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NZQA Geometry

NZQA Geometry. Excellence. Sample 2001. Read the detail. Line KM forms an axis of symmetry. Length QN = Length QK . Angle NQM = 120°. Angle NMQ = 30°. Read the detail. Line KM forms an axis of symmetry. Symmetry is a reason Length QN = Length QK . Isosceles triangle

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NZQA Geometry

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  1. NZQA Geometry Excellence

  2. Sample 2001

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

  4. 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°.

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

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

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

  8. 2002

  9. 2002

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

  11. 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. 

  12. 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)

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

  14. 2003

  15. 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

  16. 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

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

  18. 2004

  19. 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

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

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

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

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

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

  25. 2005

  26. 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)

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

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

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

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

  31. 2006

  32. 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°.

  33. BEC = x (base ’s isos ∆) x

  34. EBA = 2x (ext  ∆) x x 2x

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

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

  37. 2007

  38. 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.

  39. Calculate the size of angle ACB.

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

  41. 256 38 104 128 14 38 Now match reasons 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 )

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

  43. 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.

  44. Similar triangles

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

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