Will return graded Unit 2 tests on Friday for reflection.

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Will return graded Unit 2 tests on Friday for reflection. 11/30: Classify a triangle by its sides and angles. Do Now: Have Out: - Today’s Handouts Begin your Warm-Up Agenda: Warm-Up & Check Guided Notes Guided Practice Independent Practice Closing & HW (…maybe…) Homework Handout (5 problems + 2 EC Problems) Will return graded Unit 2 tests on Friday for reflection. 1st Period

Use the linear pair theorem to find the missing triangle angle measures: Z Y X 30° 90° 30° 90° 60° 180° 60°

Triangle Sum Theorem

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Graphic Organizer Use the examples and non-examples at your table to fill out your graphic organizer. Once you have finished, check with Mr. G BEFORE trying Examples 1 & 2

All 3 sides are congruent At least 2 congruent sides No sides are congruent

All 3 angles are congruent One right angle One obtuse angle All 3 angles are acute

How do we know it is obtuse? Example 1: Classifying a Triangle by its Sides & Angles Right Triangle Isosceles Triangle Obtuse Triangle How do we know it is obtuse?

Example 2: Apply Properties of Isosceles & Equilateral Triangles Check! 3x – 3 = 12 3(5) – 3 = 12 15 – 3 = 12 12 = 12

AB = BC 2x + 7 = 3x – 2 2x + 7 3x – 2 7 = x – 2 9 = x AB = 25 units Example 2: Apply Properties of Isosceles & Equilateral Triangles AB = BC 2x + 7 = 3x – 2 2x + 7 3x – 2 7 = x – 2 9 = x 64° AB = 2x + 7 AB = 2(9) + 7 AB = 18 + 7 AB = 25 units

Example 2: Apply Properties of Isosceles & Equilateral Triangles ∠BAC ≅ ∠BCA 52° 2x + 7 3x – 2 m∠BCA = 64° Triangle Sum Theorem 64° 64° m∠BAC + m∠BCA + m∠ABC = 180° ______ + ______ + ________ = 180° 64° 64° m∠BAC m∠ABC = 52°

Triangle Sum Theorem 6x + 18 = 180° 6x = 162 x = 27 Example 3: Apply Triangle Sum Theorem to Classify Triangles Triangle Sum Theorem (2x + 10)° m∠ABC + m∠ACB + m∠BAC = 180° (x + 20)° (3x - 12)° ______ + _______ + _______ = 180° (x + 20)° (3x – 12 )° (2x + 10)° 6x + 18 = 180° 6x = 162 x = 27

Example 3: Apply Triangle Sum Theorem to Classify Triangles m∠ABC = (x + 20)° (2x + 10)° m∠ABC = 47° (x + 20)° (3x - 12)° m∠ACB = (3x – 12)° m∠ACB = 69° m∠BAC = (2x + 10)° m∠BAC = 64°