51 Perpendicular and Angle Bisectors


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1 51 Perpendicular and Angle Bisectors 51 Perpendicular and Angle Bisectors Warm Up Lesson Presentation Lesson Quiz Holt
2 51 Perpendicular and Angle Bisectors Warm Up Construct each of the following. 1. A perpendicular bisector. 2. An angle bisector. 3. Find the midpoint and slope of the segment (2, 8) and ( 4, 6).
3 51 Perpendicular and Angle Bisectors Objectives Prove and apply theorems about perpendicular bisectors. Prove and apply theorems about angle bisectors.
4 51 Perpendicular and Angle Bisectors equidistant locus Vocabulary
5 51 Perpendicular and Angle Bisectors When a point is the same distance from two or more objects, the point is said to be equidistant from the objects. Triangle congruence theorems can be used to prove theorems about equidistant points.
6 51 Perpendicular and Angle Bisectors
7 51 Perpendicular and Angle Bisectors A locus is a set of points that satisfies a given condition. The perpendicular bisector of a segment can be defined as the locus of points in a plane that are equidistant from the endpoints of the segment.
8 51 Perpendicular and Angle Bisectors Example 1A: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. MN MN = LN MN = 2.6 Bisector Thm. Substitute 2.6 for LN.
9 51 Perpendicular and Angle Bisectors Example 1B: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. BC Since AB = AC and, is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem. BC = 2CD BC = 2(12) = 24 Def. of seg. bisector. Substitute 12 for CD.
10 51 Perpendicular and Angle Bisectors TU Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. TU = UV 3x + 9 = 7x 17 9 = 4x = 4x 6.5 = x Bisector Thm. Substitute the given values. Subtract 3x from both sides. Add 17 to both sides. Divide both sides by 4. So TU = 3(6.5) + 9 = 28.5.
11 51 Perpendicular and Angle Bisectors Find the measure. Check It Out! Example 1a Given that line l is the perpendicular bisector of DE and EG = 14.6, find DG. DG = EG DG = 14.6 Bisector Thm. Substitute 14.6 for EG.
12 51 Perpendicular and Angle Bisectors Find the measure. Check It Out! Example 1b Given that DE = 20.8, DG = 36.4, and EG =36.4, find EF. Since DG = EG and, is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem. DE = 2EF 20.8 = 2EF Def. of seg. bisector. Substitute 20.8 for DE = EF Divide both sides by 2.
13 51 Perpendicular and Angle Bisectors Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line.
14 51 Perpendicular and Angle Bisectors
15 51 Perpendicular and Angle Bisectors Based on these theorems, an angle bisector can be defined as the locus of all points in the interior of the angle that are equidistant from the sides of the angle.
16 51 Perpendicular and Angle Bisectors Example 2A: Applying the Angle Bisector Theorem Find the measure. BC BC = DC BC = 7.2 Bisector Thm. Substitute 7.2 for DC.
17 51 Perpendicular and Angle Bisectors Example 2B: Applying the Angle Bisector Theorem Find the measure. m EFH, given that m EFG = 50. Since EH = GH, and, bisects EFG by the Converse of the Angle Bisector Theorem. Def. of bisector Substitute 50 for m EFG.
18 51 Perpendicular and Angle Bisectors Example 2C: Applying the Angle Bisector Theorem Find m MKL. Since, JM = LM, and, bisects JKL by the Converse of the Angle Bisector Theorem. m MKL = m JKM Def. of bisector 3a + 20 = 2a + 26 a + 20 = 26 a = 6 Substitute the given values. Subtract 2a from both sides. Subtract 20 from both sides. So m MKL = [2(6) + 26] = 38
19 51 Perpendicular and Angle Bisectors Check It Out! Example 2a Given that YW bisects XYZ and WZ = 3.05, find WX. WX = WZ WX = 3.05 Bisector Thm. Substitute 3.05 for WZ. So WX = 3.05
20 51 Perpendicular and Angle Bisectors Check It Out! Example 2b Given that m WYZ = 63, XW = 5.7, and ZW = 5.7, find m XYZ. m WYZ + m WYX = m XYZ m WYZ = m WYX m WYZ + m WYZ = m XYZ 2m WYZ = m XYZ 2(63 ) = m XYZ 126 = m XYZ Bisector Thm. Substitute m WYZ for m WYX. Simplify. Substitute 63 for m WYZ. Simplfiy.
21 51 Perpendicular and Angle Bisectors Example 3: Application John wants to hang a spotlight along the back of a display case. Wires AD and CD are the same length, and A and C are equidistant from B. How do the wires keep the spotlight centered? It is given that. So D is on the perpendicular bisector of by the Converse of the Angle Bisector Theorem. Since B is the midpoint of, is the perpendicular bisector of. Therefore the spotlight remains centered under the mounting.
22 51 Perpendicular and Angle Bisectors Check It Out! Example 3 S is equidistant from each pair of suspension lines. What can you conclude about QS? QS bisects PQR.
