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 \( \triangle BCD \) is equal to:

• A. \( \sqrt{340} \)
• B. 12
• C. \( \sqrt{110} \)
• D. 7 \( \sqrt{3} \)

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 are asked to find the area of the face BCD of a tetrahedron ABCD. We are given that the edges AB, AC, and AD are mutually perpendicular, and the areas of the faces ABC, ACD, and ADB are 5, 6, and 7 square units, respectively.

Concept Used:

This problem can be solved using De Gua's Theorem, which is a generalization of the Pythagorean theorem to three dimensions. For a tetrahedron that has a right-angled vertex (like the corner of a cube), the square of the area of the face opposite the right-angled vertex is the sum of the squares of the areas of the other three faces.

Let A be the right-angled vertex. Let \(A_{ABC}\), \(A_{ACD}\), and \(A_{ADB}\) be the areas of the three mutually orthogonal faces. Let \(A_{BCD}\) be the area of the face opposite the right-angled vertex. The theorem states:

\[ (A_{BCD})^2 = (A_{ABC})^2 + (A_{ACD})^2 + (A_{ADB})^2 \]

Step-by-Step Solution:

Step 1: Identify the given information based on the problem statement.

The edges AB, AC, and AD are mutually perpendicular, which means the vertex A is a right-angled vertex. The faces \( \triangle ABC \), \( \triangle ACD \), and \( \triangle ADB \) are the three orthogonal faces. Their areas are given as:

\[ \text{Area}(\triangle ABC) = 5 \] \[ \text{Area}(\triangle ACD) = 6 \] \[ \text{Area}(\triangle ADB) = 7 \]

We need to find the area of the face \( \triangle BCD \).

Step 2: Apply De Gua's Theorem to the given tetrahedron.

Using the formula from the concept, we can write:

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

Step 3: Substitute the given area values into the equation.

\[ (\text{Area of } \triangle BCD)^2 = 5^2 + 6^2 + 7^2 \]

Final Computation & Result:

Step 4: Perform the calculation to find the squared area.

\[ (\text{Area of } \triangle BCD)^2 = 25 + 36 + 49 \] \[ (\text{Area of } \triangle BCD)^2 = 110 \]

Step 5: Take the square root to find the final area.

\[ \text{Area of } \triangle BCD = \sqrt{110} \]

Therefore, the area of the \( \triangle BCD \) is \( \sqrt{110} \) square units.