Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular.

Question

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 $	riangle BCD$ is equal to:

Accepted Answer

Let $AB=b,\ AC=c,\ AD=d$ (mutually perpendicular). Then
$$\frac12bc=5\Rightarrow bc=10,\qquad \frac12cd=6\Rightarrow cd=12,\qquad \frac12bd=7\Rightarrow bd=14.$$
With $A$ at the origin and $B,C,D$ along the axes: $\vec{BC}=(-b,c,0)$, $\vec{BD}=(-b,0,d)$.
$$|\vec{BC}\times\vec{BD}|=\sqrt{c^2d^2+b^2d^2+b^2c^2}=\sqrt{12^2+14^2+10^2}=\sqrt{144+196+100}=\sqrt{440}.$$
$$\text{ar}(\triangle BCD)=\frac12\sqrt{440}=\sqrt{110}.$$

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