Find: AreaABC [ft2] C A B C’ 43,560 44,600 44,630 45,000 LAB@105 F = 438.16 [ft] Find the area of Triangle A B C, in square feet. [pause] In this problem, ---- o LCC’@105 F = 203.56 [ft] o using a steel tape o calibrated at 53 F
Find: AreaABC [ft2] C A B C’ LAB@105 F 43,560 44,600 44,630 45,000 o 43,560 44,600 44,630 45,000 LAB@105 F = 438.16 [ft] the length from Point A to Point B was measured at a temperature of 105 degrees Fahrenheit. o LCC’@105 F = 203.56 [ft] o using a steel tape o calibrated at 53 F
Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630 o A B C’ LAB@105 F o 43,560 44,600 44,630 45,000 LAB@105 F = 438.16 [ft] The length from Point C to Point C prime was also measured at a temperature of 105 degrees Fahrenheit. o LCC’@105 F = 203.56 [ft] o using a steel tape o calibrated at 53 F
Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630 o A B C’ LAB@105 F o 43,560 44,600 44,630 45,000 LAB@105 F = 438.16 [ft] Both of these measurement were made using a steel tape, --- o LCC’@105 F = 203.56 [ft] o using a steel tape o calibrated at 53 F
Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630 o A B C’ LAB@105 F o 43,560 44,600 44,630 45,000 LAB@105 F = 438.16 [ft] which was calibrated at 53 degrees Fahrenheit. Also note that angles --- o LCC’@105 F = 203.56 [ft] o using a steel tape o calibrated at 53 F
Find: AreaABC [ft2] C LCC’@105 F A B C’ LAB@105 F 43,560 44,600 44,630 o A B C’ LAB@105 F o 43,560 44,600 44,630 45,000 LAB@105 F = 438.16 [ft] C C prime A, and C C prime B, measure to be 90 degrees. Therefore, we’ve effectively been given --- o LCC’@105 F = 203.56 [ft] o using a steel tape o calibrated at 53 F
Find: AreaABC [ft2] C height LCC’@105 F A B C’ LAB@105 F width 43,560 o A B C’ LAB@105 F width o 43,560 44,600 44,630 45,000 LAB@105 F = 438.16 [ft] the dimensions for the width and height, of the triangle. [pause] To find the area of the oblique Triangle --- o LCC’@105 F = 203.56 [ft] o using a steel tape o calibrated at 53 F
Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] steel tape A B (53 F) o =203.56 [ft] steel tape o A B (53 F) LAB@105 F =438.16 [ft] o A B C, we first need to determine the appropriate area equation to use. ---
Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] steel tape A B (53 F) o =203.56 [ft] steel tape o A B (53 F) LAB@105 F =438.16 [ft] o # [pause] We’ll begin by defining Capital A subscript A, B and C ---- known variables area equation 1 2 3
Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] AC AA AB steel tape A B o =203.56 [ft] AC AA AB steel tape o A B (53 F) LAB@105 F =438.16 [ft] o # as the interior angles of the triangle, at Points A, B and C respectively. Also, --- known variables area equation 1 2 3
Find: AreaABC [ft2] C LCC’@105 F =203.56 [ft] AC AA AB steel tape A B o =203.56 [ft] AC AA AB steel tape o A B (53 F) LAB@105 F =438.16 [ft] o # lower-case letters a, b and c will refer to ---- known variables area equation 1 2 3
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
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
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
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
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
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
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
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
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
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
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
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
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)
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)
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)
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)
Find: AreaABC [ft2] C height width A B # known variables area equation 1 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
Find: AreaABC [ft2] ? ? C LCC’@105 F height =203.56 [ft] width o =203.56 [ft] width steel tape o A B (53 F) LAB@105 F =438.16 [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 ? ?
Find: AreaABC [ft2] ? ? C LCC’@105 F height =203.56 [ft] width o =203.56 [ft] width steel tape o A B (53 F) LAB@105 F =438.16 [ft] o for the temperature difference. [pause] In this problem, the actual height of Triangle A B C equals ---- AreaABC=0.5 * height * width ? ?
