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1 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 5APROBLEM 5B END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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4 A TRIANGLE is a three-sided polygon Standards 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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5 55° 64°61° 21° 110° 49° RIGHT TRIANGLE ACUTE TRIANGLE OBTUSE TRIANGLE ANGLES CLASSIFYING TRIANGLES BY ANGLES STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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6 Parts of a RIGHT TRIANGLE Leg HYPOTENUSE Right Angle PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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SIDES CLASSIFYING TRIANGLES BY SIDES 10 4 9 22 13 13 13 SCALENE TRIANGLE ISOSCELES TRIANGLE EQUILATERAL TRIANGLE 9 Standards 4 and 5 Also EQUIANGULAR PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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8 Parts of an ISOSCELES TRIANGLE Base Angles Base Leg Vertex Angle Standards 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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9 B C A REFLEXIVE CONGRUENCE of triangles is REFLEXIVE: ABC ABC Standards 4 and 5 b c a PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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10 SYMMETRIC: CONGRUENCE of triangles is SYMMETRIC: A CB M K L IFTHEN ABC KLM KLM ABC Standards 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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11 TRANSITIVE CONGRUENCE of triangles is TRANSITIVE A CBM K L IFAND R TS THEN ABC RST KLM RST ABC KLM Standards 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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). + 1 m 2 m = 3 m Exterior Angle Theorem Remote Interior Angles Exterior Angle Standards 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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13 Standards 4 and 5 (3X+4)° (7X+5)° (6X+69)° Z Find in the figure below: Z m PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 +69 10X + 9 = 6X +69 -9 10X = 6X + 60 -6X 4X = 60 4 X=15 By Linear Pair: Z m + (6X+69)° = 180° Z m + 6X+69 = 180 Z m + 6( )+69 = 180 15 Z m + 90+69 = 180 Z m + 159 = 180 -159 Z m = 21° Find in the figure below: Z m Substituting the value for X: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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15 Standards 4 and 5 (3Z+8)° (9Z+6)° (5Z+84)° X Find in the figure below: X m PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 +84 12Z + 14 = 5Z +84 -14 12Z = 5Z + 70 -5Z 7Z = 70 7 Z=10 By Linear Pair: X m + (5Z+84)° = 180° X m + 5Z + 84 = 180 X m + 5( )+84 = 180 10 X m + 50+84 = 180 X m + 134 = 180 -134 X m = 46° Find in the figure below: X m Substituting the value for Z: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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17 CONGRUENT TRIANGLES CPCTC Corresponding Parts of Congruent Triangles are Congruent CPCTC ABC KLM by CPCTC BC A L M K Standards 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 56 2X – 2 6X + 8 = 56 Since both triangles are congruent, corresponding sides are congruent as well: -8 6X = 48 6 X = 8 Now finding RT: RT=2X – 2 =2( ) – 2 8 = 16 – 2 RT= 14 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 48 3X + 1 5X + 3 = 48 Since both triangles are congruent, corresponding sides are congruent as well: -3 5X = 45 5 X = 9 Now finding DF: DF=3X + 1 =3( ) + 1 9 = 27 + 1 DF= 28 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 Since both triangles are congruent, corresponding sides are congruent as well: -3 5X =3X + 10 2 Now finding ST: ST=6X + 9 =6( ) +9 5 = 30 + 9 6X + 9 -3X 2X = 10 X = 5 ST= 39 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 Since both triangles are congruent, corresponding sides are congruent as well: -2 9X =4X + 15 5 Now finding JI: JI=2X + 3 =2( ) +3 3 = 6 + 3 2X + 3 -4X 5X = 15 X = 3 JI= 9 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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22 35° 58° 87° ++ C C m B B m A A m = 180° 35° + 87° + 58° = 180° The sum of the interior angles of a triangle is always 180° ANGLE SUM THEOREM: Standards 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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23 A A m = 180° B B m + C C m + If and A = 90 ° m then = 180° B m + C m + 90° - 90° -90° = 90° B m C m + 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? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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24 Standards 4 and 5 If QRT UVW, with Q m = 100°, V m = 30°, and, find the value for Z. W m = 4Z +10 Q T RW V U Since both triangles are congruent then their corresponding angles are congruent: 4Z + 10 100° 30° Q = m U m Q U 100° = 100° Now from the Angle Sum Theorem: V m W m + =180° U m + 100° + 30° + = 180° W m 130 + = 180° W m -130° W = 50° m W m = 4Z +10 If then 4Z + 10 = 50 -10 100° + 30° + = 180° W m 4Z = 40 4 Z = 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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25 Standards 4 and 5 If DEF JKL, with D m = 110°, K m = 20°, and, find the value for Z. L m = 3Z +23 D F EL K J Since both triangles are congruent then their corresponding angles are congruent: 3Z + 23 110° 20° D = m J m D J 110° = 110° Now from the Angle Sum Theorem: K m L m + =180° J m + 110° + 20° + = 180° L m 130 + = 180° L m -130° L = 50° m L m = 3Z + 23 If then 3Z + 23 = 50 -23 110° + 20° + = 180° L m 3Z = 27 3 Z = 9 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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26 Standards 4 and 5 If FGH IJK with, and GFH= m 3X+28 JKI= m 2X – 8 KJI= m 2X+20, what is the value for x and for each angle? F H GK 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 7X = 140 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° 20 X = 20 Checking solution: 32° + 88° +60° = 180° 180°=180° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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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 SP 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 12X = 144 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° 12 X = 12 Checking solution: 66° + 68° +46° = 180° 180°=180° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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