1. Find: AreaABCD [acres]
North D C
course
bearing
length [ft]
0.50
0.67
1.00
1.35
Find the area of the polygon A B C D, in acres. [pause] In this problem, ---- o
AB
N80 45’E
250.71 o
BC
S5 22’E
147.66 o
CD
N85 13’W
257.04 o
DA
N3 26’W
85.43

2. Find: AreaABCD [acres]
North D C
course
bearing
length [ft]
0.50
0.67
1.00
1.35
Points A, B, C and D are shown. And the bearings and lengths of the ---- o
AB
N80 45’E
250.71 o
BC
S5 22’E
147.66 o
CD
N85 13’W
257.04 o
DA
N3 26’W
85.43

3. Find: AreaABCD [acres]
North D C
course
bearing
length [ft]
0.50
0.67
1.00
1.35
connecting courses, are provided in the data table. [pause] To solve this problem, --- o
AB
N80 45’E
250.71 o
BC
S5 22’E
147.66 o
CD
N85 13’W
257.04 o
DA
N3 26’W
85.43

4. Find: AreaABCD [acres]
North D C
course
bearing
length [ft]
coordinate
we’ll use the coordinate method for finding areas, where the area equals ---- o
AB
N80 45’E
250.71
method o
BC
S5 22’E
147.66 o
CD
N85 13’W
257.04 o
DA
N3 26’W
85.43

5. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
coordinate
method D C
course
bearing
length [ft]
one half times the absolute value of the sum of the product of the y coordinate of a point, times the difference between the x coordinates of the preceding point and the x coordinate of the subsequent points, for all n points. o
AB
N80 45’E
250.71 o
BC
S5 22’E
147.66 o
CD
N85 13’W
257.04 o
DA
N3 26’W
85.43

6. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
coordinate
method D C
course
bearing
length [ft] x y
Therefore we’ll have to solve for the x and y coordinates of Points A, B, C and D, using the given data. o
AB
N80 45’E
250.71 A Ax Ay o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

7. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
coordinate
method D C
course
bearing
length [ft] x y
Since no coordinates are given for any of the points, we’ll arbitrarily set Point A to the origin, ---- o
AB
N80 45’E
250.71 A Ax Ay o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

8. Find: AreaABCD [acres]y A x n Σ
Area=0.5* {yi*(xi-1-xi+1)} D C
i=1
course
bearing
length [ft] x y
of an x-y coordinate system, where A X equals 0 feet, ---- o
AB
N80 45’E
250.71 A Ax Ay o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

9. Find: AreaABCD [acres]y
A (0,0) x n Σ
Area=0.5* {yi*(xi-1-xi+1)} D C
i=1
course
bearing
length [ft] x y
and A Y equals 0 feet. Next we’ll use the bearing and length values, --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

10. Find: AreaABCD [acres]y
A (0,0) x n Σ
Area=0.5* {yi*(xi-1-xi+1)} D C
i=1
course
bearing
length [ft] x y
given for the 4 courses, to determine the x and y coordinates for Points B, ---- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

11. Find: AreaABCD [acres]y
A (0,0) x D C
course
bearing
length [ft] x y
C and D. From trigonometry, we can equate the x value of Point B to --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

12. Find: AreaABCD [acres]
Bx = Ax + LAB * sin (BAB) D C
course
bearing
length [ft] x y
the x value of Point A, plus the length of line segment A B, times sine of the bearing of course A B. o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

13. Find: AreaABCD [acres]
Bx = Ax + LAB * sin (BAB)
0 [ft] D C
course
bearing
length [ft] x y
If we plug in the x coordinate of A, length A B, and the bearing of Course A B, the x coordinate of B equals, --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

14. Find: AreaABCD [acres]
Bx = Ax + LAB * sin (BAB)= [ft]
0 [ft] D C
course
bearing
length [ft] x y
feet. [pause] We can find the y coordinate of Point B by adding the y coordinate of Point A to --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

15. Find: AreaABCD [acres]
Bx = Ax + LAB * sin (BAB)= [ft]
By = Ay + LAB * cos (BAB)
0 [ft] D C
course
bearing
length [ft] x y
to the length of line segment A B, times the cosine of the bearing of course A B, which equals, --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

16. Find: AreaABCD [acres]
Bx = Ax + LAB * sin (BAB)= [ft]
By = Ay + LAB * cos (BAB)= [ft]
0 [ft] D C
course
bearing
length [ft] x y
40.30 feet. [pause] The coordinates for Points C and D are calculated the same way, --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

