This is a very hard problem, lets see if ANYONE (Even Moderators) Can try to figure this out.

Both Circles are the same size, the box is made of perfect 90° angles,

Here is an update, left something out

I think there is still something missing because you have not told us how big the overlap is.

I think to get a number answer we would need that, or something that would allow us to find it..

I suppose I can assume there is a 1.5inch border around the circles which would mean that the circles are tight inside a rectangle  12 inches high and 17inches long 

Hi ProMagma,

I have had to make some assumptions here. They are stated in my last post.

I have drawn this to scale.

$cos\theta=\frac{2.5}{6}=\frac{5}{12}\\ \theta \approx 65.376^\circ\\ \therefore \angle AOB\approx 130.751^\circ$

$Area\;\;of\;\; \triangle AOB\approx 0.5*6*6*sin130.751^\circ\approx 13.636 u^2\\ Area\;\;of\;\; sector \;\;AOB\approx \frac{130.751}{360}\pi *6^2 \approx 41.077u^2\\ Area\;\;of\;\; minor \;\; segment \;\;AB\approx 41.077-13.636 = 27.441u^2$

This is the same as the Red region inside the intersection of the circles because both are half of the total intersection.

Now the area of the union of the 2 circles is   $2\pi r^2 - 2*27.441 \approx 226.195 - 54.882\approx 171.313$

Half of this area (because the rectangle is split in half) is $\approx 85.857$

The area of the whole rectangle is  $20*15=300 inch^2$

The are of half the rectangle is  $150inch^2$

So

$Original\;red\;shaded\:area\;\;\approx  150 - 85.857 + 27.441 \approx 91.584 \;inch^2$

I think that is right but my logic, and arithemetic does need to be double checked.

This appears to be a variation of the problem posed here: "Mind Your Decisions- Two Circles within a Rectangle.", on YouTube.

Sorry Melody: I did forget to give the link. Here it is:

https://www.youtube.com/watch?v=xnE_sO7PbBs&list=PLDZcGqoKA84E2a0L6IS68hswD4iiUN2Cv