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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
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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.
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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
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A TRIANGLE is a three-sided polygon
Standards 4 and 5
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55°
64° 61°
21°
110° 49°
RIGHT TRIANGLE
ACUTE TRIANGLE
OBTUSE TRIANGLE
CLASSIFYING TRIANGLES BY ANGLESANGLES
STANDARDS 4 and 5
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Parts of a RIGHT TRIANGLE
Leg
Leg
HYPOTENUSE
Right Angle
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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
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Parts of an ISOSCELES TRIANGLE
Base Angles
Base
LegLeg
Vertex Angle
Standards 4 and 5
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9
B C
A
CONGRUENCE of triangles is REFLEXIVEREFLEXIVE:
ABC ABC
Standards 4 and 5
bc
a
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CONGRUENCE of triangles is SYMMETRIC:SYMMETRIC:
A
CB M
K
L
IF THEN ABC KLM KLM ABC
Standards 4 and 5
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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
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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
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Standards 4 and 5
(3X+4)°
(7X+5)°
(6X+69)°Z
Find in the figure below:Zm
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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:
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Standards 4 and 5
(3Z+8)°
(9Z+6)°
(5Z+84)° X
Find in the figure below:Xm
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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
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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
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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
Since both triangles are congruent, corresponding sides are congruent as well:
-3 -35X = 455 5
X = 9
Now finding DF:
DF=3X + 1
=3( ) + 1 9
= 27 + 1
DF= 28
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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
Since both triangles are congruent, corresponding sides are congruent as well:
-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
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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
Since both triangles are congruent, corresponding sides are congruent as well:
-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
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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?
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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
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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
Since both triangles are congruent then their corresponding angles are congruent:
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
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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
Since both triangles are congruent then their corresponding angles are congruent:
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
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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
Since both triangles are congruent then their corresponding angles are congruent:
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