A plane mirror `50 cm` long, is hung on a vetical wall of a room, with its lower edge `50 cm` above the ground. `A` man stands of the mirror at a distance `2 m` away from the mirror. If his eyes are at a height `1.8 m` above the ground, then the length (distance between the extreme points of the visible region perpendicular to the mirror ) of the floor visible to him due to refliection from the mirror is `(x)/(26)m` . Find the value of `x`

Explanation

To solve the problem step by step, we will analyze the situation involving the plane mirror, the man, and the heights involved.

### Step 1: Understand the setup
- The plane mirror is 50 cm long and its lower edge is 50 cm above the ground.
- The man stands 2 m away from the mirror, and his eyes are at a height of 1.8 m above the ground.

### Step 2: Identify the relevant points
- Let’s denote:
- The bottom edge of the mirror as point O (50 cm above the ground).
- The top edge of the mirror as point S (50 cm + 50 cm = 100 cm above the ground).
- The height of the man's eyes as point E (1.8 m = 180 cm above the ground).

### Step 3: Calculate the height of the visible region
- The visible region on the floor due to the reflection in the mirror can be determined by finding the points on the ground that correspond to the reflection of the man's eyes in the mirror.

### Step 4: Use similar triangles
- The triangles formed by the eye level and the mirror can be analyzed.
- The height of the eye level (E) is 1.8 m (or 180 cm), and the height of the bottom of the mirror (O) is 0.5 m (or 50 cm).
- The height of the top of the mirror (S) is 1 m (or 100 cm).

### Step 5: Calculate the distances
- The distance from the eye level to the bottom of the mirror (O) is:
\[ EM = 1.8 \, \text{m} - 0.5 \, \text{m} = 1.3 \, \text{m} \]
- The distance from the eye level to the top of the mirror (S) is:
\[ ES = 1.8 \, \text{m} - 1.0 \, \text{m} = 0.8 \, \text{m} \]

### Step 6: Set up the ratios using similar triangles
- For the triangle formed by the eye level and the bottom of the mirror:
\[ \frac{EM}{PO} = \frac{MP}{OS} \]
where \( PO = 2 \, \text{m} \) (distance from the man to the mirror).

- For the triangle formed by the eye level and the top of the mirror:
\[ \frac{EN}{QR} = \frac{OQ}{RO} \]

### Step 7: Solve for the lengths
- Using the ratios, we can find the lengths:
- For \( OS \):
\[ OS = \frac{MP \cdot PO}{EM} = \frac{2 \cdot 1}{1.3} = \frac{20}{13} \, \text{m} \]

- For \( OR \):
\[ OR = \frac{OQ \cdot QN}{EN} = \frac{0.5 \cdot 2}{1.3} = \frac{10}{13} \, \text{m} \]

### Step 8: Calculate the visible length
- The visible length of the floor is given by:
\[ Visible \, Length = OS - OR = \frac{20}{13} - \frac{10}{13} = \frac{10}{13} \, \text{m} \]

### Step 9: Convert to the required format
- The problem states that the visible length is given as \( \frac{x}{26} \, \text{m} \).
- Setting \( \frac{10}{13} = \frac{x}{26} \):
\[ x = \frac{10 \cdot 26}{13} = 20 \]

### Final Answer
The value of \( x \) is **20**.