Asked by Anonymous
                The area of a rectangular piece of land is 280 square meters. If the length of the land was 5 meters less and the width was 1 meter more, the shape of the land would be a square.
Part A: Write an equation to find the width (x) of the land. Show the steps of your work. (5 points)
Part B: What is the width of the land in meters? Show the steps of your work. (5 points)
            
        Part A: Write an equation to find the width (x) of the land. Show the steps of your work. (5 points)
Part B: What is the width of the land in meters? Show the steps of your work. (5 points)
Answers
                    Answered by
            Reiny
            
    original width --- x
original length --- y
xy= 280
y = 280/x
new width --- x+1
new length --- y-5
new area
= (x+1)(y-5)
= (x+1)(280/x - 5)
= 280 - 5x + 280/x - 5
= 275 - 5x + 280/x
= (275x - 5x^2 + 280)/x
= -5( x^2 - 55x - 56)/x
perfect squares in the neighbourhood of 280 are
169 , 196 , 225 , 256 , 289, 324 ...
So I tried values of x starting at x = 13
Lo, and behold, when x = 14
new area = -5(14^2 - 55(14) - 56)/14 = 225 which is a perfect square.
so the original was 14 by 20 for an area of 280
check:
new width = 15
new length = 15
new area = 225
YEAHHH!
    
original length --- y
xy= 280
y = 280/x
new width --- x+1
new length --- y-5
new area
= (x+1)(y-5)
= (x+1)(280/x - 5)
= 280 - 5x + 280/x - 5
= 275 - 5x + 280/x
= (275x - 5x^2 + 280)/x
= -5( x^2 - 55x - 56)/x
perfect squares in the neighbourhood of 280 are
169 , 196 , 225 , 256 , 289, 324 ...
So I tried values of x starting at x = 13
Lo, and behold, when x = 14
new area = -5(14^2 - 55(14) - 56)/14 = 225 which is a perfect square.
so the original was 14 by 20 for an area of 280
check:
new width = 15
new length = 15
new area = 225
YEAHHH!
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