Let ABCD be a tetrahedron such that the edges | vedclass

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(C) Let the lengths of the edges $AB, AC$ and $AD$ be $c, b$ and $d$ respectively. Since the edges are mutually perpendicular,the areas of the triangl

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$\sqrt{110}$

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(C) Let the lengths of the edges $AB, AC$ and $AD$ be $c, b$ and $d$ respectively. Since the edges are mutually perpendicular,the areas of the triangles are given by: $Ar(	riangle ABC) = \frac{1}{2} bc = 5 \implies bc = 10$ $Ar(	riangle ACD) = \frac{1}{2} bd = 6 \implies bd = 12$ $Ar(	riangle ADB) = \frac{1}{2} cd = 7 \implies cd = 14$ By the projection theorem for a tetrahedron with a right-angled corner at $A$,the area of the face opposite to the right-angled corner is given by: $Ar(	riangle BCD) = \sqrt{(Ar(	riangle ABC))^2 + (Ar(	riangle ACD))^2 + (Ar(	riangle ADB))^2}$ $Ar(	riangle BCD) = \sqrt{5^2 + 6^2 + 7^2}$ $Ar(	riangle BCD) = \sqrt{25 + 36 + 49}$ $Ar(	riangle BCD) = \sqrt{110}$

Let $ABCD$ be a tetrahedron such that the edges $AB, AC$ and $AD$ are mutually perpendicular. Let the areas of the triangles $ABC, ACD$ and $ADB$ be $5, 6$ and $7$ square units respectively. Then the area (in square units) of the $\triangle BCD$ is equal to :

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