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To convert 100 inches wide by 84 inches long into feet, you divide each dimension by 12 (since there are 12 inches in a foot). This results in 100 inches being approximately 8.33 feet wide and 84 inches being 7 feet long. Therefore, the dimensions in feet are approximately 8.33 feet by 7 feet.
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The perimeter of a rectangle is 200 feet and the area is 2400 feet squared What is the length and width?
Suppose the length and width are L and W Perimeter = 2(L+W) = 200 so L+W = 100 so L= 100-W Area = L*W = 2400 so (100-W)*W = 2400 W2 - 100W + 2400 = 0 W = (100 +/- sqrt(10000 - 4*2400))/2 where the minus sign will give W and the plus sign will give L So W = (100 - sqrt(400))/2 = (100 - 20)/2 = 80/2 = 40 feet So L = 100-40 = 60 feet
What are the dimensions of a rectangle with a perimeter of 401 feet and an area of 100 feet?
The rectangle cannot have an area of 100 feet, since that is a measure of distance and not area. Assume, therefore, that the area is 100 SQUARE feet. Suppose the rectagle has length L and width W then 2(L + W) = 401 and L*W = 100 So L = 100/W Then 2(100/W + W) = 401 Multiply by W: 2(100 + W2) = 401W or 2W2 - 401W + 200 = 0 or (2W - 1)(W-200) = 0 So W = 0.5 ft or W = 200 ft which imply that L = 200 ft or 0.5 ft respectively. So, the dimensions are 200 ft by 0.5 ft (or 6 inches).
What is the perimeter of a rectangle with an area of 100 ft?
First of all, an area of 100 feet is not possible since area is measured in square feet - not in feet. But suppose we overlook that fundamental error. The area of a rectangle is not sufficient to determine its perimeter. It can have any value greater than 40 ft. Let L be ANY length ≥ 10 ft and let W = 100/L ft. Then, a rectangle with length L and width W has an area of L*W = L*100/L = 100 square feet. But since there are infinitely many choices for L, there are infinitely many perimeters (P). For example, L = 10 => W = 10 => P = 40 L = 100 => W = 1 => P = 202 L = 1000=> W = 0.1 => P = 2000.2 L = 10000=> W = 0.01 => P = 20000.02 You should see from that, there is no limit to how large L can be and so no limit on how large P can be.
How many square feet is a 10 x 10 feet room?
100 square feet. A= lxw A= area l=length w=width 10x10= 100.
How long are the sides of a 876 square feet carpet?
Any value that you like, as long as the length and breadth multiply to 876. Pick ANY positive number L and let W = 876/L Then a carper with length L feet and width W feet will have an area of L*W = L*876/L = 876 square feet. For example, 1 ft * 876 feet or 10 ft * 87.6 ft or 100 ft * 8.76 ft or 1000000 ft * 0.000876 feet
The perimeter of a rectangle is 200 feet and the area is 2400 feet squared What is the length and width?
Suppose the length and width are L and W Perimeter = 2(L+W) = 200 so L+W = 100 so L= 100-W Area = L*W = 2400 so (100-W)*W = 2400 W2 - 100W + 2400 = 0 W = (100 +/- sqrt(10000 - 4*2400))/2 where the minus sign will give W and the plus sign will give L So W = (100 - sqrt(400))/2 = (100 - 20)/2 = 80/2 = 40 feet So L = 100-40 = 60 feet
What are the dimensions of a rectangle with a perimeter of 401 feet and an area of 100 feet?
The rectangle cannot have an area of 100 feet, since that is a measure of distance and not area. Assume, therefore, that the area is 100 SQUARE feet. Suppose the rectagle has length L and width W then 2(L + W) = 401 and L*W = 100 So L = 100/W Then 2(100/W + W) = 401 Multiply by W: 2(100 + W2) = 401W or 2W2 - 401W + 200 = 0 or (2W - 1)(W-200) = 0 So W = 0.5 ft or W = 200 ft which imply that L = 200 ft or 0.5 ft respectively. So, the dimensions are 200 ft by 0.5 ft (or 6 inches).
