[The ratio of the areas of triangles with a common base or common height] [The area of a figure is equal to the sum of the areas of the figures into which it is divided

Points $M$ and $N$ are located on side $A C$ of triangle $A B C$, and points $K$ and $L$ are on side $A B$, such that $A M: M N: N C=1: 3: 1$ and $A K=K L=L B$. It is known that the area of triangle $A B C$ is 1. Find the area of quadrilateral $K L N M$.
For triangles $ABN$ and $ABC$, they share a height from vertex $C$, so their areas are in the ratio of their bases, which means, 
\$\$S_\{ABN $\}=\backslash$ frac $\{$ AN $\}\{A B\} \backslash$ cdot S_\{ABC $\}=\backslash$ frac45 $\backslash$ cdot $1=\backslash$ frac45.\$\$

Similarly, 
\$\$S_\{ANL $\}=\backslash f r a c\{A L\}\{A B\} \backslash c d o t S_\{ABN $\}=\backslash f r a c 23 \backslash c d o t \backslash f r a c 45=\backslash f r a c\{8\}\{15\}, \$ \$ \$ \$ S_\{ANK $\}=$ $\backslash$ frac $\{$ AK $\}\{A B\} \backslash$ cdot S_\{ABN $\}=\backslash$ frac $13 \backslash$ cdot $\backslash$ frac $45=\backslash$ frac4\{15\},\$\$\$S_\{AKM $\}=\backslash f r a c\{A M\}\{A N\} \backslash c d o t$ S_ $\{$ ANK $\}=\backslash$ frac14 $\backslash$ cdot $\backslash$ frac $4\{15\}=\backslash$ frac $1\{15\} . \$ \$$

Therefore, 
\$\$S_\{KLNM\} = S_\{ALN\} - S_\{AKM $\}=|f r a c 8\{15\}-| f r a c 1\{15\}=\backslash f r a c 7\{15\} . \$ \$$

## Answer

\$frac $7\{15\} \$$.
