Question

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:

• A. \( \sqrt{30} \)
• B. 12
• C. \( \sqrt{10} \)
• D. 7 \( \sqrt{5} \)

Hint:

For finding the volume of a tetrahedron, use the areas of the faces in the formula involving the cross-product and areas.

Answer 1

The Correct Option is C

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: \[ V = \sqrt{10} \]