Question

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 triangle $BCD$ is

• A. $5\sqrt{2}$
• B. $5$
• C. $\frac{\sqrt{5}}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is A $5\sqrt{2}$

Area of $\Delta BCD=\frac{1}{2}|\overrightarrow{BC}\times \overrightarrow{BD}|$

$=\frac{1}{2}\left|\right(b\hat{i}-c\hat{j})\times (b\hat{i}-d\hat{k}\left)\right|$

$=\frac{1}{2}|bd\hat{j}+bc\hat{k}+dc\hat{i}|$

$=\frac{1}{2}\sqrt{b^2{c}^2+c^2{d}^2+d^2{b}^2}$ ...$\left(1\right)$

Now,

$6=bc;8=cd;10=bd$ (Given)

$b^2{c}^2+c^2{d}^2+d^2{b}^2=200$

Substituting the value in $\left(1\right)$, we get
$A=\frac{1}{2}\sqrt{200}=5\sqrt{2}$

Hence, option $A$.