SOLUTION: A rectangle has sides with integral lengths. The number of units in its perimeter is the same as the number of square units in its area. What are the dimensions of the rectangle

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Question 68969: A rectangle has sides with integral lengths. The number of units in its perimeter is the same as the number of square units in its area. What are the dimensions of the rectangle
Found 2 solutions by ankor@dixie-net.com, Edwin McCravy:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle has sides with integral lengths. The number of units in its perimeter is the same as the number of square units in its area. What are the dimensions of the rectangle;
:
It think it will be a low value, create an equation:
:
2L + 2W = L*W
:
2L = LW - 2W
:
2L - LW = -2W
:
L(2 - W) = - 2W
:
L = -%282W%29%2F%28%282-W%29%29
:
A value of W=2 puts 0 in the denominator so
Substitute 3 for W and see if you get an integer value for L:
L = -%282%2A3%29%2F%28%282-3%29%29
:
L = %28-6%29%2F%28%282-3%29%29
:
L = -6/-1
L = 6
:
The dimensions of the rectangle is 6 by 3:
2(6) + 2(3) = 6*3



Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle has sides with integral lengths.
The number of units in its perimeter is the
same as the number of square units in its 
area. What are the dimensions of the 
rectangle?

Assume L > W

Perimeter = 2L + 2W
Area = LW

so 

Perimeter = Area

or

2L + 2W = LW

Solve for W

LW - 2W = 2L

W(L - 2) = 2L

      2L 
W = -------
     L - 2

Divide by long division:
           
           2 + 4/(L-2)
L - 2)2L + 0
      2L - 4
           4
so

    W = 2 + 4/(L-2)

W - 2 = 4/(L-2)

The left side is an integer, so
the right side must be too.

For 4/(L-2) to be an integer,

L-2 must divide evenly into 4.

The only positive integers which
divide evenly into 4 are 1, 2, and 4

So L - 2 = 1 or 2 or 4, and therefore
       L = 3 or 4 or 6

If L = 3 then

      2L        2(3)
W = ------- = ------- = 6 
     L - 2     3 - 2

We will discard this answer since 
we assumed L > W

If L = 4 then

      2L        2(4)
W = ------- = ------- = 4 
     L - 2     4 - 2

So one solution is L = 4 and W = 4, (a 4×4 square).

If L = 6 then

      2L        2(6)
W = ------- = ------- = 3 
     L - 2     6 - 2

So one solution is L = 6 and W = 3

So there are two solutions,
a 6×3 rectangle and a 4×4 square.

Edwin