9.35 Given the sides of an acute triangle are $2, 3, x$, then the range of the third side $x$ is
(A) $1<x<5$.
(B) $\sqrt{5}<x<\sqrt{13}$.
(C) $\sqrt{13}<x<5$.
(D) $1<x<\sqrt{5}$.
(E) $\sqrt{5}<x<5$.
(Chongqing, Sichuan Province, China Junior High School Mathematics Competition, 1983)
[Solution] By the cosine theorem, we know that in an acute triangle, the sum of the squares of any two sides is greater than the square of the third side, i.e.,
$$
\left\{\begin{array}{l}
2^{2}+3^{2}>x^{2}, \\
2^{2}+x^{2}>3^{2}, \\
3^{2}+x^{2}>2^{2} .
\end{array} \text { Solving, we get } \sqrt{5}<x<\sqrt{13} .\right.
$$

Therefore, the correct choice is (B).