find. the measure of the length if a rectangle

Question

Find the length of a rectangle that is 2 less than 5 times the measure  of the width. The perimeter of the rectangle is 32

Answer 1

The perimeter of a rectangle is 2 times the length plus 2 times the width: P = 2w + 2l. Here, the length is 2 less than 5 times the width: l = 5w - 2. Additionally, the perimeter is 32: P = 32. We need to plug both of these values into the perimeter equation and solve for width. Once we have the width, we can then solve for the length:

P = 2w + 2l

32 = 2w + 2(5w - 2) <--- Place 32 for P and 5w - 2 for l

32 = 2w + 10w - 4 <--- Foil the 2

32 = 12w - 4 <--- Combine the w terms

36 = 12w <--- Add 4 to both sides

3 = w  <--- Divide by 12

 

Now that we have the width, we can find the length:

l = 5w - 2

l = 5(3) - 2 <---- Plug in the width we found earlier

l = 15 - 2 = 13

 

The length is 13 units.

Answer 2

width= x

length= 5x - 2

 

perimeter = 2(x+5x-2)

 

 32            =2(6x-2)

 

 16        = 6x -2

 

18          =6x

 

3          =x

 

width = x= 3

length=5x-2=5(3)-2 =15-2 =13