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 $1,$ $3$ and $\sqrt{90}$ sq. units respectively. Then the area of triangle $BCD$ is sq. units.

Answer 1

Area of $\Delta ABC=\frac{1}{2}ab=1$
Area of $\Delta ACD=\frac{1}{2}bc=3$
Area of $\Delta ADB=\frac{1}{2}ac=\sqrt{90}$
Area of $\Delta BCD$
$=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$
$=\frac{1}{2}\times 2\sqrt{1+9+90}=10$