A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the re... A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
Understand the Problem
The question is asking for the radius of a circle that shares the center with a square of side length 2, given that the area of regions inside the circle and outside the square is equal to the area of regions outside the circle and inside the square.
Answer
The radius of the circle is \( r = \frac{2}{\sqrt{\pi}} \).
Answer for screen readers
The radius of the circle is ( r = \frac{2}{\sqrt{\pi}} ).
Steps to Solve
- Determine the area of the square
The area of the square can be calculated using the formula for the area of a square:
$$ \text{Area}_{\text{square}} = \text{side length}^2 = 2^2 = 4 $$
- Determine the area of the circle
The area of the circle is given by the formula:
$$ \text{Area}_{\text{circle}} = \pi r^2 $$
- Set up the equation based on the problem's condition
According to the problem, the area of the regions inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. This can be expressed as:
$$ (\text{Area}{\text{circle}} - \text{Area}{\text{square}}) = (\text{Area}{\text{square}} - \text{Area}{\text{circle}}) $$
This simplifies to:
$$ \pi r^2 - 4 = 4 - \pi r^2 $$
- Solve for r
Combine like terms:
$$ \pi r^2 + \pi r^2 = 4 + 4 $$
This can be rewritten as:
$$ 2\pi r^2 = 8 $$
Now, divide both sides by $2\pi$:
$$ r^2 = \frac{8}{2\pi} = \frac{4}{\pi} $$
Taking the square root of both sides gives:
$$ r = \sqrt{\frac{4}{\pi}} = \frac{2}{\sqrt{\pi}} $$
The radius of the circle is ( r = \frac{2}{\sqrt{\pi}} ).
More Information
This result shows that the relationship between the areas constrained by the circle and the square leads to a specific radius for the circle that balances the two areas evenly.
Tips
- Misunderstanding Areas: Sometimes, students confuse the areas inside and outside the shapes. It's crucial to correctly identify which areas are being compared.
- Algebra Errors: When setting up and simplifying equations, watch for errors in sign or combining like terms.
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