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Warm Up (On Separate Sheet & Pass Back Papers) Solve for x . 1. 16 x – 3 = 12 x + 13

Warm Up (On Separate Sheet & Pass Back Papers) Solve for x . 1. 16 x – 3 = 12 x + 13 2. 2 x – 4 = 90 ABCD is a parallelogram. Find each measure. 3. CD 4. m  C. 4. 47. 104°. 14. 6-4. Properties of Special Parallelograms. Holt Geometry.

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Warm Up (On Separate Sheet & Pass Back Papers) Solve for x . 1. 16 x – 3 = 12 x + 13

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  1. Warm Up(On Separate Sheet & Pass Back Papers) Solve for x. 1.16x – 3 = 12x + 13 2. 2x – 4 = 90 ABCD is a parallelogram. Find each measure. 3.CD4. mC 4 47 104° 14

  2. 6-4 Properties of Special Parallelograms Holt Geometry

  3. A rectangleis a quadrilateral with four right angles.

  4.  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.

  5. A rhombusis a quadrilateral with four congruent sides.

  6. Example 2B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find a. 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

  7. A square is a quadrilateral with four right angles and four congruent sides.

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

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

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

  11. 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.

  12. 6-5 Conditions for Special Parallelograms Holt Geometry

  13. 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

  14. 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.

  15. Example 3B: 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. W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3) Step 1 Graph WXYZ.

  16. Since , WXYZ is not a rectangle. Example 3B Continued Step 2 Find WY and XZ to determine is WXYZ is a rectangle. Thus WXYZ is not a square.

  17. Since (–1)(1) = –1, , PQRS is a rhombus. Example 3B Continued Step 3 Determine if WXYZ is a rhombus.

  18. Lesson Quiz: Part I 1. Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square?

  19. 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

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

  21. 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.

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

  23. Warm-Up 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

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