6. Given a right triangle $K L M$ with legs $L M=60$ and $K M=80$. On the sides $K L, L M$, and $M K$, points $M_{1}, K_{1}, L_{1}$ are chosen, respectively, such that $K M_{1}=L K_{1}=\mathrm{ML}_{1}=2$. Find the area of triangle $K_{1} L_{1} M_{1}$.

ANSWER: 2216.8
Solution: Let $S(K L M)=2400=S$. Then $S\left(K L_{1} M_{1}\right)=2 / 100 \times 78 / 80 \times S$; $S\left(K_{1} L M_{1}\right)=98 / 100 \times 2 / 60 \times S ; S\left(K_{1} L_{1} M\right)=58 / 60 \times 2 / 80 \times S$. We get $S\left(K_{1} L_{1} M_{1}\right)=S(1-$ $468 / 24000-580 / 24000-784 / 24000)=22168 / 24000 \times 2400=2216.8$.