23 51 Perpendicular and Angle Bisectors Example 4: Writing Equations of Bisectors in the Coordinate Plane Write an equation in pointslope form for the perpendicular bisector of the segment with endpoints C(6, 5) and D(10, 1). Step 1 Graph. The perpendicular bisector of is perpendicular to at its midpoint.
24 51 Perpendicular and Angle Bisectors Example 4 Continued Step 2 Find the midpoint of. Midpoint formula. mdpt. of =
25 51 Perpendicular and Angle Bisectors Example 4 Continued Step 3 Find the slope of the perpendicular bisector. Slope formula. Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is
26 51 Perpendicular and Angle Bisectors Example 4 Continued Step 4 Use pointslope form to write an equation. The perpendicular bisector of has slope and passes through (8, 2). y y 1 = m(x x 1 ) Pointslope form Substitute 2 for y 1, for m, and 8 for x 1.
27 51 Perpendicular and Angle Bisectors Example 4 Continued
28 51 Perpendicular and Angle Bisectors Check It Out! Example 4 Write an equation in pointslope form for the perpendicular bisector of the segment with endpoints P(5, 2) and Q(1, 4). Step 1 Graph PQ. The perpendicular bisector of is perpendicular to at its midpoint.
29 51 Perpendicular and Angle Bisectors Check It Out! Example 4 Continued Step 2 Find the midpoint of PQ. Midpoint formula.
30 51 Perpendicular and Angle Bisectors Check It Out! Example 4 Continued Step 3 Find the slope of the perpendicular bisector. Slope formula. Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is.
31 51 Perpendicular and Angle Bisectors Check It Out! Example 4 Continued Step 4 Use pointslope form to write an equation. The perpendicular bisector of PQ has slope and passes through (3, 1). y y 1 = m(x x 1 ) Pointslope form Substitute.
32 51 Perpendicular and Angle Bisectors Lesson Quiz: Part I Use the diagram for Items Given that m ABD = 16, find m ABC Given that m ABD = (2x + 12) and m CBD = (6x 18), find m ABC. 54 Use the diagram for Items Given that FH is the perpendicular bisector of EG, EF = 4y 3, and FG = 6y 37, find FG Given that EF = 10.6, EH = 4.3, and FG = 10.6, find EG. 8.6
33 51 Perpendicular and Angle Bisectors Lesson Quiz: Part II 5. Write an equation in pointslope form for the perpendicular bisector of the segment with endpoints X(7, 9) and Y( 3, 5).
34 54 The Triangle Midsegment Theorem 54 The Triangle Midsegment Theorem Warm Up Lesson Presentation Lesson Quiz Holt
35 54 The Triangle Midsegment Theorem Warm Up Use the points A(2, 2), B(12, 2) and C(4, 8) for Exercises 1 5. (3, 5), (8, 5) 1. Find X and Y, the midpoints of AC and CB. 2. Find XY. 3. Find AB Find the slope of AB. 5. Find the slope of XY. 6. What is the slope of a line parallel to 3x + 2y = 12? 0 0
36 54 The Triangle Midsegment Theorem Objective Prove and use properties of triangle midsegments.
37 54 The Triangle Midsegment Theorem Vocabulary midsegment of a triangle
38 54 The Triangle Midsegment Theorem A midsegment of a triangle is a segment that joins the midpoints of two sides of the triangle. Every triangle has three midsegments, which form the midsegment triangle.
39 54 The Triangle Midsegment Theorem Example 1: Examining Midsegments in the Coordinate Plane The vertices of XYZ are X( 1, 8), Y(9, 2), and Z(3, 4). M and N are the midpoints of XZ and YZ. Show that and. Step 1 Find the coordinates of M and N.
40 54 The Triangle Midsegment Theorem Example 1 Continued Step 2 Compare the slopes of MN and XY. Since the slopes are the same,
41 54 The Triangle Midsegment Theorem Example 1 Continued Step 3 Compare the heights of MN and XY.
42 54 The Triangle Midsegment Theorem Check It Out! Example 1 The vertices of RST are R( 7, 0), S( 3, 6), and T(9, 2). M is the midpoint of RT, and N is the midpoint of ST. Show that and Step 1 Find the coordinates of M and N.
43 54 The Triangle Midsegment Theorem Check It Out! Example 1 Continued Step 2 Compare the slopes of MN and RS. Since the slopes are equal.
44 54 The Triangle Midsegment Theorem Check It Out! Example 1 Continued Step 3 Compare the heights of MN and RS. The length of MN is half the length of RS.
45 54 The Triangle Midsegment Theorem The relationship shown in Example 1 is true for the three midsegments of every triangle.
46 54 The Triangle Midsegment Theorem Example 2A: Using the Triangle Midsegment Theorem Find each measure. BD Midsegment Thm. Substitute 17 for AE. BD = 8.5 Simplify.
47 54 The Triangle Midsegment Theorem Example 2B: Using the Triangle Midsegment Theorem Find each measure. m CBD m CBD = m BDF m CBD = 26 Midsegment Thm. Alt. Int. s Thm. Substitute 26 for m BDF.
