Question

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

Answer 1

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**