6 4 Properties of Special Parallelograms Warm UpOn

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

6 -4 Properties of Special Parallelograms Holt Geometry

6 -4 Properties of Special Parallelograms A rectangle is a quadrilateral with four right angles. Holt Geometry

6 -4 Properties of Special Parallelograms Holt Geometry

6 -4 Properties of Special Parallelograms 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. diags. bisect each other Substitute and simplify. Holt Geometry

6 -4 Properties of Special Parallelograms A rhombus is a quadrilateral with four congruent sides. Holt Geometry

6 -4 Properties of Special Parallelograms Holt Geometry

6 -4 Properties of Special Parallelograms Example 2 B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find a. m VZT = 90° 14 a + 20 = 90° a=5 Holt Geometry Rhombus diag. Substitute 14 a + 20 for m VTZ. Subtract 20 from both sides and divide both sides by 14.

6 -4 Properties of Special Parallelograms A square is a quadrilateral with four right angles and four congruent sides. Holt Geometry

6 -4 Properties of Special Parallelograms Example 3: Verifying Properties of Squares Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other. Holt Geometry

6 -4 Properties of Special Parallelograms Example 3 Continued Step 1 Show that EG and FH are congruent. Since EG = FH, Holt Geometry

6 -4 Properties of Special Parallelograms Example 3 Continued Step 2 Show that EG and FH are perpendicular. Since Holt Geometry ,

6 -4 Properties of Special Parallelograms Example 3 Continued Step 3 Show that EG and FH are bisect each other. Since EG and FH have the same midpoint, they bisect each other. The diagonals are congruent perpendicular bisectors of each other. Holt Geometry

6 -4 Properties of Special Parallelograms 6 -5 Conditions for Special Parallelograms Holt Geometry

6 -4 Properties of Special Parallelograms Holt Geometry

6 -4 Properties of Special Parallelograms Example 1: Carpentry Application 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. Holt Geometry

6 -4 Properties of Special Parallelograms Holt Geometry

6 -4 Properties of Special Parallelograms Example 2 A: 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. Holt Geometry

6 -4 Properties of Special Parallelograms Example 3 B: 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 Holt Geometry WXYZ.

6 -4 Properties of Special Parallelograms Example 3 B Continued Step 2 Find WY and XZ to determine is WXYZ is a rectangle. Since , WXYZ is not a rectangle. Thus WXYZ is not a square. Holt Geometry

6 -4 Properties of Special Parallelograms Example 3 B Continued Step 3 Determine if WXYZ is a rhombus. Since (– 1)(1) = – 1, rhombus. Holt Geometry , PQRS is a

6 -4 Properties of Special Parallelograms Lesson Quiz: Part I 1. Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square? Holt Geometry

6 -4 Properties of Special Parallelograms 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. TR 35 ft Holt Geometry 2. CE 29 ft

6 -4 Properties of Special Parallelograms Lesson Quiz: Part II PQRS is a rhombus. Find each measure. 3. QP 42 Holt Geometry 4. m QRP 51°

6 -4 Properties of Special Parallelograms 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. Holt Geometry

6 -4 Properties of Special Parallelograms Lesson Quiz: Part IV 6. Given: ABCD is a rhombus. Prove: Holt Geometry

6 -4 Properties of Special Parallelograms 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. TR 35 ft Holt Geometry 2. CE 29 ft