The length is 15ft longer than the width and enclosed an area covering 1350 square feet. What are the dimensions of the dog's play area?

Question

The manager of a park enclosed an area for small dogs to play. He made the length 15ft longer than the width and enclosed an area covering 1350 square feet. What are the dimensions of the dog's play area?

Answer 1

The length (L) is equal to the Width (W) plus 15 feet or:  L=W+15.  Now the area (A) of the rectangle (dog's play area) is:

 

L x W or by substitution we have A = (W+15) x W = 1350 sq ft (given),

 

now we have W² + 15W = 1350, subtracting 1350 from both sides give us W² + 15W - 1350 = 0

                                                                                                   

Solving for W using the quadratic formula we have that W=(-15 ± Sqrt (15² - (4 x -1350))) /2

 

We're only interested in the positive root of W= 30ft.    So now we substitute 30 for W into the equation

 

L=W+15, and we get L=30+15=45 ft.  And the dimensions are L=45 ft, W=30 ft, and A = 45 x 30 = 1350 sq ft.

 

 

Answer 2

Hi, Emily,

 

    1.  Draw a model of the story problem.  Assume the area is a rectangular shape because most animal pens are rectangular.

 

                   ____________

                   |                   |   Let x=width

                   |___________| 

                L=Length=x+15 ft.

 

Note:  Area of a rectangle=L*W                      A=1,350 ft.²

 

2.  Use substitution to solve the problem.        A=L*W         1,350= (x+15)(x)

 

                                                  1,350=x² + 15x   Subtract 1,350 from both sides of the Quadratic Equation. 

 

                                                      0=x ² + 15x -1,350        

 

  If you find this hard to factor, I suggest you use the Quadratic Formula.  You will get two answers, but use the positive one when dealing with the length of a side of something, which would never have a negative length.  

If I have helped you, please give me a thumbs up.

                                                         Susan C.