# Surveyor

Calculate the area of ​​what may vary rectangular, if it focused by surveyor and found the dimensions 18 x 15 m while in each of the four joint points can be position deviation 25 cm?

Result

ΔS =  23.3 m2

#### Solution:

$\ \\ 25^2 = c^2 + c^2 \ \\ c = \dfrac{ 25}{100 \cdot \sqrt2} = 0.177 \ m \ \\ S = ab = 270 \ m^2 \ \\ S_1 = (a+c)(b+c) = 281.79 \ m^2 \ \\ S_2 = (a-c)(b-c) = 258.46 \ m^2 \ \\ \Delta S = S_1 - S_2 = 23.3 \ \text{m}^2$

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