Answer:
1.5 m
Explanation:
Length. L = 12 m
Width, W = 16 m
Area, A = 12 x 16 = 192 m^2
Let the width of pavement be d.
The new length, L' = 12 + 2d
the new width, W' = 16 + 2d
New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2
Difference in area = A' - A
285 = 192 + 56 d + 4d^2 - 192
93 = 56 d + 4d^2
4d^2 + 56 d - 93 = 0

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d = 1.5 m
Thus, the width of the pavement is 1.5 m.
Magnets facing the same way <span />
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
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