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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. Objectives. Classify triangles by their angle measures and side lengths.

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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. Objectives Classify triangles by their angle measures and side lengths. Use triangle classification to find angle measures and side lengths. Find the measures of interior and exterior angles of triangles. Apply theorems about the interior and exterior angles of triangles.

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

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

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

  7. By Angle Measures Acute Triangle Three acute angles

  8. By Angle Measures Equiangular Triangle Three congruent acute angles

  9. By Angle Measures Right Triangle One right angle

  10. By Angle Measures Obtuse Triangle One obtuse angle

  11. DBC is an obtuse angle. So BDC is an obtuse triangle. Example 1A: Classifying Triangles by Angle Measures Classify BDC by its angle measures. DBC is an obtuse angle.

  12. Therefore mABD + mCBD = 180°. By substitution, mABD + 100°= 180°. SomABD = 80°. ABD is an acute triangle by definition. Example 1B: Classifying Triangles by Angle Measures Classify ABD by its angle measures. ABD andCBD form a linear pair, so they are supplementary.

  13. FHG is an equiangular triangle by definition. Check It Out! 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°.

  14. By Side Lengths Equilateral Triangle Three congruent sides

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

  16. By Side Lengths Scalene Triangle No congruent sides

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

  18. From the figure, . So HF = 10, and EHF is isosceles. Example 2A: Classifying Triangles by Side Lengths Classify EHF by its side lengths.

  19. By the Segment Addition Postulate, EG = EF + FG = 10 + 4 = 14. Since no sides are congruent, EHG is scalene. Example 2B: Classifying Triangles by Side Lengths Classify EHGby its side lengths.

  20. From the figure, . So AC = 15, and ACD is isosceles. Check It Out! Example 2 Classify ACD by its side lengths.

  21. 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.

  22. 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

  23. Check It Out! 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

  24. Check It Out! 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

  25. 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

  26. 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.

  27. 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

  28. Sum. Thm Example 1A: 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.

  29. 118° Example 1B: 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YWZ. Step 1 Find mWXY. mYXZ + mWXY = 180° Lin. Pair Thm. and  Add. Post. 62 + mWXY = 180 Substitute 62 for mYXZ. mWXY = 118° Subtract 62 from both sides.

  30. 118° Sum. Thm Example 1B: Application Continued 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YWZ. Step 2 Find mYWZ. mYWX + mWXY + mXYW = 180° Substitute 118 for mWXY and 12 for mXYW. mYWX + 118+ 12= 180 mYWX + 130= 180 Simplify. Subtract 130 from both sides. mYWX = 50°

  31. Sum. Thm Check It Out! Example 1 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°

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

  33. Acute s of rt. are comp. Example 2: 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.

  34. Acute s of rt. are comp. Check It Out! Example 2a 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.

  35. Acute s of rt. are comp. Check It Out! Example 2b 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.

  36. 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 Check It Out! Example 2c 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°

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

  38. 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.

  39. 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.

  40. Example 3: 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°

  41. Check It Out! Example 3 Find mACD. mACD = mA + mB Ext.  Thm. Substitute 6z – 9 for mACD, 2z + 1 for mA, and 90 for mB. 6z – 9 = 2z + 1+ 90 6z – 9= 2z + 91 Simplify. Subtract 2z and add 9 to both sides. 4z = 100 z = 25 Divide by 4. mACD = 6z – 9 = 6(25) – 9 = 141°

  42. Example 4: Applying the Third Angles Theorem Find mK and mJ. K  J Third s Thm. mK = mJ Def. of s. 4y2= 6y2 – 40 Substitute 4y2 for mK and 6y2 – 40 for mJ. –2y2 = –40 Subtract 6y2 from both sides. y2 = 20 Divide both sides by -2. So mK = 4y2 = 4(20) = 80°. Since mJ = mK,mJ =80°.

  43. Check It Out! Example 4 Find mP and mT. P  T Third s Thm. mP = mT Def. of s. 2x2= 4x2 – 32 Substitute 2x2 for mP and 4x2 – 32 for mT. –2x2 = –32 Subtract 4x2 from both sides. x2 = 16 Divide both sides by -2. So mP = 2x2 = 2(16) = 32°. Since mP = mT,mT =32°.

  44. 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

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