A square is circumscribed about a circle which is turn circumscribes another square. The ratio of the area of outer square to the inner square

Options

A. `1 , 2`
B. `2 : 1`
C. `1 : 3 `
D. `1 : 1`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Configuration
We have a square (let's call it Square ABCD) that is circumscribed about a circle. The circle is inscribed within this square. Furthermore, this circle also circumscribes another square (let's call it Square MNOP) inside it.

### Step 2: Define Variables
Let the side length of the outer square (Square ABCD) be \( s \). The diameter of the circle inscribed in this square will be equal to the side length of the square, so:
\[ \text{Diameter of circle} = s \]

### Step 3: Find the Radius of the Circle
The radius \( r \) of the circle is half of the diameter:
\[ r = \frac{s}{2} \]

### Step 4: Relate the Inner Square to the Circle
The inner square (Square MNOP) is inscribed in the circle. The diagonal of the inner square is equal to the diameter of the circle. If we denote the side length of the inner square as \( a \), then using the properties of squares, we know:
\[ \text{Diagonal of inner square} = a\sqrt{2} \]
Setting this equal to the diameter of the circle, we have:
\[ a\sqrt{2} = s \]

### Step 5: Solve for the Side Length of the Inner Square
From the equation above, we can solve for \( a \):
\[ a = \frac{s}{\sqrt{2}} = \frac{s\sqrt{2}}{2} \]

### Step 6: Calculate the Areas of the Squares
Now we can calculate the areas of both squares:
- Area of the outer square:
\[ \text{Area of outer square} = s^2 \]
- Area of the inner square:
\[ \text{Area of inner square} = a^2 = \left(\frac{s}{\sqrt{2}}\right)^2 = \frac{s^2}{2} \]

### Step 7: Find the Ratio of the Areas
Now, we find the ratio of the area of the outer square to the area of the inner square:
\[ \text{Ratio} = \frac{\text{Area of outer square}}{\text{Area of inner square}} = \frac{s^2}{\frac{s^2}{2}} = \frac{s^2 \cdot 2}{s^2} = 2 \]

### Conclusion
Thus, the ratio of the area of the outer square to the inner square is:
\[ \text{Ratio} = 2 \]

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