A ladder 5 m long leans against a vertical wall. The bottom of the ladder is 3m from the wall. If the bottom of the ladder is pulled 1 m farther from the wall, how much does the top of the ladder slide down the wall

Options

A. 1 m
B. 4 m
C. 2 m
D. 3 m

Answer

To solve the problem, we will use the Pythagorean theorem, which relates the lengths of the sides of a right triangle.

### Step-by-Step Solution:

1. **Identify the initial setup**:
- Let the length of the ladder be \( L = 5 \) m.
- The distance from the wall to the bottom of the ladder is \( OB = 3 \) m.
- The height of the ladder on the wall is \( OA \) (which we need to find).

2. **Apply the Pythagorean theorem to the initial position**:
- According to the Pythagorean theorem:
\[ AB^2 = OA^2 + OB^2 \]
- Substituting the known values:
\[ 5^2 = OA^2 + 3^2 \]
- This simplifies to:
\[ 25 = OA^2 + 9 \]
- Rearranging gives:
\[ OA^2 = 25 - 9 = 16 \]
- Taking the square root:
\[ OA = 4 \text{ m} \]

3. **Identify the new setup after pulling the ladder**:
- The bottom of the ladder is pulled 1 m farther from the wall, so the new distance from the wall is:
\[ OB' = 3 + 1 = 4 \text{ m} \]
- Let the new height of the ladder on the wall be \( OA' \).

4. **Apply the Pythagorean theorem to the new position**:
- Again using the Pythagorean theorem:
\[ AB^2 = OA'^2 + OB'^2 \]
- Substituting the known values:
\[ 5^2 = OA'^2 + 4^2 \]
- This simplifies to:
\[ 25 = OA'^2 + 16 \]
- Rearranging gives:
\[ OA'^2 = 25 - 16 = 9 \]
- Taking the square root:
\[ OA' = 3 \text{ m} \]

5. **Calculate the distance the top of the ladder slides down**:
- The distance the top of the ladder slides down the wall is:
\[ x = OA - OA' = 4 - 3 = 1 \text{ m} \]

### Final Answer:
The top of the ladder slides down the wall by **1 meter**.

---