Question

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

What is the approximate area of the courtyard? (100m and 50 m)

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

Answer 1

To find the area of the courtyard that is in the shape of a rectangle and a half-circle, we can break it down into two parts: the rectangular area and the area of the half-circle.

1. **Calculate the area of the rectangle:**
The formula for the area of a rectangle is:
\[
\text{Area of rectangle} = \text{length} \times \text{width}
\]
Given dimensions:
- Length = 100 m
- Width = 50 m

\[
\text{Area of rectangle} = 100 \, \text{m} \times 50 \, \text{m} = 5000 \, \text{m}^2
\]

2. **Calculate the area of the half-circle:**
The formula for the area of a circle is:
\[
\text{Area of circle} = \pi r^2
\]
Since we have a half-circle, we need to divide this area by 2. The radius of the half-circle is half the width of the rectangle, which is:
\[
r = \frac{50 \, \text{m}}{2} = 25 \, \text{m}
\]

So, the area of the half-circle is:
\[
\text{Area of half-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (25 \, \text{m})^2
\]
\[
= \frac{1}{2} \pi (625 \, \text{m}^2) \approx \frac{1}{2} \times 3.14 \times 625 \approx 981.25 \, \text{m}^2
\]

3. **Total area of the courtyard:**
Now, we add both areas together:
\[
\text{Total Area} = \text{Area of rectangle} + \text{Area of half-circle}
\]
\[
\text{Total Area} = 5000 \, \text{m}^2 + 981.25 \, \text{m}^2 \approx 5981.25 \, \text{m}^2
\]

Rounding this value gives approximately \( 5982 \, \text{m}^2 \).

Based on the options provided, the closest answer would be:
**C. 5,982 m²**