Question

What is the value of y?

Answer 1

Note: You forgot to add the diagram, so I found the missing diagram and attached below, based on which I am solving your query. Hopefully, it would clear your concepts anyways.

Answer:

Option B is correct, as  $\boxed{y=6\sqrt{3}}$  

Step-by-step explanation:

From the attached figure,

Δ$MTU$ is a right triangle.

Applying Pythagoras theorem,

$TM^2=TU^2+MU^2$

$\left(6\right)^2=TU^2+\left(3\right)^2$

$\left(6\right)^2-\left(3\right)^2\:=TU^2$

switching sides

$TU^2=\left(6\right)^2-\left(3\right)^2\:$

$TU^2=36-9$

$TU^2=27$

Note that Δ$NTU$ s also right triangle.

$\:NT^2\:=\:TU^2\:+\:NU^2$

         $=27\:+\:9^2$           ∵ $TU^2=27$

         $=108$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$NT=\sqrt{108},\:NT=-\sqrt{108}$

But length of a side can not be negative, so ignore the negative solution.

so

$NT=\sqrt{108}$

    $y=6\sqrt{3}$                ∵ $NT = y$

Therefore, option B is correct, as  $\boxed{y=6\sqrt{3}}$  

Answer 2

Answer:

B

Step-by-step explanation:

I know its right