<s>[INST] Find the relationship of the length of triangle's sides. 
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
  Now determine what kind of triangle $\triangle ABC$ is.
  A.Isosceles triangle which its leg and base is not equal.
  B.equilateral triangle
  C.Right triangle
  D.Isosceles Right triangle

The only information I got is from the number in the radical need to be greater than $0$. Then $b\ge4$ and $c\ge 1$. Also $10a+2\sqrt{b-4}-22\ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
 [/INST] Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|\sqrt{c-1} - 2|=(a-5)^2 + b+ |\sqrt{c-1} -2|= 2\sqrt{b-4} + 3$
And, well this seems a bit weird but that is an even number in front of the $\sqrt{b-4}$ and we have $\sqrt{b-4}$ and $b$ variables to deal with so we can complete the square again with $v = \sqrt{b-4}$ and $v^2 = b-4$.  
(In the back of my mind I'm worrying about the $\sqrt{c-1}$ which is just a single term; I'm not sure at this point what will happen.)
$(a -5)^2 + (b-4) - 2\sqrt{b-4} + 1 +|\sqrt{c-1} - 2|=3-4 + 1$
$(a-5)^2 + (\sqrt{b-4} - 1)^2 +|\sqrt{c-1} - 2| = 0$.
Oh......
We have three things that can't be negative adding up to $0$.  So they must each equal $0$.
So $(a-5)^2 = 0$ and $a =5$.  And $(\sqrt{b-4}-1)^2=0$ and $b=5$.  And $|\sqrt{c-1} -2| = 0$ so $c=5$.
Well.... okay then......
</s>