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Questions  

 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 : 

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a
52
b
5
c
52
d
52

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detailed solution

Correct option is A

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


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