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Classifying Quadrilaterals

Find the distance between the points. 1. M (2, –5), N (–7, 1) 2. P (–1, –3), Q (–6, –9) 3. C (–4, 6), D (5, –3) 4. X (0, 6), Y (4, 9) 5. R (3, 8), S (6, 0) 6. A (4, 3), B (2, 1). Find the slope of the line through each pair of points. Classifying Quadrilaterals.

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Classifying Quadrilaterals

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  1. Find the distance between the points. 1.M (2, –5), N (–7, 1) 2.P (–1, –3), Q (–6, –9) 3.C (–4, 6), D (5, –3) 4.X (0, 6), Y (4, 9) 5.R (3, 8), S (6, 0) 6.A (4, 3), B (2, 1) Find the slope of the line through each pair of points. Classifying Quadrilaterals Lesson 6-1 Check Skills You’ll Need (For help, go to Lesson 1-8 and page 165.)+ Check Skills You’ll Need 6-1

  2. Classifying Quadrilaterals Lesson 6-1 Notes A quadrilateralis a polygon with four sides. A parallelogram ()is a quadrilateral with both pairs of opposite sides parallel. A rhombusis a  with four congruent sides. A rectangleis a  with four right angles. A squareis a  with four congruent sides and four right angles. 6-1

  3. Classifying Quadrilaterals Lesson 6-1 Notes A kiteis a quadrilateral with two pairs of adjacent sides congruent and no opposite sides congruent. A trapezoidis a quadrilateral with exactlyone pair of parallel sides. An isosceles trapezoidis a trapezoid whose nonparallel sides are congruent. 6-1

  4. Classifying Quadrilaterals Lesson 6-1 Notes 6-1

  5. It is a trapezoid because AB and DC appear parallel and AD and BC appear nonparallel. Classifying Quadrilaterals Lesson 6-1 Additional Examples Classifying a Quadrilateral Judging by appearance, classify ABCD in as many ways as possible. ABCD is a quadrilateral because it has four sides. Quick Check 6-1

  6. Graph quadrilateral QBHA. First, find the slope of each side. slope of QB = slope of BH = slope of HA = slope of QA = 4 – 4 –4 – 10 9 – 9 8 – (–2) 4 – 9 10 – 8 9 – 4 –2 – (–4) 5 2 5 2 = = = = – 0 0 BH is parallel to QA because their slopes are equal. QB is not parallel to HA because their slopes are not equal. Classifying Quadrilaterals Lesson 6-1 Additional Examples Classifying by Coordinate Methods Determine the most precise name for the quadrilateral with vertices Q(–4, 4), B(–2, 9), H(8, 9), and A(10, 4). 6-1

  7. 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 Classifying Quadrilaterals Lesson 6-1 Additional Examples (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. Because QB = HA, QBHA is an isosceles trapezoid. Quick Check 6-1

  8. In parallelogram RSTU, mR = 2x – 10 and m S = 3x + 50. Find x. If lines are parallel, then interior angles on the same side of a transversal are supplementary. m R + m S = 180 Draw quadrilateral RSTU. Label R and S. RSTU is a parallelogram. Given ST || RU Definition of parallelogram Classifying Quadrilaterals Lesson 6-1 Additional Examples Using the Properties of Special Quadrilaterals 6-1

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

  10. Classifying Quadrilaterals Lesson 6-1 Lesson Quiz Judging by appearance, classify the quadrilaterals in Exercises 1 and 2 in as many ways as possible. 1. 2. quadrilateral, parallelogram, rectangle, rhombus, square quadrilateral, kite 4. What is the most precise name for the figure in Exercise 2? 3. What is the most precise name for the figure in Exercise 1? square kite 5. Find the values of the variables in the rhombus to the right. a = 60, x = 6, y = 2 6-1

  11. y2 – y1 x2– x1 y2 – y1 x2– x1 y2 – y1 x2– x1 Classifying Quadrilaterals Lesson 6-1 Check Skills You’ll Need Solutions 1.d = (x2 – x1)2 + (y2 – y1)2 = ( –7 – 2)2 + (1 – (–5))2 = ( –9)2 + 62 = 81 + 36 = 117 10.8 2.d = (x2 – x1)2 + (y2 – y1)2 = ( –6 – (–1))2 + (–9 – (–3))2 = ( –5)2 + (–6)2 = 25 + 36 = 61 7.8 3.d = (x2 – x1)2 + (y2 – y1)2 = ( 5 – (– 4))2 + (– 3 – 6)2 = 92 + (–9)2 = 81 + 81 = 162 12.7 3 4 9 – 6 4 – 0 4.m = = = 5.m = = = = – 6.m = = = = 1 0 – 8 6 – 3 –8 3 8 3 –2 –2 1 – 3 2 – 4 6-1

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