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 … 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.

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

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 = 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

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 = 4 + 25 = 29 HA = (10 – 8)2 + (4 – 9)2 = 4 + 25 = 29 BH = (8 – (–2))2 + (9 – 9)2 = 100 + 0 = 10 QA = (– 4 – 10)2 + (4 – 4)2 = 196 + 0 = 14 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 R = 2x – 10 and m S = 3x + 50. 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

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

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.

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

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 = 60 + 15 = 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

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 – 32 + 5 = 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

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

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