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how many rectangles can be there with equal area and perimeter if its sides are to be...
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(Level 1) Associate Educator, Expert, Newton
Call the width of the rectangle w and the height of the rectangle h.
The area of the rectangle is wh.
The perimenter of the rectangle is w+h+w+h=2(w+h).
So, we want wh=2w+2h, and we want w and h positive integers.
Try to "factor" this:
the -2w term suggests we want:
similarly, the -2h term suggests we want:
expand out to find what that last term must be:
We must have:
w, h are integers, so w-2, h-2 are integers. The only integers that multiply to 4 are: 4*1, 2*2, -4*-1, -2*-2
we try each of these:
w-2 = 4, h-2 = 1, means that w=6, h=3
(alternatively, we could have h-6, w=3)
w-2= 2, h-2 = 2, means that w=4, h=4
w-2=-4, h-2=-1, means that w=-2, h=1 (our dimensions must be positive, so this can't happen)
w-2=-2, h-2=-2, means that w=0, h=0 (this doesn't really describe a rectangle)
Thus, the only rectangles that can be made with integer lengths and area equal to perimenter are:
4x4, and 3x6
Posted by mlehuzzah on June 26, 2012 at 4:41 PM (Answer #1)
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