Unit 2B Parallelograms
Day 4 Friday, Nov 4, 2016
Agenda 11/4/2016
1)
2)
3)
4)
5)
6)
Bulletin
Switch and grade homework.
Constructing with Diagonals - wrap up
Pythagorean theorem problem
Tutoring Thursdays and Homework
Note: Tuesday review Part 2, and Thursday
Unit 2B test (45mins).
Homework, 11/2/2-16
6.4 Rectangles p. 344
J
K
Q
2pt
Draw the diagram.
N
7. NQ = 5x - 3
QM = 4x + 6
NK = NQ + QK = 2(5x - 3) Why?
M
Find NK.
Homework, 11/2/2-16
6.4 Rectangles p. 344
J
K
Q
Draw the diagram.
N
M
7. NQ = 5x - 3
QM = 4x + 6
Find NK.
NK = NQ + QK = 2(5x - 3) = 10x - 6
JM = JQ +QM = 2(4x+6) = 8x + 12
J
K
Q
Draw the diagram.
N
M
7. NQ = 5x - 3
QM = 4x + 6
Find NK.
NK = NQ + QK = 2(5x - 3) = 10x - 6
JM = JQ +QM = 2(4x+6) = 8x + 12
Diagonals of a rectangle are congruent
So 10x - 6 = 8x + 12
2x = 18
therefore x = 9
J
K
Q
Draw the diagram.
N
M
7. NQ = 5x - 3
QM = 4x + 6
Find NK.
2x = 18
therefore x = 9
And
NK = 10x - 6 = 10(9) - 6 = 84
NK = 84
2pt
J
K
Q
Same diagram.
N
M
9. NM = 8x - 14
JK = x2 + 1
Find JK.
Opposite sides of a rectangle are congruent so
8x - 14 = x2 + 1
0 = x2 + 1 - 8x + 14 = x2 - 8x +15 factors?
J
K
Q
Same diagram.
M
N
9. NM = 8x - 14
JK = x2 + 1
Find JK.
Opposite sides of a rectangle are congruent so
8x - 14 = x2 + 1
0 = x2 + 1 - 8x + 14 = x2 - 8x +15
0 = (x - 5) (x - 3)
2 solutions: x = 5, x = 3
So JK = 52 + 1 = 26 OR JK = 32 + 1 = 10
2pt
2pt
J
K
Q
Same diagram.
N
M
11. Find m∠JKN
if m∠NKM = x2 + 4 and m∠KNM = x +30
90° = x2 + 4 + x + 30
How do I know?
J
K
Q
Same diagram.
N
M
11. Find m∠JKN
if m∠NKM = x2 + 4 and m∠KNM = x +30
90° = x2 + 4 + x + 30
0 = x2 + x - 56
Factors?
J
K
Q
Same diagram.
N
M
11. Find m∠JKN
if m∠NKM = x2 + 4 and m∠KNM = x +30
90° = x2 + 4 + x + 30
0 = x2 + x - 56
Factors?
0 = (x + 8)(x - 7)
x = -8 or x = 7
J
K
Q
Same diagram.
N
M
11. Find m∠JKN
if m∠NKM = x2 + 4 and m∠KNM = x +30
90° = x2 + 4 + x + 30
0 = x2 + x - 56
0 = (x + 8)(x - 7)
x = - 8 or x = 7
m∠JKN = x + 30 = -8 + 30 = 22°
OR m∠JKN = 7 + 30 = 37°
2pt
2pt
W
30°
X
2
Z
13. m∠2 = (90 - 30)° = 60°
2pt
Y
1pt
W
30°
X
60°
60°
Z
15. m∠4 = (90 - 60)° = 30°
4
Y
1pt
30°
W
X
60°
6
Z
60°
30°
17. m∠6 = (90 - 30)° = 60°
30°
Y
1pt
30°
W
X
8
60°
60°
Z
30°
19. m∠8 = 30°
30°
Y
equal to angle 1
1pt
30°
W
X
11
12
Z
30°
21. m∠12 = m∠11 = 120°
60°
60°
30°
Y
2pt
Homework, 11/2/2-16
6.4 Rectangles p. 344
J
K
/20
CB
your
initials
Q
Draw the diagram.
