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1. 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 3 5/8 inches by 2 7/8 inches. How many inches does the ant travel? Show your work
2. 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.
3. A second ant walks all the way around a triangle drawn in the dirt. Two of the dimensions of the triangle are 5 1/10 centimeters and 7 1/5 centimeters. the total distance traveled by the ant is 20 centimeters. What is the length of the third side of the triangle?
4. 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. write as a 7th grader

1 answer

Sure! Let's break down each question step by step.

Question 1

To find out how far the ant travels around the perimeter of the rectangular sheet of paper with dimensions of \(3 \frac{5}{8}\) inches by \(2 \frac{7}{8}\) inches, we first need to calculate the perimeter.

  1. Convert Mixed Numbers to Improper Fractions:

    • For \(3 \frac{5}{8}\): \[ 3 \frac{5}{8} = \frac{3 \times 8 + 5}{8} = \frac{24 + 5}{8} = \frac{29}{8} \]
    • For \(2 \frac{7}{8}\): \[ 2 \frac{7}{8} = \frac{2 \times 8 + 7}{8} = \frac{16 + 7}{8} = \frac{23}{8} \]

  2. Calculate the Perimeter: The formula for the perimeter \(P\) of a rectangle is: \[ P = 2 \times (length + width) \] So we plug in the values: \[ P = 2 \times \left(\frac{29}{8} + \frac{23}{8}\right) \] First, add the fractions: \[ \frac{29}{8} + \frac{23}{8} = \frac{29 + 23}{8} = \frac{52}{8} = \frac{13}{2} \text{ (by simplifying)} \] Now multiply by 2: \[ P = 2 \times \frac{13}{2} = 13 \text{ inches} \]

So, the ant travels 13 inches.

Question 2

Now, if each dimension is increased by 1 inch, the new dimensions become:

  • New length: \[ 3 \frac{5}{8} + 1 = 4 \frac{5}{8} = \frac{37}{8} \text{ (as calculated earlier)} \]

  • New width: \[ 2 \frac{7}{8} + 1 = 3 \frac{7}{8} = \frac{31}{8} \]

Now, calculate the new perimeter: \[ P = 2 \times \left( \frac{37}{8} + \frac{31}{8} \right) \] First, we add the new dimensions: \[ \frac{37}{8} + \frac{31}{8} = \frac{68}{8} = \frac{17}{2} \] So we multiply by 2: \[ P = 2 \times \frac{17}{2} = 17 \text{ inches} \]

Thus, the ant will now travel 17 inches.

Question 3

For the triangle where two sides are \(5 \frac{1}{10}\) centimeters and \(7 \frac{1}{5}\) centimeters, and the total distance around the triangle is 20 centimeters, we need to find the third side.

  1. Convert Mixed Numbers to Improper Fractions:

    • For \(5 \frac{1}{10}\): \[ 5 \frac{1}{10} = \frac{51}{10} \]
    • For \(7 \frac{1}{5}\): \[ 7 \frac{1}{5} = \frac{36}{5} = \frac{72}{10} \text{ (by converting to a common denominator)} \]

  2. Add the two sides: \[ \frac{51}{10} + \frac{72}{10} = \frac{123}{10} \]

  3. Subtract from the total distance: The total perimeter is 20 cm, which is \(\frac{200}{10}\). Now we can find the third side: \[ \text{Third side} = \frac{200}{10} - \frac{123}{10} = \frac{77}{10} = 7.7 \text{ centimeters} \]

So, the length of the third side is 7.7 centimeters.

Question 4

We know the ant travels between 18 and 19 centimeters around a rectangular sheet of paper. To find dimensions that satisfy this, we can apply the perimeter formula again:

Let the length be \(L\) and width be \(W\). We know: \[ 18 < 2(L + W) < 19 \]

Dividing the whole inequality by 2 gives: \[ 9 < L + W < 9.5 \]

Now, let's choose an example. Assume \(L = 5\) and solve for \(W\): \[ 5 + W > 9 \implies W > 4 \] \[ 5 + W < 9.5 \implies W < 4.5 \]

Choosing \(W = 4.25\): So the dimensions can be:

  • Length: \(5 \text{ cm}\)
  • Width: \(4.25 \text{ cm}\)

Total perimeter check: \[ 2(5 + 4.25) = 2 \times 9.25 = 18.5 \text{ centimeters} \]

This is between 18 and 19.

So one possibility for the dimensions of the rectangular sheet of paper is \(5 \text{ cm} \times 4.25 \text{ cm}\).