In \( \Large \triangle BCF \), by Pythagoras theorem,

\( \Large (5)^{2}=(3)^{2}+(BF)^{2} \)

BF=4 cm

AB=2+4+4=10 cm

Now, in \( \Large \triangle ACF \), \( \Large AC^{2}=CF^{2}+FA^{2} \)

=> \( \Large AC^{2}=3^{2}+6^{2} \) => AC=\( \Large \sqrt{45} \)cm

Similarly, BD = \( \Large \sqrt{45} \)cm

Sum of diagonals

=AC+BD= \( \Large \sqrt{45}+\sqrt{45}=2\sqrt{45}=6\sqrt{5} \)cm