3

These are examples of POLYGONS:

These are NOT POLYGONS:

A POLYGON is a closed figure in a plane which is made up of line segments, called sides, that intersect only at their endpoints, named vertices.

Standards 4 and 5



4

A TRIANGLE is a three-sided polygon

Standards 4 and 5



5

55°

64° 61°

21°

110° 49°

RIGHT TRIANGLE

ACUTE TRIANGLE

OBTUSE TRIANGLE

CLASSIFYING TRIANGLES BY ANGLESANGLES

STANDARDS 4 and 5


6

Parts of a RIGHT TRIANGLE

Leg

Leg

HYPOTENUSE

Right Angle



CLASSIFYING TRIANGLES BY SIDESSIDES

10

4

9 22 22

13 13

13

SCALENE TRIANGLE

ISOSCELES TRIANGLE

EQUILATERAL TRIANGLE

9

Standards 4 and 5

Also EQUIANGULAR



8

Parts of an ISOSCELES TRIANGLE

Base Angles

Base

LegLeg

Vertex Angle

Standards 4 and 5



9

B C

A

CONGRUENCE of triangles is REFLEXIVEREFLEXIVE:

ABC ABC

Standards 4 and 5

bc

a



10

CONGRUENCE of triangles is SYMMETRIC:SYMMETRIC:

A

CB M

K

L

IF THEN ABC KLM KLM ABC

Standards 4 and 5



11

CONGRUENCE of triangles is TRANSITIVETRANSITIVE

A

CB M

K

L

IF AND

R

TS

THEN ABC RST KLM RST

ABC KLM

Standards 4 and 5



12

1

2 3

The measure of an exterior angle (angle 3) of a triangle is equal to the sum of the measures of the two remote interior angles (angles 1 and 2).

+1m 2m = 3m

Exterior Angle Theorem

Remote Interior Angles

Exterior Angle

Standards 4 and 5



13

Standards 4 and 5

(3X+4)°

(7X+5)°

(6X+69)°Z

Find in the figure below:Zm



14

Standards 4 and 5

(3X+4)°

(7X+5)°

(6X+69)°Z

(7X+5)°+(3X+4)° = (6X+69)°

By Exterior Angle Theorem:

7X+5 + 3X+4 = 6X +6910X + 9 = 6X +69

-9 -910X = 6X + 60

-6X -6X

4X = 604 4

X=15

By Linear Pair:

Zm + (6X+69)° = 180°

Zm + 6X+69 = 180

Zm + 6( )+69 = 18015

Zm + 90+69 = 180

Zm + 159 = 180

-159 -159

Zm = 21°

Find in the figure below:Zm

Substituting the value for X:



15

Standards 4 and 5

(3Z+8)°

(9Z+6)°

(5Z+84)° X

Find in the figure below:Xm



16

Standards 4 and 5

(3Z+8)°

(9Z+6)°

(5Z+84)° X

(9Z+6)°+(3Z+8)° = (5Z+84)°

By Exterior Angle Theorem:

9Z+6 + 3Z+8 = 5Z +8412Z + 14 = 5Z +84

-14 -1412Z = 5Z + 70

-5Z -5Z

7Z = 707 7

Z=10

By Linear Pair:

Xm + (5Z+84)° = 180°

Xm + 5Z + 84 = 180

Xm + 5( )+84 = 18010

Xm + 50+84 = 180

Xm + 134 = 180

-134 -134

Xm = 46°

Find in the figure below:Xm

Substituting the value for Z:



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CONGRUENT TRIANGLES

CCorresponding PParts of CCongruent TTriangles are CCongruent

ABC KLM by CPCTCCPCTC

B C

A

L M

K

Standards 4 and 5



18

Standards 4 and 5

If RST UVW with RS = 6X + 8, UV = 56, RT = 2X – 2, What is the value of X, and RT.

R

T S

U

W V

6X+8 562X – 2

6X + 8 = 56

Since both triangles are congruent, corresponding sides are congruent as well:

-8 -86X = 486 6

X = 8

Now finding RT:

RT=2X – 2

=2( ) – 2 8

= 16 – 2

RT= 14



19

Standards 4 and 5

If DEF JKL with DE = 5X + 3, JK = 48, DF = 3X + 1, What is the value of X, and DF.

D

F E

J

L K

5X+3 483X + 1

5X + 3 = 48


-3 -35X = 455 5

X = 9

Now finding DF:

DF=3X + 1

=3( ) + 1 9

= 27 + 1

DF= 28



20

Standards 4 and 5

Given that LMN RST with sides LM=5X+3, RS=3X+13, and ST = 6X + 9. Find the value of X and ST.

