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12/02/2014 Conditions for Parallelograms

12/02/2014 Conditions for Parallelograms. Grade 9 ASP. Warm Up Justify each statement. 1. 2. Evaluate each expression for x = 12 and y = 8.5. 3. 2 x + 7 4. 16 x – 9 5. (8 y + 5)°. Reflex Prop. of . Conv. of Alt. Int. s Thm. 31. 183. 73°. Objective.

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12/02/2014 Conditions for Parallelograms

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  1. 12/02/2014 Conditions for Parallelograms Grade 9 ASP

  2. Warm Up Justify each statement. 1. 2. Evaluate each expression for x = 12 and y = 8.5. 3. 2x + 7 4. 16x –9 5.(8y + 5)° Reflex Prop. of  Conv. of Alt. Int. s Thm. 31 183 73°

  3. Objective Prove that a given quadrilateral is a parallelogram.

  4. You have learned to identify the properties of a parallelogram. Now you will be given the properties of a quadrilateral and will have to tell if the quadrilateral is a parallelogram. To do this, you can use the definition of a parallelogram or the conditions below.

  5. The two theorems below can also be used to show that a given quadrilateral is a parallelogram.

  6. Example 1A: Verifying Figures are Parallelograms Show that JKLM is a parallelogram for a = 3 and b = 9. Step 1 Find JK and LM. Given LM = 10a + 4 JK = 15a – 11 Substitute and simplify. LM = 10(3)+ 4 = 34 JK = 15(3) – 11 = 34

  7. Example 1A Continued Step 2 Find KL and JM. Given KL = 5b + 6 JM = 8b – 21 Substitute and simplify. KL = 5(9) + 6 = 51 JM = 8(9) – 21 = 51 Since JK = LM and KL = JM, JKLM is a parallelogram by Theorem 6-3-2.

  8. Example 1B: Verifying Figures are Parallelograms Show that PQRS is a parallelogram for x = 10 and y = 6.5. mQ = (6y + 7)° Given Substitute 6.5 for y and simplify. mQ = [(6(6.5) + 7)]° = 46° mS = (8y – 6)° Given Substitute 6.5 for y and simplify. mS = [(8(6.5) – 6)]° = 46° mR = (15x – 16)° Given Substitute 10 for x and simplify. mR = [(15(10) – 16)]° = 134°

  9. Example 1B Continued Since 46° + 134° = 180°, R is supplementary to both Q and S. PQRS is a parallelogram by Theorem 6-3-4.

  10. PQ = RS = 16.8, so Therefore, Check It Out! Example 1 Show that PQRS is a parallelogram for a = 2.4 and b = 9. mQ = 74°, and mR = 106°, so Q and R are supplementary. So one pair of opposite sides of PQRS are || and . By Theorem 6-3-1, PQRS is a parallelogram.

  11. Example 2A: Applying Conditions for Parallelograms Determine if the quadrilateral must be a parallelogram. Justify your answer. Yes. The 73° angle is supplementary to both its corresponding angles. By Theorem 6-3-4, the quadrilateral is a parallelogram.

  12. Example 2B: Applying Conditions for Parallelograms Determine if the quadrilateral must be a parallelogram. Justify your answer. No. One pair of opposite angles are congruent. The other pair is not. The conditions for a parallelogram are not met.

  13. 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.

  14. 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.

  15. Helpful Hint To say that a quadrilateral is a parallelogram by definition, you must show that both pairs of opposite sides are parallel.

  16. Example 3A: Proving Parallelograms in the Coordinate Plane Show that quadrilateral JKLM is a parallelogram by using the definition of parallelogram. J(–1, –6), K(–4, –1), L(4, 5), M(7, 0). Find the slopes of both pairs of opposite sides. Since both pairs of opposite sides are parallel, JKLM is a parallelogram by definition.

  17. AB and CD have the same slope, so . Since AB = CD, . So by Theorem 6-3-1, ABCD is a parallelogram. Example 3B: Proving Parallelograms in the Coordinate Plane Show that quadrilateral ABCD is a parallelogram by using Theorem 6-3-1.A(2, 3), B(6, 2), C(5, 0), D(1, 1). Find the slopes and lengths of one pair of opposite sides.

  18. Both pairs of opposite sides have the same slope so and by definition, KLMN is a parallelogram. Check It Out! Example 3 Use the definition of a parallelogram to show that the quadrilateral with vertices K(–3, 0), L(–5, 7), M(3, 5), and N(5, –2) is a parallelogram.

  19. 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.

  20. Helpful Hint To show that a quadrilateral is a parallelogram, you only have to show that it satisfies one of these sets of conditions.

  21. 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.

  22. Since ABRS is a parallelogram, it is always true that . Since AB stays vertical, RS also remains vertical no matter how the frame is adjusted. Check It Out! Example 4 The frame is attached to the tripod at points A and B such that AB = RS and BR = SA. So ABRS is also a parallelogram. How does this ensure that the angle of the binoculars stays the same? Therefore the viewing never changes.

