Can you find the width of a rectangle if you only know the perimeter?

Answer

If the perimeter of a rectangle is known, finding the width is only possible if the length is also known. Infinite combinations of lengths and widths of a rectangle exist with any given perimeter.

The perimeter of a rectangle is given by the formula P = 2L + 2W, where P represents the perimeter, L represents the length and W refers to the width. If a given rectangle has a perimeter of 100 inches, the rectangle might have a width of 20 inches and a length of 30 inches. Two times 20 plus two times 30 does give a total perimeter of 100 inches. However, it is also possible that the width is 10 inches and the length is 40 inches.

Reference:
Q&A Related to "Can you find the width of a rectangle if you..."
if you have the length of one side - plus the perimeter. Divide the perimeter by 2, then subtract the known side - this leaves you with the unknown dimension. For example. Say you
http://wiki.answers.com/Q/How_do_you_find_the_widt...
60+60 will give you the length of two sides of the rectangle. This = 120. 280-120 will give you the length of the outside thats left over after measuring the width. This =160. divide
http://answers.yahoo.com/question/index?qid=201109...
Yes, if you know that it's a right, isosceles or equilateral triangle, you can compute the perimeter. However, if your triangle is a scalene, as most triangles are, you're out of
http://nz.answers.yahoo.com/question/index?qid=200...
This will be an exercise in creating a couple of equations from the text and then solving simultaneously. The area of a rectangle is L x W. So our first equation is simply: L x W
http://answers.yahoo.com/question/index?qid=201309...
Explore this Topic
According to my calculations, To find the perimeter of a rectangle you use Pythagorean theorem. A squared + B squared = C squared. Your C is the diagonal and your ...
In algebra, you use some amount of known information to solve for an unknown variable. If you are given the perimeter of a rectangle, you might be asked to solve ...
About -  Privacy -  Careers -  Ask Blog -  Mobile -  Help -  Feedback  -  Sitemap  © 2014 Ask.com