Consider the following statements.

I) In triangle ABC, if $\mathrm{c}=6$ and $\cos \mathrm{C}=\frac{11}{25}$ then $\mathrm{R}=\frac{25}{2\sqrt{14}}$

II) In triangle ABC, if a = 3, b = 4, c = 6 then ABC is acute angled triangle.
Which of the above statments is/are true?

In the statement I, Given that $\cos C=\frac{11}{25}$

Hence, 

 $\begin{array}{rcl}\sin C& =& \sqrt{1-\frac{121}{625}}\\ & =& \sqrt{\frac{504}{625}}\\ & =& \frac{6\sqrt{14}}{25}\end{array}$

Sine rule states that 

$\begin{array}{rcl}2R& =& \frac{c}{\sin C}\\ & =& \frac{6}{{\displaystyle \frac{6\sqrt{14}}{25}}}\\ & =& \frac{25}{\sqrt{14}}\end{array}$

Therefore, the circum Radius is $R=\frac{25}{2\sqrt{14}}$

Hence, the statement I is true. 

Consider the statement II

Given that $a=3,b=4,c=6$

let us find the value of $\cos C$

Use cosine rule 

$\begin{array}{rcl}\cos C& =& \frac{a^2+b^2-c^2}{2ab}\\ & =& \frac{9+16-36}{2\left(3\right)\left(4\right)}\\ & =& -\frac{11}{24}\end{array}$

The above value is negative, it means the measure of angle C is obtuse

The triangle is not acute angled triangle

Therefore, the statement II is not true.