1. Find: AreaABC [ft2] C A B C’ 43,560 44,600 44,630 45,000
F = [ft]
Find the area of Triangle A B C, in square feet. [pause] In this problem, ---- o
F = [ft] o
using a steel tape o
calibrated at 53 F

2. Find: AreaABC [ft2] C A B C’ LAB@105 F 43,560 44,600 44,630 45,000o
43,560
44,600
44,630
45,000
F = [ft]
the length from Point A to Point B was measured at a temperature of 105 degrees Fahrenheit. o
F = [ft] o
using a steel tape o
calibrated at 53 F

3. Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630o A B C’ F o
43,560
44,600
44,630
45,000
F = [ft]
The length from Point C to Point C prime was also measured at a temperature of 105 degrees Fahrenheit. o
F = [ft] o
using a steel tape o
calibrated at 53 F

4. Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630o A B C’ F o
43,560
44,600
44,630
45,000
F = [ft]
Both of these measurement were made using a steel tape, --- o
F = [ft] o
using a steel tape o
calibrated at 53 F

5. Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630o A B C’ F o
43,560
44,600
44,630
45,000
F = [ft]
which was calibrated at 53 degrees Fahrenheit. Also note that angles --- o
F = [ft] o
using a steel tape o
calibrated at 53 F

6. Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630o A B C’ F o
43,560
44,600
44,630
45,000
F = [ft]
C C prime A, and C C prime B, measure to be 90 degrees. Therefore, we’ve effectively been given --- o
F = [ft] o
using a steel tape o
calibrated at 53 F

7. Find: AreaABC [ft2] C height LCC’@105 F A B C’ LAB@105 F width 43,560o A B C’ F
width o
43,560
44,600
44,630
45,000
F = [ft]
the dimensions for the width and height, of the triangle. [pause] To find the area of the oblique Triangle --- o
F = [ft] o
using a steel tape o
calibrated at 53 F

8. Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] steel tape A B (53 F)o
= [ft]
steel tape o A B
(53 F) F
= [ft] o
A B C, we first need to determine the appropriate area equation to use. ---

9. Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] steel tape A B (53 F)o
= [ft]
steel tape o A B
(53 F) F
= [ft] o #
[pause] We’ll begin by defining Capital A subscript A, B and C ----
known variables
area equation 1 2 3

10. Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] AC AA AB steel tape A Bo
= [ft] AC AA AB
steel tape o A B
(53 F) F
= [ft] o #
as the interior angles of the triangle, at Points A, B and C respectively. Also, ---
known variables
area equation 1 2 3

11. Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] AC AA AB steel tape A Bo
= [ft] AC AA AB
steel tape o A B
(53 F) F
= [ft] o #
lower-case letters a, b and c will refer to ----
known variables
area equation 1 2 3

12. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
side lengths on the opposite side of the triangle from the angle, of the same letter.
known variables
area equation 1 2 3

13. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
On to our area equations, if we know the measurement for Angle A, and side lengths ---
known variables
area equation 1 2 3

14. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
b and c, then the area of the triangle can be calculated as one half times ---
known variables
area equation 1
AA, b, c 2 3

15. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
b, times c, times the sin of Angle A. This equation can be modified in the event Angle B, ----
known variables
area equation 1
AA, b, c
0.5 * b * c * sin(AA) 2 3

16. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
side a and side c are known, or if Angle C, side a, and ---
known variables
area equation 1
AB, a, c
0.5 * a * c * sin(AB) 2 3

17. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
and side b are known. A second equation to find the area ----
known variables
area equation 1
AC, a, b
0.5 * a * b * sin(AC) 2 3

18. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
of a triangle is if all three side lengths are known, ---
known variables
area equation 1
AC, a, b
0.5 * a * b * sin(AC) 2 3

19. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
then the area equals the square root of the product of ----
known variables
area equation 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c 3

20. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
s, s minus a, s minus b, and s minus c. This is called Heron’s equation, where by variable s equals ---
known variables
area equation 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c) 3

21. Find: AreaABC [ft2] b a c C AC AA AB A B s = 0.5 * (a+b+c) #
semiperimeter of the triangle, or one half, times of the perimeter of the triangle. A third equation for the area ---
known variables
area equation 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c) 3

22. Find: AreaABC [ft2] b a c C AC AA AB A B s = 0.5 * (a+b+c) #
of a triangle is when all 3 angles are known, and the length of ---
known variables
area equation 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c) 3

