Question

The diagram shows a track composed of a rectangle with a semicircle on each end. The area of the rectangle is 6 comma 600 square meters. What is the perimeter of the​ track? Use 3.14 for pi.
110 m
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Part 1
The perimeter of the track is
  
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square meters.
meters.

Answer 1

To find the perimeter of the track, we need to calculate both the rectangular and semicircular parts of the track.

1. **Identify the dimensions of the rectangle**:
The area of the rectangle is given as \(6600\) square meters. Let's denote the length of the rectangle as \(L\) and the width as \(W\). The area is calculated as:

\[
\text{Area} = L \times W = 6600 \quad \text{(1)}
\]

2. **Determine the semicircular parts**:
The width of the rectangle will also be the diameter of the semicircles on each end of the track, hence \(W\) is the diameter. The radius \(r\) of the semicircles will be:

\[
r = \frac{W}{2}
\]

The perimeter contributes the lengths of the two semicircles, which can be calculated as the circumference of a full circle with radius \(r\) divided by 2. The circumference \(C\) of a full circle is given by:

\[
C = 2\pi r
\]

Therefore, the combined length of the two semicircles is:

\[
\text{Length of two semicircles} = \pi r = \pi \left(\frac{W}{2}\right) = \frac{\pi W}{2} \quad \text{(2)}
\]

3. **Calculate the perimeter of the track**:
The total perimeter \(P\) of the track is then:

\[
P = \text{Length of rectangle} + \text{Length of two semicircles}
\]

\[
P = L + \frac{\pi W}{2} + L
\]

\[
P = 2L + \frac{\pi W}{2}
\]

4. **Calculate \(L\) and \(W\)**:
To move forward, we need to assume values for \(W\) since there are many combinations possible. Without loss of generality, let's assume \(W = 60\) m, then from equation (1):

\[
L \times 60 = 6600 \implies L = \frac{6600}{60} = 110 \text{ m}
\]

5. **Plug in the values into the perimeter formula**:
Since we have found \(L\) and \(W\):

- \(L = 110\) m
- \(W = 60\) m

Now, putting these values into the formula for perimeter:

\[
P = 2(110) + \frac{3.14 \times 60}{2}
\]

First, calculate the semicircle part:

\[
= \frac{3.14 \times 60}{2} = 94.2
\]

Thus, the perimeter is:

\[
P = 220 + 94.2 = 314.2
\]

Therefore, rounding to the nearest meter, the perimeter of the track is:

\[
\boxed{314 \text{ meters}}
\]