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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 Excellence
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 • Angle NQM = 120°. • Angle NMQ = 30°.
Read the detail • To prove KLMN is cyclic, you must prove that the opposite angles sum to 180 degrees.
Read the detail • QKN = 60 • (Ext. isos ∆) 60
Read the detail • QKN + QMN = 90 • LKN + LMN = 180 • (Symmetry) • Therefore KLMN is cyclic. • (Opp. ’s sum to 180) 60
Read the information • The logo is based on two regular pentagons and a regular hexagon. • AB and AC are straight lines.
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.
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)
Interior angles in a hexagon Therefore DAF = 72 (Sum interior angles of a hexagon = 720)
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
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
Let DAC= x and CFB= y CAB + CBA = 90 • Therefore ACB = 90 • (sum ∆) • Therefore AB is the diameter • ( in a semi-circle)
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
x Let EBC = x ADB = EBC = x (alt. ’s // lines) x
x x ADB = DAE = x (base ’s isos ∆) x
x x 2x AEB = DAE + ADE = 2x (ext. ∆) x
x x 2x AEB = ABE (base ’s isos. ∆) 2x x
x x 2x AEB = 2CBE 2x x = therefore
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)
Read and interpret the information Let BED=x BOD =2x ( at centre) 2x x
Read and interpret the information Let OBD=90-x (base isos. ∆) 90 - x 2x x
Read and interpret the information Let DBC = x (rad tang.) x 90 - x 2x x
Read and interpret the information CBD =BED = x x 90 - x 2x x
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°.
BEC = x (base ’s isos ∆) x
EBA = 2x (ext ∆) x x 2x
EAB = 2x (base ’s isos. ∆) x x 2x 2x
AEB = 180 - 4x ( sum ∆) x 180-4x x 2x 2x
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.
Put in everything you know. 256 38 104 128 14 38
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 )
Question 2c • Tony’s model bridge uses straight lines. • The diagram shows the side view of Tony’s model bridge.
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.
Question 2b • Kim’s model bridge uses a circular arc. • The diagram shows the side view of Kim’s model bridge.