17. Find: AreaABCD [acres]
Cx = Bx + LBC * sin (BBC)
Cy = By - LBC * cos (BBC)
Dx = Cx - LCD * sin (BCD)
Dy = Cy + LCD * cos (BCD) C
course
bearing
length [ft] x y
as was coordinate B. We’ll plug in the x and y coordinate values from the previous point on the traverse, plus or minus --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

18. Find: AreaABCD [acres]
Cx = Bx + LBC * sin (BBC)
Cy = By - LBC * cos (BBC)
Dx = Cx - LCD * sin (BCD)
Dy = Cy + LCD * cos (BCD) C
course
bearing
length [ft] x y
the length between that Point and the previous Point, times, the sine or cosine of the bearing ---- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

19. Find: AreaABCD [acres]
Cx = Bx + LBC * sin (BBC)
Cy = By - LBC * cos (BBC)
Dx = Cx - LCD * sin (BCD)
Dy = Cy + LCD * cos (BCD) C
course
bearing
length [ft] x y
of the course from the preceding point, to that point. [pause] o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

20. Find: AreaABCD Bx = 247.45 [ft] By = 40.30 [ft]
Cx = Bx + LBC * sin (BBC) B
Cy = By - LBC * cos (BBC)
Dx = Cx - LCD * sin (BCD)
Dy = Cy + LCD * cos (BCD) C
course
bearing
length [ft] x y
After plugging in the appropriate coordinate values, lengths and bearings, the coordinates of Point C and ---- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

21. Find: AreaABCD Bx = 247.45 [ft] By = 40.30 [ft]
Cx = Bx + LBC * sin (BBC) = [ft] B
Cy = By - LBC * cos (BBC) = [ft]
Dx = Cx - LCD * sin (BCD) = 5.12 [ft]
Dy = Cy + LCD * cos (BCD) = [ft] C
course
bearing
length [ft] x y
and Point D are determined. [pause] Lastly, we’ll confirm the traverse closes ---- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

22. Find: AreaABCD Ax = Dx - LDA * sin (BDA) B Ay = Dy + LDA * cos (BDA)
course
bearing
length [ft] x y
by calculating the coordinates of Point A. Using the coordinates of Point D, the length of D A, and the bearing ---- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

23. Find: AreaABCD Dx = 5.12 [ft] Dy = -85.28 [ft]
Ax = Dx - LDA * sin (BDA) B
Ay = Dy + LDA * cos (BDA) D C
course
bearing
length [ft] x y
of Course D A, we calculate Point A is indeed located at, --- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

24. Find: AreaABCD Dx = 5.12 [ft] Dy = -85.28 [ft]
Ax = Dx - LDA * sin (BDA) = 0.00 [ft] B
Ay = Dy + LDA * cos (BDA) = 0.00 [ft] D C
course
bearing
length [ft] x y
the origin, 0, 0. Which means the traverse closes, and doesn’t need to be corrected. o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

25. Find: AreaABCD Dx = 5.12 [ft] Dy = -85.28 [ft]
Ax = Dx - LDA * sin (BDA) = 0.00 [ft] B
Ay = Dy + LDA * cos (BDA) = 0.00 [ft]
no correction D
necessary C
course
bearing
length [ft] x y
Next we’ll set up a table to summarize the x and ---- o
AB
N80 45’E
250.71 A o
BC
S5 22’E
147.66 B Bx By o
CD
N85 13’W
257.04 C Cx Cy o
DA
N3 26’W
85.43 D Dx Dy

26. Find: AreaABCD [acres]
Point
x [ft]
y [ft]
y coordinate values for Points A, B, C and D, and we recall our equation for the area only involves --- A
0.00
0.00
North B
247.45
40.30 C
261.26 D
5.12
-85.28

27. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
coordinate
method
Point
x [ft]
y [ft]
these x, and y, coordinate values. Next we’ll add a column to the table, for the variable i. A
0.00
0.00
North B
247.45
40.30 C
261.26 D
5.12
-85.28

28. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
coordinate
method
Point i
x [ft]
y [ft]
Variable i corresponds to the iteration in the summation. --- A 1
0.00
0.00 B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28

29. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
coordinate
method
Point i
x [ft]
y [ft]
[pause] To avoid confusion, the term y sub i, refers to --- A 1
0.00
0.00 B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28

30. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
coordinate
method
Point i
x [ft]
y [ft]
the y coordinate of the Point whose value is i. For example, if i equals 2, then --- A 1
0.00
0.00 B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28

31. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
Example #1:
i=1
if i=2, then
coordinate
method
yi=40.30 [ft]
Point i
x [ft]
y [ft]
y sub 2 equals feet, as shown in the data table. [pause] As a second example, if i equals 1, ---- A 1
0.00
0.00 B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28

32. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
Example #1:
i=1
if i=2, then
coordinate
method
yi=40.30 [ft]
Example #2:
Point i
x [ft]
y [ft]
if i=1, then
then, x sub i minus 1, will equal x sub 0. But but there are no points ---- A 1
0.00
0.00
xi-1=x0 B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28

33. ? Find: AreaABCD [acres] Σ Area=0.5* {yi*(xi-1-xi+1)} Example #1:
if i=2, then
coordinate
method
yi=40.30 [ft]
Example #2:
Point i
x [ft]
y [ft]
if i=1, then
whose i value equals zero. In this case, we’ll loop back around ---- A 1
0.00
0.00
xi-1=x0 B 2
247.45
40.30 ? C 3
261.26 D 4
5.12
-85.28

34. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
Example #1:
i=1
if i=2, then
coordinate
method
yi=40.30 [ft]
Example #2:
Point i
x [ft]
y [ft]
if i=1, then
and use the coordinates for i equals 4, which are the coordinates for Point D. A similar situation occurs for x sub i plus 1, --- A 1
0.00
0.00
xi-1=x0=x4 B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28

35. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
Example #1:
i=1
if i=2, then
coordinate
method
yi=40.30 [ft]
Example #2:
Point i
x [ft]
y [ft]
if i=1, then
when i equals 4. To avoid this confusion, we’ll add a row for i equals 0 and --- A 1
0.00
0.00
xi-1=x0=x4 B 2
247.45
40.30
Example #3: C 3
261.26
if i=4, then D 4
5.12
-85.28
xi+1=x5=x1

36. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
Point i
x [ft]
y [ft]
new row D
5.12
-85.28 A 1
0.00
0.00
and a row for I equals 5, which are just copies of coordinate data from rows i = 4 and i =1, respectively. Returning to our equation for area, --- B 2
247.45
40.30
existing
rows C 3
261.26 D 4
5.12
-85.28
new row A 5
0.00
0.00

37. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
Point i
x [ft]
y [ft] D
5.12
-85.28 A 1
0.00
0.00
there are 4 verticies to the closed traverse, so we’ll assign --- B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28 A 5
0.00
0.00

38. Find: AreaABCD [acres]Σ
Area=0.5* {yi*(xi-1-xi+1)}
i=1
n=4
Point i
x [ft]
y [ft] D
5.12
-85.28 A 1
0.00
0.00
the value of n to equal 4. Next we’ll write out the summation equation --- B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28 A 5
0.00
0.00

39. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Point i
x [ft]
y [ft] D
5.12
-85.28 A 1
0.00
0.00
for values of i ranging from 1 to 4. Then we’ll plug in the appropriate ---- B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28 A 5
0.00
0.00

40. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Point i
x [ft]
y [ft] D
5.12
-85.28 A 1
0.00
0.00
values for x, and the appropriate values for y --- B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28 A 5
0.00
0.00

41. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Point i
x [ft]
y [ft] D
5.12
-85.28 A 1
0.00
0.00
[pause] And the area of A B C D equals --- B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28 A 5
0.00
0.00

42. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Area = 29,334 [ft2]
Point i
x [ft]
y [ft] D
5.12
-85.28 A 1
0.00
0.00
29,334 feet squared. Since the problem asks to find the area in acres, we’ll divide this quantity by ---- B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28 A 5
0.00
0.00

43. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Area = 29,334 [ft2] 1
acre
Point i
x [ft]
y [ft] *
43,560
ft2 D
5.12
-85.28 A 1
0.00
0.00
43,560 square feet per acre, and the final area equals --- B 2
247.45
40.30 C 3
261.26 D 4
5.12
-85.28 A 5
0.00
0.00

44. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Area = 29,334 [ft2] 1
acre *
Area = [acres]
43,560
ft2
0.673 acres. [pause]

45. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Area = 29,334 [ft2] 1
acre *
Area = [acres]
43,560
ft2
0.50
0.67
1.00
1.35
When reviewing the possible solutions, ---

46. Find: AreaABCD [acres]
y1*(x0-x2)+y2*(x1-x3)
Area=0.5*
+y3*(x2-x4)+y4*(x3-x5)
Area = 29,334 [ft2] 1
acre *
Area = [acres]
43,560
ft2
AnswerB
0.50
0.67
1.00
1.35
the answer is B.

47. d