8. (4 points) Given a triangle $ABC$ with sides $AB=14$, $BC=13$, and $AC=15$. On side $AB$, point $K$ is marked; on side $AC$, point $L$ is marked; and on side $BC$, point $N$ is marked. It is known that $AK=\frac{15}{14}$, $BN=9$, and $AL=1$. A line parallel to $KL$ is drawn through point $N$, intersecting side $AC$ at point $M$. Find the area of quadrilateral $KLMN$.
Answer: $\frac{36503}{1183}=30 \frac{1013}{1183}$.