Sectors in a Square

Let the radii be $r_1$ and $r_2$ . Then $r_1+r_2 = 4\sqrt{2}$ .

Then evaluating the total perimeter, we have: $\frac{\pi}{2} (r_1+r_2)+2(r_1+r_2) = 4\sqrt{2}(\frac{\pi}{2} + 2) = \sqrt{2}(2\pi+8)$ which gives the answer $\boxed{12}$ as required.

diagonal of the square, k= sqrt(4^2+4^2)=4*sqrt(2).

Radius of a single sector, r=k/2=2*sqrt(2)

perimeter of a single sector,p =2 pi r/4 + 2r =2 pi 2 sqrt(2)/4+2 2 sqrt(2) =pi sqrt(2)+4sqrt(2)

total perimeters of the sectors=p 2= =2 pi sqrt(2)+8sqrt(2) =sqrt(2) (2 pi+8)=sqrt(c) (a*pi+b)

now simplifying we get the values of a=2,b=8,c=2.

a+b+c=2+8+2=12

I assumed that these sectors are same, then I calculated the diagonal of the square and divided it by two. That's how I came to know the radius of the circle they are part of. Then I calculated the one fourth of the circumference of the circle they are part of. Then I added twice it's radius to get the perimeter of one sector,I multiplied it with 2 to get perimeter of both the sectors which gave me a,b,c.

we know, S=rθ . Here, r=2√2, θ=π/2, so S=π√2, Now total area of one shaded area is 2√2+2√2+π√2. So Finally total area of both shaded area is 2(2√2+2√2+π√2) So here the calculation: 2(4√2+π√2) =2√2(4+π)=√2(2π+8) which is resemble to the expressed form. so the answer is 12

lol just by luck

same concept. taking x & y as radii of two quadrants & x+y=4*2^(1/2).