G2.1 It is given that the lengths of the sides of a triangle are 6,8 , and 10 .
G2.2 Given that $f\left(x+\frac{1}{x}\right)=x^{3}+\frac{1}{x^{3}}$ and $f(4)=b$, find the value of $b$.
G2.3 Given that $2002^{2}-2001^{2}+2000^{2}-1999^{2}+\ldots+4^{2}-3^{2}+2^{2}-1^{2}=c$, find the value of $c$.
G2.4 $P Q R S$ is a square, $P T U$ is an isosceles triangle, and $\angle T P U=30^{\circ}$. Points $T$ and $U$ lie on $Q R$ and $R S$ respectively. The area of $\triangle P T U$ is 1 . If the area of $P Q R S$ is $d$, find the value of $d$.
If the area of the triangle is $a$, find the value of $a$.
$$
6^{2}+8^{2}=36+64=100=10^{2}
$$

It is a right angled triangle.
The area of the triangle $=6 \times 8 \div 2=24$
$$
a=24
$$

Reference: 1987 FG8.2, 2002 HI10
Let $y=x+\frac{1}{x}$
$$
y^{2}-2=x^{2}+\frac{1}{x^{2}}
$$
$$
\begin{array}{l}
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right) \\
\quad=y\left(y^{2}-3\right)=y^{3}-3 y \\
f(y)=y^{3}-3 y \\
b=f(4)=4^{3}-3(4)=52
\end{array}
$$

Reference: 1997 HI5, 2004 HI1, 2015 FI3.2, 2015 FG4.1
$$
\begin{aligned}
c & =(2002+2001)(2002-2001)+(2000+1999)(2000-1999)+\ldots+(4+3)(4-3)+(2+1)(2-1) \\
c & =4003+3999+\ldots+7+3 \\
& =\frac{4003+3}{2} \times 1001=2005003
\end{aligned}
$$

Let $P T=a=P U$
$$
\begin{array}{l}
\frac{1}{2} a^{2} \sin 30^{\circ}=1 \\
\Rightarrow a=2 \\
\triangle P S U \cong \triangle P Q T \text { (RHS) }
\end{array}
$$

Let $P S=x=P Q ; S U=y=Q T$
$\angle S P U=\angle Q P T=30^{\circ}$ (corr. $\angle \mathrm{s} \cong \Delta$ )
$$
\begin{array}{l}
x=P U \cos 30^{\circ}=2 \frac{\sqrt{3}}{2}=\sqrt{3} \\
d=\text { area of } P Q R S=\sqrt{3}^{2}=3
\end{array}
$$