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6-3 6-3 Conditions Conditionsfor forParallelograms Parallelograms Warm Up Lesson Presentation Lesson Quiz Holt Geometry Holt Geometry 6-3 Conditions for Parallelograms Warm Up Justify each statement. 1. 2. Reflex Prop. of Conv. of Alt. Int. s Thm. Evaluate each expression for x = 12 and y = 8.5. 3. 2x + 7 31 4. 16x – 9 183 5. (8y + 5)° 73° Holt Geometry 6-3 Conditions for Parallelograms Objective Prove that a given quadrilateral is a parallelogram. Prove and apply properties of rectangles, rhombuses, and squares. Use properties of rectangles, rhombuses, and squares to solve problems. Holt Geometry 6-4 Properties of Special Parallelograms Vocabulary rectangle rhombus square Holt Geometry 6-3 Conditions for Parallelograms Holt Geometry 6-3 Conditions for Parallelograms The two theorems below can also be used to show that a given quadrilateral is a parallelogram. Holt Geometry 6-3 Conditions for Parallelograms You have learned several ways to determine whether a quadrilateral is a parallelogram. You can use the given information about a figure to decide which condition is best to apply. Holt Geometry 6-3 Conditions for Parallelograms Example 1A: Verifying Figures are Parallelograms Show that JKLM is a parallelogram for a = 3 and b = 9. Step 1 Find JK and LM. JK = 15a – 11 Given LM = 10a + 4 Substitute JK = 15(3) – 11 = 34 and simplify. LM = 10(3)+ 4 = 34 Holt Geometry 6-3 Conditions for Parallelograms Example 1A Continued Step 2 Find KL and JM. KL = 5b + 6 KL = 5(9) + 6 = 51 Given JM = 8b – 21 Substitute and simplify. JM = 8(9) – 21 = 51 Since JK = LM and KL = JM, JKLM is a parallelogram by Theorem 6-3-2. Holt Geometry 6-3 Conditions for Parallelograms Example 1B: Verifying Figures are Parallelograms Show that PQRS is a parallelogram for x = 10 and y = 6.5. mQ = (6y + 7)° mQ = [(6(6.5) + 7)]° = 46° Given Substitute 6.5 for y and simplify. mS = (8y – 6)° Given Substitute 6.5 for y and simplify. mR = (15x – 16)° Given Substitute 10 for x mR = [(15(10) – 16)]° = 134° and simplify. mS = [(8(6.5) – 6)]° = 46° Holt Geometry 6-3 Conditions for Parallelograms Example 1B Continued Since 46° + 134° = 180°, R is supplementary to both Q and S. PQRS is a parallelogram by Theorem 6-3-4. Holt Geometry 6-3 Conditions for Parallelograms Check It Out! Example 2a Determine if the quadrilateral must be a parallelogram. Justify your answer. Yes The diagonal of the quadrilateral forms 2 triangles. Two angles of one triangle are congruent to two angles of the other triangle, so the third pair of angles are congruent by the Third Angles Theorem. So both pairs of opposite angles of the quadrilateral are congruent . By Theorem 6-3-3, the quadrilateral is a parallelogram. Holt Geometry 6-3 Conditions for Parallelograms Check It Out! Example 2b Determine if each quadrilateral must be a parallelogram. Justify your answer. No. Two pairs of consective sides are congruent. None of the sets of conditions for a parallelogram are met. Holt Geometry 6-3 Conditions for Parallelograms Helpful Hint To say that a quadrilateral is a parallelogram by definition, you must show that both pairs of opposite sides are parallel. Holt Geometry 6-3 Conditions for Parallelograms Helpful Hint To show that a quadrilateral is a parallelogram, you only have to show that it satisfies one of these sets of conditions. Holt Geometry 6-3 Conditions for Parallelograms Example 4: Application The legs of a keyboard tray are connected by a bolt at their midpoints, which allows the tray to be raised or lowered. Why is PQRS always a parallelogram? Since the bolt is at the midpoint of both legs, PE = ER and SE = EQ. So the diagonals of PQRS bisect each other, and by Theorem 6-3-5, PQRS is always a parallelogram. Holt Geometry 6-4 Properties of Special Parallelograms Holt Geometry 6-4 Properties of Special Parallelograms A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles. Holt Geometry 6-4 Properties of Special Parallelograms Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2. Holt Geometry 6-4 Properties of Special Parallelograms Example 1: Craft Application A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM. Rect. diags. KM = JL = 86 Def. of segs. diags. bisect each other Substitute and simplify. Holt Geometry 6-4 Properties of Special Parallelograms A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides. Holt Geometry 6-4 Properties of Special Parallelograms Holt Geometry 6-4 Properties of Special Parallelograms Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses. Holt Geometry 6-4 Properties of Special Parallelograms Example 2B Continued mVTZ = mZTX Rhombus each diag. bisects opp. s mVTZ = (5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ = [5(5) – 5)]° Substitute 5 for a and simplify. = 20° Holt Geometry 6-4 Properties of Special Parallelograms A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three. Holt Geometry 6-4 Properties of Special Parallelograms Helpful Hint Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms. Holt Geometry 6-4 Properties of Special Parallelograms Example 3: Verifying Properties of Squares Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other. Holt Geometry 6-4 Properties of Special Parallelograms Example 3 Continued Step 1 Show that EG and FH are congruent. Since EG = FH, Holt Geometry 6-4 Properties of Special Parallelograms Example 3 Continued Step 2 Show that EG and FH are perpendicular. Since Holt Geometry , 6-4 Properties of Special Parallelograms Example 3 Continued Step 3 Show that EG and FH are bisect each other. Since EG and FH have the same midpoint, they bisect each other. The diagonals are congruent perpendicular bisectors of each other. Holt Geometry 6-4 Properties of Special Parallelograms Check It Out! Example 4 Given: PQTS is a rhombus with diagonal Prove: Holt Geometry 6-4 Properties of Special Parallelograms Check It Out! Example 4 Continued Statements 1. PQTS is a rhombus. 2. 3. QPR SPR 4. 5. 6. 7. Holt Geometry Reasons 1. Given. 2. Rhombus → each diag. bisects opp. s 3. Def. of bisector. 4. Def. of rhombus. 5. Reflex. Prop. of 6. SAS 7. CPCTC 6-4 Properties of Special Parallelograms Lesson Quiz: Part I A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length. 1. TR 35 ft Holt Geometry 2. CE 29 ft 6-4 Properties of Special Parallelograms Lesson Quiz: Part II PQRS is a rhombus. Find each measure. 3. QP 42 Holt Geometry 4. mQRP 51° 6-3 Conditions for Parallelograms Class work problems Page 402: 10, 12,14,16,18,20 Page 412 :10, 14, 16, 17, 22, 24 Holt Geometry