L

N M

R

T S

5X+3 3X+13

5X +3 = 3X + 13


-3 -3

5X =3X + 10

2 2

Now finding ST:

ST=6X + 9

=6( ) +9 5

= 30 + 9

6X + 9

-3X -3X

2X = 10

X = 5

ST= 39



21

Standards 4 and 5

Given that EFG HIJ with sides EF =9X+2, HI=4X+17, and JI = 2X + 3. Find the value of X and JI.

E

G F

H

J I

9X+2 4X+17

9X +2 = 4X + 17


-2 -2

9X =4X + 15

5 5

Now finding JI:

JI=2X + 3

=2( ) +3 3

= 6 + 3

2X + 3

-4X -4X

5X = 15

X = 3

JI= 9



22

35° 58°

87°

+ +

C

Cm

B

Bm

A

Am =180°

35° + 87° + 58° = 180°

The sum of the interior angles of a triangle is always 180°

ANGLE SUM THEOREM:Standards 4 and 5



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A

Am =180°

B

Bm+

C

Cm+If and A = 90°m then

=180°Bm+ Cm+90°- 90° -90°

=90°Bm Cm+

Conclusion: In a right triangle, both non- right angles are acute and complementary!

What kind of angles are the non-right angles in a right triangle?


24

Standards 4 and 5

If QRT UVW, with Qm = 100°, Vm = 30°,

and , find the value for Z.Wm = 4Z +10

Q

T R W V

U

Since both triangles are congruent then their corresponding angles are congruent:

4Z + 10

100°

30°

Q =m Um

Q U

100°

= 100°

Now from the Angle Sum Theorem:

Vm Wm+ =180°Um +

100° + 30° + = 180°Wm

130 + = 180°Wm-130° -130°

W = 50°m

Wm = 4Z +10If then 4Z + 10 = 50

-10 -10

100° + 30° + = 180°Wm

4Z = 404 4

Z = 10PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

25

Standards 4 and 5

If DEF JKL, with Dm = 110°, Km = 20°,

and , find the value for Z.Lm = 3Z +23

D

F E L K

J


3Z + 23

110°

20°

D =m Jm

D J

110°

= 110°

Now from the Angle Sum Theorem:

Km Lm+ =180°Jm +

110° + 20° + = 180°Lm

130 + = 180°Lm-130° -130°

L = 50°m

Lm = 3Z + 23If then 3Z + 23 = 50

-23 -23

110° + 20° + = 180°Lm

3Z = 273 3

Z = 9PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

26

Standards 4 and 5

If FGH IJK with , andGFH=m 3X+28 JKI=m 2X – 8

KJI=m 2X+20 , what is the value for x and for each angle?

F

H G K J

I

3X+28

2X – 8 2X + 20°

3X + 28


HFG KIJ

HFG=m KIJ=m 3X+28

JKI +m KIJ+m KJI=m 180°

(2X – 8 ) + (3X + 28) + (2X+20) = 180

Now from the Angle Sum Theorem in KIJ:

2X – 8 + 3X + 28 + 2X+20 = 180

2X + 3X + 2X – 8 + 28+20 = 180

7X + 40 = 180

-40 -40

7X = 1407 7

JKI =m 2X – 8

= 2( ) – 8

= 40 - 8

= 32°

20

KIJ =m 3X+28

= 3( ) +28

= 60+28

= 88°

20

KJI =m 2X+20

= 2( ) +20

= 40+20

= 60°

20X = 20

Checking solution:

32° + 88° +60° = 180°

180°=180°PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

27

Standards 4 and 5

If RST NOP with , and

TRS=m 4X+20 NPO=m 5X + 6

PON=m 3X+10 , what is the value for x and for each angle?

R

T S P O

N

4X+20

5X +6 3X + 10°

4X + 20


TRS PNO

TRS =m PNO=m 4X+20

NPO+m PNO+m PON=m 180°

(5X + 6 ) + (4X + 20) + (3X+10) = 180

Now from the Angle Sum Theorem in NOP:

5X + 6 + 4X + 20 + 3X+10 = 180

5X + 4X + 3X + 6 + 20+10 = 180

12X + 36 = 180

-36 -36

12X = 14412 12

NPO=m 5X + 6

= 5( ) +6

=60 + 6

= 66°

12

PNO=m 4X+20

= 4( ) +20

= 48+20

= 68°

12

PON=m 3X+10

= 3( ) +10

= 36+10

= 46°

12X = 12

Checking solution:

66° + 68° +46° = 180°

180°=180°PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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