1. 6.1 Classifying Quadrilaterals page 288
Obj 1: To define & classify special types of quadrilaterals

2. And why…
To use the properties of special quadrilaterals with a kite, as in Example 3.

3. Seven important types of quadrilaterals …
Parallelogram-has both pairs of opposite sides parallel
Rhombus-has four congruent sides
Rectangle-has four right angles
Square-has four congruent sides and four right angles
Kite-has two pairs of adjacent sides congruent and no opposite opposite sides congruent.

4. Continued….
Trapezoid-has exactly one pair of parallel sides. (you have same side interior angles)
Isosceles trapezoid-is a trapezoid whose non-parallel opposite sides are congruent

5. 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

6. You try one
Turn to page 289 and complete check understanding 1 (top of page)

7. Classifying by Coordinate Method
Do you remember the slope formula?
Do you remember the distance formula that finds the distance between two points?

8. Do you remember how to tell if two lines are parallel?
Do you remember how to tell if two lines are perpendicular?

9. 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 = slope of BH = slope of HA = slope of QA =
9 – 4
–2 – (–4) 5 2 =
9 – 9
8 – (–2)
4 – 9
10 – 8
= –
4 – 4
–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

10. 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 = = 29
BH = (8 – (–2))2 + (9 – 9)2 = = 10
QA = (– 4 – 10)2 + (4 – 4)2 = = 14
Because QB = HA, QBHA is an isosceles trapezoid.
6-1

11. You try one
Turn to page 289 and complete check understanding 2 (bottom of page).

12. Classifying Quadrilaterals
In parallelogram RSTU, m R = 2x – 10 and
m S = 3x Find x.
Draw quadrilateral RSTU. Label R and S.
RSTU is a parallelogram.
Given
Definition of parallelogram
ST || RU
m R + m S = 180
If lines are parallel, then interior
angles on the same side of a
transversal are supplementary.
6-1

13. Classifying Quadrilaterals
(continued)
(2x – 10) + (3x + 50) = 180
Substitute 2x – 10 for m R and
3x + 50 for m S.
5x + 40 = 180
Simplify.
Subtract 40 from each side.
5x = 140
x = 28
Divide each side by 5.
6-1

14. You try one
Turn to page 290 and complete check understanding 3 (middle of page)

15. 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?

16. 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

17. You can use what you know about parallel lines & transversals to prove some theorems about parallelograms
Theorem 6.1 p Opposite sides of a parallelogram are congruent

18. 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.

19. Properties of Parallelograms
GEOMETRY LESSON 6-2
Use KMOQ to find m O.
Q and O are consecutive angles of
KMOQ, so they are supplementary.
Definition of supplementary angles
m O + m Q = 180
Substitute 35 for m Q.
m O + 35 = 180
Subtract 35 from each side.
m O = 145
6-2

20. Properties of Parallelograms
GEOMETRY LESSON 6-2
Find the value of x in ABCD. Then find m A.
x + 15 = 135 – x
Opposite angles of a are congruent.
2x + 15 = 135
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 = = 75
Consecutive angles of a
parallelogram are supplementary.
m A + m B = 180
m A + 75 = 180
Substitute 75 for m B.
Subtract 75 from each side.
m A = 105
6-2

21. Find the values of x and y in
KLMN.
x = 7y – 16
The diagonals of a parallelogram
bisect each other.
2x + 5 = 5y
2(7y – 16) + 5 = 5y
Substitute 7y – 16 for x in the
second equation to solve for y.
14y – = 5y
Distribute.
14y – 27 = 5y
Simplify.
–27 = –9y
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
Simplify.
So x = 5 and y = 3.
6-2

22. Summary 6.2 What are the properties of parallelograms? Theorem 6.1-

23. Homework
6.1 page 290
2-26 E, 37-42