Why can't the width, length, and diagonal of a rectangle be consecutive odd integers?

Answer 1

Because of Pythagoras' theorem,

OK, here comes the proof. I recommend you understand it... don't just copy it down for your homework.

OK. there are three consecutive integers. let the first one equal x.
the second one must be x+2, and the third one must be x+4. (if this doesn't make sense, pick any odd integer, and plug it into x. you'll get 3 consecutive odd integers.)

Since the diagonal of a rectangle is always the largest number, it must be x+4.

therefore by Pythagoras' theorem...
(x+4)^2=(x+2)^2 + x^2
x^2+8x+16=2x^2+4x+4
-x^2+4x+12=0
x^2-4x-12=0
so, we get that
(x-6)(x+2)=0
x=6,-2

Since x HAS to be positive (imagine a triangle with a negative length... not going to happen)
the only numbers that this works for are 6,8,10.
Since these are not odd... you get the drift.

I hope this helps.

Answer 2

A rectangle has four corners, all of which are at a right angle. Thus, we can use Pythagoras' Theorem which states that a^2 + b^2 = c^2, where a & b are adjacent sides & c the diagonal connecting the opposite corners to form a triangle.

This works for a = 3, b = 4, c = 5, but not for consecutive odd integers.
3^2 + 4^2 = 5^2
9 + 16 = 25
25 = 25 :)

Unfortunately, consecutive odd integers just don't fit the theorem or the reality of the situation. You can try to solve it algebraically, but will note that the solution (for those boundaries) is impossible:

x^2 + (x+2) ^2 = (x+4)^2
x^2 + x^2 + 4x + 4 = x^2 + 8x + 16
2 x^2 + 4x + 4 = x^2 + 8x + 16
x^2 - 4x - 12 = 0
(x-6)(x+2) = 0
So x = 6 or -2, neither of which is a positive odd integer.

I hope this helps.