A geometry problem

In the above image if

 \[r=20\]

Find the Area of 

\(L+G\)

:Solution

First i will write the equations of the three circles

\[A- \ x^{2}+y^{2}=400 \\B- \ (x-10)^{2}+y^2=100 \\ C- x^2+(y-10)^2=100\]

Now the area

\[L=G\]

?how

 \[G=\ the \ area \ of \ quarter \ ABC -\ the\ area \ of \ the\ semi \ circle \ BC \ - \ the\ semi\ circle \ AC+L\]

we add \(L\) because we substract it twice when we substract the two semicircles

\[G=\pi \frac{20^{2}}{4}-\pi \frac{10^{2}}{2}-\pi \frac{10^{2}}{2}+L\]

do the calculation

 \[=> G=L\]

now we need to find the points of intersection from eq.\(B\) and eq. \(C\)

\[x^{2}-20x+100+y^{2}=x^{2}+y^{2}-20y+100\]

then  \(\boxed{x=y}\)     substitute in eq.  \(B\) we get

 \[y=0 \  \ x=0 \ or \boxed{y=10 \ ,  x=10}\]


the points of intersection leads us to drow the square with side length  \(10\) as shown in the photo

 now the area of
  \(G=\ the \ area \ of \ quarter \ ABC -\ the\ area \ of \ the\ square \\ -2 \ the\ area \ of \ the \ quarter \ of \ circle \ with \ r=10\) 


 \(G=\pi \frac{20^{2}}{4}-10^{2}-2\pi\frac{10^{2}}{4}\)
 \(50(\pi-2)\)=
,Finally 
 \(L+G=2*50(\pi-2)=\boxed{100(\pi-2)}\)