Page 4. Line, segment and angle Postulates Line and Segment Postulates Line – a line contains at...
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Transcript of Page 4. Line, segment and angle Postulates Line and Segment Postulates Line – a line contains at...
Page 4
(a) and
(b) and
(c) and
(d)
(e) and
Line, segment and angle Postulates
Line and Segment Postulates
Line – a line contains at least 2 points.
Plane – contains at least 3 noncollinear points.
Line and Segment Postulates
Intersecting planes
- The intersection of 2 planes is a line.
Line and Segment Postulates
Segment addition postulate – given 3 collinear points, AB+BC=AC
A B C
A BB C A C+ =
Angle Postulates
AOCmBOCmAOBm Angle addition postulate – if B is in the interior of angle AOC , then
A
O C
B20º45º
𝑚∠𝐴𝑂𝐶=𝑚∠𝐴𝑂𝐵+𝑚∠𝐵𝑂𝐶
𝑚∠𝐴𝑂𝐶=20+45=65
Angle Postulates
Linear pair postulate – if 2 angles form a linear pair, they are supplementary
OA
B
30º C
B
150ºO
A C
B
30º 150º
O
𝑚∠𝐴𝑂𝐶=𝑚∠𝐴𝑂𝐵+𝑚∠𝐵𝑂𝐶
𝑚∠𝐴𝑂𝐶=30+150=180
Page 6
∠1+∠2=18032+∠2=180−32−32
∠2=148
Page 6
L NM
Page 6
R TS
Page 6
121+59=180
Page 7
Page 7
W YX
2𝑚+2 𝑚+3
4𝑚
2𝑚+2+𝑚+3=4𝑚3𝑚+5=4𝑚−3𝑚−3𝑚
5=𝑚
Page 8
Two angles are adjacent if they share a common vertex and
common side, but have no interior points in common.
𝑌 𝑒𝑠They share a common vertex and common side, but have no interior
points in common.
𝑁𝑜They do
not share a common vertex.
𝑌 𝑒𝑠They share a
common vertex and common
side, but have no interior points in
common.
𝑁𝑜They have
interior points in common.
Page 8
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝐴𝑛𝑔𝑙𝑒𝑠 :
∠𝐿𝑅𝑄+∠𝑄𝑅𝑀∠𝑄𝑅𝑀+∠𝑀𝑅𝑃∠𝑀𝑅𝑃+∠ 𝑃𝑅𝐿∠𝑃𝑅𝐿+∠𝐿𝑅𝑄
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐴𝑛𝑔𝑙𝑒𝑠:
∠𝐿𝑅𝑄+∠𝑀𝑅𝑃∠𝑄𝑅𝑀+∠𝑃𝑅𝐿
∠𝐷𝐴𝐶+∠𝐶𝐴𝐵
Page 8
45
𝐿𝑒𝑡 𝑥=𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝐿𝑒𝑡 𝑥+60=𝑎𝑛𝑔𝑙𝑒𝑥+𝑥+60=902 𝑥+60=90−60−602 𝑥=30
2𝑥2
=302
𝑥=15𝑥+60=15+60=75
Page 8
Perpendicular Lines and Planes
Page 9
Perpendicular Lines and Planes
Page 8
Perpendicular Lines and Planes
Page 10
Perpendicular Lines and Planes
Page10
Perpendicular Lines and Planes
Page 10
Perpendicular Lines and Planes
HomeworkPage 8#2-12 evens, 6 do only parts a,b,f
∠𝐶𝐵𝐷𝑎𝑛𝑑∠ 𝐴𝐵𝐷∠𝐶𝐷𝐵𝑎𝑛𝑑∠ 𝐴𝐷𝐵
𝑎 ¿∠𝐴𝐵𝐸𝑎𝑛𝑑∠𝐶𝐵𝐸𝑏¿∠𝐵𝐸𝐶𝑎𝑛𝑑∠ 𝐴𝐸𝐷
𝑐 ¿∠𝐴𝐸𝐵𝑎𝑛𝑑∠𝐵𝐸𝐶
𝑎 ¿90−50=40𝑏¿90−27=63𝑓 ¿90−𝑥
𝐿𝑒𝑡 𝑥= h𝑡 𝑒𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑎𝑛𝑔𝑙𝑒𝐿𝑒𝑡 5 𝑥= h𝑡 𝑒 𝑙𝑎𝑟𝑔𝑒𝑟 𝑎𝑛𝑔𝑙𝑒
5 𝑥+𝑥=906 𝑥=906 𝑥6
=906
𝑥=155 𝑥=5 ∙15=75
𝐿𝑒𝑡 𝑥= h𝑡 𝑒 𝑙𝑎𝑟𝑔𝑒𝑟 𝑎𝑛𝑔𝑙𝑒𝐿𝑒𝑡 𝑥−50= h𝑡 𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑎𝑛𝑔𝑙𝑒
𝑥+𝑥−50=902 𝑥−50=90
+50+502 𝑥=1402𝑥2
=1402
𝑥=70
𝑥=70𝑥−50=70−50=20
𝐿𝑒𝑡 𝑥=𝑡 h𝑒𝑎𝑛𝑔𝑙𝑒𝐿𝑒𝑡 𝑥+24=𝑡 h𝑒𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡
𝑥+𝑥+24=902 𝑥+24=90−24−242 𝑥=662𝑥2
=662
𝑥=33
𝑥=33𝑥+24=33+24=57
