Select the procedure that can be used to show the converse of the Pythagorean theorem using side lengths chosen from 6 feet, 8 feet, 10 feet, and 11 feet.1. Knowing that 62 + 82 = 102, draw the 6-foot side and the 8-foot side with a right angle between them. The 10-foot side will fit to form a right triangle.
2. Knowing that 82 + 102 < 112, draw the 4-foot side and the 10-foot side with a right angle between them. The 11-foot side will fit to form a right triangle.
3. Knowing that 62 + 102 ≠ 112, draw the 6-foot side and the 10-foot side with a right angle between them. The 11-foot side will fit to form a right triangle.
4. Knowing that 62 + 82 = 102, draw any two of the sides with a right angle between them. The third side will fit to form a right triangle
Complete your answers in the spaces below and submit this word file or print it, fill it by hand and scan it.Most of the tables have extra lines that you may not need to use so don’t worry if there are unused spaces.
1) Determine the scale of the plan (3 Marks)
Line
Distance on Plan (mm)
Surveyed Dist (m)
CALCULATED PLAN SCALE
Bar Scale
A – Y
Y – X
A – Start of Road
PLAN SCALE ADOPTED BY YOU (rounded to nearest 10 – no units or decimals) …. Please give a brief reason why this scale was adopted
1) and 3) Both your plan and longsection will need to be scanned then pasted to the end of this word document. (Presentation and data for Plan 2 Marks; and Longsection 16 Marks)
2) HORIZONTAL ALIGNMENT (4 Marks)
Bearing (nearest degree) (Ch00 – IP1) …
Bearing (nearest degree) (IP1 – IP2) …
Bearing (nearest degree) (IP2 – End of Rd.) .
Deflection Angle at IP1 (nearest degree) …
Deflection Angle at IP2 (nearest degree) ….
Lengths as scaled from the plan and computed using your plan scale (Round to the nearest metre and use that value for the precise length of each straight in all further calculations.) 00 – IP1 …
IP1- IP2 …..
IP2 – End ……
Curve Elements (6 Marks)
Tangent Distance for Curve 1 …
Arc Length of Curve 1 …
Tangent Distance for Curve 2 …
Arc Length of Curve 2 …
Chainages along the road (16 Marks)
Location
Chainage
Start of Road
00
First Bank of River (by scale to nearest metre)
Second Bank of River (by scale to nearest metre)
Chainage of Hor. Curve TP 1 (to 2 decimal places)
Chainage of Hor. Curve TP 2 (to 2 decimal places)
Chainage of Hor. Curve TP 3 (to 2 decimal places)
Chainage of Hor. Curve TP 4 (to 2 decimal places)
Chainage of end of Road (to 2 decimal places)
2) VERTICAL ALIGNMENT (20 Marks; 15 for Design Levels, 5 for Low Point) Complete this table by calculating the design level at every 30m of running chainage where there is no vertical curve and every 10m inside each vertical curve. Please indicate the TPs and IPs of each VC in the chainage column with the chainage. (e.g. 320 TP, 360 IP) Your longsection must be pasted at the end of this answer file. Where possible please scan it. (You may not fill every line in this table).
CHAINAGE GRADE GRADE LEVEL ORDINATE DESIGN R.L.
00
Show the chainage and RL of the Low Point in the row below.
Shows calculations for Low Point here.
4a) Traverse and Coordinate Calculations (8 Marks)
Survey Party’s Traverse from SSM to point X
Horiz.
E
N
CO-ORD
INATES
LINE
Bearing
Dist
E (+)
W (-)
N (+)
S(-)
E
N
PT.
300.000
600.000
A
A – Y
Y
Y – X
X
Calculate the Coordinates of the end of your road
Horiz.
E
N
CO-ORD
INATES
LINE
Bearing
Dist
E (+)
W (-)
N (+)
S(-)
E
N
PT.
300.000
600.000
A
A – Ch 0
Ch 0
Ch 0 – IP1
IP1
IP1 -IP2
IP2
IP2 – END
End of Rd
Calculate the bearing and distance to set out the end of your road from point X.
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