Chapter 5 Section 4

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