#1**+7 **

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

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

xerxes
May 11, 2015

#3**-1 **

I don't know, but there must be a way. I never posted a question.

Some people obviously disagree with me answer. Can someone please state why?

xerxes
May 11, 2015

#4**+8 **

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. :)

Melody
May 11, 2015

#5**+12 **

Best Answer

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

xerxes
May 11, 2015

#6**+8 **

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.

Melody
May 11, 2015