Diagonal Road

Let $x$ be the base of the road's parallelogram. Recognize two similar right triangles:

We get the proportion

$\frac{\sqrt{x^2-100}}{10} = \frac{100-x}{100}.$

This leads to the quadratic equation

$99x^2+200x-20\,000=0,$

which has the positive solution

$x = \frac{-200+\sqrt{200^2+4\times 99\times 20\,000}}{2\times 99}\simeq 13.24\,\text{m}.$

The road's area proportion is thus

$\frac{x\times 100}{100^2}=\frac{x}{100} = 0.1324 = 13.24\,\%.$