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

The objective is to calculate the area of face BCD of a tetrahedron ABCD. It is given that edges AB, AC, and AD are mutually orthogonal. The areas of faces ABC, ACD, and ADB are provided as 5, 6, and 7 square units, respectively.

Concept Used:

This problem is solvable using De Gua's Theorem, a three-dimensional extension of the Pythagorean theorem. For a tetrahedron with a vertex forming a right angle (similar to a cube's corner), the square of the area of the face opposite this vertex equals 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 \(A_{ADB}\) denote the areas of the three mutually orthogonal faces. Let \(A_{BCD}\) represent the area of the face opposite the right-angled vertex. The theorem is stated as:

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

Step-by-Step Solution:

Step 1: Extract the given information.

The mutual perpendicularity of edges AB, AC, and AD signifies that vertex A is a right-angled vertex. The faces \( \triangle ABC \), \( \triangle ACD \), and \( \triangle ADB \) are the orthogonal faces with the following given areas:

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

The target is to find the area of face \( \triangle BCD \).

Step 2: Apply De Gua's Theorem.

Using the theorem's formula:

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

Step 3: Substitute the given area values.

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

Final Computation & Result:

Step 4: Calculate the squared area.

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

Step 5: Determine the final area by taking the square root.

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

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