Warm up
Notes
Preliminary Activity
Activity
Surveying
For Fun
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1. Find the area of the following triangles:
a)
Area = ½absinC
= ½ x 9 x 8 x sin115o
= 32.6cm2
b)
Area = ½absinC
= ½ x 6.9 x 4.8 x
sin108o
= 15.7cm2
c)
Area = ½absinC
= ½ x 19 x 18 x
sin105o
= 165.2cm2
d)
Area = ½absinC
= ½ x 6.9 x 7.8 x
sin118o
= 23.8cm2
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Offset survey - this was looked at in Year 11 as a
field diagram. The longest diagonal of a field is used as a
traverse (to cross) line, and the perpendicular distances from
this line to each corner of the field are measured - these are
called offsets.
The plane table radial survey and the compass
bearing radial survey methods result in similar diagrams.
The cosine rule is used to find the length of each
boundary and hence the perimeter. The area of each part is
found using
A = ½absinC.
The main difference in the methods is that the angles
must be measured from the plane table diagram using a
protractor, while the angles in the compass survey are
calculated from the compass bearings.
The perimeter of the plane table survey is usually
found using measurement and calculation while the compass
survey uses the cosine rule to find the perimeter.
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Next
A
Example 1: Plane table radial survey
The following information was the result of a
plane table radial survey.
i) Find the perimeter of each field
ii) Calculate the area of each field
B
D
i) Use the cosine rule to find the length of the missing side.
c2 = a2 + b2 - 2ab cosC
AB2 = 542 + 482 - 2 x 54 x 48 x cos110o
AB2 = 6993.03....
AB = √6993.03....
AB = 83.6m
BC = 57.2m;
CD = 47.7m;
∴ Perimeter = 257m
DA = 68.5m
ii) The area of each part is found by using A = ½absinC
ΔAPB = ½ x 54 x 48 x sin110o
= 1217.8m2
ΔBPC = 943.1m2
ΔCPD = 730.2m2
ΔDPA = 995.6m2
∴
Area = 3886m2
C
E
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A
Example 2: The compass radial survey
A field was surveyed, all measurements in metres.
i) Calculate the perimeter using the cosine rule.
ii) Calculate the area.
D
B
C
i) <APB = 125o - 53o = 72o
<BPC = 195o - 125o = 70o
<CPD = 269o - 195o = 74o
<DPE = 345o - 269o = 76o
<EPA = 35o + 53o = 68o
AB2 = 7.92 + 12.12 - 2 x 7.9 x 12.1 x cos 72o
AB2 = 149.7....
AB = √149.7...
AB = 12.2m
BC = 8.6m;
CD = 8.9m;
DE = 13.4m;
14.1m
∴ Perimeter = 57m
ii) A = ½absinC
ΔAB = ½ x 7.9 x 12.1 x sin72o
Δ
AB = 45.5m2
BC = 26.4m2;
ΔCD = 26.3m2;
72.9m2
∴ Area = 220m2
Δ
EA =
ΔDE = 48.6m2;
ΔEA =
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Complete exercise 5-10
Questions 1, 2, 4, 7, 8, 10, 12, 14
41.7%
56.3%
75.7%
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