ABCD is a tetrahedron such that each of the `triangleABC`, `triangleABD` and `triangleACD` has a right angle at A. If `ar(triangleABC) = k_(1). Ar(triangleABD)= k_(2), ar(triangleBCD)=k_(3)` then `ar(triangleACD)` is

Options

A. `sqrt(k_(1)^(2)+k_(2)^(2)+k_(3)^(2))`
B. `sqrt((k_(1)k_(2)k_(3))/(k_(1)+k_(2)+k_(3))`
C. `sqrt(|k_(1)^(2)+k_(2)^(2)-k_(3)^(2)|)`
D. `sqrt(|k_(2)^(2)-k_(1)^(2)-k_(3)^(2)|)`

Explanation

To find the area of triangle ACD in the tetrahedron ABCD, given that triangles ABC, ABD, and ACD have right angles at A, we can use the relationship between the areas of these triangles.

### Step-by-step Solution:

1. **Identify the Areas of Triangles**:
- Let the area of triangle ABC be \( \text{ar}(ABC) = k_1 \).
- Let the area of triangle ABD be \( \text{ar}(ABD) = k_2 \).
- Let the area of triangle BCD be \( \text{ar}(BCD) = k_3 \).
- We need to find the area of triangle ACD, denoted as \( \text{ar}(ACD) = k_4 \).

2. **Use the Relationship of Areas**:
- In a tetrahedron, if three faces are right triangles at a common vertex, the areas of these triangles satisfy the following relationship:
\[ \text{ar}(ABC)^2 + \text{ar}(ABD)^2 + \text{ar}(ACD)^2 = \text{ar}(BCD)^2 \]
- Substituting the known areas into this equation gives:
\[ k_1^2 + k_2^2 + k_4^2 = k_3^2 \]

3. **Rearranging the Equation**:
- To isolate \( k_4^2 \), we rearrange the equation:
\[ k_4^2 = k_3^2 - k_1^2 - k_2^2 \]

4. **Taking the Square Root**:
- To find \( k_4 \), we take the square root of both sides:
\[ k_4 = \sqrt{k_3^2 - k_1^2 - k_2^2} \]

5. **Conclusion**:
- Therefore, the area of triangle ACD is:
\[ \text{ar}(ACD) = \sqrt{k_3^2 - k_1^2 - k_2^2} \]

### Final Answer:
The area of triangle ACD is \( \sqrt{k_3^2 - k_1^2 - k_2^2} \).