Gch04 l1
- 1. Holt Geometry 4-1 Classifying Triangles4-1 Classifying Triangles Holt Geometry Warm UpWarm Up Lesson PresentationLesson Presentation Lesson QuizLesson Quiz
- 2. Holt Geometry 4-1 Classifying Triangles Warm Up Classify each angle as acute, obtuse, or right. 1. 2. 3. 4. If the perimeter is 47, find x and the lengths of the three sides. right acute x = 5; 8; 16; 23 obtuse
- 3. Holt Geometry 4-1 Classifying Triangles Classify triangles by their angle measures and side lengths. Use triangle classification to find angle measures and side lengths. Objectives
- 4. Holt Geometry 4-1 Classifying Triangles acute triangle equiangular triangle right triangle obtuse triangle equilateral triangle isosceles triangle scalene triangle Vocabulary
- 5. Holt Geometry 4-1 Classifying Triangles Recall that a triangle ( ) is a polygon with three sides. Triangles can be classified in two ways: by their angle measures or by their side lengths.
- 6. Holt Geometry 4-1 Classifying Triangles B A C AB, BC, and AC are the sides of ABC. A, B, C are the triangle's vertices.
- 7. Holt Geometry 4-1 Classifying Triangles Acute Triangle Three acute angles Triangle Classification By Angle Measures
- 8. Holt Geometry 4-1 Classifying Triangles Equiangular Triangle Three congruent acute angles Triangle Classification By Angle Measures
- 9. Holt Geometry 4-1 Classifying Triangles Right Triangle One right angle Triangle Classification By Angle Measures
- 10. Holt Geometry 4-1 Classifying Triangles Obtuse Triangle One obtuse angle Triangle Classification By Angle Measures
- 11. Holt Geometry 4-1 Classifying Triangles Classify BDC by its angle measures. Example 1A: Classifying Triangles by Angle Measures ∠B is an obtuse angle. ∠B is an obtuse angle. So BDC is an obtuse triangle.
- 12. Holt Geometry 4-1 Classifying Triangles Classify ABD by its angle measures. Example 1B: Classifying Triangles by Angle Measures ∠ABD and ∠CBD form a linear pair, so they are supplementary. Therefore m∠ABD + m∠CBD = 180°. By substitution, m∠ABD + 100° = 180°. So m∠ABD = 80°. ABD is an acute triangle by definition.
- 13. Holt Geometry 4-1 Classifying Triangles Classify FHG by its angle measures. Check It Out! Example 1 ∠EHG is a right angle. Therefore m∠EHF +m∠FHG = 90°. By substitution, 30°+ m∠FHG = 90°. So m∠FHG = 60°. FHG is an equiangular triangle by definition.
- 14. Holt Geometry 4-1 Classifying Triangles Equilateral Triangle Three congruent sides Triangle Classification By Side Lengths
- 15. Holt Geometry 4-1 Classifying Triangles Isosceles Triangle At least two congruent sides Triangle Classification By Side Lengths
- 16. Holt Geometry 4-1 Classifying Triangles Scalene Triangle No congruent sides Triangle Classification By Side Lengths
- 17. Holt Geometry 4-1 Classifying Triangles Remember! When you look at a figure, you cannot assume segments are congruent based on appearance. They must be marked as congruent.
- 18. Holt Geometry 4-1 Classifying Triangles Classify EHF by its side lengths. Example 2A: Classifying Triangles by Side Lengths From the figure, . So HF = 10, and EHF is isosceles.
- 19. Holt Geometry 4-1 Classifying Triangles Classify EHG by its side lengths. Example 2B: Classifying Triangles by Side Lengths By the Segment Addition Postulate, EG = EF + FG = 10 + 4 = 14. Since no sides are congruent, EHG is scalene.
- 20. Holt Geometry 4-1 Classifying Triangles Classify ACD by its side lengths. Check It Out! Example 2 From the figure, . So AC = 15, and ACD is isosceles.
- 21. Holt Geometry 4-1 Classifying Triangles Find the side lengths of JKL. Example 3: Using Triangle Classification Step 1 Find the value of x. Given. JK = KL Def. of ≅ segs. 4x – 10.7 = 2x + 6.3 Substitute (4x – 10.7) for JK and (2x + 6.3) for KL. 2x = 17.0 x = 8.5 Add 10.7 and subtract 2x from both sides. Divide both sides by 2.
