Answer :
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 [tex]\boxed{y=6\sqrt{3}}[/tex]
Step-by-step explanation:
From the attached figure,
Δ[tex]MTU[/tex] is a right triangle.
Applying Pythagoras theorem,
[tex]TM^2=TU^2+MU^2[/tex]
[tex]\left(6\right)^2=TU^2+\left(3\right)^2[/tex]
[tex]\left(6\right)^2-\left(3\right)^2\:=TU^2[/tex]
switching sides
[tex]TU^2=\left(6\right)^2-\left(3\right)^2\:[/tex]
[tex]TU^2=36-9[/tex]
[tex]TU^2=27[/tex]
Note that Δ[tex]NTU[/tex] s also right triangle.
[tex]\:NT^2\:=\:TU^2\:+\:NU^2[/tex]
[tex]=27\:+\:9^2[/tex] ∵ [tex]TU^2=27[/tex]
[tex]=108[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]NT=\sqrt{108},\:NT=-\sqrt{108}[/tex]
But length of a side can not be negative, so ignore the negative solution.
so
[tex]NT=\sqrt{108}[/tex]
[tex]y=6\sqrt{3}[/tex] ∵ [tex]NT = y[/tex]
Therefore, option B is correct, as [tex]\boxed{y=6\sqrt{3}}[/tex]