1. NZQA Geometry Excellence

2. Sample 2001

4. Read the detail Line KM forms an axis of symmetry. Length QN = Length QK. Angle NQM = 120°. Angle NMQ = 30°.

5. Read the detail Line KM forms an axis of symmetry. Symmetry is a reason Length QN = Length QK. Isosceles triangle Angle NQM = 120°. Angle NMQ = 30°.

6. Read the detail To prove KLMN is cyclic, you must prove that the opposite angles sum to 180 degrees.

7. Read the detail  QKN = 60 (Ext.  isos ∆) 60

8. Read the detail  QKN +  QMN = 90  LKN +  LMN = 180 (Symmetry) Therefore KLMN is cyclic. (Opp.  ’s sum to 180) 60

9. 2002

11. Read the information The logo is based on two regular pentagons and a regular hexagon. AB and AC are straight lines.

12. Interior angles in a hexagon Interior  ’s sum to (6-2) x 180 = 720 Exterior angles in regular figures are 360/no. of sides. Interior angle is 180 minus the ext. 

13. Interior angles in a hexagon  ADG = HFA = 360/5= 72 (ext.  regular pentagon)  DGE=  EHF = 132 (360-108-120) (Interior angles regular figures) (  ’s at a point) Reflex  GEH = 240 (360-120) (Interior angles regular figures) (  ’s at a point)

14. Interior angles in a hexagon Therefore  DAF = 72 (Sum interior angles of a hexagon = 720)

15. 2003

17. Read the information and absorb what this means The lines DE and FG are parallel. Coint  ’s sum to 180 AC bisects the angle DAB.  DAC=  CAB BC bisects the angle FBA.  CBF=  CBA

18. Let  DAC= x and  CFB= y  DAB = 2x (  DAC=  CAB)  FBA= 2y (  FBC=  CBA) 2x + 2y = 180 (coint  ’s // lines) X + y = 90 I.e.  CAB +  CBA = 90

19. Let  DAC= x and  CFB= y  CAB +  CBA = 90 Therefore  ACB = 90 (  sum ∆) Therefore AB is the diameter (  in a semi-circle)

20. 2004

22. Read and interpret the information In the figure below AD is parallel to BC. Coint  s sum to 180 Corr.  s are equal Alt.  s are equal A is the centre of the arc BEF. ∆ABE is isos E is the centre of the arc ADG. ∆AED is isos

23. x x Let  EBC = x  ADB =  EBC = x (alt.  ’s // lines)

24. x x  ADB =  DAE = x (base  ’s isos ∆) x

25. x x  AEB =  DAE +  ADE = 2x (ext.  ∆) x 2x

26. x x  AEB =  ABE (base  ’s isos. ∆) x 2x

27. x x  AEB = 2  CBE x 2x  = therefore

28. 2005

30. Read and interpret the information The circle, centre O, has a tangent AC at point B. ∆BOD isos. AB  OB (rad  tang) The points E and D lie on the circle.  BOD=2  BED (  at centre)

31. Read and interpret the information x 2x Let  BED=x  BOD =2x (  at centre)

32. Read and interpret the information x 2x Let  OBD=90-x (base  isos. ∆) 90 - x

33. Read and interpret the information x 2x Let  DBC = x (rad  tang.) 90 - x x

34. Read and interpret the information x 2x  CBD =  BED = x 90 - x x

35. 2006

37. Read and interpret In the above diagram, the points A, B, D and E lie on a circle. Angles same arc Cyclic quad AE = BE = BC. AEB, EBC Isos ∆s The lines BE and AD intersect at F. Angle DCB = x°.

38. x  BEC = x (base  ’s isos ∆)

39. x  EBA = 2x (ext  ∆) 2x x

40. x  EAB = 2x (base  ’s isos. ∆) 2x x

41. x  AEB = 180 - 4x (  sum ∆) 2x x 180-4x

42. 2007

43. Question 3 A, B and C are points on the circumference of the circle, centre O. AB is parallel to OC. Angle CAO = 38°. Calculate the size of angle ACB. You must give a geometric reason for each step leading to your answer.

44. Calculate the size of angle ACB.

45. Put in everything you know. 38 104 256 128 38 14

46. Now match reasons 38 104 256 128 38 14  ACO =38 (base  ’s isos   AOC = 104 (angle sum  )  AOC = 256 (  ’s at a pt)  ABC=128 (  at centre)  BAC=38 (alt  ’s // lines)  ACB= 14 (  sum  )

47. Question 2c Tony ’ s model bridge uses straight lines. The diagram shows the side view of Tony ’ s model bridge.

48. BCDE is an isosceles trapezium with CD parallel to BE. AC = 15 cm, BE = 12 cm, CD = 20 cm. Calculate the length of DE. You must give a geometric reason for each step leading to your answer.

49. Similar triangles

50. Question 2b Kim ’ s model bridge uses a circular arc. The diagram shows the side view of Kim ’ s model bridge.

51. WX = WY = UV = VX. UX = XY. U, V, W and Y lie on the circumference of the circle. Angle VXW = 132°.

52. Calculate the size of angle WYZ. You must give a geometric reason for each step leading to your answer.

53. Write in the angles and give reasons as you go.  WXY=48 (adj  on a line)

54. Write in the angles and give reasons as you go.  WXY=48 (adj  on a line)  XYZ=48 (base  ’s isos  )

55. Write in the angles and give reasons as you go.  WXY=48 (adj  on a line)  XYZ=48 (base  ’s isos  )  XWY=84 (  sum  )

56. Write in the angles and give reasons as you go.  WXY=48 (adj  on a line)  XYZ=48 (base  ’s isos  )  XWY=84 (  sum  )  WYZ=132 (ext  )