Area Computation_Mock Exam
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8/9/2019 Area Computation_Mock Exam
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1. Sketch: (5%)
6
8
4 10 16
2 11 18
14
20
1 3 5 7 9 12
13 15 17 19
21
Offsets:
0.96 2.16 4.16 3.7 1.46 0.46
2.5 2.66 5.16
4.31 2.16
2. Using Area By Trapezoid (points 1-12): (20%)
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8/9/2019 Area Computation_Mock Exam
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Area = 10[((0.96+0.46)/2) + 2.16 + 4.16 + 3.7 + 1.46]
Area = 121.9 m2
3. Sum of Area by triangles: (40%)
Triangle Side (a) Side (b) Side(C)
s Area (m2)
123 10.000 0.960 10.046
10.503
4.800
234 10.046 2.160 10.072
11.139
10.800
345 10.000 2.160 10.231
11.196
10.800
456 10.231 4.160 10.198
12.295
20.800
567 10.000 4.160 10.831
12.496
20.800
678 10.831 3.700 10.011
12.271
18.501
789 10.000 3.700 10.66
3
12.18
2
18.500
8910 10.6631.460
10.248
11.186
7.300
91012 10.000 1.460 10.106
10.783
7.300
101112 10.106 0.460 10.050
10.308
2.300
Total Area=121.901
m2
Therefore: are by Trapezoidal Rule = Area by Triangles
4. Using Area By Simpsons 1/3 Rule (points 11-21): 20%
Area = 12.5/3 (2.96 + 2.16 + (2x 5.16) + (4x6.97))
Area = 180.500 m
2
5. New area of the excavated land:
++++
+= 1n32
n1hhh
2
hhdArea
( )evenoddn1 h4h2hh3
dArea +++=
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8/9/2019 Area Computation_Mock Exam
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Area = 121.900 + (50x2.5) = 246.900 m2
Total volume of the excavated land:
Volume = Area x Width = (246.900 + 180.500) x 15 m =6411 m3
a. Scale:
X=Xhighest Xlowest = 100 0 = 100
Y=Yhighest Ylowest = 505 498.34 = 6.66
Since: X > Y then use X = 100
Scale = 1:400 since max. Difference in coordinates = 90-120
