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Special Parallelograms

Special Parallelograms. Warm Up What is the sum of the interior angles of 11-gon. Given Parallelogram ABCD, what is the value of y? Explain why the quadrilateral , JKLM, is a Parallelogram . Objectives. Prove and apply properties of rectangles, rhombuses, and squares.

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Special Parallelograms

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  1. Special Parallelograms

  2. Warm Up • What is the sum of the interior angles of 11-gon. • Given Parallelogram ABCD, what is the value of y? • Explain why the quadrilateral, JKLM, is a Parallelogram

  3. Objectives Prove and apply properties of rectangles, rhombuses, and squares. Use properties of rectangles, rhombuses, and squares to solve problems.

  4. A second type of special quadrilateral is a rectangle. A rectangleis a quadrilateral with four right angles.

  5. 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.

  6.  diags. bisect each other 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. Substitute and simplify.

  7. Check It Out! Example 1a Carpentry The rectangular gate has diagonal braces. Find HJ. Rect.  diags.  HJ = GK = 48 Def. of  segs.

  8. Check It Out! Example 1b Carpentry The rectangular gate has diagonal braces. Find HK. Rect.  diags.  Rect.  diagonals bisect each other JL = LG Def. of  segs. JG = 2JL = 2(30.8) = 61.6 Substitute and simplify.

  9. When you are given a parallelogram with certain properties, you can use the theorems below to determine whether the parallelogram is a rectangle.

  10. When you are given a parallelogram with certain properties, you can use the theorems below to determine whether the parallelogram is a rectangle.

  11. A manufacture builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle? Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1. Example 1: Carpentry Application

  12. A rhombus is another special quadrilateral. A rhombusis a quadrilateral with four congruent sides.

  13. Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

  14. Example 2A: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find TV. WV = XT Def. of rhombus 13b – 9=3b + 4 Substitute given values. 10b =13 Subtract 3b from both sides and add 9 to both sides. b =1.3 Divide both sides by 10.

  15. Example 2A Continued TV = XT Def. of rhombus Substitute 3b + 4 for XT. TV =3b + 4 TV =3(1.3)+ 4 = 7.9 Substitute 1.3 for b and simplify.

  16. Example 2B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find mVTZ. mVZT =90° Rhombus  diag.  Substitute 14a + 20 for mVTZ. 14a + 20=90° Subtract 20 from both sides and divide both sides by 14. a=5

  17. Example 2B Continued Rhombus  each diag. bisects opp. s mVTZ =mZTX mVTZ =(5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ =[5(5) – 5)]° = 20° Substitute 5 for a and simplify.

  18. Check It Out! Example 2a CDFG is a rhombus. Find CD. CG = GF Def. of rhombus 5a =3a + 17 Substitute a =8.5 Simplify GF = 3a + 17=42.5 Substitute CD = GF Def. of rhombus CD = 42.5 Substitute

  19. Check It Out! Example 2b CDFG is a rhombus. Find the measure. mGCH if mGCD = (b + 3)° and mCDF = (6b – 40)° Def. of rhombus mGCD + mCDF = 180° b + 3 + 6b –40 = 180° Substitute. 7b = 217° Simplify. b = 31° Divide both sides by 7.

  20. Check It Out! Example 2b Continued mGCH + mHCD = mGCD Rhombus  each diag. bisects opp. s 2mGCH = mGCD Substitute. 2mGCH = (b + 3) Substitute. 2mGCH = (31 + 3) Simplify and divide both sides by 2. mGCH = 17°

  21. Below are some conditions you can use to determine whether a parallelogram is a rhombus.

  22. Example 2A: Applying Conditions for Special Parallelograms Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a rhombus. The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.

  23. 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. • One special characteristic of a square is that the diagonals are • Congruent • Perpendicular • Bisect one another

  24. Example 3: Verifying Properties of Squares Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other.

  25. Step 1 Show that EG and FH are congruent. Since EG = FH, Example 3 Continued

  26. Step 2 Show that EG and FH are perpendicular. Since , Example 3 Continued

  27. Step 3 Show that EG and FH are bisect each other. Since EG and FH have the same midpoint, they bisect each other. Example 3 Continued The diagonals are congruent perpendicular bisectors of each other.

  28. SV = TW = 122 so, SV @ TW . 1 slope of SV = 11 slope of TW = –11 SV ^ TW Check It Out! Example 3 The vertices of square STVW are S(–5, –4), T(0, 2), V(6, –3) , and W(1, –9) . Show that the diagonals of square STVW are congruent perpendicular bisectors of each other.

  29. Step 1 Show that SV and TW are congruent. Since SV = TW, Check It Out! Example 3 Continued

  30. Step 2 Show that SV and TW are perpendicular. Since Check It Out! Example 3 Continued

  31. Step 3 Show that SV and TW bisect each other. Since SV and TW have the same midpoint, they bisect each other. Check It Out! Example 3 Continued The diagonals are congruent perpendicular bisectors of each other.

  32. Caution In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram. To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus. You will explain why this is true in Exercise 43.

  33. Quad. with diags. bisecting each other  Example 2B: Applying Conditions for Special Parallelograms Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a square. Step 1 Determine if EFGH is a parallelogram. Given EFGH is a parallelogram.

  34. with diags.   rect. with one pair of cons. sides  rhombus Example 2B Continued Step 2 Determine if EFGH is a rectangle. Given. EFGH is a rectangle. Step 3 Determine if EFGH is a rhombus. EFGH is a rhombus.

  35. Example 2B Continued Step 4 Determine is EFGH is a square. Since EFGH is a rectangle and a rhombus, it has four right angles and four congruent sides. So EFGH is a square by definition. The conclusion is valid.

  36. Check It Out! Example 2 Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given:ABC is a right angle. Conclusion: ABCD is a rectangle. The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram .

  37. Example 3A: Identifying Special Parallelograms in the Coordinate Plane Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)

  38. Example 3A Continued Step 1 Graph PQRS.

  39. Since , the diagonals are congruent. PQRS is a rectangle. Example 3A Continued Step 2 Find PR and QS to determine is PQRS is a rectangle.

  40. Since , PQRS is a rhombus. Example 3A Continued Step 3 Determine if PQRS is a rhombus. Step 4 Determine if PQRS is a square. Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.

  41. Check It Out! Example 4 Given: PQTS is a rhombus with diagonal Prove:

  42. 2. 4. 5. 6. 7. Check It Out! Example 4 Continued 1. PQTSis a rhombus. 1. Given. 2. Rhombus → each diag. bisects opp. s 3. QPR  SPR 3. Def. of  bisector. 4. Def. of rhombus. 5. Reflex. Prop. of  6. SAS 7. CPCTC

  43. 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.TR2.CE 35 ft 29 ft

  44. Lesson Quiz: Part II PQRS is a rhombus. Find each measure. 3.QP4. mQRP 42 51°

  45. Lesson Quiz: Part III 5. The vertices of square ABCD are A(1, 3), B(3, 2), C(4, 4), and D(2, 5). Show that its diagonals are congruent perpendicular bisectors of each other.

  46. 6.Given:ABCD is a rhombus. Prove:  Lesson Quiz: Part IV

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