SOLUTION: The length of a rectangle is 8cm longer than the width. If the width is increased by 4cm and the length is decreased by 5cm, the area of the rectangle remains the same. Find the or

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 8cm longer than the width. If the width is increased by 4cm and the length is decreased by 5cm, the area of the rectangle remains the same. Find the or      Log On


   

Question 1089299: The length of a rectangle is 8cm longer than the width. If the width is increased by 4cm and the length is decreased by 5cm, the area of the rectangle remains the same. Find the original dimensions of the rectangle.
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let w be the width.

Then your equation is

w*(w+8) = (w+4)*((w+8)-5),    or, which is the same,

w*(w+8) = (w+4)*(w+3),

w^2 + 8w = w^2 + 4w + 3w + 12,

w = 12.


Check.  12*(12+8) = 240.

        (12+4)(12+3) = 16*15 = 240.  ! Correct !


Answer.  the original dimensions of the rectangle are 12 x 8 cm.

Solved.