48 54 The Triangle Midsegment Theorem Check It Out! Example 2a Find each measure. JL 2(36) = JL 72 = JL Midsegment Thm. Substitute 36 for PN and multiply both sides by 2. Simplify.
49 54 The Triangle Midsegment Theorem Check It Out! Example 2b Find each measure. PM Midsegment Thm. Substitute 97 for LK. PM = 48.5 Simplify.
50 54 The Triangle Midsegment Theorem Check It Out! Example 2c Find each measure. m MLK Midsegment Thm. m MLK = m JMP Similar triangles m MLK = 102 Substitute.
51 54 The Triangle Midsegment Theorem Example 3: Indirect Measurement Application In an Aframe support, the distance PQ is 46 inches. What is the length of the support ST if S and T are at the midpoints of the sides? Midsegment Thm. Substitute 46 for PQ. ST = 23 Simplify. The length of the support ST is 23 inches.
52 54 The Triangle Midsegment Theorem Check It Out! Example 3 What if? Suppose Anna s result in Example 3 (p. 323) is correct. To check it, she measures a second triangle. How many meters will she measure between H and F? Midsegment Thm. Substitute 1550 for AE. HF = 775 m Simplify.
53 54 The Triangle Midsegment Theorem Lesson Quiz: Part I Use the diagram for Items 1 3. Find each measure. 1. ED 2. AB m BFE 44
54 54 The Triangle Midsegment Theorem Lesson Quiz: Part II 4. Find the value of n XYZ is the midsegment triangle of WUV. What is the perimeter of XYZ? 11.5
55 Indirect Proof Proof and and Inequalities in in One One Triangle Triangle Warm Up Lesson Presentation Lesson Quiz Holt
56 55 Indirect Proof and Inequalities in One Triangle Warm Up 1. Write a conditional from the sentence An isosceles triangle has two congruent sides. If a is isosc., then it has 2 sides. 2. Write the contrapositive of the conditional If it is Tuesday, then John has a piano lesson. If John does not have a piano lesson, then it is not Tuesday. 3. Show that the conjecture If x > 6, then 2x > 14 is false by finding a counterexample. x = 7
57 55 Indirect Proof and Inequalities in One Triangle Write indirect proofs. Objectives Apply inequalities in one triangle.
58 55 Indirect Proof and Inequalities in One Triangle indirect proof Vocabulary
59 55 Indirect Proof and Inequalities in One Triangle So far you have written proofs using direct reasoning. You began with a true hypothesis and built a logical argument to show that a conclusion was true. In an indirect proof, you begin by assuming that the conclusion is false. Then you show that this assumption leads to a contradiction. This type of proof is also called a proof by contradiction.
60 55 Indirect Proof and Inequalities in One Triangle
61 55 Indirect Proof and Inequalities in One Triangle Helpful Hint When writing an indirect proof, look for a contradiction of one of the following: the given information, a definition, a postulate, or a theorem.
62 55 Indirect Proof and Inequalities in One Triangle Example 1: Writing an Indirect Proof Write an indirect proof that if a > 0, then Step 1 Identify the conjecture to be proven. Given: a > 0 Prove: Step 2 Assume the opposite of the conclusion. Assume
63 55 Indirect Proof and Inequalities in One Triangle Example 1 Continued Step 3 Use direct reasoning to lead to a contradiction. Given, opposite of conclusion Zero Prop. of Mult. Prop. of Inequality 1 0 Simplify. However, 1 > 0.
64 55 Indirect Proof and Inequalities in One Triangle Example 1 Continued Step 4 Conclude that the original conjecture is true. The assumption that is false. Therefore
65 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 1 Write an indirect proof that a triangle cannot have two right angles. Step 1 Identify the conjecture to be proven. Given: A triangle s interior angles add up to 180. Prove: A triangle cannot have two right angles. Step 2 Assume the opposite of the conclusion. An angle has two right angles.
66 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 1 Continued Step 3 Use direct reasoning to lead to a contradiction. m 1 + m 2 + m 3 = m 3 = m 3 = 180 m 3 = 0 However, by the Protractor Postulate, a triangle cannot have an angle with a measure of 0.
67 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 1 Continued Step 4 Conclude that the original conjecture is true. The assumption that a triangle can have two right angles is false. Therefore a triangle cannot have two right angles.
68 55 Indirect Proof and Inequalities in One Triangle The positions of the longest and shortest sides of a triangle are related to the positions of the largest and smallest angles.
69 55 Indirect Proof and Inequalities in One Triangle Example 2A: Ordering Triangle Side Lengths and Angle Measures Write the angles in order from smallest to largest. The shortest side is, so the smallest angle is F. The longest side is, so the largest angle is G. The angles from smallest to largest are F, H and G.
