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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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spsav | Student, Grade 10 | eNoter

Posted June 26, 2012 at 4:12 PM via web

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how many rectangles can be there with equal area and perimeter if its sides are to be integers

explain with full method to solve it plz plz plz plz !!!!!! also the sides must be integers

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted June 26, 2012 at 4:41 PM (Answer #1)

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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.

wh-2w-2h=0

Try to "factor" this:

(w+?)(h+?)+?=0

the -2w term suggests we want:

(w+?)(h-2)+?=0

similarly, the -2h term suggests we want:

(w-2)(h-2)+?=0

expand out to find what that last term must be:

wh-2w-2h+4+?=0

We must have:

(w-2)(h-2)-4=0

(w-2)(h-2)=4

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



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