#1**+3 **

We can use some symmetry here....this pic might help :

Locate the center of the top circle and draw two radial lines to each end of the shaded figure in ths circle

Let the center of the circle be A....let one of the radial lines intersect the "bottom" of the shaded area in this circle at B and Let the other radial line intersect the other end of the shaded area in this circle at C

Connect BC....and this will divide the shaded area in two equal parts

Now we can take the circle sector ABC as the quarter area of this circle.....its area will be (1/4)pi(4^2) = 4pi (1)

And we can take the area of triangle ABC as (1/2) (4^2) sin(90°) = 8 (2)

Now the difference in (1) - (2) will give us half of this shaded area =

(4pi - 8) units ^2 = 4 (pi - 2) units^2

So...the whole shaded area in one circle is twice this = 2 * 4 (pi - 2) units^2 = 8(pi - 2) units^2

And since we have 4 of these areas - one in each circle - the total of the shaded areas is just

4* 8(pi - 2) units^2 = 32 (pi - 2) units^2

CPhill Apr 2, 2018