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 the triangle $BCD$, is

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

Answer 1

Answer: (A)

$5\sqrt{2}$

Place $A,B,C,$ and $D$ at $(0,0,0),(b,0,0),(0,c,0),$ and $(0,0,d)$ in Cartesian coordinate space, with $b,c,$and $d$ positive.

Given that $bc/2=3,cd/2=4$ and $bd/2=5$, as the area of triangles are given in the question.

Area of triangle $BCD$, will be half of the cross product of $\overrightarrow{BC}$ and $\overrightarrow{BD}$.

This cross product is $(-cd,-db,-bc)$, which has the length $=2\sqrt{3^2+4^2+5^2}=2\sqrt{50}$

$\therefore$ area $=\frac{1}{2}\times 2\sqrt{50}=5\sqrt{2}$ sq. unit