23. In the figure below, $A B C$ is a triangle with $A B=5 \mathrm{~cm}, B C=13 \mathrm{~cm}$ and $A C=10 \mathrm{~cm}$. Points $P$ and $Q$ lie on sides $A B$ and $A C$ respectively such that $\frac{\text { area of } \triangle A P Q}{\text { area of } \triangle A B C}=\frac{1}{4}$. Given that the least possible length of $P Q$ is $k \mathrm{~cm}$, find the value of $k$.
23. Answer: 6

We have $\cos A=\frac{A B^{2}+A C^{2}-B C^{2}}{2(A B)(A C)}=\frac{5^{2}+10^{2}-13^{2}}{2(5)(10)}=-\frac{11}{25}$.
Let $A P=x \mathrm{~cm}$ and $A Q=y \mathrm{~cm}$. Since area of $\triangle A P Q=\frac{1}{2} x y \sin A$ and area of $\triangle A B C=\frac{1}{2}(A B)(A C) \sin A=\frac{1}{2}(5)(10) \sin A$, we obtain $\frac{x y}{50}=\frac{1}{4}$, that is, $x y=\frac{25}{2}$.
Hence
$$
\begin{aligned}
P Q^{2} & =x^{2}+y^{2}-2 x y \cos A=x^{2}+\left(\frac{25}{2 x}\right)^{2}-25\left(-\frac{11}{25}\right) \\
& =x^{2}+\frac{625}{4 x^{2}}+11 \geq 2 \sqrt{x^{2} \cdot \frac{625}{4 x^{2}}}+11=25+11=36 .
\end{aligned}
$$

Consequently, $P Q \geq 6$, with the equality attained when $x=y=\frac{5}{\sqrt{2}}$.