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What is 100 w by 84 L IN FEET?

Updated: 9/27/2023
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Q: What is 100 w by 84 L IN 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 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+

Related questions

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 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 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+


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


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.


The perimeter of a rectangle is 14 feet and the area is 12 square feet What is the width of the rectangle?

Suppose the length is L feet and the width is W feet. Then, Perimeter = 14 => 2*(L+W) = 14 or L+W=7 Also, Area = 12 => L*W=12 You could try trial and error to solve these two equations and in this particular case you will get to the answer quite quickly but for a more general method: L+W=7 => W=7-L Substituting in the second equation, L*(7-L)=12 or L^2 - 7L + 12 = 0 L^2 - 3L - 4L + 12 = 0 (L-3)*(L-4) = 0 so L=3 or L=4 But L>W and L+W = 7 so L > 3.5 so L= 4 => W=3 feet


What is the formula for square feet including ceiling?

' L ' = length of the room ' W ' = width of the room ' H ' = height of the ceiling Floor = ( L W ) square feet Ceiling = ( L W ) square feet 2 walls = ( L H ) square feet, each other 2 walls = ( W H ) square feet, each


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