Question

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of $\Delta$ABC, $\Delta$ACD and $\Delta$ADB be 3, 4 and 5 sq units, respectively. Then, the area of the $\Delta$BCD, is

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

Answer 1

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

$AB,AC,AD$ are mutually perpendicular

area of $\Delta ABC=\left|AB\right|\left|AC\right|=6$

area of $\Delta ACD=\left|AC\right|\left|AD\right|=8$

area of $\Delta ADB=\left|AB\right|\left|AD\right|=10$

Then the area of $△BCD=\frac{1}{2}|\overrightarrow{BC}\times \overrightarrow{BD}|$

$\Rightarrow area=\frac{1}{2}\left|\right(\overrightarrow{AC}-\overrightarrow{AB})\times (\overrightarrow{AD}-\overrightarrow{AB}\left)\right|$

$\Rightarrow area=\frac{1}{2}|\overrightarrow{AC}\times \overrightarrow{AD}-\overrightarrow{AC}\times \overrightarrow{AB}-\overrightarrow{AB}\times \overrightarrow{AD}+\overrightarrow{AB}\times \overrightarrow{AB}|$

$\Rightarrow area=\frac{1}{2}|\overrightarrow{AC}\times \overrightarrow{AD}-\overrightarrow{AC}\times \overrightarrow{AB}-\overrightarrow{AB}\times \overrightarrow{AD}+0|$

$\Rightarrow area=\frac{1}{2}|\overrightarrow{AC}\times \overrightarrow{AD}-\overrightarrow{AC}\times \overrightarrow{AB}-\overrightarrow{AB}\times \overrightarrow{AD}|$

$\Rightarrow area=\frac{1}{2}|\overrightarrow{AC}\times \overrightarrow{AD}-\overrightarrow{AC}\times \overrightarrow{AB}-\overrightarrow{AB}\times \overrightarrow{AD}|$

$\Rightarrow area=\sqrt{(\frac{\overrightarrow{AC}\times \overrightarrow{AD}}{2}{)}^2+(\frac{\overrightarrow{AC}\times \overrightarrow{AB}}{2}{)}^2+(\frac{\overrightarrow{AB}\times \overrightarrow{AD}}{2}{)}^2}$

$\Rightarrow area=\sqrt{(areaof\Delta ACD{)}^2+(areaof\Delta ABC{)}^2+(areaof\Delta ABD{)}^2}$

$\Rightarrow area=\sqrt{3^2+4^2+5^2}$

$\Rightarrow area=\sqrt{9+16+25}$

$\Rightarrow area=\sqrt{50}$

$\Rightarrow area=5\sqrt{2}$