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Warm Up Classify each angle as acute, obtuse, or right. 1. 2. 3.

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. obtuse. x = 5; 8; 16; 23. 1. In the figure, n is a whole number. What is the smallest possible value for n?. 2.

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Warm Up Classify each angle as acute, obtuse, or right. 1. 2. 3.

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  1. 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 obtuse x = 5; 8; 16; 23

  2. 1. In the figure, n is a whole number. What is the smallest possible value for n? 2. A sewing club is making a quilt consisting of 25 squares with each side of the square measuring 30 centimeters. If the quilt has five rows and five columns, what is the perimeter of the quilt?

  3. 1. In the figure, n is a whole number. What is the smallest possible value for n?

  4. 2. A sewing club is making a quilt consisting of 25 squares with each side of the square measuring 30 centimeters. If the quilt has five rows and five columns, what is the perimeter of the quilt?

  5. Objectives Classify triangles by their angle measures and side lengths. Use triangle classification to find angle measures and side lengths.

  6. Vocabulary acute triangle equiangular triangle right triangle obtuse triangle equilateral triangle isosceles triangle scalene triangle

  7. California Standards 12.0 Students find and use measures of sides and of interior and exterior angles of triangles and polygons to classify figures and solve problems. 15.0 Students use the Pythagorean theorem to determine distance and find missing lengths of sides of right triangles.

  8. 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.

  9. C A AB, BC, and AC are the sides of ABC. B A, B, C are the triangle's vertices.

  10. By Angle Measures Acute Triangle Three acute angles

  11. By Angle Measures Equilateral (Equiangular) Triangle Three congruent acute angles

  12. By Angle Measures Right Triangle One right angle

  13. By Angle Measures Obtuse Triangle One obtuse angle

  14. FHG is an equilateral (Equiangular) triangle by definition. Teach! Example 1 Classify FHG by its angle measures. EHG is a right angle. Therefore mEHF +mFHG = 90°. By substitution, 30°+ mFHG = 90°. SomFHG = 60°.

  15. By Side Lengths Equilateral Triangle Three congruent sides

  16. By Side Lengths Isosceles Triangle At least two congruent sides

  17. By Side Lengths Scalene Triangle No congruent sides

  18. Remember! When you look at a figure, you cannot assume segments are congruent based on appearance. They must be marked as congruent.

  19. From the figure, . So HF = 10, and EHF is isosceles. Example 1: Classifying Triangles by Side Lengths Classify EHF by its side lengths.

  20. By the Segment Addition Postulate, EG = EF + FG = 10 + 4 = 14. Since no sides are congruent, EHG is scalene. TEACH! : Classifying the Triangle by Side Lengths Classify EHGby its side lengths.

  21. From the figure, . So AC = 15, and ACD is scalene and probably obtuse. TEACH! Example 3 Classify ACD by its side lengths.

  22. Example 3: Using Triangle Classification Find the side lengths of JKL. Step 1 Find the value of x. Given. JK = KL Def. of  segs. Substitute (4x – 10.7) for JK and (2x + 6.3) for KL. 4x – 10.7 = 2x + 6.3 Add 10.7 and subtract 2x from both sides. 2x = 17.0 x = 8.5 Divide both sides by 2.

  23. Example 3 Continued Find the side lengths of JKL. 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

  24. TEACH! Example 3 Find the side lengths of equilateral FGH. Step 1 Find the value of y. Given. FG = GH = FH Def. of  segs. Substitute (3y – 4) for FG and (2y + 3) for GH. 3y – 4 = 2y + 3 Add 4 and subtract 2y from both sides. y = 7

  25. TEACH! Example 3 Continued Find the side lengths of equilateral FGH. 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

  26. Example 4: Application 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? 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

  27. 420  54 = 7 triangles 7 9 Example 4: Application Continued 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? 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. 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.

  28. TEACH! Example 4 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. 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.

  29. 100  7 = 14 triangles 2 7 TEACH! Example 4 Continued 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. 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. There is not enough steel to complete a fifteenth triangle. So the manufacturer can make 14 triangles from a 100 in. piece of steel.

