2. (a) In the diagram, $\triangle P Q R$ is right-angled at $R$. If $P Q=25$ and $R Q=24$, determine the perimeter and area of $\triangle P Q R$.
(b) In the diagram, $\triangle A B C$ is right-angled at $C$ with $A B=c, A C=b$, and $B C=a$. Also, $\triangle A B C$ has perimeter 144 and area 504. Determine all possible values of $c$.
(You may use the facts that, for any numbers $x$ and $y$, $(x+y)^{2}=x^{2}+2 x y+y^{2}$ and $(x-y)^{2}=x^{2}-2 x y+y^{2}$.)
2. (a) Since $\triangle P Q R$ is right-angled at $R$, we can apply the Pythagorean Theorem to say that $P R^{2}+R Q^{2}=P Q^{2}$.
Since $R Q=24$ and $P Q=25$, then $P R^{2}=25^{2}-24^{2}=625-576=49$.
Since $P R>0$, then $P R=7$.
The perimeter of $\triangle P Q R$ is $P R+R Q+P Q=7+24+25=56$.
Since $\triangle P Q R$ is right-angled at $R$, its area equals $\frac{1}{2}(P R)(R Q)=\frac{1}{2}(7)(24)=\frac{1}{2}(168)=84$.
(b) Since the perimeter of $\triangle A B C$ is 144 and its side lengths are $a, b$ and $c$, then $a+b+c=144$. Since $\triangle A B C$ is right-angled at $C$ and its area is 504 , then $\frac{1}{2}(C B)(A C)=504$ or $\frac{1}{2} a b=504$ or $a b=1008$.
Since $a+b+c=144$, then
$$
\begin{aligned}
a+b & =144-c \\
(a+b)^{2} & =(144-c)^{2} \quad \text { (squaring both sides) } \\
a^{2}+2 a b+b^{2} & =c^{2}-288 c+144^{2} \quad \text { (expanding) } \\
\left(a^{2}+b^{2}\right)+2 a b & =c^{2}-288 c+144^{2} \\
2 a b & \left.=-288 c+144^{2} \quad \text { (since } a^{2}+b^{2}=c^{2}\right) \\
2(1008) & \left.=-288 c+144^{2} \quad \text { (since } a b=1008\right) \\
288 c & =144^{2}-2(1008) \\
144(2 c) & =144^{2}-2(7)(144) \\
2 c & =144-2(7) \\
c & =65
\end{aligned}
$$

Therefore, $c=65$.
(We note that when $c=65$, we obtain $a+b=144-65=79$ and $a b=1008$ which can be solved to obtain $a=16$ and $b=63$ or $a=63$ and $b=16$. We can verify that $16^{2}+63^{2}=65^{2}$ so we do obtain a right-angled triangle.)