Math and Arithmetic
Algebra
Geometry

# What is the length and width of a1380 sq ft house?

012

###### 2013-01-22 15:37:56

length = 1380/width or width = 1380/length

Unless a length or width is given it's impossible to give an accuratee answer.

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Length*Width = Area so Width = Area/Length = 40.64 sq ft/12.7 ft = 3.2 ft.

Perimeter = 2(length + width) &rarr; width = perimeter &divide; 2 - length length = 1 foot &rarr; width = 70 ft &divide; 2 - 1 ft = 34 ft; length = 2 feet &rarr; width = 70 ft &divide; 2 - 2 ft = 33 ft.

area = width x length width = area/length width = 925 sq ft/37 ft width = 25 ft

Runway: 25R Length x Width: 12923 ft x 150 ft Runway: 07L Length x Width: 12923 ft x 150 ft Runway: 25L Length x Width: 11095 ft x 200 ft Runway: 24R Length x Width: 8926 ft x 150 ft Runway: 24L Length x Width: 10885 ft x 150 ft Runway: 07R Length x Width: 11095 ft x 200 ft Runway: 06R Length x Width: 10885 ft x 150 ft Runway: 06L Los Angeles Ramp/Taxi: 131.45

From here, it appears that the length is double the width.

The perimeter of a standard- sized rectangular rug is 94 ft. the length is 3 ft more than the width. Find the length and the width.

The area formula is: Area = length x width Let: length = 40 ft width = 50 ft Then: A = 40 ft x 50 ft = 2000 ft&sup2;

The length of the rectangle that has the area of 120 ft and the width of 8 ft is 960 ft that is the correct answer 960 ft

Depends length-25ft width-10ft length - 125ft width - 2ft length - 50ft width- 5ft

If the average square footage of houses is about 2700 ft^2 and the average length is about 60 ft, then it will be around 45 ft.

Assuming Width &lt;= Length then the width can be any positive value less than sqrt(270) feet = 16.43 feet (approx). Width = 15 ft goes with Length = 270/15 = 18 feet Width = 10 ft with Length = 270/10 = 27 ft Width = 0.0001 ft with Length = 270/0.0001 = 2700000 ft etc.

Area = Length * Width = 13 ft * 6 ft = 13*6 sq ft = 78 sq ft

Length is 12ft, width is 4 ft. 2 x (length + width) = perimeter = 32 ft &rArr; length = 16ft - width [1] length x width = area = 48 sq ft Using [1] above: &rArr; (16ft - width) x width = 48 sq ft multiplying out and rearranging: &rArr; width2 - 16 x width + 48 = 0 factorizing: &rArr; (width - 4)x(width - 12) = 0 Thus width = 4 or 12, so using [1] above: &rArr; width = 4ft, length = 16 - 4 = 12ft or width = 12ft, length = 16 - 12 = 4ft Convention is that the length is larger than the width, thus rectangle is 12ft long by 4 ft wide.

Let the length be 4x and the width be x: 2(length)+width = 45 feet of fence used 2(4x)+x = 45 feet 8x+x = 45 9x = 45 x = 5 length = 20 ft and the width = 5 ft Check: 2(20)+5 = 45 ft

Not at all.Rectangle A:-- Length = 20-ft, Width = 10-ft.-- Area = 200 square ft-- Perimeter = 60-ftRectangle B:-- Length = 16-ft, Width = 12.5-ft-- Area = 200 square ft-- Perimeter = 57-ftRectangle C:-- Length = 40-ft, Width = 5-ft-- Area = 200 square ft-- Perimeter = 90-ftRectangle D:-- Length = 50-ft, Width = 4-ft-- Area = 200 square ft-- Perimeter = 108-ftRectangle E:-- Length = 100-ft, Width = 2-ft-- Area = 200 square ft-- Perimeter = 204-ftRectangle F:-- Length = 800-ft, Width = 3 inches-- Area = 200 square ft-- Perimeter = 1600-ft 6-in

By multiplying its length to its width. Area= length x width

You can't without one or the other. If you know length then width is the area divided by the length.

Let the width be 2x and the length be x: width*length = area 2x*x = 200 sq ft 2x2 = 200 Divide both sides by 2 and then square root both sides: x = 10 Therefore: length = 10 ft and width = 20 ft

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