SOLUTION: I was doing an IQ game ("challenging" level) and was faced with a problem, which I solved by logical thinking and trial/error. However, before that I tried to figure out a way to s

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Question 466398: I was doing an IQ game ("challenging" level) and was faced with a problem, which I solved by logical thinking and trial/error. However, before that I tried to figure out a way to solve the problem with algebra. I'm still not sure if it is possible or not.
The problem/question:
"Charlie removed a one-metre wide strip of grass from all the way around a square lawn, which was made info a flower bed. The remaining grass was four-fifths the area of the new flower bed.
What were the dimensions of the lawn originally?"
It seemed to me that we are missing a value (a length), but then I thought that it might be possible to figure it out by using equations, and canceling out values against each other. I tried different approaches, but I haven't done equations for a long time and got stuck.
This was my approach:
a = length or height of square lawn
b = new flower bed
c = remaining grass
d = dimensions of original lawn (area)
c = 4/5b
d = b + c
(a - 2)x2 = (4/5b) + (5/5b)
Is it just not possible, or am I missing something?
Any help with pointing out if and how this can be done would be greatly appreciated!
Kind regards,
Rob

Found 2 solutions by ankor@dixie-net.com, scott8148:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Charlie removed a one-metre wide strip of grass from all the way around a square lawn, which was made into a flower bed.
The remaining grass was four-fifths the area of the new flower bed.
What were the dimensions of the lawn originally?
:
Let the original dimension of the square lawn = x by x
Area of original lawn = x^2
:
After removing the 1 meter strip all the way around, the dimensions would be:
(x-2) by (x-2)
Area of the remaining grass = (x-2)^2, FOIL this and you have:
A = x^2 - 4x + 4; the remaining grass
:
then find removed strip area:
A = x^2 - (x^2 - 4x + 4)
A = x^2 - x^2 + 4x - 4
A = 4x - 4, the area of the flowerbed
:
x^2 - 4x + 4 = 4%2F5(4x - 4)

:

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
if the original square lawn had a side length of S, then the remaining grass square had a side length of S-2 (one meter removed from each end)

the new grass square is 4/5 the area of the one meter border (flower bed), whose area is 5/5 of itself
___ this means that the grass is 4/9 (4 / [4+5]) of the total area

(4/9)S^2 = (S - 2)^2 ___ 4S^2 = 9S^2 - 36 S + 36 ___ 0 = 5S^2 - 36 S + 36

factoring ___ (5S - 6)(S - 6) = 0

5S - 6 = 0 ___ S = 6/5 ___ not realistic considering the one meter border reduction

S - 6 = 0 ___ S = 6