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

[@matthieudutriaux]

Great job ! We can't do better. I just did like you : * PQRS is a cyclic quadrilateral (inscribed quadrilateral) : Brahmagupta's formula : area=sqrt((s-a)*(s-b)*(s-c)*(s-d)) with s=(a+b+c+d)/2 * PQRS is a tangent(ial) quadrilateral : s=(a+b+c+d)/2=a+d=b+c Then : (s-a)=d ; (s-b)=c ; (s-c)=b ; (s-d)=a area=sqrt(a*b*c*d) area^2=a*b*c*d=a*b*c*(-a+b+c) with a=10*sqrt(3) and b=40*sqrt(3) area^2/(10*sqrt(3))^4=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))^2/(300^2)=1*4*C*(-1+4+C) (6*sqrt(6))^2=4*C*(C+3) 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/(10*sqrt(3))=1/2*(A+B+C+D)*r 1800*sqrt(6)/(10*sqrt(3))=1/2*(1+4+6+9)*r 180*sqrt(2)=10*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

[@jasongoldman3850]

Where are you getting 2x and 2y from?

[@MathandEngineering]

We assumed the lengths of PS to be 2x and RS = 2y

[@jasongoldman3850]

@MathandEngineering But why?