All the vertices of a rectangle lie on a circle of radius $$ R $$. If the perimeter of the rectangle is $$ P $$, then the area of the rectangle is:

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To solve this problem:

1. **Define Dimensions**:  
   Let the sides of the rectangle be $$ a $$ and $$ b $$. The area of the rectangle is given by:  
   $$
   	ext{Area} = a 	imes b
   $$

2. **Perimeter Relation**:  
   The perimeter of the rectangle is:  
   $$
   P = 2(a + b) \implies a + b = \frac{P}{2}.
   $$

3. **Diagonal Relation**:  
   Since all vertices lie on a circle, the diagonal of the rectangle equals the diameter of the circle:  
   $$
   \sqrt{a^2 + b^2} = 2R.
   $$

4. **Square of Sum Relation**:  
   Expand $$(a + b)^2$$ using the perimeter relation:  
   $$
   (a + b)^2 = a^2 + b^2 + 2ab = \left(\frac{P}{2}ight)^2.
   $$

5. **Substitute Diagonal Relation**:  
   Replace $$ a^2 + b^2 $$ with $$ (2R)^2 = 4R^2 $$:  
   $$
   4R^2 + 2ab = \left(\frac{P}{2}ight)^2.
   $$

6. **Solve for Area**:  
   Rearrange to find $$ ab $$ (the area):  
   $$
   ab = \frac{P^2}{8} - 2R^2.
   $$

Thus, the area of the rectangle is **$$ \frac{P^2}{8} - 2R^2 $$**.

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