70 55 Indirect Proof and Inequalities in One Triangle Example 2B: Ordering Triangle Side Lengths and Angle Measures Write the sides in order from shortest to longest. m R = 180 ( ) = 48 The smallest angle is R, so the shortest side is. The largest angle is Q, so the longest side is. The sides from shortest to longest are
71 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 2a Write the angles in order from smallest to largest. The shortest side is, so the smallest angle is B. The longest side is, so the largest angle is C. The angles from smallest to largest are B, A, and C.
72 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 2b Write the sides in order from shortest to longest. m E = 180 ( ) = 68 The smallest angle is D, so the shortest side is. The largest angle is F, so the longest side is. The sides from shortest to longest are
73 55 Indirect Proof and Inequalities in One Triangle A triangle is formed by three segments, but not every set of three segments can form a triangle.
74 55 Indirect Proof and Inequalities in One Triangle A certain relationship must exist among the lengths of three segments in order for them to form a triangle.
75 55 Indirect Proof and Inequalities in One Triangle Example 3A: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain. 7, 10, 19 No by the Triangle Inequality Theorem, a triangle cannot have these side lengths.
76 55 Indirect Proof and Inequalities in One Triangle Example 3B: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain. 2.3, 3.1, 4.6 Yes the sum of each pair of lengths is greater than the third length.
77 55 Indirect Proof and Inequalities in One Triangle Example 3C: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain. n + 6, n 2 1, 3n, when n = 4. Step 1 Evaluate each expression when n = 4. n n 2 1 (4) n 3(4) 12
78 55 Indirect Proof and Inequalities in One Triangle Example 3C Continued Step 2 Compare the lengths. Yes the sum of each pair of lengths is greater than the third length.
79 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3a Tell whether a triangle can have sides with the given lengths. Explain. 8, 13, 21 No by the Triangle Inequality Theorem, a triangle cannot have these side lengths.
80 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3b Tell whether a triangle can have sides with the given lengths. Explain. 6.2, 7, 9 Yes the sum of each pair of lengths is greater than the third side.
81 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3c Tell whether a triangle can have sides with the given lengths. Explain. t 2, 4t, t 2 + 1, when t = 4 Step 1 Evaluate each expression when t = 4. t t 4(4) 16 t (4)
82 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 3c Continued Step 2 Compare the lengths. Yes the sum of each pair of lengths is greater than the third length.
83 55 Indirect Proof and Inequalities in One Triangle Example 4: Finding Side Lengths The lengths of two sides of a triangle are 8 inches and 13 inches. Find the range of possible lengths for the third side. Let x represent the length of the third side. Then apply the Triangle Inequality Theorem. x + 8 > 13 x > 5 x + 13 > 8 x > > x 21 > x Combine the inequalities. So 5 < x < 21. The length of the third side is greater than 5 inches and less than 21 inches.
84 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 4 The lengths of two sides of a triangle are 22 inches and 17 inches. Find the range of possible lengths for the third side. Let x represent the length of the third side. Then apply the Triangle Inequality Theorem. x + 22 > 17 x > 5 x + 17 > 22 x > > x 39 > x Combine the inequalities. So 5 < x < 39. The length of the third side is greater than 5 inches and less than 39 inches.
85 55 Indirect Proof and Inequalities in One Triangle Example 5: Travel Application The figure shows the approximate distances between cities in California. What is the range of distances from San Francisco to Oakland? Let x be the distance from San Francisco to Oakland. x + 46 > 51 x > 5 x + 51 > 46 x > 5 5 < x < 97 Combine the inequalities > x 97 > x Inequal. Thm. Subtr. Prop. of Inequal. The distance from San Francisco to Oakland is greater than 5 miles and less than 97 miles.
86 55 Indirect Proof and Inequalities in One Triangle Check It Out! Example 5 The distance from San Marcos to Johnson City is 50 miles, and the distance from Seguin to San Marcos is 22 miles. What is the range of distances from Seguin to Johnson City? Let x be the distance from Seguin to Johnson City. x + 22 > 50 x > 28 x + 50 > 22 x > > x 28 < x < 72 Combine the inequalities. 72 > x Inequal. Thm. Subtr. Prop. of Inequal. The distance from Seguin to Johnson City is greater than 28 miles and less than 72 miles.
87 55 Indirect Proof and Inequalities in One Triangle Lesson Quiz: Part I 1. Write the angles in order from smallest to largest. C, B, A 2. Write the sides in order from shortest to longest.
88 55 Indirect Proof and Inequalities in One Triangle Lesson Quiz: Part II 3. The lengths of two sides of a triangle are 17 cm and 12 cm. Find the range of possible lengths for the third side. 5 cm < x < 29 cm 4. Tell whether a triangle can have sides with lengths 2.7, 3.5, and 9.8. Explain. No; is not greater than Ray wants to place a chair so it is 10 ft from his television set. Can the other two distances shown be 8 ft and 6 ft? Explain. Yes; the sum of any two lengths is greater than the third length.
