Transcript 6.1 Classifying Quadrilaterals page 288

6.1
Classifying
Quadrilaterals
page 288
Obj 1: To define &
classify special types of
quadrilaterals
And why…
• To use the properties of special
quadrilaterals with a kite, as in
Example 3.
Seven important types of
quadrilaterals …
1) Parallelogram-has both pairs of
opposite sides parallel
2) Rhombus-has four congruent sides
3) Rectangle-has four right angles
4) Square-has four congruent sides and
four right angles
5) Kite-has two pairs of adjacent sides
congruent and no opposite opposite
sides congruent.
Continued….
6. Trapezoid-has exactly one pair of
parallel sides. (you have same side
interior angles)
7. Isosceles trapezoid-is a trapezoid
whose non-parallel opposite sides
are congruent
Classifying Quadrilaterals
Judging by appearance, classify ABCD in as many
ways as possible.
ABCD is a quadrilateral because it has
four sides.
It is a trapezoid because AB and DC appear
parallel and AD and BC appear nonparallel.
6-1
You try one
• Turn to page 289 and complete
check understanding 1 (top of page)
Classifying by Coordinate
Method
• Do you remember the slope formula?
• Do you remember the distance
formula that finds the distance
between two points?
• Do you remember how to tell if two
lines are parallel?
• Do you remember how to tell if two
lines are perpendicular?
Classifying Quadrilaterals
Determine the most precise name for the quadrilateral with
vertices Q(–4, 4), B(–2, 9), H(8, 9), and A(10, 4).
Graph quadrilateral QBHA.
First, find the slope of each side.
slope of QB =
9–4
5
=
–2 – (–4)
2
slope of BH =
slope of HA =
9–9
=0
8 – (–2)
4–9
5
=
–
10 – 8
2
slope of QA =
4–4
=0
–4 – 10
BH is parallel to QA because their slopes are equal. QB is not parallel
to HA because their slopes are not equal.
6-1
Classifying Quadrilaterals
(continued)
One pair of opposite sides are parallel, so QBHA is a trapezoid.
Next, use the distance formula to see whether any pairs
of sides are congruent.
QB =
( –2 – ( –4))2 + (9 – 4)2 =
HA =
(10 – 8)2 + (4 – 9)2 =
BH =
(8 – (–2))2 + (9 – 9)2 =
100 + 0 = 10
QA =
(– 4 – 10)2 + (4 – 4)2 =
196 + 0 = 14
4 + 25 =
4 + 25 =
29
29
Because QB = HA, QBHA is an isosceles trapezoid.
6-1
You try one
• Turn to page 289 and complete
check understanding 2 (bottom of
page).
Classifying Quadrilaterals
In parallelogram RSTU, m
m
R = 2x – 10 and
S = 3x + 50. Find x.
Draw quadrilateral RSTU. Label
RSTU is a parallelogram.
ST || RU
m R + m S = 180
R and
S.
Given
Definition of parallelogram
If lines are parallel, then interior
angles on the same side of a
transversal are supplementary.
6-1
Classifying Quadrilaterals
(continued)
(2x – 10) + (3x + 50) = 180
5x + 40 = 180
5x = 140
x = 28
Substitute 2x – 10 for m R and
3x + 50 for m S.
Simplify.
Subtract 40 from each side.
Divide each side by 5.
6-1
You try one
• Turn to page 290 and complete
check understanding 3 (middle of
page)
Summary 6.1



What are the seven types of quadrilaterals
we have described today?
How do you tell if two lines are parallel?
How do you tell if two lines are
perpendicular?
6.2 Properties of
Parallelograms (page 294)
• Obj 1: to use relationships among
sides & among angles of
parallelograms
• Obj 2: to use relationships involving
diagonals of parallelograms &
transversals
• You can use what you know about
parallel lines & transversals to prove
some theorems about parallelograms
• Theorem 6.1 p. 294---Opposite
sides of a parallelogram are
congruent
•
Theorems Continued…
• Theorem 6.2 page 295 –Opposite angles
of a parallelogram are congruent
• Theorem 6.3 page 296—the diagonals
of a parallelogram bisect each other
• Theorem 6.4 page 297—If three of
more parallel lines cut off congruent
segments on one transversal, then they
cut off congruent segments on every
transveral.
GEOMETRY LESSON 6-2
Properties of Parallelograms
Use
KMOQ to find m O.
Q and O are consecutive angles of
KMOQ, so they are supplementary.
m O + m Q = 180
m O + 35 = 180
m O = 145
Definition of supplementary angles
Substitute 35 for m Q.
Subtract 35 from each side.
6-2
GEOMETRY LESSON 6-2
Properties of Parallelograms
Find the value of x in
x + 15 = 135 – x
2x + 15 = 135
ABCD. Then find m A.
Opposite angles of a
are congruent.
Add x to each side.
2x = 120
Subtract 15 from each side.
x = 60
Divide each side by 2.
Substitute 60 for x.
m B = 60 + 15 = 75
m A + m B = 180
Consecutive angles of a
parallelogram are supplementary.
m A + 75 = 180
Substitute 75 for m B.
m A = 105
Subtract 75 from each side.
6-2
Find the values of x and y in
KLMN.
x = 7y – 16
The diagonals of a parallelogram
2x + 5 = 5y
bisect each other.
2(7y – 16) + 5 = 5y
14y – 32 + 5 = 5y
14y – 27 = 5y
–27 = –9y
Substitute 7y – 16 for x in the
second equation to solve for y.
Distribute.
Simplify.
Subtract 14y from each side.
3=y
Divide each side by –9.
x = 7(3) – 16
Substitute 3 for y in the first
equation to solve for x.
x=5
So x = 5 and y = 3.
Simplify.
6-2
Summary 6.2
What are the properties of
parallelograms?
Theorem 6.1Theorem 6.2Theorem 6.3Theorem 6.4-
Homework
6.1 page 290
2-26 E, 37-42