the area of the triangle whose vertices are the incentre,centroid, and circumcentre of a triangle with sides 14,50,48 is??

Dear Student,

we know, A=1/4 sqrt((a + b - c) (a - b + c) (-a + b + c) (a + b + c))
Substituting a=14, b=50, c=48:
A=1/4 sqrt((14 + 50 - 48) (14 - 50 + 48) (-14 + 50 + 48) (14 + 50 + 48))
=>A=(sqrt((14 + 50 - 48) (14 - 50 + 48) (-14 + 50 + 48) 112))/(4)
=>A=(sqrt((14 + 50 - 48) (14 - 50 + 48) 84×112))/(4)
14 - 50 + 48=12:
=>A=(sqrt((14 + 50 - 48) 12×84×112))/(4)
14 + 50 - 48=16:
=>A=(sqrt(16×12×84×112))/(4)16×12 =192
=>sqrt(1806336) = sqrt(2^12×21^2) = 2^6×21:
=>A=(2^6×21)/4
=>A=((2×4)^2 21)/4
A=16×21
16×21 = 336 (ans)