89 56 Inequalities in Two Triangles 56 Inequalities in Two Triangles Warm Up Lesson Presentation Lesson Quiz Holt
90 56 Inequalities in Two Triangles Warm Up 1. Write the angles in order from smallest to largest. X, Z, Y 2. The lengths of two sides of a triangle are 12 cm and 9 cm. Find the range of possible lengths for the third side. 3 cm < s < 21 cm
91 56 Inequalities in Two Triangles Objective Apply inequalities in two triangles.
92 56 Inequalities in Two Triangles
93 56 Inequalities in Two Triangles Example 1A: Using the Hinge Theorem and Its Converse Compare m BAC and m DAC. Compare the side lengths in ABC and ADC. AB = AD AC = AC BC > DC By the Converse of the Hinge Theorem, m BAC > m DAC.
94 56 Inequalities in Two Triangles Example 1B: Using the Hinge Theorem and Its Converse Compare EF and FG. Compare the sides and angles in EFH angles in GFH. m GHF = = 98 EH = GH FH = FH m EHF > m GHF By the Hinge Theorem, EF < GF.
95 56 Inequalities in Two Triangles Example 1C: Using the Hinge Theorem and Its Converse Find the range of values for k. Step 1 Compare the side lengths in MLN and PLN. LN = LN LM = LP MN > PN By the Converse of the Hinge Theorem, m MLN > m PLN. 5k 12 < 38 k < 10 Substitute the given values. Add 12 to both sides and divide by 5.
96 56 Inequalities in Two Triangles Example 1C Continued Step 2 Since PLN is in a triangle, m PLN > 0. 5k 12 > 0 k < 2.4 Substitute the given values. Add 12 to both sides and divide by 5. Step 3 Combine the two inequalities. The range of values for k is 2.4 < k < 10.
97 56 Inequalities in Two Triangles Check It Out! Example 1a Compare m EGH and m EGF. Compare the side lengths in EGH and EGF. FG = HG EG = EG EF > EH By the Converse of the Hinge Theorem, m EGH < m EGF.
98 56 Inequalities in Two Triangles Compare BC and AB. Check It Out! Example 1b Compare the side lengths in ABD and CBD. AD = DC BD = BD m ADB > m BDC. By the Hinge Theorem, BC > AB.
99 56 Inequalities in Two Triangles Example 2: Travel Application John and Luke leave school at the same time. John rides his bike 3 blocks west and then 4 blocks north. Luke rides 4 blocks east and then 3 blocks at a bearing of N 10º E. Who is farther from school? Explain.
100 56 Inequalities in Two Triangles Example 2 Continued The distances of 3 blocks and 4 blocks are the same in both triangles. The angle formed by John s route (90º) is smaller than the angle formed by Luke s route (100º). So Luke is farther from school than John by the Hinge Theorem.
101 56 Inequalities in Two Triangles Check It Out! Example 2 When the swing ride is at full speed, the chairs are farthest from the base of the swing tower. What can you conclude about the angles of the swings at full speed versus low speed? Explain. The of the swing at full speed is greater than the at low speed because the length of the triangle on the opposite side is the greatest at full swing.
102 56 Inequalities in Two Triangles Example 3: Proving Triangle Relationships Write a twocolumn proof. Given: Prove: AB > CB Proof: Statements Reasons 1. Given 2. Reflex. Prop. of 3. Hinge Thm.
103 56 Inequalities in Two Triangles Check It Out! Example 3a Write a twocolumn proof. Given: C is the midpoint of BD. m 1 = m 2 m 3 > m 4 Prove: AB > ED
104 56 Inequalities in Two Triangles Proof: 1. C is the mdpt. of BD m 3 > m 4, m 1 = m Statements Reasons 1. Given 2. Def. of Midpoint 3. Def. of s 4. Conv. of Isoc. Thm. 5. AB > ED 5. Hinge Thm.
105 56 Inequalities in Two Triangles Write a twocolumn proof. Given: SRT STR TU > RU Prove: m TSU > m RSU Check It Out! Example 3b Statements 1. SRT STR TU > RU 1. Given Reasons 2. Conv. of Isoc. Thm. 3. Reflex. Prop. of 4. m TSU > m RSU 4. Conv. of Hinge Thm.
106 56 Inequalities in Two Triangles Lesson Quiz: Part I 1. Compare m ABC and m DEF. m ABC > m DEF 2. Compare PS and QR. PS < QR
107 56 Inequalities in Two Triangles Lesson Quiz: Part II 3. Find the range of values for z. 3 < z < 7
108 56 Inequalities in Two Triangles Lesson Quiz: Part III 4. Write a twocolumn proof. Prove: m XYW < m ZWY Given: Proof: Statements Reasons 1. Given 2. Reflex. Prop. of 3. m XYW < m ZWY 3. Conv. of Hinge Thm.
