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Lesson 6–5 Objectives

Lesson 6–5 Objectives. Be able to prove that a given quadrilateral is a rectangle, rhombus, or square. The following can be used to prove that a parallelogram is a rectangle.

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Lesson 6–5 Objectives

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  1. Lesson 6–5 Objectives • Be able to prove that a given quadrilateral is a rectangle, rhombus, or square.

  2. The following can be used to prove that a parallelogram is a rectangle.

  3. 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. Proving a figure is a rectangle A manufacturer makes a desktop mold so that AB  CD and BC  AD and mABC = 90°. Explain why ABCD must be a rectangle.

  4. Remember! You can also prove that a given quadrilateral is a rectangle, rhombus, or square by using the definitions of the special quadrilaterals. To prove that a given quadrilateral is a square, it is sufficient to show that the figure is BOTH a rectangle AND a rhombus.

  5. Applying special parallelogram conditions Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: EFFG, EGFH Conclusion: EFGH is a rhombus.

  6. Applying special parallelogram conditions Given: EFFG, EGFH 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.

  7. Applying special parallelogram conditions Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: EBBG, FBBH, EGFH, ∆EBF∆EBH Conclusion: EFGH is a square.

  8. Quad. with diags. bisecting each other  Applying special parallelogram conditions Given: EBBG, FBBH, EGFH, ∆EBF∆EBH Conclusion: EFGH is a square. Step 1 Determine if EFGH is a parallelogram. Given EFGH is a parallelogram.

  9. with diags.   rect. with one pair of cons. sides  rhombus Applying special parallelogram conditions 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.

  10. Applying special parallelogram conditions 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.

  11. Applying special parallelogram conditions 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.

  12. Applying special parallelogram conditions 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 .

  13. Identifying special parallelograms 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)

  14. Lesson 6–5 Assignment

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