6. In $\triangle A B C, A B=20, B C=21$, and $C A=29$. Point $M$ is on side $A B$ with $\frac{A M}{M B}=\frac{3}{2}$, while point $N$ is on side $B C$ with $\frac{C N}{N B}=2 . P$ and $Q$ are points on side $A C$ such that the line $M P$ is parallel to $B C$ and the line $N Q$ is parallel to $A B$. Suppose that $M P$ and $N Q$ intersect at point $R$. Find the area of $\triangle P Q R$.
Answer: $\frac{224}{15}$
Solution: We use similar triangles here. Note that triangles $A B C, A M P, Q N C$ and $Q R P$ are all similar right (by the Pythagorean theorem, since $20^{2}+21^{2}=29^{2}$ ) triangles by AA similarity and corresponding angle theorem. We see that $A P=\frac{3}{5} \cdot 29=\frac{87}{5}$ and $C Q=\frac{2}{3} \cdot 29=\frac{58}{3}$.
Hence, $P Q=A P+Q C-A C=\frac{87}{5}+\frac{58}{3}-29=\frac{116}{15}=29\left(\frac{4}{15}\right)$. Thus, the ratio of similitude between $Q R P$ and $A B C$ is $\frac{4}{15}$, and the area of triangle $Q R P$ is $\left(\frac{16}{225}\right)\left(\frac{1}{2}\right)(20)(21)=\frac{224}{15}$.