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Transcript
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
• 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
60
• QKN = 60
• (Ext.  isos ∆)
Read the detail
• QKN + QMN = 90
60
• LKN + LMN = 180
• (Symmetry)
• Therefore KLMN is
cyclic.
• (Opp. ’s sum to 180)
2002
2002
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)
2003
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)
2004
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
x
ADB = EBC = x
(alt. ’s // lines)
x
x
ADB = DAE = x
x
(base ’s isos ∆)
x
x
2x
x
AEB = DAE +
ADE = 2x
(ext.  ∆)
x
x
2x
AEB = ABE
2x
x
(base ’s isos. ∆)
x
x
2x
AEB = 2CBE
2x
x
 = therefore
2005
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)
x
2x
Read and interpret the information
Let OBD=90-x
(base  isos. ∆)
90 - x
x
2x
Read and interpret the information
Let DBC = x
(rad  tang.)
90 - x
x
2x
x
Read and interpret the information
CBD =BED = x
90 - x
x
2x
x
2006
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
2x
x
EAB = 2x
(base ’s isos. ∆)
x
2x
2x
x
AEB = 180 - 4x
( sum ∆)
x
180-4x
2x
2x
x
2007
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.
Calculate the size of angle ACB.
Put in everything you know.
256
104
38
128
14
38
Now match reasons
ACO =38 (base ’s isos 
AOC = 104 (angle sum )
256
104
38
ABC=128 ( at centre)
BAC=38 (alt ’s // lines)
128
14
AOC = 256 (’s at a pt)
38
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.
12 AB

20 15
AB  9
Similar triangles
CB  6
ED  6 isos trapezium

Question 2b
• Kim’s model bridge uses a circular arc.
• The diagram shows the side view of Kim’s model
bridge.
WX = WY = UV = VX .
UX = XY.
U, V, W and Y lie on the circumference of the circle.
Angle VXW = 132°.
Calculate the size of angle WYZ.
You must give a geometric reason for each step
leading to your answer.
Write in the angles and give reasons as you go.
WXY=48 (adj  on a line)
Write in the angles and give reasons as you go.
WXY=48 (adj  on a line)
XYZ=48 (base
’s isos )
Write in the angles and give reasons as you go.
WXY=48 (adj  on a line)
XYZ=48 (base ’s isos )
XWY=84 (sum )
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)