Is it possible to have an odd perimeter without a half in the area?

Answer 1

Yes. Consider, if you will, the triangle with sides of 3, 5, and 7 centimeters. Or, perhaps, the trapezoid with sides of 3, 5, 3, and 8 inches. Or even a circle with a perimeter of 55 feet.

However, I believe you're asking if you can have a rectangle with an odd perimeter, without a 1/2 in two of its sides.

The area, you see, is (side)*(adjacent side), so in order for there to be a 1/2 in the area, there would have to be a 1/2 in one of the factors, and the other would have to be odd. And if we imagine a square with sides of 2.25 each, then that makes an odd perimeter of 9, with an area of 81/16 (or 5.0625), which I would consider odd.

And, of course, as you saw above, the square with sides of 2.25 each had no halves in each individual side, and had an odd perimeter. There are many ways to slice the fractions so that there is no 1/2 in any side.

It cannot, however, be done with a whole number, for in a rectangle, two sides absolutely must be equal, such that it is as if you multiply any side length by two. When you multiply a whole number by two, it will inevitably turn out even, regardless of the other factor's oddness or evenness.