A lotus flower is 5 cm up to the water level in a lake. Due to strong wind it drown in water 10 cm away from it original position. What is the depth of water at position of flower?

Options

A. 5 cm
B. 5 sqrt(5) cm
C. 7.5 cm
D. 10 cm

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Problem
We have a lotus flower that is 5 cm above the water level. It is blown by the wind to a position that is 10 cm away horizontally from its original position. We need to find the depth of the water at the position of the flower after it has been moved.

### Step 2: Define Variables
Let:
- \( x \) = depth of the water at the position of the flower.
- The height of the flower above the water = 5 cm.
- The horizontal distance moved = 10 cm.

### Step 3: Set Up the Relationship
When the flower is at its original position, the total distance from the bottom of the lake to the top of the flower is \( x + 5 \) cm.

### Step 4: Use the Pythagorean Theorem
We can visualize this situation as a right triangle where:
- One leg (vertical) is the depth of the water, \( x \).
- The other leg (horizontal) is the distance the flower moved, which is 10 cm.
- The hypotenuse is the total distance from the bottom of the lake to the top of the flower, which is \( x + 5 \).

According to the Pythagorean theorem:
\[ (OC)^2 = (OA)^2 + (AC)^2 \]
Substituting the values we have:
\[ (x + 5)^2 = x^2 + 10^2 \]

### Step 5: Expand and Simplify the Equation
Expanding the left side:
\[ (x + 5)^2 = x^2 + 10x + 25 \]
So, we have:
\[ x^2 + 10x + 25 = x^2 + 100 \]

### Step 6: Eliminate \( x^2 \) from Both Sides
Subtract \( x^2 \) from both sides:
\[ 10x + 25 = 100 \]

### Step 7: Solve for \( x \)
Now, isolate \( x \):
\[ 10x = 100 - 25 \]
\[ 10x = 75 \]
\[ x = \frac{75}{10} = 7.5 \text{ cm} \]

### Conclusion
The depth of the water at the position of the flower is **7.5 cm**.

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