- 22. Holt Geometry 4-1 Classifying Triangles Find the side lengths of JKL. Example 3 Continued Step 2 Substitute 8.5 into the expressions to find the side lengths. JK = 4x – 10.7 = 4(8.5) – 10.7 = 23.3 KL = 2x + 6.3 = 2(8.5) + 6.3 = 23.3 JL = 5x + 2 = 5(8.5) + 2 = 44.5
- 23. Holt Geometry 4-1 Classifying Triangles Find the side lengths of equilateral FGH. Check It Out! Example 3 Step 1 Find the value of y. Given. FG = GH = FH Def. of ≅ segs. 3y – 4 = 2y + 3 Substitute (3y – 4) for FG and (2y + 3) for GH. y = 7 Add 4 and subtract 2y from both sides.
- 24. Holt Geometry 4-1 Classifying Triangles Find the side lengths of equilateral FGH. Check It Out! Example 3 Continued Step 2 Substitute 7 into the expressions to find the side lengths. FG = 3y – 4 = 3(7) – 4 = 17 GH = 2y + 3 = 2(7) + 3 = 17 FH = 5y – 18 = 5(7) – 18 = 17
- 25. Holt Geometry 4-1 Classifying Triangles The amount of steel needed to make one triangle is equal to the perimeter P of the equilateral triangle. P = 3(18) P = 54 ft A steel mill produces roof supports by welding pieces of steel beams into equilateral triangles. Each side of the triangle is 18 feet long. How many triangles can be formed from 420 feet of steel beam? Example 4: Application
- 26. Holt Geometry 4-1 Classifying Triangles A steel mill produces roof supports by welding pieces of steel beams into equilateral triangles. Each side of the triangle is 18 feet long. How many triangles can be formed from 420 feet of steel beam? Example 4: Application Continued To find the number of triangles that can be made from 420 feet of steel beam, divide 420 by the amount of steel needed for one triangle. 420 ÷ 54 = 7 triangles 7 9 There is not enough steel to complete an eighth triangle. So the steel mill can make 7 triangles from a 420 ft. piece of steel beam.
- 27. Holt Geometry 4-1 Classifying Triangles The amount of steel needed to make one triangle is equal to the perimeter P of the equilateral triangle. P = 3(7) P = 21 in. Each measure is the side length of an equilateral triangle. Determine how many 7 in. triangles can be formed from a 100 in. piece of steel. Check It Out! Example 4a
- 28. Holt Geometry 4-1 Classifying Triangles To find the number of triangles that can be made from 100 inches of steel, divide 100 by the amount of steel needed for one triangle. 100 ÷ 7 = 14 triangles 2 7 There is not enough steel to complete a fifteenth triangle. So the manufacturer can make 14 triangles from a 100 in. piece of steel. Each measure is the side length of an equilateral triangle. Determine how many 7 in. triangles can be formed from a 100 in. piece of steel. Check It Out! Example 4a Continued
- 29. Holt Geometry 4-1 Classifying Triangles The amount of steel needed to make one triangle is equal to the perimeter P of the equilateral triangle. P = 3(10) P = 30 in. Each measure is the side length of an equilateral triangle. Determine how many 10 in. triangles can be formed from a 100 in. piece of steel. Check It Out! Example 4b
- 30. Holt Geometry 4-1 Classifying Triangles To find the number of triangles that can be made from 100 inches of steel, divide 100 by the amount of steel needed for one triangle. 100 ÷ 10 = 10 triangles The manufacturer can make 10 triangles from a 100 in. piece of steel. Each measure is the side length of an equilateral triangle. Determine how many 10 in. triangles can be formed from a 100 in. piece of steel. Check It Out! Example 4b Continued
- 31. Holt Geometry 4-1 Classifying Triangles Lesson Quiz Classify each triangle by its angles and sides. 1. MNQ 2. NQP 3. MNP 4. Find the side lengths of the triangle. acute; equilateral obtuse; scalene acute; scalene 29; 29; 23