Surveyor

Calculate the area of ​​what may vary rectangular if it was focused by a surveyor and found the dimensions 10 x 16 m while in each of the four joint points can be position deviation 8 cm?

Final Answer:

Δ =  2.94 m²

Step-by-step explanation:

$$\begin{gathered} a=10\text{m} \\ b=16\text{m} \\ o=8\text{cm}\to \text{m}=8:100\text{m}=0.08\text{m} \\ {o}^2=c^2+c^2 \\ c=o/\sqrt{2}=0.08/\sqrt{2}\doteq 0.0566\text{m} \\ S=a\cdot b=10\cdot 16=160{\text{m}}^2 \\ {S}_1=(a+c)\cdot (b+c)=(10+0.0566)\cdot (16+0.0566)\doteq 161.474{\text{m}}^2 \\ {S}_2=(a-c)\cdot (b-c)=(10-0.0566)\cdot (16-0.0566)\doteq 158.5324{\text{m}}^2 \\ \Delta =S_1-S_2=161.474-158.5324=2.94{\text{m}}^2 \end{gathered}$$

Try another example