Pearson Unit 1 Topic 6: Polygons and Quadrilaterals 6-5: Conditions for Rhombuses, Rectangles, and Squares Pearson Texas Geometry ©2016 Holt Geometry.

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Presentation transcript:

Pearson Unit 1 Topic 6: Polygons and Quadrilaterals 6-5: Conditions for Rhombuses, Rectangles, and Squares Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (6)(E) Prove a quadrilateral is a parallelogram, rectangle, square, or rhombus using opposite sides, opposite angles, or diagonals and apply these relationships to solve problems. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication.

Remember:

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 can also prove that a given quadrilateral is a rectangle, rhombus, or square by using the definitions of the special quadrilaterals. Remember!

Example: 1 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. You must first know that the quadrilateral is a parallelogram.

Example: 2 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 Quad. with diags. bisecting each other  EFGH is a parallelogram.

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

Example: 3 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. If one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this, you need to know that ABCD is a parallelogram .

Example: 4 For what values of x and y is quadrilateral QRST a rhombus? 2y = y+ 25 4x – 32 = 2x + 4 y = 25 2x – 32 = 4 2x = 36 x = 18

Example: 5 Can you conclude that each parallelogram is a rectangle, rhombus or a square? Explain. Rectangle, because diagonals are congruent Rhombus, because diagonals are perpendicular

Example: 6 m2 = 28 m3 = 28 m4 = 28

Example: 7 For what value of x is quadrilateral ABCD a rectangle? A B D C 5x + 2 + 3x = 90 8x + 2 = 90 8x = 88 x = 11 (5x + 2) (3x)