  23. Lesson Quiz: Part I 1. Show that JKLM is a parallelogram for a = 4 and b = 5. 2. Determine if QWRT must be a parallelogram. Justify your answer. JN = LN = 22; KN = MN = 10; so JKLM is a parallelogram by Theorem 6-3-5. No; One pair of consecutive s are , and one pair of opposite sides are ||. The conditions for a parallelogram are not met.

  24. Lesson Quiz: Part II 3. Show that the quadrilateral with vertices E(–1, 5), F(2, 4), G(0, –3), and H(–3, –2) is a parallelogram. Since one pair of opposite sides are || and , EFGH is a parallelogram by Theorem 6-3-1.

  25. L.O. • Application in Racing • Using Properties in Parallelograms to find Measures. • Parallelograms in Coordinate Planes. • Using Properties of Parallelograms in Proof.

  26. Lesson Quiz 1. Name the polygon by the number of its sides. Then tell whether the polygon is regular or irregular, concave or convex. nonagon; irregular; concave 2. Find the sum of the interior angle measures of a convex 11-gon. 1620° 3. Find the measure of each interior angle of a regular 18-gon. 4. Find the measure of each exterior angle of a regular 15-gon. 160° 24°

  27. CLASSWORK AND HOMEWORK CLASSWORK HOMEWORK See Homework booklet. Week 3 • (Pages 407 to 409 ) 1, 2, 3 to 13, 14, 21 to 24, 25, 26, 27 to 30, 32 to 43, 46, 47, 50, 51, 52, 53.

  28. Objectives Prove and apply properties of parallelograms. Use properties of parallelograms to solve problems.

  29. Any polygon with four sides is a quadrilateral. However, some quadrilaterals have special properties. These special quadrilaterals are given their own names.

  30. Helpful Hint Opposite sides of a quadrilateral do not share a vertex. Opposite angles do not share a side.

  31. A quadrilateral with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you use the symbol .

  32. In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find CF. opp. sides Example 1A: Properties of Parallelograms CF = DE Def. of segs. CF = 74 mm Substitute 74 for DE.

  33. In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find mEFC. cons. s supp. Example 1B: Properties of Parallelograms mEFC + mFCD = 180° mEFC + 42= 180 Substitute 42 for mFCD. mEFC = 138° Subtract 42 from both sides.

  34. In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find DF. diags. bisect each other. Example 1C: Properties of Parallelograms DF = 2DG DF = 2(31) Substitute 31 for DG. DF = 62 Simplify.

  35. opp. sides Check It Out! Example 1a In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find KN. LM = KN Def. of segs. LM = 28 in. Substitute 28 for DE.

  36. opp. s  Check It Out! Example 1b In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find mNML. NML  LKN mNML = mLKN Def. of  s. mNML = 74° Substitute 74° for mLKN. Def. of angles.

  37. diags. bisect each other. Check It Out! Example 1c In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find LO. LN = 2LO 26 = 2LO Substitute 26 for LN. LO = 13 in. Simplify.

  38. opp. s  Example 2A: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find YZ. YZ = XW Def. of  segs. 8a – 4 = 6a + 10 Substitute the given values. Subtract 6a from both sides and add 4 to both sides. 2a = 14 a = 7 Divide both sides by 2. YZ = 8a – 4 = 8(7) – 4 = 52

  39. cons. s supp. Example 2B: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find mZ. mZ + mW = 180° (9b + 2)+ (18b –11) = 180 Substitute the given values. Combine like terms. 27b – 9 = 180 27b = 189 b = 7 Divide by 27. mZ = (9b + 2)° = [9(7) + 2]° = 65°

  40. diags. bisect each other. Check It Out! Example 2a EFGH is a parallelogram. Find JG. EJ = JG Def. of  segs. 3w = w + 8 Substitute. 2w = 8 Simplify. w = 4 Divide both sides by 2. JG = w + 8 = 4 + 8 = 12

  41. diags. bisect each other. Check It Out! Example 2b EFGH is a parallelogram. Find FH. FJ = JH Def. of  segs. 4z – 9 = 2z Substitute. 2z = 9 Simplify. z = 4.5 Divide both sides by 2. FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18

  42. Remember! When you are drawing a figure in the coordinate plane, the name ABCD gives the order of the vertices.

  43. L K J Example 3: Parallelograms in the Coordinate Plane Three vertices of JKLM are J(3, –8), K(–2, 2), and L(2, 6). Find the coordinates of vertex M. Since JKLM is a parallelogram, both pairs of opposite sides must be parallel. Step 1 Graph the given points.

  44. Step 2 Find the slope of by counting the units from K to L. • The rise from 2 to 6 is 4. • The run of –2 to 2 is 4. L K J Example 3 Continued • Step 3 Start at J and count the • same number of units. • A rise of 4 from –8 is –4. • A run of 4 from 3 is 7. Label (7, –4) as vertex M. M

  45. Step 4 Use the slope formula to verify that L K M J Example 3 Continued The coordinates of vertex M are (7, –4).

  46. Q S P Check It Out! Example 3 Three vertices of PQRS are P(–3, –2), Q(–1, 4), and S(5, 0). Find the coordinates of vertex R. Since PQRS is a parallelogram, both pairs of opposite sides must be parallel. Step 1 Graph the given points.

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