1. Stating the problem. A wall 10 m long, 20 cm thick and 1.2 m high is constructed as a partition between two plots of land. Find the cost of cementing it on five sides at the rate of 25 per sq. m. 2. Known values and unit conversion. Length: $l = 10$ m. Thickness: $t = 0.2$ m (20 cm converted to meters). Height: $h = 1.2$ m. 3. Formula and important rule. The part to be cemented is the sum of areas of five faces of the rectangular block: two long faces, two ends and the top. The formula is: $$A = 2(lh) + 2(th) + lt$$ This counts both faces of length and height, both ends of thickness and height, and the top of length and thickness. 4. Intermediate calculations with full simplification. Compute each product first. $$l h = 10 \times 1.2 = 12$$ $$t h = 0.2 \times 1.2 = 0.24$$ $$l t = 10 \times 0.2 = 2$$ Substitute into the formula and simplify. $$A = 2 \times 12 + 2 \times 0.24 + 2 = 24 + 0.48 + 2 = 26.48$$ So the area to be cemented is 26.48 sq. m. 5. Cost calculation and final answer. Let the rate be $r = 25$ per sq. m. Cost equals area times rate. $$\text{Cost} = A \times r = 26.48 \times 25 = 662$$ Therefore the cost of cementing is 662.