Var(X+X) = Var(X) + Var(X) or $2^2$ Var(X)? "We want to measure the perimeter of rectangle ABCD. The measurement AB and CD have standard deviation of 3cm, whereas the measurement of BC and AD have St. deviation of 2cm. 
b) What is the standard deviation for the perimeter if we measure each side and add them up? " 
Answer: $\sqrt{26}$
Here is what I did so far:
Let X: Random variable of AB = CD measurement
    Y: RV of AD = BC measurement
$Var(X + X + Y + Y) = Var(2X + 2Y) = 2^2Var(X+Y) = 2^2Var(X) + 2^2 Var(Y) = 52 $
or 
$Var(X + X + Y + Y) = Var(X) + Var(X) + Var(Y) + Var(Y) = 26$
Then of course, the standard deviation would be the square root of the variances. 
I have 2 questions: 


*

*Is it true to say that $Var(X+X) \mathrel{{=}\llap{/\,}} Var(X) + Var(X)$ because X is not independent of X? (I learnt that $Var(X+Y) = Var(X) + Var(Y)$ if X independent of Y) ? 
If yes, then the second calculation is wrong.

*So the only way I see is if I assign each side of the rectangle a different random variable, but why should I? If two sides are the same, why can't I use the same random variable? 
 A: The bilinearity of covariance states: $$\mathsf {Var}(X+Y)=\mathsf {Var}(X)+\mathsf {Var}(Y)+2\mathsf {Cov}(X,Y)$$
Now, when $X,Y$ are independent, then their covariance is $0$, so $$\mathsf {Var}(X+Y)=\mathsf {Var}(X)+\mathsf {Var}(Y)$$
Similarly the bilineatity of covariance states.
$$\mathsf {Var}(X+X)=\mathsf {Var}(X)+\mathsf {Var}(X)+2\mathsf {Cov}(X,X)$$
Now, $X$ is not independent from $X$. Indeed, the covariance of $X$ and itself is the variance of $X$, so we have the following
$$\begin{split}\mathsf {Var}(X+X)&=\mathsf {Var}(X)+\mathsf {Var}(X)+2~\mathsf {Var}(X)\\&=4~\mathsf {Var}(X)\end{split}$$
And in general $\mathsf {Var}(aX)=a^2\mathsf {Var}(X)$ for any constant $a$, as derived from the definition for variance $\mathsf {Var}(aX)=\mathsf E(a^2X^2)-[\mathsf E(aX)]^2$ and the linearity of expectation.


So the only way I see is if I assign each side of the rectangle a different random variable, but why should I? If two sides are the same, why can't I use the same random variable?

The opposing sides of a rectangle have equal lengths, the adjacent sides need not be.   $\lozenge\rm ABCD$ was specified as a rectangle, rather than a square.
So let the measurement of $\overline{\rm AB}$ and $\overline{\rm CD}$ be random variable $X$, and the measurement of $\overline{\rm AD}$ and  $\overline{\rm BC}$ be random variable $Y$.
$$\mathsf {Var}(2X+2Y)=4~(\mathsf{Var}(X)+\mathsf{Var}(Y))$$
A: Your answer to "question 1" at the end of your post is correct.
The answer to the original problem might be different than your approach, due to an interpretation of the question.
The problem says to "measure each side and add them up." This means you should measure AB and CD individually (possibly getting two different measurements even though they are the same length) and same for the other pair of sides. Then all four measurements are independent, and you may add up the four variances individually.
What you have done is measured one side (say, AB) once and assumed the opposite side (say, CD) has has the same length, and the same for the other pair of sides. If this is allowed, then your first computation (with answer $\sqrt{52}$) is correct. But this is not what the question is asking.
