Let ABCD be a tetrahedron such that the edges AB,AC and AD are  mutually perpendicular. Let the area of triangles ABC,ACD and ADB be 3,4 and 5 sq. units respectively. Then the area of the triangle BCD is :

Area of ABCD=12|BC→XBD→| =12  cj^ -bi∧  dk^-bi∧ =12 |dci∧ + bck∧ + bdj∧  | =12 b2c2 + c2d2 + d2b2                                               X ( Now  6=bc;  8=cd;  10=bd b2c2+c2d2=d2b2=200  Substituting the value in eqn. (i),  A=12200 =  52