Question is the image, I am stuck of how to do this question

Thanks Melody, I appreciate your support :) 

However, I do not like to do the homework for other people. Helping is fine tough. If someone asks a short question, he or she gets probably a short answer back 

$${\mathtt{AG}} = {\sqrt{{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{8}}}^{{\mathtt{2}}}}}$$

Please mark this post as "answer" if it helped you! 

Thanks, how do I post this as answer?

Triangle EGH is a right angled triangle.  EG is the hypotenuse.   SO

$$EG=\sqrt{6^2+8^2}=\sqrt{100}=10cm$$

Now triangle AEG is a right angled triangle and  AG is the hypotenuse

so

$$\\AG=\sqrt{10^2+2^2}=\sqrt{104}=\sqrt{4*26}=2\sqrt{26}cm\\
AG\approx 10.2cm\qquad $correct to 2 decimal places$$$

Lets compare this to xerxes's answer.

$$\\AG^2=AE^2+EG^2\\
\\AG^2=2^2+\left(\sqrt{(EH^2+HG^2}\right)^2\\
\\AG^2=2^2+(EH^2+HG^2)\\
\\AG^2=2^2+(8^2+6^2)\\
\\AG^2=2^2+8^2+6^2\\
\\AG=\sqrt{2^2+8^2+6^2}\\$$

So you see, Xerxes wass correct in the fist place.  Maybe he should have explained him/herself better but I think he should get positive points not negative ones.

I only have 3 points that i can give you, I am sorry that i cannot give you any more.  :)

Thanks you xerxes for answering questions and for getting so involved in the forum :)

I take your point but you did not give a hint on how to do the question.  

You gave a final answer that I do not think any student could have learned from.  

I really like it when people give hints without giving an answer but giving an answer without any explanation is usually pretty pointless.