In `DeltaABC,angleC=90^(@)` and CD is perpendicular to AB at D. If `(AD)/(BD)=sqrtk`, then `(AC)/(BC)=?`

Options

A. a)k
B. b)`sqrtk`
C. c)`1/(sqrtk)`
D. d)`root(4)(k)`

Answer

To solve the problem, we will use the properties of similar triangles and the given ratio of segments.

### Step-by-Step Solution:

1. **Identify the Triangles**: In triangle \( \Delta ABC \), we know that \( \angle C = 90^\circ \). The point \( D \) is where the perpendicular \( CD \) meets \( AB \). This creates two smaller triangles, \( \Delta ACD \) and \( \Delta BCD \).

2. **Use the Given Ratio**: We are given that \( \frac{AD}{BD} = \sqrt{k} \). Let's denote \( AD = x \) and \( BD = y \). Thus, we can write:
\[ \frac{x}{y} = \sqrt{k} \]

3. **Express \( AB \)**: The total length of \( AB \) can be expressed as:
\[ AB = AD + BD = x + y \]

4. **Relate the Sides**: From the properties of similar triangles \( \Delta ACD \) and \( \Delta BCD \), we can write:
\[ AC^2 = AD \cdot AB \quad \text{and} \quad BC^2 = BD \cdot AB \]

5. **Substitute for \( AB \)**: Since \( AB = x + y \), we can rewrite the equations:
\[ AC^2 = AD \cdot (AD + BD) = x(x + y) \]
\[ BC^2 = BD \cdot (AD + BD) = y(x + y) \]

6. **Form the Ratio**: Now, we can form the ratio \( \frac{AC^2}{BC^2} \):
\[ \frac{AC^2}{BC^2} = \frac{x(x + y)}{y(x + y)} = \frac{x}{y} \]

7. **Substitute the Known Ratio**: Since \( \frac{x}{y} = \sqrt{k} \), we have:
\[ \frac{AC^2}{BC^2} = \sqrt{k} \]

8. **Take the Square Root**: To find \( \frac{AC}{BC} \), we take the square root of both sides:
\[ \frac{AC}{BC} = \sqrt{\sqrt{k}} = k^{1/4} \]

### Final Answer:
\[ \frac{AC}{BC} = k^{1/4} \]