Daily logic 7

Label the path as follows and draw the width of the path intersecting A perpendicular to the path's edges:

Let $x = AD$ . Then $AC = 55 - x$ , and since $BC = 40$ , by Pythagorean's Theorem $AB = \sqrt{40^2 + (55 - x)^2}$ .

Since $DF || AB$ , $\angle ADE \cong \angle CAB$ ; and since $\angle AED$ and $\angle ACB$ are right angles, $\angle AED \cong \angle ACB$ . Therefore, $\triangle AED \sim \triangle ACB$ by AA similarity, and $\frac{AD}{AE} = \frac{AB}{BC}$ , or $\frac{x}{1} = \frac{\sqrt{40^2 + (55 - x)^2}}{40}$ , which solves to $x = \frac{5}{3}$ .

The shape of the path is a parallelogram, so when $x = \frac{5}{3}$ its area is $AB \cdot AE = \sqrt{40^2 + (55 - \frac{5}{3})^2} \cdot 1 = \frac{200}{3} \approx \boxed{66.67 \text{ m}^2}$ .

Longer side of the parallelogram shaped road is 66.66666666...m. Therefore it's area is (1)(66.6666666...)=66.66666..sq. m.

$40 \csc (\theta )\text{/.}\, \text{Solve}\left[\cot ^{-1}\left(\frac{1}{40} (55-\csc (\theta ))\right)=\theta \land \frac{\pi }{8}\leq \theta \leq \frac{\pi }{2}\right] \approx 66\frac23$ .

I say approximately as I do not know to simplify the expression solved above to the necessary $\csc ^{-1}\left(\frac{5}{3}\right)$ .