Since, m1 and m2 are the roots of the equation x2+(√3+2)x+(√3−1)=0 then m1+m2=−(√3+2), ∴m1−m2=√(m1+m2)2−4m1m2 =√(3+4+4√3−4√3+4) =√11 and coordinates of the vertices of the given triangles are (0,0),(c∕m1,c) and (c∕m2,c). Hence, the required area of triangle =