the length of a rectangle is 19 centimeters less than five times its width. Its area is 30 centimeters. Find the dimensions of the rectangle.

Question

the length of a rectangle is 19 centimeters less than five times its width. Its area is 30 centimeters. Find the dimensions of the rectangle.

Answer 1

w = width

l = length = 5w - 19            (5 times the w less 19 in)

 

A = l * w

 

A = w * (5w -19)

 

A = 5W² - 19w

 

30 = 5w² - 19w

 

0 = 5w² - 19w - 30

 

0 = (5w + 6) (w - 5)

 

set the factors equal to zero and solve for w

 

5w + 6 = 0

5w = -6

w = -6/5   (can't be right because our width must be a positive number)

 

w - 5 =0

w = 5

 

l = 5w -19

l = 25 - 19

l = 6

 

 

So width is 5 and length is 6

 

and that would give us an area of 30

 

 

Answer 2

Let w = width
l = 5w - 19

Area of a rectangle is length * width:
Area = l * w = 30

Substitute 5w-19 for l:

Area = (5w - 19)(w) = 0
30 = 5w^2 - 19w


5w^2 - 19w = 30

5w^2 - 19w - 30 = 0

(5w + 6) (w - 5) = 0


5w + 6 = 0

5w = -6

w = -6/5 No negative widths



w - 5 = 0
w = 5 cm

l = 5w - 19 = 5(5) - 19 = 25-19 = 6 cm



Check:

A = l * w = 6 * 5 = 30