Section 1.1 Angle Measure and Arc Length Section 1.1 Angle Measure and Arc Length.
NZQA Geometry Excellence. Sample 2001 Read the detail Line KM forms an axis of symmetry. Length QN =...
-
Upload
barnard-phelps -
Category
Documents
-
view
222 -
download
2
Transcript of NZQA Geometry Excellence. Sample 2001 Read the detail Line KM forms an axis of symmetry. Length QN =...
![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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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
![Page 9: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/9.jpg)
2002
![Page 10: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/10.jpg)
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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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)
![Page 14: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/14.jpg)
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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/15.jpg)
2003
![Page 16: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/16.jpg)
![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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/20.jpg)
2004
![Page 21: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/21.jpg)
![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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/24.jpg)
x
x
ADB = DAE = x
(base ’s isos ∆)
x
![Page 25: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/25.jpg)
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.fdocuments.net/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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/27.jpg)
x
x
AEB = 2CBE
x
2x
2x
= therefore
![Page 28: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/28.jpg)
2005
![Page 29: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/29.jpg)
![Page 30: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/30.jpg)
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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/31.jpg)
Read and interpret the information
x2x
Let BED=x
BOD =2x
( at centre)
![Page 32: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/32.jpg)
Read and interpret the information
x2x
Let OBD=90-x
(base isos. ∆)
90 - x
![Page 33: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/33.jpg)
Read and interpret the information
x2x
Let DBC = x
(rad tang.)
90 - x x
![Page 34: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/34.jpg)
Read and interpret the information
x2x
CBD =BED = x
90 - x x
![Page 35: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/35.jpg)
2006
![Page 36: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/36.jpg)
![Page 37: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/37.jpg)
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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/42.jpg)
2007
![Page 43: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/43.jpg)
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.
![Page 44: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/44.jpg)
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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/45.jpg)
Put in everything you know.
38
104
256
128
38
14
![Page 46: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/46.jpg)
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.fdocuments.net/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.
![Page 48: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/48.jpg)
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.
![Page 49: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/49.jpg)
Similar triangles
€
1220
=AB15
AB=9CB=6ED=6 isos trapezium
![Page 50: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/50.jpg)
Question 2b
• Kim’s model bridge uses a circular arc.• The diagram shows the side view of Kim’s model
bridge.
![Page 51: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/51.jpg)
WX = WY = UV = VX .UX = XY.
U, V, W and Y lie on the circumference of the circle.Angle VXW = 132°.
![Page 52: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/52.jpg)
Calculate the size of angle WYZ.You must give a geometric reason for
each step leading to your answer.
![Page 53: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/53.jpg)
Write in the angles and give reasons as you go.
WXY=48 (adj on a line)
![Page 54: 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.fdocuments.net/reader036/viewer/2022062422/56649e725503460f94b708df/html5/thumbnails/54.jpg)
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.fdocuments.net/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.fdocuments.net/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)