Holt Geometry
6-4 Properties of Special Parallelograms
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. CD 4. mC
4
47
14 104°
Holt Geometry
6-4 Properties of Special Parallelograms
6-4 Properties of Special Parallelograms
Holt Geometry
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.
Def. of segs.
Substitute and simplify.
KM = JL = 86
diags. bisect each other
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
Rhombus diag.
Example 2B: Using Properties of Rhombuses to Find Measures
TVWX is a rhombus. Find a.
Substitute 14a + 20 for mVTZ.
Subtract 20 from both sides and divide both sides by 14.
mVZT = 90°
14a + 20 = 90°
a = 5
Holt Geometry
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
The diagonals are congruent perpendicular bisectors of each other.
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.
Holt Geometry
6-4 Properties of Special Parallelograms
6-5 Conditions for Special Parallelograms
Holt Geometry
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 mABC = 90°. Why must ABCD be a rectangle?
Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 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 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.
Holt Geometry
6-4 Properties of Special Parallelograms
Example 3B: Identifying Special Parallelograms in the Coordinate Plane
W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3)
Step 1 Graph WXYZ.
Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.
Holt Geometry
6-4 Properties of Special Parallelograms
Step 2 Find WY and XZ to determine is WXYZ is a rectangle.
Thus WXYZ is not a square.
Example 3B Continued
Since , WXYZ is not a rectangle.
Holt Geometry
6-4 Properties of Special Parallelograms
Step 3 Determine if WXYZ is a rhombus.
Example 3B Continued
Since (–1)(1) = –1, , PQRS is a rhombus.
Holt Geometry
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 2. CE
35 ft 29 ft
Holt Geometry
6-4 Properties of Special Parallelograms
Lesson Quiz: Part II
PQRS is a rhombus. Find each measure.
3. QP 4. mQRP
42 51°
Holt Geometry
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
ABE CDF
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 2. CE
35 ft 29 ft