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6-3
6-3 Conditions
Conditionsfor
forParallelograms
Parallelograms
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Geometry
Holt
Geometry
6-3 Conditions for Parallelograms
Warm Up
Justify each statement.
1.
2.
Reflex Prop. of 
Conv. of Alt. Int. s Thm.
Evaluate each expression for x = 12 and
y = 8.5.
3. 2x + 7 31
4. 16x – 9 183
5. (8y + 5)° 73°
Holt Geometry
6-3 Conditions for Parallelograms
Objective
Prove that a given quadrilateral is a
parallelogram.
Prove and apply properties of
rectangles, rhombuses, and squares.
Use properties of rectangles,
rhombuses, and squares to solve
problems.
Holt Geometry
6-4 Properties of Special Parallelograms
Vocabulary
rectangle
rhombus
square
Holt Geometry
6-3 Conditions for Parallelograms
Holt Geometry
6-3 Conditions for Parallelograms
The two theorems below can also be used to show that
a given quadrilateral is a parallelogram.
Holt Geometry
6-3 Conditions for Parallelograms
You have learned several ways to determine whether a
quadrilateral is a parallelogram. You can use the given
information about a figure to decide which condition is
best to apply.
Holt Geometry
6-3 Conditions for Parallelograms
Example 1A: Verifying Figures are Parallelograms
Show that JKLM is
a parallelogram for
a = 3 and b = 9.
Step 1 Find JK and LM.
JK = 15a – 11
Given
LM = 10a + 4
Substitute
JK = 15(3) – 11 = 34 and simplify. LM = 10(3)+ 4 = 34
Holt Geometry
6-3 Conditions for Parallelograms
Example 1A Continued
Step 2 Find KL and JM.
KL = 5b + 6
KL = 5(9) + 6 = 51
Given
JM = 8b – 21
Substitute
and simplify. JM = 8(9) – 21 = 51
Since JK = LM and KL = JM, JKLM is a parallelogram
by Theorem 6-3-2.
Holt Geometry
6-3 Conditions for Parallelograms
Example 1B: Verifying Figures are Parallelograms
Show that PQRS is a
parallelogram for x = 10
and y = 6.5.
mQ = (6y + 7)°
mQ = [(6(6.5) + 7)]° = 46°
Given
Substitute 6.5 for y
and simplify.
mS = (8y – 6)°
Given
Substitute 6.5 for y
and simplify.
mR = (15x – 16)°
Given
Substitute 10 for x
mR = [(15(10) – 16)]° = 134°
and simplify.
mS = [(8(6.5) – 6)]° = 46°
Holt Geometry
6-3 Conditions for Parallelograms
Example 1B Continued
Since 46° + 134° = 180°, R is supplementary to
both Q and S. PQRS is a parallelogram by
Theorem 6-3-4.
Holt Geometry
6-3 Conditions for Parallelograms
Check It Out! Example 2a
Determine if the
quadrilateral must be a
parallelogram. Justify
your answer.
Yes
The diagonal of the quadrilateral forms 2 triangles.
Two angles of one triangle are congruent to two
angles of the other triangle, so the third pair of
angles are congruent by the Third Angles Theorem.
So both pairs of opposite angles of the quadrilateral
are congruent .
By Theorem 6-3-3, the quadrilateral is a parallelogram.
Holt Geometry
6-3 Conditions for Parallelograms
Check It Out! Example 2b
Determine if each
quadrilateral must be a
parallelogram. Justify
your answer.
No. Two pairs of consective sides
are congruent.
None of the sets of conditions for a
parallelogram are met.
Holt Geometry
6-3 Conditions for Parallelograms
Helpful Hint
To say that a quadrilateral is a parallelogram by
definition, you must show that both pairs of
opposite sides are parallel.
Holt Geometry
6-3 Conditions for Parallelograms
Helpful Hint
To show that a quadrilateral is a parallelogram,
you only have to show that it satisfies one of
these sets of conditions.
Holt Geometry
6-3 Conditions for Parallelograms
Example 4: Application
The legs of a keyboard tray are
connected by a bolt at their
midpoints, which allows the tray to
be raised or lowered. Why is PQRS
always a parallelogram?
Since the bolt is at the midpoint of both legs, PE = ER
and SE = EQ. So the diagonals of PQRS bisect each
other, and by Theorem 6-3-5, PQRS is always a
parallelogram.
Holt Geometry
6-4 Properties of Special Parallelograms
Holt Geometry
6-4 Properties of Special Parallelograms
A second type of special quadrilateral is a rectangle.
A rectangle is a quadrilateral with four right angles.
Holt Geometry
6-4 Properties of Special Parallelograms
Since a rectangle is a parallelogram by Theorem 6-4-1,
a rectangle “inherits” all the properties of
parallelograms that you learned in Lesson 6-2.
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 another special quadrilateral. 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
Like a rectangle, a rhombus is a parallelogram. So you
can apply the properties of parallelograms to
rhombuses.
Holt Geometry
6-4 Properties of Special Parallelograms
Example 2B Continued
mVTZ = mZTX
Rhombus  each diag.
bisects opp. s
mVTZ = (5a – 5)°
Substitute 5a – 5 for mVTZ.
mVTZ = [5(5) – 5)]° Substitute 5 for a and simplify.
= 20°
Holt Geometry
6-4 Properties of Special Parallelograms
A square is a quadrilateral with four right angles and
four congruent sides. In the exercises, you will show
that a square is a parallelogram, a rectangle, and a
rhombus. So a square has the properties of all three.
Holt Geometry
6-4 Properties of Special Parallelograms
Helpful Hint
Rectangles, rhombuses, and squares are
sometimes referred to as special parallelograms.
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
Check It Out! Example 4
Given: PQTS is a rhombus with diagonal
Prove:
Holt Geometry
6-4 Properties of Special Parallelograms
Check It Out! Example 4 Continued
Statements
1. PQTS is a rhombus.
2.
3. QPR  SPR
4.
5.
6.
7.
Holt Geometry
Reasons
1. Given.
2. Rhombus → each
diag. bisects opp. s
3. Def. of  bisector.
4. Def. of rhombus.
5. Reflex. Prop. of 
6. SAS
7. CPCTC
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°
6-3 Conditions for Parallelograms
Class work problems
Page 402: 10, 12,14,16,18,20
Page 412 :10, 14, 16, 17, 22, 24
Holt Geometry