109 57 The Pythagorean Theorem 57 The Pythagorean Theorem Warm Up Lesson Presentation Lesson Quiz Holt
110 57 The Pythagorean Theorem Warm Up Classify each triangle by its angle measures acute right 3. Simplify If a = 6, b = 7, and c = 12, find a 2 + b 2 and find c 2. Which value is greater? 85; 144; c 2
111 57 The Pythagorean Theorem Objectives Use the Pythagorean Theorem and its converse to solve problems. Use Pythagorean inequalities to classify triangles.
112 57 The Pythagorean Theorem Pythagorean triple Vocabulary
113 57 The Pythagorean Theorem The Pythagorean Theorem is probably the most famous mathematical relationship. As you learned in Lesson 16, it states that in a right triangle, the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse. a 2 + b 2 = c 2
114 57 The Pythagorean Theorem Example 1A: Using the Pythagorean Theorem Find the value of x. Give your answer in simplest radical form. a 2 + b 2 = c 2 Pythagorean Theorem = x 2 Substitute 2 for a, 6 for b, and x for c. 40 = x 2 Simplify. Find the positive square root. Simplify the radical.
115 57 The Pythagorean Theorem Example 1B: Using the Pythagorean Theorem Find the value of x. Give your answer in simplest radical form. a 2 + b 2 = c 2 Pythagorean Theorem (x 2) = x 2 Substitute x 2 for a, 4 for b, and x for c. x 2 4x = x 2 Multiply. 4x + 20 = 0 Combine like terms. 20 = 4x Add 4x to both sides. 5 = x Divide both sides by 4.
116 57 The Pythagorean Theorem Check It Out! Example 1a Find the value of x. Give your answer in simplest radical form. a 2 + b 2 = c 2 Pythagorean Theorem = x 2 Substitute 4 for a, 8 for b, and x for c. 80 = x 2 Simplify. Find the positive square root. Simplify the radical.
117 57 The Pythagorean Theorem Check It Out! Example 1b Find the value of x. Give your answer in simplest radical form. a 2 + b 2 = c 2 x = (x + 4) 2 x = x 2 + 8x + 16 Pythagorean Theorem Substitute x for a, 12 for b, and x + 4 for c. Multiply. 128 = 8x Combine like terms. 16 = x Divide both sides by 8.
118 57 The Pythagorean Theorem Example 2: Crafts Application Randy is building a rectangular picture frame. He wants the ratio of the length to the width to be 3:1 and the diagonal to be 12 centimeters. How wide should the frame be? Round to the nearest tenth of a centimeter. Let l and w be the length and width in centimeters of the picture. Then l:w = 3:1, so l = 3w.
119 57 The Pythagorean Theorem Example 2 Continued a 2 + b 2 = c 2 (3w) 2 + w 2 = w 2 = 144 Pythagorean Theorem Substitute 3w for a, w for b, and 12 for c. Multiply and combine like terms. Divide both sides by 10. Find the positive square root and round.
120 57 The Pythagorean Theorem Check It Out! Example 2 What if...? According to the recommended safety ratio of 4:1, how high will a 30 foot ladder reach when placed against a wall? Round to the nearest inch. Let x be the distance in feet from the foot of the ladder to the base of the wall. Then 4x is the distance in feet from the top of the ladder to the base of the wall.
121 57 The Pythagorean Theorem Check It Out! Example 2 Continued a 2 + b 2 = c 2 (4x) 2 + x 2 = x 2 = 900 Pythagorean Theorem Substitute 4x for a, x for b, and 30 for c. Multiply and combine like terms. Since 4x is the distance in feet from the top of the ladder to the base of the wall, 4(7.28) 29 ft 1 in.
122 57 The Pythagorean Theorem A set of three nonzero whole numbers a, b, and c such that a 2 + b 2 = c 2 is called a Pythagorean triple.
123 57 The Pythagorean Theorem Example 3A: Identifying Pythagorean Triples Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. a 2 + b 2 = c 2 Pythagorean Theorem = c 2 Substitute 14 for a and 48 for b = c 2 Multiply and add. 50 = c Find the positive square root. The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2, so they form a Pythagorean triple.
124 57 The Pythagorean Theorem Example 3B: Identifying Pythagorean Triples Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. a 2 + b 2 = c 2 Pythagorean Theorem b 2 = 12 2 Substitute 4 for a and 12 for c. b 2 = 128 Multiply and subtract 16 from both sides. Find the positive square root. The side lengths do not form a Pythagorean triple because is not a whole number.
125 57 The Pythagorean Theorem Check It Out! Example 3a Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. a 2 + b 2 = c 2 Pythagorean Theorem = c 2 Substitute 8 for a and 10 for b. 164 = c 2 Multiply and add. Find the positive square root. The side lengths do not form a Pythagorean triple because is not a whole number.
126 57 The Pythagorean Theorem Check It Out! Example 3b Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. The side lengths are nonzero whole numbers that satisfy the equation a 2 + b 2 = c 2, so they form a Pythagorean triple. a 2 + b 2 = c 2 Pythagorean Theorem b 2 = 26 2 Substitute 24 for a and 26 for c. b 2 = 100 Multiply and subtract. b = 10 Find the positive square root.
