word problem

Question

 

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the perimeter of a badminton court is 128 feet. after a game of badminton the coach estimates that the players have run a total of 444 feet which is equivalent to six times the courts length plus nine times its width. what are the dimensions of a standard badminton court?

Answer 1

The perimeter of a rectangle is

     

     Perimeter = 2L + 2W    ...

         

         where

 

            W = width

            L = length

 

Since the perimeter is 128 feet,  we can construct the following equation:

 

     2L + 2W = 128       (equation 1)

 

It it is given that "the players have run a total pf 444 feet which is equivalent to six times the courts length plus nine times its width.'  Translating this to math gives us the following:

 

     6L + 9W = 444      (equation 2)

 

We bow have a system of 3 equations.

Wr can multiply equation 1 by -3.  This gives us opposite coefficients for "L".  We can then add the resulting equations, eliminating L and solving for W.

 

     -6L  - 6W = -384

      6L + 9W =  188

     --------------------------

              3W = 60

                W = 20

 

Substitute W = 20 into equation 1 and solve for L

 

    2L + 2W = 128

    2L + 2(20) = 128

    2L + 40 = 128

    2L = 128 - 40

    2L = 88

    L = 44

 

So for a standard badminton court,

 

   Length = 44 feet

   Width = 20 feet