Two vertical poles AL and BM of height 25 m and 100 m respectively stand apart on a horizontal plane. If A, B be the feet of the poles and AM and BL intersect at P, then the height of P from the horizontal plane is equal to

Options

A. 20 m
B. 18 m
C. 16 m
D. 15 m

Answer

To solve the problem, we will use the concept of similar triangles and the Basic Proportionality Theorem (also known as Thales' theorem).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two vertical poles: AL (height = 25 m) and BM (height = 100 m).
- Let A and B be the feet of the poles, and AM and BL intersect at point P.

2. **Assigning Variables**:
- Let the height of point P from the horizontal plane be denoted as \( z \).
- Let \( x \) be the distance from A to P along line AM.
- Let \( y \) be the distance from B to P along line BL.

3. **Applying Basic Proportionality Theorem**:
- In triangle APL (where AL is the pole of height 25 m):
\[ \frac{y}{z} = \frac{AB}{AL} = \frac{x + y}{25} \]
- Rearranging gives us:
\[ y = z \cdot \frac{x + y}{25} \quad \text{(Equation 1)} \]

4. **Applying Basic Proportionality Theorem to Triangle BMP**:
- In triangle BMP (where BM is the pole of height 100 m):
\[ \frac{x}{z} = \frac{AB}{BM} = \frac{x + y}{100} \]
- Rearranging gives us:
\[ x = z \cdot \frac{x + y}{100} \quad \text{(Equation 2)} \]

5. **Adding Equations 1 and 2**:
- From Equation 1:
\[ y = z \cdot \frac{x + y}{25} \]
- From Equation 2:
\[ x = z \cdot \frac{x + y}{100} \]
- Adding these two equations:
\[ x + y = z \left( \frac{x + y}{25} + \frac{x + y}{100} \right) \]
- Let \( S = x + y \):
\[ S = z \left( \frac{4S}{100} + \frac{S}{100} \right) \]
- Simplifying gives:
\[ S = z \cdot \frac{5S}{100} \]
- Canceling \( S \) (assuming \( S \neq 0 \)):
\[ 1 = \frac{5z}{100} \]
- Thus:
\[ z = \frac{100}{5} = 20 \text{ m} \]

6. **Conclusion**:
- The height of point P from the horizontal plane is \( z = 20 \) m.