127 57 The Pythagorean Theorem Check It Out! Example 3c Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. No. The side length 2.4 is not a whole number.
128 57 The Pythagorean Theorem Check It Out! Example 3d Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. Yes. The three side lengths are nonzero whole numbers that satisfy Pythagorean's Theorem. a 2 + b 2 = c 2 Pythagorean Theorem = c 2 Substitute 30 for a and 16 for b. c 2 = 1156 c = 34 Multiply. Find the positive square root.
129 57 The Pythagorean Theorem The converse of the Pythagorean Theorem gives you a way to tell if a triangle is a right triangle when you know the side lengths.
130 57 The Pythagorean Theorem You can also use side lengths to classify a triangle as acute or obtuse. c B a A b C
131 57 The Pythagorean Theorem To understand why the Pythagorean inequalities are true, consider ABC.
132 57 The Pythagorean Theorem Remember! By the Triangle Inequality Theorem, the sum of any two side lengths of a triangle is greater than the third side length.
133 57 The Pythagorean Theorem Example 4A: Classifying Triangles Tell if the measures can be the side lengths of a triangle. If so, classify the triangle as acute, obtuse, or right. 5, 7, 10 Step 1 Determine if the measures form a triangle. By the Triangle Inequality Theorem, 5, 7, and 10 can be the side lengths of a triangle.
134 57 The Pythagorean Theorem Step 2 Classify the triangle. Example 4A Continued c 2? = a 2 + b ? = ? 100 = > 74 Compare c 2 to a 2 + b 2. Substitute the longest side for c. Multiply. Add and compare. Since c 2 > a 2 + b 2, the triangle is obtuse.
135 57 The Pythagorean Theorem Example 4B: Classifying Triangles Tell if the measures can be the side lengths of a triangle. If so, classify the triangle as acute, obtuse, or right. 5, 8, 17 Step 1 Determine if the measures form a triangle. Since = 13 and 13 > 17, these cannot be the side lengths of a triangle.
136 57 The Pythagorean Theorem Check It Out! Example 4a Tell if the measures can be the side lengths of a triangle. If so, classify the triangle as acute, obtuse, or right. 7, 12, 16 Step 1 Determine if the measures form a triangle. By the Triangle Inequality Theorem, 7, 12, and 16 can be the side lengths of a triangle.
137 57 The Pythagorean Theorem Check It Out! Example 4a Continued Step 2 Classify the triangle. c 2? = a 2 + b ? = ? 256 = > 193 Compare c 2 to a 2 + b 2. Substitute the longest side for c. Multiply. Add and compare. Since c 2 > a 2 + b 2, the triangle is obtuse.
138 57 The Pythagorean Theorem Check It Out! Example 4b Tell if the measures can be the side lengths of a triangle. If so, classify the triangle as acute, obtuse, or right. 11, 18, 34 Step 1 Determine if the measures form a triangle. Since = 29 and 29 > 34, these cannot be the sides of a triangle.
139 57 The Pythagorean Theorem Check It Out! Example 4c Tell if the measures can be the side lengths of a triangle. If so, classify the triangle as acute, obtuse, or right. 3.8, 4.1, 5.2 Step 1 Determine if the measures form a triangle. By the Triangle Inequality Theorem, 3.8, 4.1, and 5.2 can be the side lengths of a triangle.
140 57 The Pythagorean Theorem Check It Out! Example 4c Continued Step 2 Classify the triangle. c 2? = a 2 + b ? = ? = < Compare c 2 to a 2 + b 2. Substitute the longest side for c. Multiply. Add and compare. Since c 2 < a 2 + b 2, the triangle is acute.
141 57 The Pythagorean Theorem 1. Find the value of x. Lesson Quiz: Part I An entertainment center is 52 in. wide and 40 in. high. Will a TV with a 60 in. diagonal fit in it? Explain.
142 57 The Pythagorean Theorem Lesson Quiz: Part II 3. Find the missing side length. Tell if the side lengths form a Pythagorean triple. Explain. 4. Tell if the measures 7, 11, and 15 can be the side lengths of a triangle. If so, classify the triangle as acute, obtuse, or right. yes; obtuse 13; yes; the side lengths are nonzero whole numbers that satisfy Pythagorean s Theorem.
143 58 Applying Special Right Triangles 58 Applying Special Right Triangles Warm Up Lesson Presentation Lesson Quiz Holt
144 58 Applying Special Right Triangles Warm Up For Exercises 1 and 2, find the value of x. Give your answer in simplest radical form Simplify each expression
145 58 Applying Special Right Triangles Objectives Justify and apply properties of triangles. Justify and apply properties of triangles.
146 58 Applying Special Right Triangles A diagonal of a square divides it into two congruent isosceles right triangles. Since the base angles of an isosceles triangle are congruent, the measure of each acute angle is 45. So another name for an isosceles right triangle is a triangle. A triangle is one type of special right triangle. You can use the Pythagorean Theorem to find a relationship among the side lengths of a triangle.
