Question

1. westview school’s track is shown below. determine the area of the field within the track as well as the perimeter of the track. (image of a track with length 60 m and semicircle diameter 30 m)

Explanation:

Step1: Analyze the track shape

The track consists of a rectangle and two semicircles (which form a full circle). The length of the rectangle is \( 60 \, \text{m} \), and the diameter of the semicircles (and thus the width of the rectangle) is \( 30 \, \text{m} \), so the radius \( r \) of the circle is \( \frac{30}{2} = 15 \, \text{m} \).

Step2: Calculate the area of the rectangle

The area of a rectangle is given by \( A_{\text{rectangle}} = \text{length} \times \text{width} \). Here, length \( = 60 \, \text{m} \) and width \( = 30 \, \text{m} \), so \( A_{\text{rectangle}} = 60 \times 30 = 1800 \, \text{m}^2 \).

Step3: Calculate the area of the circle

The area of a circle is given by \( A_{\text{circle}} = \pi r^2 \). With \( r = 15 \, \text{m} \), we have \( A_{\text{circle}} = \pi \times 15^2 = 225\pi \, \text{m}^2 \approx 225 \times 3.14 = 706.5 \, \text{m}^2 \).

Step4: Calculate the total area of the track

The total area is the sum of the area of the rectangle and the area of the circle: \( A_{\text{total}} = A_{\text{rectangle}} + A_{\text{circle}} = 1800 + 706.5 = 2506.5 \, \text{m}^2 \).

(For the perimeter, the perimeter of the track consists of the two lengths of the rectangle and the circumference of the circle. The circumference of the circle is \( C = \pi d = 30\pi \approx 94.2 \, \text{m} \), and the two lengths of the rectangle are \( 2 \times 60 = 120 \, \text{m} \). So the perimeter \( P = 120 + 94.2 = 214.2 \, \text{m} \). But since the question mentions "area" (and maybe a typo for perimeter? But we'll go with area as per the initial calculation for area, or perimeter if needed. Wait, the original question: "Determine the area of the field within the track as well as the perimeter of the track." So we did area, now perimeter:

Wait, let's recheck. The track's perimeter: the two straight sides (each 60m) and the curved part (circumference of the circle with diameter 30m). So perimeter \( P = 2 \times 60 + \pi \times 30 = 120 + 30\pi \approx 120 + 94.2 = 214.2 \, \text{m} \).

But the initial problem says "Determine the area of the field within the track as well as the perimeter of the track." So we have two parts:

Area: \( 1800 + 225\pi \approx 2506.5 \, \text{m}^2 \)

Perimeter: \( 120 + 30\pi \approx 214.2 \, \text{m} \)

Assuming the question might have wanted perimeter (maybe a translation error), but let's confirm the shape. The track has two semicircles (top and bottom) which make a full circle, and two straight sections (each 60m). So perimeter is \( 2 \times 60 + \pi \times 30 \).

Answer:

• Area of the field within the track: \( \boldsymbol{2506.5 \, \text{square meters}} \) (or \( 1800 + 225\pi \, \text{m}^2 \))
• Perimeter of the track: \( \boldsymbol{214.2 \, \text{meters}} \) (or \( 120 + 30\pi \, \text{m} \))