I have been given the length $L$ and the width $W$ of a rectangle and the radius $R$ of circle which is situated in the center of the rectangle . I need to find the area of the red marked portion. That means the area of the four points where the circle intersects the rectangle. Is there any formula available for this?
On
Consider the area you need to calculate as being two right triangles. I'll rotate the illustration to make it take a bit less space:
(We know that $W^2 + L^2 \gt 4R^2$, and that one of $W$ or $L$ has to be less than $2 R$ for this figure to apply. I've drawn $W \lt L$. Indeed, we do not even need $L$ for anything! We only need the shorter edge length of the outer rectangle, and I'll use $W$ for that.)
Because the blue rectangle and the circle have the same center, the red rectangle (the one whose area we are interested in) is also centered on that same center. Therefore, the diagonal of the red rectangle is as long as the diameter of the circle; twice the radius, $2 R$. You can solve $X$ using the Pythagorean theorem. Note that $X$ can only be positive.
After you've solved for $X$, you can easily calculate the area of the red rectangle using $X$ and $W$.
If you start with the area, and substitute $X$ with the expression you used to solve $X$ (using the Pythagorean theorem), you do actually end up with a simple formula to calculate the area of the red rectangle, given radius $R$ and the length of the shorter side of the rectangle (as $W$).
Hint. The distance from from any corner of the red region to the center is $R$.
Now consider the right triangle which has the diagonal of the red region as hypotenuse. How long is the hypotenuse, and how long are the other sides of this right triangle (you will need the Pythagorean theorem)? Use this to calculate the area of the entire region.
This answer assumes that we have the situation presented in the image where $W<2R$ and $L>2R$.