I3.2 In figure $1, P Q R$ is a right-angled triangle and $R S T U$ is a rectangle. Let $A, B$ and $C$ be the areas of the corresponding regions. If $A: B=m: 2$ and $A: C=n: 1$, find the value of $n$.
$$
A: B=3: 2, A: C=n: 1 \Rightarrow A: B: C=3 n: 2 n: 3
$$

Let $T S=U R=x, Q U=y$
$\triangle P T S \sim \triangle T Q U \sim \triangle P Q R$ (equiangular)
$$
\begin{array}{l}
S_{\triangle P P S}: S_{\triangle T Q U}: S_{\triangle P Q R}=A: C:(A+B+C)=3 n: 3:(5 n+3) \\
x^{2}: y^{2}:(x+y)^{2}=3 n: 3:(5 n+3) \\
\frac{y}{x}=\frac{1}{\sqrt{n}} \ldots \text { (1), } \frac{x+y}{y}=\frac{\sqrt{5 n+3}}{\sqrt{3}} \ldots \text { (2) }
\end{array}
$$

From (2): $\frac{x}{y}+1=\frac{\sqrt{5 n+3}}{\sqrt{3}} \ldots$
Sub. (1) into (3): $\sqrt{n}+1=\frac{\sqrt{5 n+3}}{\sqrt{3}}$
$$
\begin{array}{l}
\sqrt{3 n}+\sqrt{3}=\sqrt{5 n+3} \\
(\sqrt{3 n}+\sqrt{3})^{2}=5 n+3 \\
3 n+6 \sqrt{n}+3=5 n+3 \\
6 \sqrt{n}=2 n \Rightarrow n=9
\end{array}
$$

Method 2
Let $S R=x, P S=z$
Join $T R$ which bisects the area of the rectangle.
$$
\begin{array}{l}
\frac{A}{B}=\frac{3}{2} \Rightarrow \frac{A}{\frac{B}{2}}=\frac{3}{1} \\
\frac{S_{\triangle T P S}}{S_{\triangle T S R}}=\frac{3}{1} \Rightarrow \frac{z}{x}=\frac{3}{1} \ldots \text { (4) } \\
\because \triangle P T S \sim \triangle T Q U \text { (equiangular) } \\
\therefore \frac{A}{C}=\frac{n}{1} \Rightarrow\left(\frac{z}{x}\right)^{2}=\frac{n}{1} \\
n=3^{2}=9 \text { (by (4)) }
\end{array}
$$