A rectangular corner lot has sidewalks on two adjacent sides for a total length of 85 feet. If the diagonal path across the lot is 64 feet, what are the length of the two sides of the walk? Round your answers to two decimal places.

A) 77.4 feet, 68.4 feet
B) 42.5 feet, 42.5 feet
C) 58.05 feet, 26.95 feet
D) 80.12 feet, 70.12 feet
E) 58.05 feet, 55.98 feet

Let the two adjacent sides of the lot be $x$ and $y$. Then we have the following system of equations:

$$\{\begin{array}{l}x+y=85\\ x^2+y^2=64^2\end{array}$$

Squaring the first equation, we get $x^2+2xy+y^2=85^2$. Subtracting the second equation from this gives $2xy=85^2-64^2$. Solving for $xy$, we get $xy=\frac{(85+64)(85-64)}{4}=1237.5$.

We know that $(x+y)^2=x^2+y^2+2xy$, so substituting in the values we have, we get $(x+y)^2=64^2+2(1237.5)$. Solving for $x+y$, we get $x+y=\sqrt{2(64^2+2(1237.5))} \approx 77.4$.

Subtracting this from 85, we get the length of the remaining side of the lot, which is approximately $85-77.4 \approx 7.6$. Therefore, the two sides of the walk are approximately $7.6$ and $\sqrt{(64^2)-(7.6^2)} \approx 26.95$. Thus, the answer is $C$.