I4.1 In the figure, the area of $P Q R S$ is $80 \mathrm{~cm}^{2}$. If the area of $\triangle Q R T$ is $A \mathrm{~cm}^{2}$, find $A$.

I4.2 If $B=\log _{2}\left(\frac{8 A}{5}\right)$, find $B$.
I4.3 Given $x+\frac{1}{x}=B$. If $C=x^{3}+\frac{1}{x^{3}}$, find $C$.
I4.4 Let $(p, q)=q D+p$. If $(C, 2)=212$, find $D$.
$\triangle Q R T$ has the same base and same height as the parallelogram $P Q R S$.
$$
\begin{array}{l}
A=\frac{1}{2} \cdot 80=40 \\
B=\log _{2}\left(\frac{8 \cdot 40}{5}\right) \\
=\log _{2} 64 \\
=\log _{2} 2^{6}=6 \\
x+\frac{1}{x}=6 \\
x^{2}+\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)^{2}-2 \\
=6^{2}-2=34 \\
C=x^{3}+\frac{1}{x^{3}} \\
=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}-1\right) \\
=6(34-1)=198 \\
2 D+C=212 \\
\Rightarrow 2 D=212-198=14 \\
\Rightarrow D=7 \\
\end{array}
$$