A cistern 6 m long and 4 m wide, contains water up to a depth of 1m 25 cm. The total area of the wet surface is

Options

A. `55 m^(2)`
B. `53.5 m^(2)`
C. `50 m^(2)`
D. `49 m^(2)`

Answer

To find the total area of the wet surface of the cistern, we need to calculate the areas of the bottom and the four sides of the cistern that are in contact with the water.

### Step-by-Step Solution:

1. **Identify the dimensions of the cistern:**
- Length (L) = 6 m
- Width (W) = 4 m
- Depth of water (H) = 1 m 25 cm = 1.25 m (convert cm to meters)

2. **Calculate the area of the bottom surface:**
- Area of the bottom = Length × Width
- Area of the bottom = 6 m × 4 m = 24 m²

3. **Calculate the area of the two longer sides:**
- Each longer side has dimensions of Height × Length.
- Area of one longer side = H × L = 1.25 m × 6 m = 7.5 m²
- Since there are two longer sides, total area for both = 2 × 7.5 m² = 15 m²

4. **Calculate the area of the two shorter sides:**
- Each shorter side has dimensions of Height × Width.
- Area of one shorter side = H × W = 1.25 m × 4 m = 5 m²
- Since there are two shorter sides, total area for both = 2 × 5 m² = 10 m²

5. **Calculate the total area of the wet surface:**
- Total area of wet surface = Area of bottom + Area of two longer sides + Area of two shorter sides
- Total area of wet surface = 24 m² + 15 m² + 10 m² = 49 m²

### Final Answer:
The total area of the wet surface is **49 m²**.