Complete the activity. An ant stands at one vertex of a rectangular sheet of paper. The ant then proceeds to walk all the way around the perimeter of the paper. The dimensions of the sheet of paper are 358 inches by 278 inches. How many inches does the ant travel? Show your work. Suppose that each dimension of the sheet of paper described in question 1 is increased by one inch. How many inches will the ant travel in this case? Explain your thinking. A second ant walks all the way around a triangle drawn in the dirt. Two of the dimensions of the triangle are 5110 centimeters and 715 centimeters. The total distance traveled by the ant is 20 centimeters. What is the length of the third side of the triangle? A third 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 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.

1 answer

To find the distance the ant travels around the rectangular sheet of paper, we can use the formula for the perimeter of a rectangle: P = 2(length + width).
Given the dimensions of the paper, length = 358 inches and width = 278 inches.
Plugging these values into the formula, we get:
P = 2(358 + 278)
P = 2(636)
P = 1272 inches
Therefore, the ant travels 1272 inches around the rectangular sheet of paper.

If each dimension of the sheet of paper is increased by one inch, the new dimensions would be length = 359 inches and width = 279 inches.
Using the same formula, we can calculate the new distance the ant travels around the paper:
P = 2(359 + 279)
P = 2(638)
P = 1276 inches
Therefore, the ant would travel 1276 inches in this case.

Let the third side of the triangle be x centimeters.
The perimeter of the triangle is the sum of all three sides, which is given as 20 centimeters.
So, we have the equation:
5110 + 715 + x = 20
5825 + x = 20
x = 20 - 5825
x = -5805 centimeters
Since negative length doesn't make sense, there seems to be an error in the information provided. It's not possible for a triangle with side lengths 5110 and 715 centimeters to have a combined perimeter of 20 centimeters.

Let the dimensions of the rectangular sheet of paper be length = a centimeters and width = b centimeters.
From the problem, we know that the perimeter is between 18 and 19 centimeters, so we can set up the inequality:
2(a + b) > 18 and 2(a + b) < 19
Dividing both sides of the inequalities by 2, we get:
a + b > 9 and a + b < 9.5
Since the dimensions must be expressed in decimal numbers to the hundredth place, and the tenths and hundredths digits are non-zero, we can have a = 4.50 and b = 4.50.
This would give us:
2(4.50 + 4.50) = 2(9) = 18 centimeters
Therefore, one possibility for the dimensions of the sheet of paper is 4.50 centimeters by 4.50 centimeters.