A square is inscribed in a quarter-circle in such a manner that two of its adjacent vertices lie on the two radii at an equal distance from the centre, while the other two vertices lie on the circular arc. If the square has sides of length x, then the radius of the circle is

Options

A. `sqrt(2)x`
B. `(16x)/(pi + 4)`
C. `(2x)/(sqrtpi)`
D. `(sqr(5)x)/(sqrt2)`

Answer

To solve the problem of finding the radius of a quarter-circle in which a square is inscribed, we can follow these steps:

### Step 1: Understand the Geometry
We have a quarter-circle with a square inscribed in it. The square has vertices A, B, C, and D. Points A and D lie on the two radii of the quarter-circle, and points B and C lie on the circular arc.

### Step 2: Define Variables
Let the side length of the square be \( x \). The distance from the center of the quarter-circle (point O) to points A and D is equal, and we denote this distance as \( AO = OD = d \).

### Step 3: Set Up the Right Triangle
In the right triangle AOD, we can apply the Pythagorean theorem:
\[ AD^2 = AO^2 + OD^2 \]
Since \( AD = x \) (the side of the square), we have:
\[ x^2 = d^2 + d^2 = 2d^2 \]
Thus, we can express \( d \) in terms of \( x \):
\[ d^2 = \frac{x^2}{2} \implies d = \frac{x}{\sqrt{2}} \]

### Step 4: Analyze Triangle BCD
Next, we look at triangle BCD. Since B and C lie on the circular arc, we can also apply the Pythagorean theorem here:
\[ BC^2 + CD^2 = BD^2 \]
Since \( BC = x \) and \( CD = x \):
\[ x^2 + x^2 = BD^2 \implies 2x^2 = BD^2 \implies BD = x\sqrt{2} \]

### Step 5: Set Up the Right Triangle OBD
Now, we analyze triangle OBD. We can again apply the Pythagorean theorem:
\[ OB^2 = OD^2 + BD^2 \]
Substituting the known lengths:
\[ OB^2 = d^2 + (x\sqrt{2})^2 \]
Substituting \( d = \frac{x}{\sqrt{2}} \):
\[ OB^2 = \left(\frac{x}{\sqrt{2}}\right)^2 + (x\sqrt{2})^2 \]
Calculating each term:
\[ OB^2 = \frac{x^2}{2} + 2x^2 = \frac{x^2}{2} + \frac{4x^2}{2} = \frac{5x^2}{2} \]
Thus,
\[ OB = \sqrt{\frac{5x^2}{2}} = \frac{x\sqrt{5}}{\sqrt{2}} \]

### Step 6: Conclusion
Since \( OB \) represents the radius \( r \) of the quarter-circle, we have:
\[ r = \frac{x\sqrt{5}}{\sqrt{2}} \]

### Final Answer
The radius of the circle is:
\[ \boxed{\frac{\sqrt{5}x}{\sqrt{2}}} \]