Squares

$Let\quad AB=r\quad and\quad EF=s\quad (so,\quad R=r^{ 2 }\quad and\quad S=s^{ 2 }).\quad Also,\quad \\ let\quad H\quad be\quad the\quad point\quad obtained\quad by\quad the\quad interception\quad of\quad \\ the\quad extention\quad of\quad the\quad segments\quad AB\quad and\quad FG\quad (clearly,\quad \\ we\quad have\quad two\quad new\quad triangles,\quad AHG\quad and\quad BHF).\quad Then:\quad \\ \\ Area_{ gray }=Area_{ AHG }-Area_{ BHF },\quad where:\quad \\ Area_{ AHG }=\frac { (r+s)\times r }{ 2 } ,\quad and\quad \\ Area_{ BHF }=\frac { (r-s)\times s }{ 2 } ;\quad \\ \\ Thus,\quad Area_{ gray }=\frac { (r+s)\times r }{ 2 } -\frac { (r-s)\times s }{ 2 } =\boxed{\frac { R+S }{ 2 } }$

$S_{grey}=(r^2+s^2)+\frac{s(r-s)}{2}-\frac{r(r+r)}{2}=\frac{r^2+s^2}{2}=\boxed{\frac{S+R}{2}}$

Add a new point H above F so that BHFE is a rectangle, and the area of triangle EFB is half of that rectangle. So the area of the shaded region is (R+S+BHFE)/2 - BHFE/2. That is just (R+S)/2.

Area of triangle ABG = 1/2 area of square ABCD = R/2

Area of triangle BFG = 1/2 area of square EFGC = S/2

By adding

Areaa of the gray region = (R + S)/2

I added an imaginary point "H".

Area of rectangle ADGH: (sqrt(R)+sqrt(S))(sqrt(R)) = R+sqrt(RS)

Area of triangle ADG: 0.5(sqrt(R)+sqrt(S))(sqrt(R)) = 0.5R+0.5sqrt(RS)

Area of triangle BEF: 0.5((sqrt(R)-sqrt(S))(sqrt(S)) = 0.5sqrt(RS)-0.5S

The area of triangle BEF is congruent to the area of triangle BHF.

Find the area of quadrilateral ABFG: (R+sqrt(RS)) - (0.5R+0.5sqrt(RS)) - (0.5sqrt(RS)-0.5S) = 0.5R+0.5S = (R+S)/2

Area of ABFG= Area of ADG - Area of BEF. let AD=a & FG=b. now area of ADG=(a (a+b)/2). area of BEF=((b (a-b)/2). so area of ABFG=(a^2+a b-b a+b^2)/2=(a^2+b^2)/2 a^2=R,b^2=S. so area of ABFG=(R+S)/2