Is there an equation or something to find out in how many ways I can build a rectangle that has the same area? Let's say the area is 5000. The simplest way would be 10 x 500. How do I find how many rectangles I can build with integer dimensions (ie. height is not decimal, only whole numbers) and what are those dimensions?
2007-02-05 06:26:59 · 7 answers · asked by n.daniel 2 in Science & Mathematics ➔ Mathematics
The formula for dimensions of a rectangle with a fixed area, say 5000, is:
5000 = h*w
Now, the domain of this function is all real numbers. For example, you could have h = 37.5 and w = 133 1/3. If you want all of the possible integer results, you will have to factor 5000:
5000 = 2*2*2*5*5*5*5
So, for any h and w which satisfy the original equation, h*w = 2^3 * 5^4. Every combination of these factors will yield an integer solution...
h = 2^0 * 5^0, w = 2^3 * 5^4
h = 2^1 * 5^0, w = 2^2 * 5^4
h = 2^2 * 5^0, w = 2^1 * 5^4
h = 2^3 * 5^0, w = 2^0 * 5^4
h = 2^0 * 5^1, w = 2^3 * 5^3
h = 2^1 * 5^1, w = 2^2 * 5^3
h = 2^2 * 5^1, w = 2^1 * 5^3
h = 2^3 * 5^1, w = 2^0 * 5^3
h = 2^0 * 5^2, w = 2^3 * 5^2
h = 2^1 * 5^2, w = 2^2 * 5^2
h = 2^2 * 5^2, w = 2^1 * 5^2
h = 2^3 * 5^2, w = 2^0 * 5^2
h = 2^0 * 5^3, w = 2^3 * 5^1
h = 2^1 * 5^3, w = 2^2 * 5^1
h = 2^2 * 5^3, w = 2^1 * 5^1
h = 2^3 * 5^3, w = 2^0 * 5^1
h = 2^0 * 5^4, w = 2^3 * 5^0
h = 2^1 * 5^4, w = 2^2 * 5^0
h = 2^2 * 5^4, w = 2^1 * 5^0
h = 2^3 * 5^4, w = 2^0 * 5^0
2007-02-05 06:47:45 · answer #1 · answered by computerguy103 6 · 1⤊ 0⤋
There is no formula
Simply take the one you have and do height 1, 2, 3, 4, 5, 6 and so on.
If you want integers on both sides, then you have to factor 5000 (in this case) to its integers - 1,2,5,10, 20, 50, 100, etc and use those.
2007-02-05 06:32:53 · answer #2 · answered by Mike1942f 7 · 0⤊ 0⤋
Area = length x height. I'd say, list all the factors of 5000. For every set, you have one possible rectangle with that area.
2007-02-05 06:30:45 · answer #3 · answered by Master Maverick 6 · 2⤊ 0⤋
You have to break down your area in primes and then choose.
Here 5000= 2^3 x 5^4 so you have 4x5 choices for one side and one choice for the other. By symmetry, that gives you only 10 solutions.
2007-02-05 06:32:54 · answer #4 · answered by gianlino 7 · 3⤊ 0⤋
It would be a matter of finding all the pairs of numbers that are factors of 5000 .
2007-02-05 06:31:49 · answer #5 · answered by Gene 7 · 1⤊ 0⤋
we would choose to be responsive to the section A Rectangle a million tells us that A = ab Rectangle 2 tells us that A = (a-3)(b+2) Rectangle 3 tells us that A = (a+3)(b-a million) are you able to verify how we could use those to verify what we could be responsive to?
2016-12-13 09:28:37 · answer #6 · answered by fette 4 · 0⤊ 0⤋
What are the factors of 5000?
That's your answer.
2007-02-05 06:32:09 · answer #7 · answered by bequalming 5 · 2⤊ 0⤋