$\text{ Let }ABCD\text{ be a tetrahedron such that the edges }AB,AC\text{ and }AD\text{ are }$

$\text{ mutually perpendicular. Let the area of triangles }\mathrm{ABC},\mathrm{ACD}\text{ and }\mathrm{ADB}\text{ be }3,4\text{ and }$

$5\text{ sq. units respectively. Then the area of the triangle }\mathrm{BCD}\text{ is : }$

$\begin{array}{rcl}\text{ Area of ∆}\mathrm{BCD}& =& \frac{1}{2}\left|\overrightarrow{BC}X\overrightarrow{BD}\right|\\ & =& \frac{1}{2}\left|\left(c\hat{j} -b\stackrel{\wedge }{i}\right) \left(d\hat{k}-b\stackrel{\wedge }{i}\right)\right|\\ & =& \frac{1}{2}|dc\stackrel{\wedge }{i}+bc\stackrel{\wedge }{k}+bd\stackrel{\wedge }{j}|\\ & =& \frac{1}{2}\sqrt{b^2{c}^2+c^2{d}^2+d^2{b}^2}\\ & & \stackrel{ }{ }\end{array}$

$$\begin{gathered} \text{ Now } 6=bc; 8=cd; 10=bd \\ \Rightarrow b^2{c}^2+c^2{d}^2+d^2{b}^2=200 \\ \text{ Substituting the value in eqn. (i), } \\ A=\frac{1}{2}\sqrt{200}=5\sqrt{2} \end{gathered}$$