Matching distribution with functions 
$A,B,C,D,E$ are independently, identically distributed normal random variables with mean $0$ and variance $1$.
  1) $$A+2B+3C+4D+5E$$
  2) $$\frac{{\sqrt3 A}}{{A^{2}+B^{2}+C^{2}}}$$
  3) $$ \frac{2E}{\sqrt{(A+B)^{2}+(C+D)^{2}}} $$
  4) $$\frac{(A+B+C+D)^{2}}{4}$$
  The choices given are :
  (i) $\chi^2$ with $1$ degree of freedom
  (ii) $N(0,55) $
  (iii) $t$-distribution with $2$ degree of freedom
  (iv) none of these distribution

Anyone can help me with this? I have already solved $6$ out of the $10$ matching questions, but unsure about these $4$, please, thank you so much :)
 A: So, you figured that 1) matches (ii) (which is right).
Here is my thought process
2) Looks suspicious, skip.
3) This is probably the $t$-distribution because of the square root.
4) The numerator is a sum of normals, so it is normal again. It is squared, so this is probably the $\chi^2$. 
So by process of elimination, 2. is "none of these".  

Formal work:
3)
\begin{align*}
\frac{2E}{\sqrt{(A+B)^{2}+(C+D)^{2}}} &= \frac{2E}{\sqrt{X^2+Y^2}}\tag a\\
&=\frac{2 E}{\sqrt{\left(\sqrt 2\frac{X}{\sqrt 2}\right)^2+\left(\sqrt 2\frac{Y}{\sqrt 2}\right)^2}}\\
&=\frac{2E}{\sqrt{2\left(\mathcal X^2+\mathcal Y^2\right)}}\tag b\\
&=\frac{2E}{\sqrt 2\sqrt{\mathcal Z}}\tag c\\
&=\frac{\sqrt 2 E}{\sqrt{\mathcal Z}}\\
&=\frac{\frac{1}{\sqrt 2}\sqrt 2 E}{\frac{1}{\sqrt 2}\sqrt{\mathcal Z}}\\
&=\frac{E}{\sqrt{\mathcal Z/2}}
\end{align*}
where in 


*

*$(a)$ $A+B = X\overset{d}{=}Y\sim N(0,2)$

*$(b)$ $X/\sqrt 2 = \mathcal X\overset{d}{=}\mathcal Y\sim N(0,1)$

*$(c)$ $\mathcal X^2\overset{d}{=}\mathcal Y^2\sim\chi_{(1)}^2$ and $\mathcal X^2+\mathcal Y^2 = \mathcal Z\sim\chi_{(2)}^2$;


Since $E$ is a standard normal, and $\mathcal Z$ follows a $\chi^2$ with $2$ degrees of freedom, then the last equality is by definition (or otherwise well-known to follow) a $t$-distribution with $2$ degrees of freedom.
4)
\begin{align*}
\frac{(A+B+C+D)^2}{4} &= \frac{\left(\sqrt 4\frac{X}{\sqrt 4}\right)^2}{4}\tag d\\
&=\frac{4\mathcal X^2}{4}\tag e\\
&=\mathcal X^2 
\end{align*}
where in $(d)$ 
$$A+B+C+D = X \sim N(0,4)$$
and in $(e)$
$$\frac{X}{\sqrt 4} = \mathcal X \sim N(0,1).$$
Since $\mathcal X$ follows a standard normal, then the final equality is by definition (or is otherwise well-known to follow) a $\chi^2$ distribution with $1$ degree of freedom.
Hence 2) is "none of these".

Note that $\mathcal X$ is not the same character as $\chi$.
