A circle is inscribed in a large square and circumscribed around a small square. If \(A_1\) is the area of the large square and \(A_2\) is the area of the small square, find the ratio \(A_1/A_2\).
Problem Statement
A circle is inscribed in one square and circumscribed around another square as shown below. Let \(A_1\) be the area of the large square and \(A_2\) the area of the small square. Determine the ratio \(A_1/A_2\).
Solution
Let \(x\) be the side length of the small square. Then:
Step 1: Area of small square
\(A_2 = x^2\)
Step 2: Diagonal of small square
Using the Pythagorean theorem:
\[ d^2 = x^2 + x^2 = 2x^2 \] \[ d = x\sqrt{2} \]Step 3: Relate to large square
The diagonal \(d\) of the small square equals the side length of the large square (as seen in the diagram).
Step 4: Area of large square
\[ A_1 = (x\sqrt{2})^2 = 2x^2 \]
Step 5: Calculate the ratio
\[ \displaystyle \frac{A_1}{A_2} = \frac{2x^2}{x^2} = 2 \]
Answer
The ratio of the areas is \(A_1/A_2 = 2\).
Explanation & Verification
Explanation
The circle acts as an intermediary: it touches the large square at the midpoints of its sides (inscribed) and passes through the vertices of the small square (circumscribed). The diagonal of the small square equals the side of the large square because both connect opposite points where the circle touches the squares.
Verification using Radius \(r\)
Let the circle have radius \(r\). Then:
- Large square side = \(2r \implies A_1 = (2r)^2 = 4r^2\)
- Small square diagonal = \(2r \implies \text{side} = r\sqrt{2} \implies A_2 = (r\sqrt{2})^2 = 2r^2\)
- Ratio: \(\frac{A_1}{A_2} = \frac{4r^2}{2r^2} = 2\)
Both methods confirm the result.