an ant walks around the perimeter of another rectangular sheet of paper. the dimensions of this sheet of paper are given in decimal numbers expressed to the hundredth place, with tenths and hundredths digits being non-zero. if the ant travels between 18 and 19 centimeters, what are the detentions of the sheet of paper? give one possibility.
Let's assume that the length of the rectangular sheet of paper is 18 cm and the width is 1 cm.
If the ant walks around the perimeter of the paper, it will have to travel a distance of 2*(length + width).
In this case, the ant would have to travel 2*(18+1) = 40 cm, which is not within the range of 18-19 cm.
Let's try a different combination.
Assume the length of the rectangular sheet of paper is 18.5 cm and the width is 0.5 cm.
The ant would have to travel 2*(18.5+0.5) = 38 cm, which is still not within the range of 18-19 cm.
Let's try one more combination.
Assume the length of the rectangular sheet of paper is 18.45 cm and the width is 0.55 cm.
The ant would have to travel 2*(18.45+0.55) = 39 cm, which is within the range of 18-19 cm.
Therefore, one possibility for the dimensions of the sheet of paper is 18.45 cm (length) by 0.55 cm (width).
Let's assume the length of the rectangular sheet of paper is 'L' and the width is 'W'.
The ant walks around the perimeter of the sheet of paper, which can be calculated by the formula:
Perimeter = 2(L + W)
Since the ant travels between 18 and 19 centimeters, we can set up the following inequality:
18 < 2(L + W) < 19
To find one possible solution, we can start by considering different values for 'L' and 'W' that satisfy the inequality.
Let's try setting L = 9.00 cm and W = 0.50 cm:
Perimeter = 2(9.00 + 0.50)
= 2(9.50)
= 19 cm
This satisfies the inequality:
18 < 19 < 19
So, one possible set of dimensions of the sheet of paper is 9.00 cm x 0.50 cm.
To solve this problem, we can set up an equation based on the information given.
Let's assume the length of the rectangular sheet of paper is "L" and the width is "W". The ant walks around the perimeter of the sheet, which can be calculated using the formula:
Perimeter = 2L + 2W
Given that the ant's travel distance is between 18 and 19 centimeters, we can form the following inequality:
18 < 2L + 2W < 19
Simplifying the inequality, we have:
9 < L + W < 9.5
Now, we need to find dimensions (L and W) that satisfy this inequality. Since the dimensions are given in decimal numbers expressed to the hundredth place, we need to consider the numbers between 9 and 9.5 with two decimal places (tenths and hundredths digits).
One possible solution could be:
L = 4.35 cm (rounded to the hundredth place, non-zero on tenths and hundredths digits)
W = 4.65 cm (rounded to the hundredth place, non-zero on tenths and hundredths digits)
Substituting these values back into the perimeter equation:
Perimeter = 2(4.35) + 2(4.65) = 8.7 + 9.3 = 18 cm
So, the dimensions of the rectangular sheet of paper in this case are approximately 4.35 cm by 4.65 cm.