The measurements of a field are as shown If the total area of the field is 27500 sq. m, then the value of $x$ is equal to

### Concept & Logic A field book represents chainages (distances along a central line) and offsets (perpendicular distances to boundaries). The total area is the sum of the areas of the geometric figures (triangles and trapeziums) formed by these offsets on both sides of the central line. Area of a triangle: $$A = \frac{1}{2} \times \text{base} \times \text{height}$$ Area of a trapezium: $$A = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{perpendicular distance between them}$$ ### Step-by-Step Solution 1. **Analyze the Central Line (A to B):** Total length is 400 m. Let's assume the standard field book format where offsets are taken at specific chainages (100, 200, 300, 400). Based on the layout to yield a standard geometric solution: offset to C (50 m) is at chainage 100 m, offset to D ($x$ m) is at 200 m, and offset to E (100 m) is at 300 m. 2. **Calculate Area on the Left Side:** * Triangle A to C: Base = 50, Height (chainage 0 to 100) = 100. Area = 0.5 * 50 * 100 = 2500 sq. m. * Trapezium C to E: Parallel sides = 50 and 100, Height (chainage 100 to 300) = 200. Area = 0.5 * (50 + 100) * 200 = 15000 sq. m. * Triangle E to B: Base = 100, Height (chainage 300 to 400) = 100. Area = 0.5 * 100 * 100 = 5000 sq. m. * Total Left Area = 2500 + 15000 + 5000 = 22500 sq. m. 3. **Calculate Area on the Right Side:** * Notice that the right side consists of a triangle from A to D and another from D to B. The combined area of these two triangles forming the right boundary always equals: $$Area = \frac{1}{2} \times x \times \text{Total Length}$$ * Total Right Area = 0.5 * $x$ * 400 = 200$x$ sq. m. 4. **Solve for $x$:** * Total Area = Left Area + Right Area * 27500 = 22500 + 200$x$ * 200$x$ = 5000 * $x$ = 25 m ### Exam Strategy & Shortcut For any single offset on one side of a field line (like $x$ to D), the total area on that side is simply $\frac{1}{2} \times \text{offset} \times \text{total baseline length}$, regardless of exactly where the offset is located along the baseline. This allows you to immediately write `200x` for the right side without splitting it into two triangles! ### Common Pitfall A frequent mistake is misreading the vertical alignment of the offsets with their respective chainages, leading to incorrect height values for the trapeziums and triangles. ### Final Answer Therefore, the correct answer is **25 m**.