It might help to assemble four quarter circles into a full one:

Let's say the doubly shaded right triangles have shorter sides $x>y$.

The outer square's side length is $2x+12$, area $4x^2+48x+144$.

We have to subtract the four white rectangles with area $1+x$ each.

Further add the doubly shaded triangle areas $2xy$ (they count twice).

Then the answer is obviously a fourth of that.

That's how far I got, but I couldn't get $x$ and $y$.

Hum ... this one looks hard.

The answer is 171.21.

Actual numbers are in the notes.

Main steps:

• Realize that one can draw a vertical line from the top to the tangent point, due to it being a quarter circle.
• Set the origin at the tangent point
• Use point (5,1) = G being part of the circle to obtain the radius of the circle R by using

• With R = 13, we can use Pythagoras to find that the length of AB = 12.
• We can find the angle t using the 5-12-13 triangle denoted by ABC corners.
• Use the $\cos (t)=12/13$ to find that $\sin (90-t)=12/13=a/12$, hence $a=12$.
• Apply Pythagoras on triangle ADE to find that $13-b=5$ and thus $b=8$.
• Use the ratios between sides of similar triangles ABC and FHG to find that the length of FG equals $91/12$.
• You have all the information to find the area of the big rectangle (136) and the remaining triangle forming the rest of the shaded region ($\sim 35.21$).
• Add up the two values to find the final answer.