The plan of an old survey, plotted to a scale of 5 m to a cm found to have shrunk so that a line originally 10 cm long was 9.6 cm. There was also a note on the plan that the 20 m c... The plan of an old survey, plotted to a scale of 5 m to a cm found to have shrunk so that a line originally 10 cm long was 9.6 cm. There was also a note on the plan that the 20 m chain used was 0.8 link too short. If the area of the plan measured now by planimeter is 48 cm², find the true area of the survey. Read more

Steps to Solve

1. Determine the true length of the line The original line length is 10 cm on the plan, but it now measures 9.6 cm. To find the scale factor for the length, we calculate: $$ \text{Scale Factor} = \frac{\text{Original Length}}{\text{Measured Length}} = \frac{10 \text{ cm}}{9.6 \text{ cm}} \approx 1.04167 $$

2. Calculate the true length adjustment The chain was originally 20 m long but was 0.8 link too short. We assume one link is a certain length (usually set for such problems). If we assume one link is about 20 m, then:

• True length of the chain = $20 - 0.8 \cdot (20/100) = 20 - 0.16 = 19.84 \text{ m}$.

3. Factor in both the length and chain adjustments To find how adjustments will affect the area measured, we revise the scale: $$ \text{True Length Scale} = \text{Scale Factor} \times \left(\frac{\text{True Length of Chain}}{\text{Measured Length}}\right) $$. Here, we adjust for both shrunk length and shortened chain accordingly.

4. Calculate the reduction factor for the area Area scales by the square of the length scale: $$ \text{Area Factor} = (\text{Scale Factor})^2 \approx (1.04167)^2 \approx 1.0865. $$

5. Determine the true area Finally, the true area can be computed by multiplying the measured area by the area factor: $$ \text{True Area} = \text{Measured Area} \times \text{Area Factor} = 48 \text{ cm}^2 \times 1.0865 \approx 52.15 \text{ cm}^2 $$.