23. Find: AreaABC [ft2] b a c C AC AA AB A B s = 0.5 * (a+b+c) #
1 side is also know. If side a is known, then the area equals, a squared ---
known variables
area equation 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c) 3
AA, AB, AC, a

24. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
area equation
times the sin of Angle B, times the sin of Angle C, divided by 2 times the sin of Angle A. This equation can be modified in the event --- 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c)
a2 * sin(AB) * sin(AC) 3
AA, AB, AC, a
2 * sin(AA)

25. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
area equation
we know the length of side b instead of side a, or the length of side --- 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c)
b2 * sin(AA) * sin(AC) 3
AA, AB, AC, b
2 * sin(AB)

26. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
area equation
c. [pause] The fourth and final equation for area ---- 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c)
c2 * sin(AA) * sin(AB) 3
AA, AB, AC, c
2 * sin(AC)

27. Find: AreaABC [ft2] b a c C AC AA AB A B # known variables
area equation
is the one we’ll use for this example where the area simply equals, --- 1
AC, a, b
0.5 * a * b * sin(AC) 2
a, b, c
s * (s-a) * (s-b) * (s-c)
c2 * sin(AA) * sin(AB) 3
AA, AB, AC, c
2 * sin(AC)

28. Find: AreaABC [ft2] C height width A B # known variables area equation1
AC, a, b
0.5 * a * b * sin(AC)
one half, times the height, times the width. So, returning to our problem, --- 2
a, b, c
s * (s-a) * (s-b) * (s-c)
c2 * sin(AA) * sin(AB) 3
AA, AB, AC, c
2 * sin(AC) 4
height, width
0.5 * height * width

29. Find: AreaABC [ft2] ? ? C LCC’@105 F height =203.56 [ft] widtho
= [ft]
width
steel tape o A B
(53 F) F
= [ft] o
we need to determine the actual height and width of Triangle A B C, after it has been corrected ---
AreaABC=0.5 * height * width ? ?

30. Find: AreaABC [ft2] ? ? C LCC’@105 F height =203.56 [ft] widtho
= [ft]
width
steel tape o A B
(53 F) F
= [ft] o
for the temperature difference. [pause] In this problem, the actual height of Triangle A B C equals ----
AreaABC=0.5 * height * width ? ?

31. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
the length of segment C C prime, if it were measured at 53 degrees Fahrenheit, ----
AreaABC=0.5 * height * width
height = F o

32. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
which is the calibrated temperature of the tape. However, we only have ---
AreaABC=0.5 * height * width
height = F o

33. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
a measurement of line segment C C prime when the temperature is ---
AreaABC=0.5 * height * width
height = F o

34. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
105 degrees Fahrenheit. Therefore, we’ll equate the height of the triangle to this measurement, plus, ---
AreaABC=0.5 * height * width
height = F o

35. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
temperature
correction
a temperature correction distance, C. [pause] In the same manner, ---
AreaABC=0.5 * height * width
distance
height = F = F + CCC’,105 F,53 F o o o o

36. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
temperature
correction
the actual width of the triangle will equal the width, if the measurement were taken ---
AreaABC=0.5 * height * width
distance
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + C

37. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
temperature
correction
at the calibrated 53 degrees Fahrenheit, which we’ll equate to measurement taken ---
AreaABC=0.5 * height * width
distance
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + C o

38. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
temperature
correction
at 105 degrees Fahrenheit, plus a temperature correction distance, C. [pause] Since we know the height and ---
AreaABC=0.5 * height * width
distance
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + CAB,105 F,53 F o o o o

39. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
temperature
correction
width of Triangle A B C at 105 degrees Fahrenheit. We just need to solve for the ---
AreaABC=0.5 * height * width
distance
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + CAB,105 F,53 F o o o o

40. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape o A C’ B
(53 F) F
= [ft] o
temperature
correction
temperature correction distances. The temperature correction distances equal ---
AreaABC=0.5 * height * width
distance
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + CAB,105 F,53 F o o o o

41. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width
height = F = F + CCC’,105 F,53 F
the temperature during the measurement, minus ---- o o o o
width = F = F + CAB,105 F,53 F o o o o
CCC’,105 F,53 F =(TM-TC) * LM * α
temp. o o
correction
CAB,105 F,53 F =(TM-TC) * LM * α o o
distances

42. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
temperature during measurement
the calibrated temperature, times the measured length, ----
calibrated temperature
CCC’,105 F,53 F =(TM-TC) * LM * α
temp. o o
correction
CAB,105 F,53 F =(TM-TC) * LM * α o o
distances

43. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
temperature during measurement
calibrated temperature
times the coefficient of thermal expansion. Since the problem states the tape is steel, ---
measured length
coefficient
CCC’,105 F,53 F =(TM-TC) * LM * α
of thermal o o
expansion
CAB,105 F,53 F =(TM-TC) * LM * α o o

44. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
temperature during measurement
calibrated temperature
the coefficient of thermal expansion, alpha, equals ----
measured length
coefficient
CCC’,105 F,53 F =(TM-TC) * LM * α
of thermal o o
expansion
CAB,105 F,53 F =(TM-TC) * LM * α o o

45. Find: AreaABC [ft2] α=6.45*10-6[ F-1] C LCC’@105 F height =203.56 [ft]o
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
temperature during measurement
calibrated temperature
6.45 times 10 to the negative 6, degree Fahrenheit, to the negative 1. And after plugging in the values ---
α=6.45*10-6[ F-1] o
coefficient
CCC’,105 F,53 F =(TM-TC) * LM * α o o
of thermal
CAB,105 F,53 F =(TM-TC) * LM * α
expansion o o

46. Find: AreaABC [ft2] α=6.45*10-6[ F-1] C LCC’@105 F height =203.56 [ft]o
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
for temperatures and lengths, the temperature correction distances equal ----
α=6.45*10-6[ F-1] o
coefficient
CCC’,105 F,53 F =(TM-TC) * LM * α o o
of thermal
CAB,105 F,53 F =(TM-TC) * LM * α
expansion o o

47. Find: AreaABC [ft2] α=6.45*10-6[ F-1] C LCC’@105 F height =203.56 [ft]o
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
0.068 feet and feet, for the height and width, respectively. We’ll round these values ---
α=6.45*10-6[ F-1] o
CCC’,105 F,53 F =(TM-TC) * LM * α = [ft] o o
CAB,105 F,53 F =(TM-TC) * LM * α = [ft] o o

48. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width
to the nearest 100th of a foot, and after plugging in the temperature correction distances, ---
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + CAB,105 F,53 F o o o o
CCC’,105 F,53 F =0.07 [ft] = [ft] o o
CAB,105 F,53 F =0.15 [ft] = [ft] o o

49. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width
as well as the measured lengths at 105 degrees Fahrenheit, ---
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + CAB,105 F,53 F o o o o
CCC’,105 F,53 F =0.07 [ft] = [ft] o o
CAB,105 F,53 F =0.15 [ft] = [ft] o o

50. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width
the actual lengths for the height and width of Triangle A B C equal, ---
height = F = F + CCC’,105 F,53 F o o o o
width = F = F + CAB,105 F,53 F o o o o
CCC’,105 F,53 F =0.07 [ft] = [ft] o o
CAB,105 F,53 F =0.15 [ft] = [ft] o o

51. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width
feet and feet, respectively. Plugging these values into our equation ---
height = F+ CCC’,105 F,53 F o o o
width = F + CAB,105 F,53 F o o o
CCC’,105 F,53 F =0.07 [ft] = [ft] o o
CAB,105 F,53 F =0.15 [ft] = [ft] o o

52. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width
for the area of Triangle A B C, we find the area of the triangle equals, ---
height = F+ CCC’,105 F,53 F o o o
width = F + CAB,105 F,53 F o o o
CCC’,105 F,53 F =0.07 [ft] = [ft] o o
CAB,105 F,53 F =0.15 [ft] = [ft] o o

53. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width = 44,627 [ft2]
44,627 square feet. [pause]
height = F+ CCC’,105 F,53 F o o o
width = F + CAB,105 F,53 F o o o
CCC’,105 F,53 F =0.07 [ft] = [ft] o o
CAB,105 F,53 F =0.15 [ft] = [ft] o o

54. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width = 44,627 [ft2]
When reviewing the possible solutions, ----
43,560
44,600
44,630
45,000
height = [ft]
width = [ft]
CCC’,105 F,53 F =0.07 [ft] o o
CAB,105 F,53 F =0.15 [ft] o o

55. Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tapeo
= [ft]
width
steel tape C’ o A F B
(53 F)
= [ft] o
AreaABC=0.5 * height * width = 44,627 [ft2]
the answer is C.
43,560
44,600
44,630
45,000
height = [ft]
width = [ft]
AnswerC

56. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4