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The diagram shows a track composed of a rectangle with a semicircle on each end. The area of the rectangle is 6 comma 600 squar...Asked by fernando
The diagram shows a track composed of a rectangle with a semicircle on each end. The area of the rectangle is 2 comma 800 square meters. What is the perimeter of the track? Use 3.14 for pi.
70 m
Question content area bottom
Part 1
The perimeter of the track is
enter your response here
▼
(Round to the nearest tenth as needed.)
70 m
Question content area bottom
Part 1
The perimeter of the track is
enter your response here
▼
(Round to the nearest tenth as needed.)
Answers
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Answered by
GPT-4o mini
To find the perimeter of the track, which consists of a rectangle and two semicircles at each end, we will follow these steps:
1. **Identify the dimensions of the rectangle:**
- Let's denote the length of the rectangle as \( L \) and the width (diameter of the semicircles) as \( W \).
- The area of the rectangle is given by:
\[
A = L \times W = 2800 \text{ m}^2
\]
2. **Calculate the perimeter of the track:**
- The perimeter consists of the lengths of the two long sides of the rectangle and the circumference of the two semicircles.
- The circumference of a full circle is \( C = \pi D \), where \( D \) is the diameter. Since we have two semicircles, the total contribution from the semicircles will be the same as the circumference of a full circle with diameter \( W \):
\[
\text{Circumference of two semicircles} = \pi W
\]
- Thus, the total perimeter \( P \) is given by:
\[
P = 2L + \pi W
\]
3. **Determine specific values:**
- Since no specific dimensions are provided for \( W \) or \( L \), let’s first express one variable in terms of the other from the area equation. If we rearrange:
\[
L = \frac{2800}{W}
\]
- Substitute \( L \) into the perimeter equation:
\[
P = 2 \left( \frac{2800}{W} \right) + \pi W
\]
- Using \( \pi \approx 3.14 \):
\[
P = \frac{5600}{W} + 3.14 W
\]
4. **Finding dimensions to calculate the perimeter:**
- If you need a specific perimeter, you would need specific dimensions \( L \) and \( W \). However, if we assume likely values based on common rectangular dimensions providing 2800 m² area, such as \( W = 40 \, \text{m} \):
- Then \( L = \frac{2800}{40} = 70 \, \text{m} \)
- Substitute into the perimeter formula:
\[
P = 2(70) + 3.14(40)
\]
\[
P = 140 + 125.6 = 265.6
\]
Thus, the perimeter of the track is approximately **265.6 meters**.
If you have specific dimensions or a different width/length ratio in mind, please provide them for a more accurate calculation.
1. **Identify the dimensions of the rectangle:**
- Let's denote the length of the rectangle as \( L \) and the width (diameter of the semicircles) as \( W \).
- The area of the rectangle is given by:
\[
A = L \times W = 2800 \text{ m}^2
\]
2. **Calculate the perimeter of the track:**
- The perimeter consists of the lengths of the two long sides of the rectangle and the circumference of the two semicircles.
- The circumference of a full circle is \( C = \pi D \), where \( D \) is the diameter. Since we have two semicircles, the total contribution from the semicircles will be the same as the circumference of a full circle with diameter \( W \):
\[
\text{Circumference of two semicircles} = \pi W
\]
- Thus, the total perimeter \( P \) is given by:
\[
P = 2L + \pi W
\]
3. **Determine specific values:**
- Since no specific dimensions are provided for \( W \) or \( L \), let’s first express one variable in terms of the other from the area equation. If we rearrange:
\[
L = \frac{2800}{W}
\]
- Substitute \( L \) into the perimeter equation:
\[
P = 2 \left( \frac{2800}{W} \right) + \pi W
\]
- Using \( \pi \approx 3.14 \):
\[
P = \frac{5600}{W} + 3.14 W
\]
4. **Finding dimensions to calculate the perimeter:**
- If you need a specific perimeter, you would need specific dimensions \( L \) and \( W \). However, if we assume likely values based on common rectangular dimensions providing 2800 m² area, such as \( W = 40 \, \text{m} \):
- Then \( L = \frac{2800}{40} = 70 \, \text{m} \)
- Substitute into the perimeter formula:
\[
P = 2(70) + 3.14(40)
\]
\[
P = 140 + 125.6 = 265.6
\]
Thus, the perimeter of the track is approximately **265.6 meters**.
If you have specific dimensions or a different width/length ratio in mind, please provide them for a more accurate calculation.
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