Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles \( ABC, ACD, \) and \( ADB \) be 5, 6 and 7 square units respectively. Then the area (in square units) of the tetrahedron ABCD is equal to:
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For finding the volume of a tetrahedron, use the areas of the faces in the formula involving the cross-product and areas.
We can calculate the volume of the tetrahedron using the areas of the triangles. The area of each triangle is given by:
\[
\text{Area of } \triangle ABC = 5, \quad \text{Area of } \triangle ACD = 6, \quad \text{Area of } \triangle ADB = 7
\]
Using the formula for the volume of a tetrahedron:
\[
V = \frac{1}{3} \sqrt{(A_{ABC} A_{ACD} A_{ADB})}
\]
Substituting the values gives the volume as:
\[
V = \sqrt{10}
\]
