Unit 4 – Lesson 1 Apply Triangle Sum Properties

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Presentation transcript:

Unit 4 – Lesson 1 Apply Triangle Sum Properties And Angle Relationships

All sides are  Type of ∆ Definition Picture Equilateral Triangle CLASSIFICATION BY SIDES

At least 2 sides  leg leg base Type of ∆ Definition Picture Isosceles Triangle At least 2 sides  CLASSIFICATION BY SIDES leg leg base

No sides are  Type of ∆ Definition Picture Scalene Triangle CLASSIFICATION BY SIDES

All angles are  Type of ∆ Definition Picture Equiangular Triangle CLASSIFICATION BY ANGLES

All angles are acute Type of ∆ Definition Picture Acute Triangle CLASSIFICATION BY ANGLES

hypotenuse leg One right angle leg Type of ∆ Definition Picture Right Triangle hypotenuse leg One right angle CLASSIFICATION BY ANGLES leg

One obtuse angle and 2 acute angles Type of ∆ Definition Picture Obtuse Triangle One obtuse angle and 2 acute angles CLASSIFICATION BY ANGLES

Interior Angles – Angles inside a shape Exterior Angles – Angles outside a shape Corollary – Statement easily proved from another statement

b c a b c a The sum of the measures of a triangle = _______ 180°

If two triangles have two angles that are _______,  If two triangles have two angles that are _______, then the third angle is also ___________.  B E 80° 80° 40° 60° 40° 60° A C D F

The measure of an exterior angle of a triangle is equal to the sum of the measures of the two remote interior angles. a + b = 180° b + c + d = 180° b + c + d = a + b c + d = a

The acute angles of a right triangle are ______________. complementary a + c + 90 = 180° a a + c = 90° c

Classify the triangle by its sides and angles. scalene right

Classify the triangle by its sides and angles. equilateral equiangular

Classify the triangle by its sides and angles. isosceles obtuse

4. Find the value of x. Then classify the triangle by its angles. 2x + 3x + 55 = 180 5x + 55 = 180 -55 -55 75° 5x = 125 5 5 50° x = 25° acute

4. Find the value of x. Then classify the triangle by its angles. 4x – 5 + 3x + 5 + 40 = 180 7x + 40 = 180 -40 -40 75° 7x = 140 65° 7 7 x = 20° acute

4. Find the value of x. Then classify the triangle by its angles. 2x + 2 + 2x + 2 + 3x + 1 = 180 52° 7x + 5 = 180 52° -5 -5 76° 7x = 175 7 7 x = 25° acute

5. Find the measure of each exterior angle shown. x + 80 = 3x – 22 -x -x . 80 = 2x – 22 +22 +22 102 = 2x 2 2 51° = x 3(51) – 22 = 131°

5. Find the measure of each exterior angle shown. 2x + 103 – x = 6x – 7 x + 103 = 6x – 7 -x -x . 103 = 5x – 7 +7 +7 110 = 5x 5 5 22° = x 6(22) – 7 = 125°

5. Find the measure of each exterior angle shown. 2x + 3 + 51 = 4x + 8 2x + 54 = 4x + 8 -2x -2x . 54 = 2x + 8 46 = 2x 23° = x 4(23) + 8 = 100°

6. Solve for the variables. 65° 25° y + 90 + 65 = 180 y + 155 = 180 y = 25°

6. Solve for the variables. 5y + 55 + 70 = 180 5y + 125 = 180 -125 -125 5y = 55 5 5 y = 11° 55°

7. Find the measure of each missing angle. 120° 30° 60° 30° 60° 120°

HW Problem 4.1 # 26 Ans: 4.1 221 229 1-6, 14-18, 21-26 (FRONT PAGE) LT Page # Assignment Due 4.1 221 229 1-6, 14-18, 21-26 (FRONT PAGE) 32, 33, 35, 15, 16 (BACK PAGE) 4.1 # 26 Ans: