A third ant walks around the perimeter of another rectangle sheet of paper. The dimensions of this sheet of paper are given in decimal numbers expressed to the hundredth place, with the tenths and hundredths digits being non-zero. If the ant travels between 18 and 19 centimeters, what are the dimensions of the sheet of paper. Give one possibility.
Let the length and width of the rectangle sheet of paper be $l$ and $w$ respectively. Then the ant walks $l+w+l+w = 2l + 2w = 2(l+w)$ centimeters around the perimeter of the rectangle sheet of paper. Since the ant travels between 18 and 19 centimeters, we have $18 \leq 2(l+w) \leq 19$. Dividing both sides by 2 gives $9 \leq l+w \leq \dfrac{19}{2}$. Since the dimensions of the sheet of paper are given in decimal numbers expressed to the hundredth place, with the tenths and hundredths digits being non-zero, we can only have $l+w = 9.00$ or $l+w = 9.50$.
If $l+w = 9.00$, then $(l,w) = (4.5,4.5)$.
If $l+w = 9.50$, then $(l,w) = (4.75,4.75)$.
Thus, one possibility for the dimensions of the sheet of paper is $\boxed{4.75}$ centimeters by $\boxed{4.75}$ centimeters.
To calculate the dimensions of the sheet of paper, we need to find the possible length and width that allow for a perimeter between 18 and 19 centimeters.
Let's assume the length of the rectangle is L and the width is W.
The formula for perimeter is: P = 2L + 2W
Given that the ant travels between 18 and 19 centimeters:
18 ≤ 2L + 2W ≤ 19
Let's find a possible solution:
Assume L = 5.4 (5.4 cm)
Substitute this value into the equation:
18 ≤ 2(5.4) + 2W ≤ 19
18 ≤ 10.8 + 2W ≤ 19
Subtract 10.8 from all sides:
7.2 ≤ 2W ≤ 8.2
Divide all sides by 2:
3.6 ≤ W ≤ 4.1
So, a possible solution could be a rectangle with dimensions 5.4 cm by 3.6 cm (length x width), where the perimeter ranges between 18 and 19 centimeters.