1. GT Geometry Unit 6: Quadrilaterals Jeopardy

2. Angles of Polygons || - ogram Properties || - ogram Tests Rhombi/Tra pezoids Area Coordinate Plane 100 200 300 400 500

3. $100 What is the sum of the interior angles of a pentagon?

4. $100 540 Degrees

5. $200 What is the measure of each exterior angle of a regular octagon?

6. $200 45 degrees

7. $300 If each interior angle of a regular polygon is 140 degrees how many sides does the polygon have?

8. $300 9 sides

9. $400 If each exterior angle of a regular polygon is 72 degrees how many sides does the polygon have?

10. $400 5 sides

11. $500 If each interior angle of a regular polygon is 150 degrees what is the measure of each exterior angle?

12. $500 30 degrees

13. $100 Find x if the quad below is a parallelogram

14. $100 X = 7

15. $200 Find x if the quad below is a parallelogram

16. $200 X = 12

17. $300 Find x if the quad below is a parallelogram

18. $300 X = 3

19. $400 Find x if the quadrilateral below is a parallelogram

20. $400 X= 33

21. $500 Find x if the quadrilateral below is a parallelogram

22. $500 X = 14.5

23. $100 Can we prove this quadrilateral is a parallelogram?

24. $100 Yes both pairs of opposite sides are congruent

25. $200 Can we prove this quadrilateral is a parallelogram?

26. $200 No, we don’t know that both pairs of opposite angles are congruent

27. $300 Can we prove this quadrilateral is a parallelogram?

28. $300 Yes, one pair of opposites sides is both congruent and parallel

29. $400 Can we prove this quadrilateral is a parallelogram?

30. $400 Yes, diagonals bisect each other

31. $500 Can we prove this quadrilateral is a parallelogram?

32. $500 Yes, the total sum of the angles of a quadrilateral is 360 degrees. Therefore x = 100. Since the opposite angles are congruent it is a parallelogram

33. $100 If the quadrilateral below is a rhombus, find x

34. $100 X = 4.5

35. $200 If the trapezoid below is an isoceles trapezoid find x.

36. $200 X = 12

37. $300 If the trapezoid below is an isosceles trapezoid, find x

38. $300 X = 14.5

39. $400 If the quadrilateral below is a rhombus find x

40. $400 X = 2 Diagonals of a rhombus bisect angles

41. $500 If the quadrilateral below is a rhombus find x

42. $500 X = 17 The diagonals of a rhombus are perpendicular so use the Pythagorean theorem

43. $100 Find the area of the polygon

44. $100 A = 70 Area of a rhombus = ½ (d 1 )(d 2 ) D 1 = 7 + 7 = 14 D 2 = 5+5 = 10

45. $200 Find the area of the quadrilateral

46. $200 A = 64 Area of a trapezoid = ½ (b 1 +b 2 )h = ½ ( 6+10) 8

47. $300 Find the area of the quadrilateral below. Hint (use the Pythagorean theorem to find the missing side.)

48. $300 A = 192 The height of the rectangle = 12. 12 x 16 = 192

49. $400 The quadrilateral has an area of 60 sq inches. Find x

50. $400 x = 8

51. $500 Find the area of the yellow region.

52. $500 X = 96. The area of the rectangle = 16 x 12. The area of the two triangles are ½ (8)(12). Subtract the two.

53. $100 JKLM is a quadrilateral with J(0,0), K (3,7), L(9,7) and M(6,0). Is JKLM a parallelogram?

54. $100 Yes opposite sides are parallel and congruent Slope: JK = 7/3 LM = 7/3 KL = 0 JM = 0

55. $200 Is ABCD a rhombus? A (3,1) B(3,-3) C(-2,-3) D (-2,1)

56. $200 No. Diagonals are not perpendicular Slope of diagonals AC = 4/5 BD = - 4/5

57. $300 Is LMNO a trapezoid? L ( 5,2) M (1,9) N (-3, 2) O (1,-5)

58. $300 Yes. 1 opposite side is parallel. It is also an isosceles trapezoid. SlopeCongruent Legs LM = - 7/4LO = sq rt 65 ON = -7/4MN = sq rt 65

59. $400 Is PQRS a square? P (5,2) Q (2,5)R( -1,2)S (2,-1)

60. $400 Yes. Diagonals are congruent and perpendicular CongruentSlope RP = 6RP = 0 QS = 6QS = undefined.

61. $500 JKLM is a quadrilateral with K(6,0) L (7,2) and M (2,8) what are the coordinates of J to make JKLM a parallelogram?

62. $500 J = (1,6) or (11,-6) The slope of LM = -6/5. Therefore the slope of JK = -6/5. The slope could also be written as 6/-5. Therefore we must solve for x and y. The following two coordinates would make this slope (1,6) or (11,-6) y – 0 = 6 x = 1, y = 6y – 0 = -6 x = 11, y = -6 x – 6 = -5x – 6 = 5 Then we find that the distance for LM = sq rt 61. Therefore, we plug in both possible coordinates to determine which one gives us a distance for JK = sq rt 61. Since they both do both answers are correct. JK when J = (1,6) = sq rt 61JK when J = 11,-6) = sq rt 61