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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+