147 58 Applying Special Right Triangles
148 58 Applying Special Right Triangles Example 1A: Finding Side Lengths in a 4545º 90º Triangle Find the value of x. Give your answer in simplest radical form. By the Triangle Sum Theorem, the measure of the third angle in the triangle is 45. So it is a triangle with a leg length of 8.
149 58 Applying Special Right Triangles Example 1B: Finding Side Lengths in a 45º 45º 90º Triangle Find the value of x. Give your answer in simplest radical form. The triangle is an isosceles right triangle, which is a triangle. The length of the hypotenuse is 5. Rationalize the denominator.
150 58 Applying Special Right Triangles Check It Out! Example 1a Find the value of x. Give your answer in simplest radical form. By the Triangle Sum Theorem, the measure of the third angle in the triangle is 45. So it is a triangle with a leg length of x = 20 Simplify.
151 58 Applying Special Right Triangles Check It Out! Example 1b Find the value of x. Give your answer in simplest radical form. The triangle is an isosceles right triangle, which is a triangle. The length of the hypotenuse is 16. Rationalize the denominator.
152 58 Applying Special Right Triangles Example 2: Craft Application Jana is cutting a square of material for a tablecloth. The table s diagonal is 36 inches. She wants the diagonal of the tablecloth to be an extra 10 inches so it will hang over the edges of the table. What size square should Jana cut to make the tablecloth? Round to the nearest inch. Jana needs a triangle with a hypotenuse of = 46 inches.
153 58 Applying Special Right Triangles Check It Out! Example 2 What if...? Tessa s other dog is wearing a square bandana with a side length of 42 cm. What would you expect the circumference of the other dog s neck to be? Round to the nearest centimeter. Tessa needs a triangle with a hypotenuse of 42 cm.
154 58 Applying Special Right Triangles A triangle is another special right triangle. You can use an equilateral triangle to find a relationship between its side lengths.
155 58 Applying Special Right Triangles Example 3A: Finding Side Lengths in a 30º60º90º Triangle Find the values of x and y. Give your answers in simplest radical form. 22 = 2x 11 = x Hypotenuse = 2(shorter leg) Divide both sides by 2. Substitute 11 for x.
156 58 Applying Special Right Triangles Example 3B: Finding Side Lengths in a 30º60º90º Triangle Find the values of x and y. Give your answers in simplest radical form. Rationalize the denominator. y = 2x Hypotenuse = 2(shorter leg). Simplify.
157 58 Applying Special Right Triangles Check It Out! Example 3a Find the values of x and y. Give your answers in simplest radical form. Hypotenuse = 2(shorter leg) Divide both sides by 2. y = 27 Substitute for x.
158 58 Applying Special Right Triangles Check It Out! Example 3b Find the values of x and y. Give your answers in simplest radical form. y = 2(5) y = 10 Simplify.
159 58 Applying Special Right Triangles Check It Out! Example 3c Find the values of x and y. Give your answers in simplest radical form. 24 = 2x 12 = x Hypotenuse = 2(shorter leg) Divide both sides by 2. Substitute 12 for x.
160 58 Applying Special Right Triangles Check It Out! Example 3d Find the values of x and y. Give your answers in simplest radical form. Rationalize the denominator. x = 2y Hypotenuse = 2(shorter leg) Simplify.
161 58 Applying Special Right Triangles Example 4: Using the 30º60º90º Triangle Theorem An ornamental pin is in the shape of an equilateral triangle. The length of each side is 6 centimeters. Josh will attach the fastener to the back along AB. Will the fastener fit if it is 4 centimeters long? Step 1 The equilateral triangle is divided into two triangles. The height of the triangle is the length of the longer leg.
162 58 Applying Special Right Triangles Example 4 Continued Step 2 Find the length x of the shorter leg. 6 = 2x Hypotenuse = 2(shorter leg) 3 = x Divide both sides by 2. Step 3 Find the length h of the longer leg. The pin is approximately 5.2 centimeters high. So the fastener will fit.
163 58 Applying Special Right Triangles Check It Out! Example 4 What if? A manufacturer wants to make a larger clock with a height of 30 centimeters. What is the length of each side of the frame? Round to the nearest tenth. Step 1 The equilateral triangle is divided into two 30º 60º90º triangles. The height of the triangle is the length of the longer leg.
164 58 Applying Special Right Triangles Check It Out! Example 4 Continued Step 2 Find the length x of the shorter leg. Rationalize the denominator. Step 3 Find the length y of the longer leg. y = 2x Hypotenuse = 2(shorter leg) Simplify. Each side is approximately 34.6 cm.
165 58 Applying Special Right Triangles Lesson Quiz: Part I Find the values of the variables. Give your answers in simplest radical form x = 10; y = 20
166 58 Applying Special Right Triangles Lesson Quiz: Part II Find the perimeter and area of each figure. Give your answers in simplest radical form. 5. a square with diagonal length 20 cm 6. an equilateral triangle with height 24 in.
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