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [ft] o the length of segment C C prime, if it were measured at 53 degrees Fahrenheit, ---- AreaABC=0.5 * height * width height = LCC’@53 F o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [ft] o which is the calibrated temperature of the tape. However, we only have --- AreaABC=0.5 * height * width height = LCC’@53 F o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [ft] o a measurement of line segment C C prime when the temperature is --- AreaABC=0.5 * height * width height = LCC’@53 F o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [ft] o 105 degrees Fahrenheit. Therefore, we’ll equate the height of the triangle to this measurement, plus, --- AreaABC=0.5 * height * width height = LCC’@53 F o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [ft] o temperature correction a temperature correction distance, C. [pause] In the same manner, --- AreaABC=0.5 * height * width distance height = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [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 = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + C
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [ft] o temperature correction at the calibrated 53 degrees Fahrenheit, which we’ll equate to measurement taken --- AreaABC=0.5 * height * width distance height = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + C o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [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 = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + CAB,105 F,53 F o o o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [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 = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + CAB,105 F,53 F o o o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape o A C’ B (53 F) LAB@105 F =438.16 [ft] o temperature correction temperature correction distances. The temperature correction distances equal --- AreaABC=0.5 * height * width distance height = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + CAB,105 F,53 F o o o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width height = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F the temperature during the measurement, minus ---- o o o o width = LAB@53 F = LAB@105 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
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [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
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [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
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [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
Find: AreaABC [ft2] α=6.45*10-6[ F-1] C LCC’@105 F height =203.56 [ft] o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [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
Find: AreaABC [ft2] α=6.45*10-6[ F-1] C LCC’@105 F height =203.56 [ft] o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [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
Find: AreaABC [ft2] α=6.45*10-6[ F-1] C LCC’@105 F height =203.56 [ft] o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o 0.068 feet and 0.147 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 = 0.068 [ft] o o CAB,105 F,53 F =(TM-TC) * LM * α = 0.147 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width to the nearest 100th of a foot, and after plugging in the temperature correction distances, --- height = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + CAB,105 F,53 F o o o o CCC’,105 F,53 F =0.07 [ft] = 0.068 [ft] o o CAB,105 F,53 F =0.15 [ft] = 0.147 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width as well as the measured lengths at 105 degrees Fahrenheit, --- height = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + CAB,105 F,53 F o o o o CCC’,105 F,53 F =0.07 [ft] = 0.068 [ft] o o CAB,105 F,53 F =0.15 [ft] = 0.147 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width the actual lengths for the height and width of Triangle A B C equal, --- height = LCC’@53 F = LCC’@105 F + CCC’,105 F,53 F o o o o width = LAB@53 F = LAB@105 F + CAB,105 F,53 F o o o o CCC’,105 F,53 F =0.07 [ft] = 0.068 [ft] o o CAB,105 F,53 F =0.15 [ft] = 0.147 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width 203.63 feet and 438.31 feet, respectively. Plugging these values into our equation --- height = 203.63 [ft]=LCC’@105 F+ CCC’,105 F,53 F o o o width = 438.31 [ft]=LAB@105 F + CAB,105 F,53 F o o o CCC’,105 F,53 F =0.07 [ft] = 0.068 [ft] o o CAB,105 F,53 F =0.15 [ft] = 0.147 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width for the area of Triangle A B C, we find the area of the triangle equals, --- height = 203.63 [ft]=LCC’@105 F+ CCC’,105 F,53 F o o o width = 438.31 [ft]=LAB@105 F + CAB,105 F,53 F o o o CCC’,105 F,53 F =0.07 [ft] = 0.068 [ft] o o CAB,105 F,53 F =0.15 [ft] = 0.147 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width = 44,627 [ft2] 44,627 square feet. [pause] height = 203.63 [ft]=LCC’@105 F+ CCC’,105 F,53 F o o o width = 438.31 [ft]=LAB@105 F + CAB,105 F,53 F o o o CCC’,105 F,53 F =0.07 [ft] = 0.068 [ft] o o CAB,105 F,53 F =0.15 [ft] = 0.147 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width = 44,627 [ft2] When reviewing the possible solutions, ---- 43,560 44,600 44,630 45,000 height = 203.63 [ft] width = 438.31 [ft] CCC’,105 F,53 F =0.07 [ft] o o CAB,105 F,53 F =0.15 [ft] o o
Find: AreaABC [ft2] C LCC’@105 F height =203.56 [ft] width steel tape o =203.56 [ft] width steel tape C’ o A LAB@105 F B (53 F) =438.16 [ft] o AreaABC=0.5 * height * width = 44,627 [ft2] the answer is C. 43,560 44,600 44,630 45,000 height = 203.63 [ft] width = 438.31 [ft] AnswerC
? 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