In six minutes the truck can drive $5$ miles on a highway or $1.4$ miles across the prairie, and an optimal route is a highway stretch followed by a straight prairie segment. Work in the first quadrant with the highways as axes. Driving to $(d, 0)$ takes $\frac{d}{50}$ hours, leaving a prairie range of $14\left(\frac{1}{10} - \frac{d}{50}\right) = 1.4\left(1 - \frac{d}{5}\right)$ miles. As $d$ runs from $0$ to $5,$ these disks shrink linearly to a point, so their union is the "cone": the convex hull of the disk of radius $1.4$ about the origin and the point $(5, 0),$ bounded by the tangent line from $(5, 0).$ The tangent length is $\sqrt{5^2 - 1.4^2} = 4.8,$ so the ratios are $7$–$24$–$25$ and the tangent line is $7x + 24y = 35.$ The $y$-axis gives the mirror-image region bounded by $24x + 7y = 35.$

The two tangent lines meet at $P = \left(\frac{35}{31}, \frac{35}{31}\right),$ which lies at distance $\frac{35\sqrt{2}}{31} \gt 1.4$ from the origin — outside the circle — so in the first quadrant the reachable set is exactly the (non-convex) quadrilateral with vertices $(0, 0),$ $(5, 0),$ $P,$ $(0, 5).$ Splitting it along the diagonal from the origin to $P$ gives two triangles, each with area $\frac{1}{2} \cdot 5 \cdot \frac{35}{31},$ for a quadrant area of $\frac{175}{31}.$

The full region is four copies, with area $\frac{700}{31}$ square miles. Since $\gcd(700, 31) = 1,$ the answer is $700 + 31 = 731.$