nzqa geometry excellence. sample 2001 read the detail line km forms an axis of symmetry. length qn =...
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![Page 1: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/1.jpg)
NZQA Geometry
Excellence
![Page 2: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/2.jpg)
Sample 2001
![Page 3: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/3.jpg)
![Page 4: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/4.jpg)
Read the detail
• Line KM forms an axis of symmetry.
• Length QN = Length QK.
• Angle NQM = 120°.
• Angle NMQ = 30°.
![Page 5: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/5.jpg)
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°.
![Page 6: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/6.jpg)
Read the detail
• To prove KLMN is cyclic, you must prove that the opposite angles sum to 180 degrees.
![Page 7: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/7.jpg)
Read the detail
QKN = 60• (Ext. isos ∆)60
![Page 8: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/8.jpg)
Read the detail
QKN + QMN = 90
LKN + LMN = 180• (Symmetry)• Therefore KLMN is
cyclic.• (Opp. ’s sum to 180)
60
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2002
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2002
![Page 11: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/11.jpg)
Read the information
• The logo is based on two regular pentagons and a regular hexagon.
• AB and AC are straight lines.
![Page 12: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/12.jpg)
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.
![Page 13: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/13.jpg)
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)
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Interior angles in a hexagon
Therefore DAF = 72(Sum interior angles of a
hexagon = 720)
![Page 15: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/15.jpg)
2003
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![Page 17: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/17.jpg)
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
![Page 18: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/18.jpg)
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
![Page 19: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/19.jpg)
Let DAC= x and CFB= y
CAB + CBA = 90
• Therefore ACB = 90
• (sum ∆)
• Therefore AB is the diameter
• ( in a semi-circle)
![Page 20: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/20.jpg)
2004
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![Page 22: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/22.jpg)
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
![Page 23: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/23.jpg)
x
x
Let EBC = x
ADB = EBC = x
(alt. ’s // lines)
![Page 24: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/24.jpg)
x
x
ADB = DAE = x
(base ’s isos ∆)
x
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x
x
AEB = DAE + ADE = 2x
(ext. ∆)
x
2x
![Page 26: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/26.jpg)
x
x
AEB = ABE
(base ’s isos. ∆)
x
2x
2x
![Page 27: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/27.jpg)
x
x
AEB = 2CBE
x
2x
2x
= therefore
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2005
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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)
![Page 31: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/31.jpg)
Read and interpret the information
x2x
Let BED=x
BOD =2x
( at centre)
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Read and interpret the information
x2x
Let OBD=90-x
(base isos. ∆)
90 - x
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Read and interpret the information
x2x
Let DBC = x
(rad tang.)
90 - x x
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Read and interpret the information
x2x
CBD =BED = x
90 - x x
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2006
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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°.
![Page 38: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/38.jpg)
x
BEC = x
(base ’s isos ∆)
![Page 39: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/39.jpg)
x
EBA = 2x
(ext ∆)
2xx
![Page 40: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/40.jpg)
x
EAB = 2x
(base ’s isos. ∆)
2xx
2x
![Page 41: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/41.jpg)
x
AEB = 180 - 4x
( sum ∆)
2xx
2x
180-4x
![Page 42: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/42.jpg)
2007
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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.
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Calculate the size of angle ACB.
![Page 45: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/45.jpg)
Put in everything you know.
38
104
256
128
38
14
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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 )
![Page 47: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/47.jpg)
Question 2c
• Tony’s model bridge uses straight lines.• The diagram shows the side view of Tony’s model
bridge.
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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.
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Similar triangles
€
1220
=AB15
AB=9CB=6ED=6 isos trapezium
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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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WX = WY = UV = VX .UX = XY.
U, V, W and Y lie on the circumference of the circle.Angle VXW = 132°.
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Calculate the size of angle WYZ.You must give a geometric reason for
each step leading to your answer.
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Write in the angles and give reasons as you go.
WXY=48 (adj on a line)
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Write in the angles and give reasons as you go.
WXY=48 (adj on a line)
XYZ=48 (base ’s isos )
![Page 55: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/55.jpg)
Write in the angles and give reasons as you go.
WXY=48 (adj on a line)XYZ=48 (base ’s isos )
XWY=84 (sum )
![Page 56: 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°](https://reader036.vdocuments.site/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/56.jpg)
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)