  30. TEACH! Example 5 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. 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.

  31. TEACH! Example 5 Continued 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. 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.

  32. 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

  33. Warm Up 1. Find the measure of exterior DBA of BCD, if mDBC = 30°, mC= 70°, and mD = 80°. 2. What is the complement of an angle with measure 17°? 3. How many lines can be drawn through N parallel to MP? Why? 150° 73° 1; Parallel Post.

  34. Objectives Find the measures of interior and exterior angles of triangles. Apply theorems about the interior and exterior angles of triangles.

  35. Vocabulary auxiliary line corollary interior exterior interior angle exterior angle remote interior angle

  36. An auxiliary line is a line that is added to a figure to aid in a proof. An auxiliary line used in the Triangle Sum Theorem

  37. Sum. Thm Example 6A: Application After an accident, the positions of cars are measured by law enforcement to investigate the collision. Use the diagram drawn from the information collected to find mXYZ. mXYZ + mYZX + mZXY = 180° Substitute 40 for mYZX and 62 for mZXY. mXYZ + 40+ 62= 180 mXYZ + 102= 180 Simplify. mXYZ = 78° Subtract 102 from both sides.

  38. Sum. Thm TEACH! Example 6 Use the diagram to find mMJK. mMJK + mJKM + mKMJ = 180° Substitute 104 for mJKM and 44 for mKMJ. mMJK + 104+ 44= 180 mMJK + 148= 180 Simplify. Subtract 148 from both sides. mMJK = 32°

  39. A corollary is a theorem whose proof follows directly from another theorem. Here are two corollaries to the Triangle Sum Theorem.

  40. Acute s of rt. are comp. Example 7: Finding Angle Measures in Right Triangles One of the acute angles in a right triangle measures 2x°. What is the measure of the other acute angle? Let the acute angles be A and B, with mA = 2x°. mA + mB = 90° 2x+ mB = 90 Substitute 2x for mA. mB = (90 – 2x)° Subtract 2x from both sides.

  41. Acute s of rt. are comp. TEACH! Example 7a The measure of one of the acute angles in a right triangle is 63.7°. What is the measure of the other acute angle? Let the acute angles be A and B, with mA = 63.7°. mA + mB = 90° 63.7 + mB = 90 Substitute 63.7 for mA. mB = 26.3° Subtract 63.7 from both sides.

  42. Acute s of rt. are comp. TEACH! Example 7b The measure of one of the acute angles in a right triangle is x°. What is the measure of the other acute angle? Let the acute angles be A and B, with mA = x°. mA + mB = 90° x+ mB = 90 Substitute x for mA. mB = (90 – x)° Subtract x from both sides.

  43. Let the acute angles be A and B, with mA = 48 . Acute s of rt. are comp. 48 + mB = 90 Substitute 48 for mA. mB = 41 Subtract 48 from both sides. 3° 5 2° 5 2° 5 2 5 2 5 2 5 TEACH! Example 7c The measure of one of the acute angles in a right triangle is 48 . What is the measure of the other acute angle? mA + mB = 90°

  44. The interior is the set of all points inside the figure. The exterior is the set of all points outside the figure. Exterior Interior

  45. An interior angle is formed by two sides of a triangle. An exterior angle is formed by one side of the triangle and extension of an adjacent side. 4 is an exterior angle. Exterior Interior 3 is an interior angle.

  46. Each exterior angle has two remote interior angles. A remote interior angle is an interior angle that is not adjacent to the exterior angle. 4 is an exterior angle. The remote interior angles of 4 are 1 and 2. Exterior Interior 3 is an interior angle.

  47. Corollary: The measure of an exterior angle of a triangle is greater than the measure of either of its remote angles.

  48. Example 8: Applying the Exterior Angle Theorem Find mB. mA + mB = mBCD Ext.  Thm. Substitute 15 for mA, 2x + 3 for mB, and 5x – 60 for mBCD. 15 + 2x + 3= 5x – 60 2x + 18= 5x – 60 Simplify. Subtract 2x and add 60 to both sides. 78 = 3x 26 = x Divide by 3. mB = 2x + 3 = 2(26) + 3 = 55°

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