Can you find area of the yellow circle? | (Fun Geometry Problem) |

[@matthieudutriaux]

Great job !
We can't do better.
I just noticed, because PQRS is a circumscribed quadrilateral and a cyclic quadrilateral, that :
area=sqrt((s-a)*(s-b)*(s-c)*(s-d)) with 2*s=a+b+c+d and a+d=b+c
area=1/4*sqrt((-a+b+c+d)*(a-b+c+d)*(a+b-c+d)*(a+b+c-d))
16*area^2=(-a+b+c+d)*(a-b+c+d)*(a+b-c+d)*(a+b+c-d)
16*area^2=(2*d)*(2*c)*(2*b)*(2*a)
16*area^2=16*a*b*c*d
area^2=a*b*c*d
area=sqrt(a*b*c*d)
area^2=a*b*c*(-a+b+c) with a=10*sqrt(3) and b=40*sqrt(3)
(area/(10*sqrt(3))^2)^2=A*B*C*(-A+B+C) with A=a/(10*sqrt(3))=1 and B=b/(10*sqrt(3))=4 and C=c/(10*sqrt(3)) and D=d/(10*sqrt(3))
(1800*sqrt(6)/300)^2=1*4*C*(-1+4+C)
216=4*C*(C+3)
54=C^2+3*C
C^2+3*C-54=0
(C-6)*(C+9)=0
C=6 (because C=-9 is impossible)
D=-A+B+C=-1+4+6=9
area=1/2*(a+b+c+d)*r
area=1/2*(A+B+C+D)*10*sqrt(3)*r
1800*sqrt(6)=1/2*(1+4+6+9)*10*sqrt(3)*r
18*sqrt(2)=r
648=r^2
A(circle)=Pi*r^2=648*Pi m²

[@vcvartak7111]

Problem was too lengthy,but some new concepts are learned