SlideServe
Télécharger Obtenir Pro
Menu
Accueil Pro
Préférences
ApparenceLight Aide et commentaires
© SlideServe · v2026

Chapter 5 Section 4

amir-clark avatar
amir-clark
12 diapositives · 311 Views · Nov 07, 2014

Chapter 5 Section 4. Inequalities for Sides and Angles of a triangle. Warm-Up. Determine whether triangle STU is congruent to triangle VUT, using the given information. Justify. 1) <S is congruent to <V 2) SU is congruent to VT 3) <STU and <VUT are right angles

Chapter 5 Section 4 - slide 1
01 / 12Chapter 5 Section 4
1 / 17

Télécharger la présentation

Chapter 5 Section 4
Une image / un lien ci-dessous est fourni (tel quel) pour télécharger la présentation

Politique de téléchargement : le contenu du site vous est fourni TEL QUEL pour votre information et votre usage personnel ; il ne peut être vendu, concédé sous licence ni partagé sur d’autres sites sans l’accord de son auteur.Le contenu vous est fourni TEL QUEL, pour votre information et votre usage personnel uniquement. Téléchargez la présentation en cliquant sur ce lien. Lors du téléchargement, si vous ne parvenez pas à télécharger une présentation, l’éditeur a peut-être supprimé le fichier de son serveur.Pendant le téléchargement, si vous ne parvenez pas à obtenir une présentation, le fichier a peut-être été supprimé par l’éditeur.

F I N

Plus de contenu associé

Transcription de la présentation

  1. Chapter 5Section 4 Inequalities for Sides and Angles of a triangle

  2. Warm-Up • Determine whether triangle STU is congruent to triangle VUT, using the given information. Justify. • 1) <S is congruent to <V • 2) SU is congruent to VT • 3) <STU and <VUT are right angles • 4) Given: <STU and <UVT are right angles; SU is congruent to VT. • Prove: <S is congruent to <V. T S V U

  3. Warm-Up • Determine whether triangle STU is congruent to triangle VUT, using the given information. Justify. • 1) <S is congruent to <V • Yes by LA • 2) SU is congruent to VT • Yes by HL • 3) <STU and <VUT are right angles • No T S V U

  4. Warm-Up 4) Given: <STU and <UVT are right angles; SU is congruent to VT. Prove: <S is congruent to <V. T S V U <STU and <UVT are right angles; SU is congruent to VT. Given Triangle STU and triangle VUT are right triangles Definition of a right triangle Congruence of segments is reflexive TU is congruent to UT Triangle STU is congruent to triangle VUT HL <S is congruent to <V CPCTC

  5. Vocabulary Theorem 5-9- If one side of a triangle is longer that another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side. Theorem 5-10- If one angle of a triangle has a greater measure than another angle, then the side opposite the greater angle is longer than the side opposite the lesser angle. Theorem 5-11- The perpendicular segment from a point to a line is the shortest segment from the point to the line. Corollary 5-1- The perpendicular segment from a point to a plane is the shortest segment from the point to the plane.

  6. D Example 1) Given: m<A is greater than m<D Prove: BD is greater than AB. E B C A m<A + m<E + m<EBA = 180, m<C + m<D + m<CBD = 180 Angle Sum Theorem • m<EBA = m<EBD Vertical angles are congruent m<A + m<E = 180 - m<EBA, m<C + m<D = 180 - m<CBD Subtraction Property of Equality m<A + m<D = m<C + m<D Substitution Property of Equality m<A is greater than m<D Given m<C is greater than m<E Subtraction Property of Equality If an angle of a triangle is greater than another, the side opposite the greater angle is longer that the side opposite the lesser angle. BD is greater than AB

