# parallelogram

```6-2 Properties of Parallelograms
A quadrilateral with two pairs of parallel sides is a
parallelogram. To write the name of a parallelogram,
you use the symbol
.
Holt Geometry
6-2 Properties of Parallelograms
Holt Geometry
6-2 Properties of Parallelograms
Holt Geometry
6-2 Properties of Parallelograms
Go to the following videos to see how to work the next problems.
http://my.hrw.com/math06_07/nsmedia/lesson_videos/geo/player.h
tml?contentSrc=6551/6551.xml
http://my.hrw.com/math06_07/nsmedia/lesson_videos/geo/player.h
tml?contentSrc=6754/6754.xml
http://my.hrw.com/math06_07/nsmedia/lesson_videos/geo/player.h
tml?contentSrc=6755/6755.xml
Holt Geometry
6-2 Properties of Parallelograms
Example 1A: Properties of Parallelograms
In
CDEF, DE = 74 mm,
DG = 31 mm, and mFCD = 42&deg;.
Find CF.
 opp. sides 
CF = DE
Def. of  segs.
CF = 74 mm
Substitute 74 for DE.
Holt Geometry
6-2 Properties of Parallelograms
Example 1B: Properties of Parallelograms
In
CDEF, DE = 74 mm,
DG = 31 mm, and mFCD = 42&deg;.
Find mEFC.
mEFC + mFCD = 180&deg;
mEFC + 42 = 180
mEFC = 138&deg;
Holt Geometry
 cons. s supp.
Substitute 42 for mFCD.
Subtract 42 from both sides.
6-2 Properties of Parallelograms
Example 1C: Properties of Parallelograms
In
CDEF, DE = 74 mm,
DG = 31 mm, and mFCD = 42&deg;. Find DF.
DF = 2DG
 diags. bisect each other.
DF = 2(31)
Substitute 31 for DG.
DF = 62
Simplify.
Holt Geometry
6-2 Properties of Parallelograms
Watch this video. It will help with your homework
worksheet.
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/geo/player.html?contentSrc=6754/6754.xml
Holt Geometry
6-2 Properties of Parallelograms
Check It Out! Example 2a
EFGH is a parallelogram.
Find JG.
 diags. bisect each other.
EJ = JG
Def. of  segs.
3w = w + 8 Substitute.
2w = 8
Simplify.
w=4
Divide both sides by 2.
JG = w + 8 = 4 + 8 = 12
Holt Geometry
6-2 Properties of Parallelograms
Check It Out! Example 2b
EFGH is a parallelogram.
Find FH.
 diags. bisect each other.
FJ = JH
4z – 9 = 2z
2z = 9
z = 4.5
Def. of  segs.
Substitute.
Simplify.
Divide both sides by 2.
FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18
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
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.html?contentSrc=6552/6552.xml
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)&deg;
mQ = [(6(6.5) + 7)]&deg; = 46&deg;
Given
Substitute 6.5 for y
and simplify.
mS = (8y – 6)&deg;
Given
Substitute 6.5 for y
and simplify.
mR = (15x – 16)&deg;
Given
Substitute 10 for x
mR = [(15(10) – 16)]&deg; = 134&deg;
and simplify.
mS = [(8(6.5) – 6)]&deg; = 46&deg;
Holt Geometry
6-3 Conditions for Parallelograms
Example 1B Continued
Since 46&deg; + 134&deg; = 180&deg;, R is supplementary to
both Q and S. PQRS is a parallelogram by
Theorem 6-3-4.
Holt Geometry
6-3 Conditions for Parallelograms
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/player.html?contentSrc=6757/6757.xml
Holt Geometry
6-3 Conditions for Parallelograms
Example 2A: Applying Conditions for Parallelograms
Determine if the quadrilateral must be a
Yes. The 73&deg; angle is
supplementary to both
its corresponding angles.
By Theorem 6-3-4, the