N
7. NQ = 5x - 3
QM = 4x + 6
NK = NQ + QK = 2(5x - 3) Why?
M
Find NK.
Constructions with Diagonals
Review 1 - 6 (pdf)
7. Complete the chart below by
identifying the quadrilateral(s) for
which the given condition is
necessary.
Now with desk partner - quickly see how to
fit the results into the chart - some ideas in
the reasons column ready to share.
Conditions
Quadrilaterals
Diagonals are perpendicular.
Rhombus
Square
kite
Diagonals are perpendicular
AND only one diagonal is
bisected.
kite
Diagonals are congruent AND
intersect, BUT are not
perpendicular.
Rectangle
Isosceles
trapezoid
Reasons
Conditions
Quadrilaterals
Diagonals bisect each other
Parallelogram,
rectangle,
rhombus, square
Diagonals are perpendicular AND
bisect each other.
rhombus
Diagonals are congruent AND
bisect each other.
rectangle
Diagonals bisect each other AND
are congruent AND are
perpendicular.
square
Reasons
8. As you add more conditions to describe
diagonals, how does it change the types of
quadrilaterals possible? Why does this make
sense.
Prepare an answer to share. (Rough draft).
8. As you add more conditions to describe
diagonals, how does it change the types of
quadrilaterals possible? Why does this make
sense.
As more conditions are placed on the
quadrilateral, fewer types of quadrilaterals meet
all the restrictions.
We are making more specific demands on the
shape with each condition we require, so it
makes sense there will not be many ways to
meet all the conditions.
9. Name each of the figures below using as
many names as possible and state as many
properties as you can about each figure
Row 1 A & E
Row 2 B & E
Row 3 C & F
Row 4 D & F
2 mins
Figu Names
re
A
B
D
D
E
F
Properties
Figu Names
re
Properties
A
Parallelogram,
rectangle
Opposite sides parallel, vertices 90°,
diagonals bisect and are congruent
B
kite
Diagonals perpendicular, angles
between non-congruent sides are
congruent, diagonal between
congruent angles is bisected by the
other diagonal
C
D
Figu Names
re
Properties
C
Parallelogram,
Diagonals bisect each other and are
rectangle,
perpendicular
square, rhombus
D
trapezoid
No specific properties
E
parallelogram
Diagonals bisect each other
F
Parallelogram,
rhombus
Diagonals bisect each other and are
perpendicular
Property
Opposite sides are
parallel
Only one pair of
opposite sides is
parallel
Opposite sides are
congruent
Only one pair of
opposite sides is
congruent
Parallelogram
Rectangle
Rhombus
Square
Isosceles
trapezoid
Kite
Property
Opposite sides are
parallel
Parallelogram
x
Rectangle
x
Rhombus
x
Square
x
Only one pair of
opposite sides is
parallel
Opposite sides are
congruent
x
x
x
x
x
Only one pair of
opposite sides is
congruent
Opposite angles
are congruent
Isosceles
trapezoid
x
x
x
x
x
Kite
Property
Parallelogram
Rectangle
Rhombus
Square
Isosceles
trapezoid
Only one pair of
opposite angles is
congruent
X
Each diagonal
forms 2 congruent
triangles
x
x
x
x
Diagonals bisect
each other
x
x
x
x
x
x
Diagonals are
perpendicular
Diagonals are
congruent
Diagonals bisect
vertex angles
Kite
x
x
x
x
x
x
x
Property
All angles are right
angles
All sides are
congruent
Two pairs of
consecutive sides
are congruent.
Parallelogram
Rectangle
Rhombus
x
Square
Isosceles
trapezoid
Kite
x
x
x
x
Polygons
Quadrilaterals
Parallelograms
rectangles
rhombi
squares
kites
triangles
trapezoids
Isosceles
trapezoids
P.
64
Homework
Complete Pythagorean theorem review on
Page 67 and 68 and “Possible Extensions for
Class Activity” on page 69.
Then complete the Unit 2B Review Part 1
Homework sheet (use your textbook as well as
the work we’ve done in class.)