  7. Example 2: Triangle JKL with vertices J(-4, 2), K(4, 3), and L(1, -3). List the angles in order from least measure to greatest measure. Find the length of each side. The distance formula is d=√((x2 – x1)2 + (y2 – y1)2) JK d=√((x2 – x1)2 + (y2 – y1)2) d=√((-4 – 4)2 + (2 – 3)2) d=√((-8)2 + (-1)2) d=√(64 + 1) d=√(65) JL d=√((x2 – x1)2 + (y2 – y1)2) d=√((-4 – 1)2 + (2 – -3)2) d=√((-1 – 1)2 + (3 + 3)2) d=√((-5)2 + (6)2) d=√(25 + 36) d=√(61) The shortest side is KL, so the smallest angle is <J. The next shortest side is JL, so the next smallest angle is <K. The greatest side is JK, so the greatest angle is <L. <J, <K, <L KL d=√((x2 – x1)2 + (y2 – y1)2) d=√((4 – 1)2 + (3 – -3)2) d=√((4 – 1)2 + (3 + 3)2) d=√((3)2 + (6)2) d=√(9 + 36) d=√(45)

  8. Example 3: Triangle JKL with vertices J(-2, 4), K(-5, -8), and L(6, 10). List the angles in order from least measure to greatest measure. Find the length of each side. The distance formula is d=√((x2 – x1)2 + (y2 – y1)2) JK d=√((x2 – x1)2 + (y2 – y1)2) d=√((-2 – -5)2+ (4 – -8)2) d=√((-2 + 5)2+ (4 + 8)2) d=√((-3)2+ (12)2) d=√(9+ 144) d=√(153) JL d=√((x2 – x1)2 + (y2 – y1)2) d=√((-2 – 6)2+ (4 – 10)2) d=√((-8)2+ (-6)2) d=√(64+ 36) d=√(100) d = 10 The shortest side is JL, so the smallest angle is <K. The next shortest side is JK, so the next smallest angle is <L. The greatest side is KL, so the greatest angle is <J. <K, <J, <L KL d=√((x2 – x1)2 + (y2 – y1)2) d=√((-5 – 6)2+ (-8 – 10)2) d=√((-11)2+ (-18)2) d=√(121 + 324) d=√(445)

  9. Example 4) Refer to the figure. A) Which is greater, m<CBD or m<CDB? <CDB because it is across from 16 which is greater than 15. B) Is m<ADB greater that m<DBA? Yes because it is across from a side that is longer. C)Which is greater, m<CDA or m<CBA? <CDA because it is across from 10 + 16 = 26 and <CBA is across from 8 + 15 = 23. 15 D C 16 8 12 A B 10

  10. Example 5) Refer to the figure. A) Which side of triangle RTU is the longest? First find all the angles in the triangle. 180 = 110 + m<RUT 70 = m<RUT 180 = 30 + 70 + m<TRU 180 = 100 + m<TRU 80 = m<TRU The greatest angle is 80 degrees so the longest side is TU. T 30 70 110 80 S R U B) Name the side of triangle UST that is the longest. Since there can be only one obtuse angle in a triangle, the longest side is across from the obtuse angle. TS is the longest side.

  11. Example 6) Find the value of x and list the sides of Triangle ABC in order from shortest to longest if <A= 3x +20, <B = 2x + 37, and <C = 4x + 15 The angles of a triangle add up to 180. 180 = m<A + m<B + m<C 180 = 3x + 20 + 2x + 37+ 4x + 15 180 = 9x + 72 108 = 9x 12 = x Plug 12 in for x in each equation. m<A = 3x + 20 m<A = 3(12) + 20 m<A = 36 + 20 m<A = 56 BC, AC, AB m<B = 2x + 37 m<B = 2(12) + 37 m<B = 24 + 37 m<B = 61 m<C = 4x + 15 m<C = 4(12) + 15 m<C = 48 + 15 m<C = 63

  12. Example 7) Find the value of x and list the sides of Triangle ABC in order from shortest to longest if <A= 9x +29, <B = 93 – 5x, and <C = 10x + 2 The angles of a triangle add up to 180. 180 = m<A + m<B + m<C 180 = 9x + 29 + 93 – 5x + 10x + 2 180 = 14x + 124 56 = 14x 4 = x Plug 4 in for x in each equation. m<A = 9x + 29 m<A = 9(4) + 29 m<A = 36 + 29 m<A = 65 AB, BC, AC m<B = 93 – 5x m<B = 93 – 5(4) m<B = 93 - 20 m<B = 73 m<C = 10x + 2 m<C = 10(4) + 2 m<C = 40 + 2 m<C = 42