The perimeter of a rectangular field is P feet The width of the field is 200 feet less than its length In terms of P what is the length of the field in Feet?
P = 2 x (L + W), so L + W = 100 ie P/2 = L + W W = L - 200 so P/2 = L + L - 200 = 2L - 200 therefore L = P/4 + 100
What are the dimensions of a rectangle with perimeter 200 feet and area less than 100 square feet?
To find the dimensions of a rectangle with a perimeter of 200 feet, we can use the formulas for perimeter (P = 2(l + w)) and area (A = l * w). Given that the perimeter is 200 feet, we have ( l + w = 100 ). However, for the area to be less than 100 square feet, the dimensions must be such that ( l * w < 100 ). Since the maximum area occurs when ( l ) and ( w ) are equal, the dimensions would need to be less than 10 feet each, which is not possible under these constraints. Therefore, no rectangle can satisfy both conditions.
What is the perimeter of a rectangle with an area of 100 ft?
First of all, an area of 100 feet is not possible since area is measured in square feet - not in feet. But suppose we overlook that fundamental error. The area of a rectangle is not sufficient to determine its perimeter. It can have any value greater than 40 ft. Let L be ANY length ≥ 10 ft and let W = 100/L ft. Then, a rectangle with length L and width W has an area of L*W = L*100/L = 100 square feet. But since there are infinitely many choices for L, there are infinitely many perimeters (P). For example, L = 10 => W = 10 => P = 40 L = 100 => W = 1 => P = 202 L = 1000=> W = 0.1 => P = 2000.2 L = 10000=> W = 0.01 => P = 20000.02 You should see from that, there is no limit to how large L can be and so no limit on how large P can be.
How many square feet is a 10 x 10 feet room?
100 square feet. A= lxw A= area l=length w=width 10x10= 100.
How long are the sides of a 876 square feet carpet?
Any value that you like, as long as the length and breadth multiply to 876. Pick ANY positive number L and let W = 876/L Then a carper with length L feet and width W feet will have an area of L*W = L*876/L = 876 square feet. For example, 1 ft * 876 feet or 10 ft * 87.6 ft or 100 ft * 8.76 ft or 1000000 ft * 0.000876 feet
How do you find the perimeter of a rectangle in feet?
you add the l+l and the w+w so it look like this l+l+w+w+
If farmer Dan has 100 feet of fencing write an inequality to find the dimensions of the rectangle with the largest perimeter that can be created using 100 feet of fencing?
To find the dimensions of a rectangle with the largest perimeter using 100 feet of fencing, we can express the perimeter ( P ) of a rectangle in terms of its length ( l ) and width ( w ) as ( P = 2l + 2w ). Since the total amount of fencing is 100 feet, we set up the inequality ( 2l + 2w \leq 100 ). Simplifying this gives ( l + w \leq 50 ). The dimensions that maximize the area (which is a related concept) would be when ( l = w = 25 ) feet, creating a square shape.
What is the width of an area of 300 square feet and a length of L feet?
A = W x L Width = 300 divided by L (or W = 300/L)
How many square feet do you need if your room measures 12ft by 7ft?
Square feet is area, or length times width (L x W), 12 ft x 7 ft = 84 square feet.
What are the dimensions of a rectangle with an area of fifty square feet and a perimeter of 201 feet?
Area of a rectangle = length(L) x width(W) Perimeter of a rectangle = 2L + 2W. 50 = LW : therefore W = 50/L 201 = 2L + 2 x (50/L) = 2L + 100/L : Multiply by L 201L = 2L2 + 100 : Therefore 2L2 - 201L + 100 = 0 This quadratic equation factorises (2L - 1)(L -100) = 0 When 2L - 1 = 0 then L = 1/2 When L - 100 = 0 then L = 100 The dimensions of the rectangle are therefore length = 100 ft, width = 1/2 foot
