ACTIVITY No.4: PROBLEM EXERCISES Hinge Theorem Hinge Theorem states that if two sides of one triangle are congruent, to two sides of another triangle, but the included angle of the rst triangle is larger than the included angle of the second, then the third side of the rst triangle is longer than the third side of the second triangle. : HE 5 BE, '13? .=:- if, ,2} and sz > 1114?), g ' then AC > DF. .2} 1. write an inequality relating the lengths of 55 and KB A 2; Two sides of ABC D are congruent m iwo sides of ABAD 3 8 0 and szCD-> mafBAD. By the Hinge Theorem, C 2. Write an inequality to describe the possible values of x. l20>115. 4x~10>30 4x>30+ IO 4x>40 53>?- 4 4 If: 3. _Which of the following statements must be true? D) AK = YU AS YU C) SK mZO SK > YU Converse of the Hinge Theorem Converse of the Hinge Theorem states that if two sides of one triangle are congruent to the other two sides of another triangle respectively, and the third side of the first triangle is longer than the third side of the second, then the angle opposite the longer side is larger than the angle opposite the third side of the second triangle. If AABC and ADEF with AC = DF, 8 AB = DE, and 5 BC > EF, then mZA > mZD. A Larger angie B D smaller angle E Examples: 1. Given the diagram below, what can be concluded regarding mau and mLE? If ABUG and APET with BU = PE, 9 UG = ET, and BC ~ PT then my( > mZE. B G 2. Write an inequality relating the mcC and mzZ. X If AABC and AXYZ with A AC = XZ 24 48 50 BC = YZ, and 8 30 30 AB > XY, then mZC > mZZ. 3. Write an inequality to describe the possible values of x and determine the possible integral values of x. 60 cm Step 1. 36 cm Find the upper limit of the value of x. 33 3x - 3 0 The angle must be greater than zero. 1 3 By Addition Property of Inequality. The possible integral values of x are 3x > 1 Then divide both sides by 3 (2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. x2 1Directions: Write an inequality or inequalities to describe the possible values of x. 23 Solution here: 15 105 115 15 5x + 8 Solution here: x+2 50 3 ACTIVITY NO.5: TABLE Listed below are the Properties of Equalities that you may use: Properties of Equalities For any real numbers a, b, c, and d: Properties Illustrations Addition Property of Equality If a = b, then a + c = B + c. or If a = band c = d, then a + c = b + d Subtraction Property of Equality If a = b, then a -c = b . c or If a = b, and c = d, then a - c = b - d Multiplication Property of Equality If a = b, then ac = bc or If a = band c = d, then ac = bd Substitution Property If a = b, then a may be replaced by b in any equation. Properties of Inequalities may also be used in this module: Properties of Equalities For any real numbers x, y, and z be real numbers: Properties Illustrations Addition Property of Inequality If x 0, then xz yz If x 0, then x + z x+ y Reversal Property of Inequality If x > y, then y EG, (6x + 1510 6 Therefore, by Converse of Hinge Theorem, mZJHE H > mZEHG. Step 1. 6x + 15 > 65 By substitution 6x > 65 - 15 Subtraction Property of Inequality 6x > 50 x > 8- Division Property of Inequality Step 2. Use the fact that the measure of any angle in a triangle is less than 180 to write a second inequality. MZJHE 8- and x ZBDA, Therefore, by Hinge Theorem, BC > BA. (4x - 4) cm 28 em 120 140 A DStep 1. 28 > 4x - 4 By substitution 28 + 4 > 4x Addition Property of Inequality 32 > 4x 8 > X Division Property of Inequality x 0, then solve for the range of values of x: 4x - 4 > 0 4x > 4 Subtraction Property of Inequality x> 1 Division Property of Inequality Step 3. Write x 1 as the compound inequality: 1 2, then 4 + 5 > 2 + 5 2. If 3x = 15, then x = 5 3, If 4 10, then 10 AH M H Proof: Statements Reasons 1. Let M be a point on a ray opposite TH such that 1. Congruent segment construction MT = AT 2. MH = MT + TH 2. Segment Addition Postulate 3. MH - AT + TH 3. Substitution Property of Equality 4. Point T is in the interior of ZMAH 4. Definition of interior point 5. myMAH = mcMAT + mcTAH 5. Angle addition postulate 6. meMAH. > MZTAM 6. If a = b + c and c > 0, then a > b 7. mcTMA = myTAM 7. Isosceles Triangle Theorem 8. mcMAH > mcTMA 8. Substitution Property 9. MH > AH 9. Angle Side Inequality theorem 10. AT + TH > AH Substitution Property 3 and 9.Unequal Side Theorem - If one side of a triangle is longer than the second side, then the angle opposite the longer side is larger than the angle opposite the second side. Example: A Given: A CAR CA > RA Prove: L ARC > C R Statements Reasons 1. Draw RE such that AR = AE 1. Ruler Postulate 2. AR = AE 2. Definition of Congruent Segment 3. m_1 = mz2 3. Isosceles Triangle Theorem 4. mz1 = mz2 4. Definition of Congruence Angles 5. mz1 = mz2 + mc3 5. Angle Addition Postulate 6. mzR > mz2 6. Comparison Property of Inequality 7. maR > mz1 7. Substitute 3 and 6 8. mzi > m &C 8. Exterior Angle Postulate 9. mcARC > MAC 9. Transitive Property of inequality Unequal Angles Theorem - If one of a triangle is larger than the second angle, then the side opposite the larger angle is longer than the side opposite the second angle. Example: Given: PQ 1 m PR is any other segment from ? to m Prove: PR > PQ 1 2 m Q Statements Reasons 1. PQ 1 m 1. Given 2. 41 = 42 2. Definition of Perpendicularity 3. my1 = mz2 3. Definition of Congruent Angles 4. mz1 > m23 4. An exterior angle is greater than any of its corresponding remote interior angles 5. m42 > PQ 5. Substitution 3 and 4 6. PR > PQ 6. The Unequal Angle Theorem Directions: Prove the given triangle. R Given: RS > QR Prove: mcRaS > mas 2. Q 3 Statements Reasons 1. PR = RQ 1. 2. 41 = 42 3, mz1 = mc2 3. 4. MARQS = + 4. Angle Addition Postulate 5. 5. mZRQS = mZ2 6. mz2 = m23 + 6. Exterior Angle Theorem 7 . 7. MARQS > MLS