8. In the figure on the right, the line $y=b-x$, where $0<b<4$, intersects the $y$-axis at $P$ and the line $x=4$ at $S$. If the ratio of the area of $\triangle Q R S$ to the area of $\triangle Q O P$ is $9: 25$, determine the value of $b$.
Answer: $b=\frac{5}{2}$
Solution: It is easy to see that $\triangle Q R S$ has legs $4-b$ and $4-b$ while $\triangle Q O P$ has legs $b$ and $b$. Since the areas have a ratio of $9: 25, b$ must therefore satisfy the equation
$$
\frac{(4-b)^{2}}{b^{2}}=\frac{9}{25} \Longrightarrow 16 b^{2}-200 b+400=0 \Longrightarrow 2 b^{2}-25 b+50=0
$$
which has roots $\frac{5}{2}$ and 10 . Since $0<b<4$, this means that $b=\frac{5}{2}$.