The distance between $J(2,7)$ and $K(5,3)$ is equal to $\sqrt{(2-5)^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = 5$. Therefore, if we consider $\triangle JKL$ as having base $JK$ and height $h$, then we want $\frac{1}{2} \cdot JK \cdot h \leq 10$ which means that $h \leq 10 \cdot \frac{2}{5} = 4$. In other words, $L(r, t)$ can be any point with $0 \leq r \leq 10$ and $0 \leq t \leq 10$ whose perpendicular distance to the line through $J$ and $K$ is at most 4. The slope of the line through $J(2,7)$ and $K(5,3)$ is equal to $\frac{7-3}{2-5} = -\frac{4}{3}$. Therefore, this line has equation $y - 7 = -\frac{4}{3}(x - 2)$. Multiplying through by 3, we obtain $3y - 21 = -4x + 8$ or $4x + 3y = 29$. We determine the equation of the line above this line that is parallel to it and a perpendicular distance of 4 from it. The equation of this line will be of the form $4x + 3y = c$ for some real number $c$, since it is parallel to the line with equation $4x + 3y = 29$. To determine the value of $c$, we determine the coordinates of one point on this line. To determine such a point, we draw a perpendicular of length 4 from $K$ to a point $P$ above the line. Since $JK$ has slope $-\frac{4}{3}$ and $KP$ is perpendicular to $JK$, then $KP$ has slope $\frac{3}{4}$. Draw a vertical line from $P$ and a horizontal line from $K$, meeting at $Q$. Since $KP$ has slope $\frac{3}{4}$, then $PQ : QK = 3 : 4$, which means that $\triangle KQP$ is similar to a 3-4-5 triangle. Since $KP = 4$, then $PQ = \frac{3}{5} KP = \frac{12}{5}$ and $QK = \frac{4}{5} KP = \frac{16}{5}$. Thus, the coordinates of $P$ are $\left(5 + \frac{16}{5}, 3 + \frac{12}{5}\right)$ or $\left(\frac{41}{5}, \frac{27}{5}\right)$. Since $P$ lies on the line with equation $4x + 3y = c$, then $c = 4 \cdot \frac{41}{5} + 3 \cdot \frac{27}{5} = \frac{164}{5} + \frac{81}{5} = \frac{245}{5} = 49$ and so the equation of the line parallel to $JK$ and 4 units above it is $4x + 3y = 49$. In a similar way, we find that the line parallel to $JK$ and 4 units below it has equation $4x + 3y = 9$. (Note that $49 - 29 = 29 - 9$.) The points $L$ that satisfy the given conditions are exactly the points within the square, below the line $4x + 3y = 49$ and above the line $4x + 3y = 9$. In other words, the region $\mathcal{R}$ is the region inside the square and between these lines. To find the area of $\mathcal{R}$, we take the area of the square bounded by the lines $x = 0, x = 10, y = 0,$ and $y = 10$ (this area equals $10 \cdot 10$ or 100) and subtract the area of the two triangles inside the square and not between the lines. The line with equation $4x + 3y = 9$ intersects the $y$-axis at $(0,3)$ (we see this by setting $x = 0$) and the $x$-axis at $\left(\frac{9}{4}, 0\right)$ (we see this by setting $y = 0$). The line with equation $4x + 3y = 49$ intersects the line $x = 10$ at $(10,3)$ (we see this by setting $x = 10$) and the line $y = 10$ at $\left(\frac{19}{4}, 10\right)$ (we see this by setting $y = 10$). The bottom triangle that is inside the square and outside $\mathcal{R}$ has area $\frac{1}{2} \cdot 3 \cdot \frac{9}{4} = \frac{27}{8}$. The top triangle that is inside the square and outside $\mathcal{R}$ has horizontal base of length $10 - \frac{19}{4}$ or $\frac{21}{4}$ and vertical height of length $10 - 3$ or 7, and thus has area $\frac{1}{2} \cdot \frac{21}{4} \cdot 7 = \frac{147}{8}$. Finally, this means that the area of $\mathcal{R}$ is $100 - \frac{27}{8} - \frac{147}{8} = 100 - \frac{174}{8} = 100 - \frac{87}{4} = \frac{313}{4}$ which is in lowest terms since the only divisors of the denominator that are larger than 1 are 2 and 4, while the numerator is odd. When we write this area in the form $\frac{300 + a}{40 - b}$ where $a$ and $b$ are positive integers, we obtain $a = 13$ and $b = 36$, giving $a + b = 49$.