The courtyard outside East Middle School is in the shape of a rectangle and half-circle.

50m x 100m
What is the approximate area of the courtyard?

Responses
A.8 , 925 m 2
B.6 , 963 m 2
C.5 , 982 m 2
D.5 , 079 m 2

1 answer

To calculate the area of the courtyard, which consists of a rectangle and a half-circle, we need to find the area of each shape and then combine them.

  1. Area of the rectangle: The area \(A\) of a rectangle is given by the formula: \[ A_{rectangle} = \text{length} \times \text{width} \] Here, the length is 100 meters and the width is 50 meters. \[ A_{rectangle} = 100 \times 50 = 5000 \text{ m}^2 \]

  2. Area of the half-circle: The area \(A\) of a full circle is given by the formula: \[ A_{circle} = \pi r^2 \] Since the half-circle is attached to the longer side of the rectangle, the diameter of the half-circle is 50 meters. Therefore, the radius \(r\) is: \[ r = \frac{50}{2} = 25 \text{ m} \] Now, we can calculate the area of the half-circle: \[ A_{half-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (25)^2 = \frac{1}{2} \pi (625) = 312.5 \pi \text{ m}^2 \] Using \( \pi \approx 3.14 \): \[ A_{half-circle} \approx 312.5 \times 3.14 \approx 980.25 \text{ m}^2 \]

  3. Total area of the courtyard: Adding the area of the rectangle and the area of the half-circle together: \[ A_{total} = A_{rectangle} + A_{half-circle} \approx 5000 + 980.25 \approx 5980.25 \text{ m}^2 \]

Now we can round that to the nearest whole number: \[ A_{total} \approx 5980 \text{ m}^2 \]

When matching this to the provided options:

  • A. 8,925 m²
  • B. 6,963 m²
  • C. 5,982 m²
  • D. 5,079 m²

The closest to our calculation of 5980 m² is option C. 5,982 m². Therefore, the approximate area of the courtyard is:

C. 5,982 m²