SOLUTION: The length of a rectangle is 2 feet less than its width. If the length is increased by 4 feet, and the width is decreased by 2 feet, the area of the new rectangle is 8 sq ft grea

Question 388631: The length of a rectangle is 2 feet less than its width. If the length is
increased by 4 feet, and the width is decreased by 2 feet, the area of the
new rectangle is 8 sq ft greater than the original. Find the area of the original rectangle.
Found 2 solutions by mananth, irshad_ahmed0@yahoo.com:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 2 feet less than its width. If the length is
increased by 4 feet, and the width is decreased by 2 feet, the area of the
new rectangle is 8 sq ft greater than the original. Find the area of the original rectangle.
width = x feet
length = x-2 feet
area = x(x-2) sq.ft
..
width x-2
length x-2+4=x+2
area = (x-2)(x+2)
..
(x-2)(x+2)-x(x-2)=8
x^2-4-x^2+2x=8
-4+2x=8
add 4
2x=12
/2
x= 6 feet the width
length = x-2 = 6-2 =4
...
area = 6*4 = 24 ft^2 --- Original rectangle
...
m.ananth@hotmail.ca
Answer by irshad_ahmed0@yahoo.com(7) (Show Source): You can put this solution on YOUR website!
lo=x-2 and wo=x so area o=(x-2)x=x^2-2x eqn 1 thrfre area of new rectangle=area of original+8 eqn 2
an=l*b=(x-2+4)(x-2)=x^2-4 as we know that an=ao+8 put values of an and ao get,x=6,put value of x in ao=24 and in an=32 so ao+8=32 proved
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