1. Inequalities for Sides and Angles of a triangle

2. 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. S T U V

3. 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 S T U V

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

5. 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. Example 1) Given: m<A is greater than m<D Prove: BD is greater than AB. C B StatementsReasons m<A + m<E + m<EBA = 180, m<C + m<D + m<CBD = 180 m<EBA = m<EBD m<A + m<E = 180 - m<EBA, m<C + m<D = 180 - m<CBD m<A is greater than m<D Angle Sum Theorem Substitution Property of Equality Subtraction Property of Equality m<A + m<D = m<C + m<D Given Vertical angles are congruent BD is greater than AB 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. A E D m<C is greater than m<E Subtraction Property of Equality

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=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) JK d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((-4 – 4) 2 + (2 – 3) 2 ) d=√((-8) 2 + (-1) 2 ) d=√(64 + 1) d=√(65) KL d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 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) JL d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 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

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=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) JK d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 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) KL d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((-5 – 6) 2 + (-8 – 10) 2 ) d=√((-11) 2 + (-18) 2 ) d=√(121 + 324) d=√(445) JL d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 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

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. B A D C 10 8 15 16 12

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. UR T S 30 70 80 110 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