A lotus is seen 5 cm above the water level of a lake. With the onset of the wind it sinks in the water 10 cm away from its place. How deep is the water in that place?

Options

A. 5cm
B. `5sqrt5 cm`
C. 7.5 cm
D. 10 cm

Answer

To solve the problem step by step, let's break it down:

1. **Understanding the Problem**:
- A lotus is 5 cm above the water level.
- When the wind blows, it sinks 10 cm away from its original position.
- We need to find the depth of the water (let's denote it as \( x \)).

2. **Visualizing the Situation**:
- Let's denote the point where the lotus is above the water as point A (the top of the lotus).
- The water surface is at point B, and the bottom of the lake is at point C.
- The lotus sinks to point D, which is 10 cm away horizontally from point A.

3. **Setting Up the Geometry**:
- The height of the lotus above the water is 5 cm, so the distance from point B (water surface) to point A (top of the lotus) is 5 cm.
- The depth of the water is \( x \) cm, which is the distance from point B to point C (the bottom of the lake).
- The total length of the lotus from the bottom to the top is \( x + 5 \) cm.

4. **Applying the Pythagorean Theorem**:
- In the right triangle ABD (where A is the top of the lotus, B is the water surface, and D is the point where the lotus sinks), we can apply the Pythagorean theorem:
\[ AD^2 = AB^2 + BD^2 \]
- Here, \( AD = x + 5 \), \( AB = x \), and \( BD = 10 \) cm (the horizontal distance).

5. **Setting Up the Equation**:
- Substituting the values into the Pythagorean theorem:
\[ (x + 5)^2 = x^2 + 10^2 \]
\[ (x + 5)^2 = x^2 + 100 \]

6. **Expanding and Rearranging**:
- Expanding the left side:
\[ x^2 + 10x + 25 = x^2 + 100 \]
- Now, subtract \( x^2 \) from both sides:
\[ 10x + 25 = 100 \]
- Rearranging gives:
\[ 10x = 100 - 25 \]
\[ 10x = 75 \]
\[ x = \frac{75}{10} = 7.5 \text{ cm} \]

7. **Conclusion**:
- The depth of the water in that place is \( 7.5 \) cm.