Let \(A B C D\) be a tetrahedron such that the edges \(A B, A C\) and \(A D\) are mutually perpendicular. Let the areas …

Since edges AB, AC, AD are mutually perpendicular,

we can place point \(A\) at the origin \((0,0,0)\), and assign coordinates along the 3 axes:

Let \(AB=x\), \(AC=y\), \(AD=z\)

Coordinates: \(A(0,0,0)\), \(B(x,0,0)\), \(C(0,y,0)\), \(D(0,0,z)\)

\(\text{Area of }\Delta ABC=\frac{1}{2}xy=5\text{ ⇒ xy=10}\)

\(\text{Area of }\Delta ACD=\frac{1}{2}yz=6\text{ ⇒ yz=12}\)

\(\text{Area of }\Delta ADB=\frac{1}{2}zx=7\text{ ⇒ zx=14}\)

For a tetrahedron with 3 mutually perpendicular edges from a vertex,

we know that the square of the area of the opposite face (\(\Delta BCD\)) is equal to the sum of squares of the areas of the three adjacent right faces.

\({(\text{Area of }\Delta BCD)}^{2}={(\text{Area of }\Delta ABC)}^{2}+{(\text{Area of }\Delta ACD)}^{2}+{(\text{Area of }\Delta ADB)}^{2}\)

\({(\text{Area})}^{2}={5}^{2}+{6}^{2}+{7}^{2}=25+36+49=110\)