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Four circles, each with a radius of 2 inches, are removed from a square. What is the remaining area of the square?

A. [tex]\((16 - 4 \pi)\)[/tex] in[tex]\(^2\)[/tex]

B. [tex]\((16 - \pi)\)[/tex] in[tex]\(^2\)[/tex]

C. [tex]\((64 - 16 \pi)\)[/tex] in[tex]\(^2\)[/tex]

D. [tex]\((64 - 4 \pi)\)[/tex] in[tex]\(^2\)[/tex]


Sagot :

To find the remaining area of the square after removing four circles each with a radius of 2 inches, we need to follow several steps:

1. Determine the area of one circle:
- The formula for the area of a circle is [tex]\( \pi r^2 \)[/tex], where [tex]\( r \)[/tex] is the radius.
- Here, the radius [tex]\( r \)[/tex] is 2 inches.
- So, the area of one circle is [tex]\( \pi \times (2^2) = 4\pi \)[/tex] square inches.

2. Calculate the total area of four circles:
- Since there are four circles, multiply the area of one circle by 4.
- The total area of four circles is [tex]\( 4 \times 4\pi = 16\pi \)[/tex] square inches.

3. Calculate the side length of the square:
- The four circles are placed perfectly within the square. Each circle has a radius of 2 inches, meaning the diameter is [tex]\( 2 \times 2 = 4 \)[/tex] inches.
- Since four circles fit perfectly in a square, one side of the square must be [tex]\( 2 \times 4 = 8 \)[/tex] inches.

4. Determine the area of the square:
- The formula for the area of a square is [tex]\( \text{side length}^2 \)[/tex].
- Here, the side length of the square is 8 inches.
- So, the area of the square is [tex]\( 8^2 = 64 \)[/tex] square inches.

5. Calculate the remaining area after removing the circles:
- The remaining area is the area of the square minus the total area of the four circles.
- Thus, the remaining area is [tex]\( 64 \)[/tex] square inches (area of the square) minus [tex]\( 16\pi \)[/tex] square inches (total area of the four circles).
- This can be represented as [tex]\( 64 - 16\pi \)[/tex] square inches.

Therefore, the remaining area of the square after removing the four circles is [tex]\( (64 - 16\pi) \)[/tex] square inches.

The correct answer is:
[tex]\[ \boxed{(64-16\pi) \text{ in}^2} \][/tex]