The perimeter of a rectangle is 52. The length of the rectangle is 7 less than twice the width.Find the dimensions of the rectangle

Question

Need help figuring out how to find dimensions and set up an equation for this problem unsure of how to do it.

 

Answer 1

Well first of all we know the perimeter is 2 times the length plus 2 times the width so 52 = 2L + 2W, which is our first equation

 

Now for the second equation you can practically take it straight from the second sentence "the Length of the rectangle is " L =   " 7 less"   -7     " twice the width"     2W;   or L = 2W -7 .   So now we have two equations and two unknowns. L is already solved for in the second equation and you just substitute L in the first equation(52 = 2L + 2W) for what L equals in the second(2W - 7), or  52 = 2(2W -7) + 2W. Solve for W then plug the result for W back into the first equation to find L.

Answer 2

 X - Width

 

 2x  - 7 - Length

 

  2X + 2 ( 2X - 7 ) = 6X - 14  perimeter

 

    6X - 14 = 52

 

   6X = 66

 

   X = 11  / Width

 

    2X - 7 = 2(11 ) - 7 = 15  / length

 

   Test:

 

     2 ( 11 ) + 2 ( 15) = 52

 

 

Answer 3

P=2L+2W

width=W

length=2W-7

substitute into the formula

52=2(2W-7)+2W

52=4W-14+2W

52=6W-14

52+14=6W

66=6W

11=W

2W-7=2(11)-7=22-7=15

L=15

W=11

check:52=2*15+2*11=30+22=52