Let `A B C D` be a tetrahedron such that the edges `A B ,A Ca n dA D` are mutually perpendicular. Let the area of triangles `A B C ,A C Da n dA D B` be 3, 4 and 5sq. units, respectively. Then the area of triangle `B C D` is `5sqrt(2)` b. `5` c. `(sqrt(5))/2` d. `5/2`

Explanation

`{AB}, {AC}, {AD}` are mutually perpendicular

area of `\triangle {ABC}=|{AB}||{AC}|=6`


area of `\triangle {ACD}=|{AC}||{AD}|=8`


area of `\Delta {ADB}=|{AB}||{AD}|=10`


Then the area of `\triangle {BCD}=\frac{1}{2}|{{BC}} \times {{BD}}|`

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