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

Problem Restatement

Find the wetted surface area (bottom plus side walls up to the water level) of a rectangular cistern.

Given data

• Length L = 6 m
• Width W = 4 m
• Water depth h = 1.25 m

Concept / Approach

Wet surface area = bottom area + area of four side walls in contact with water.

Step-by-step calculation

Bottom area = L × W = 6 × 4 = 24 m^2Side walls area = Perimeter × depth = 2(L + W) × h= 2(6 + 4) × 1.25 = 20 × 1.25 = 25 m^2Total wet surface area = 24 + 25 = 49 m^2

Common pitfalls

• Including the top surface (free surface) is incorrect.
• For sides, using Lh + Wh once; it must be twice each (four walls), or use perimeter.

Final Answer

Total wet surface area = 49 m^2.