Answer:
6[tex]\sqrt{3}[/tex]
Step-by-step explanation:
We have to use Pythagoras' identity twice to find y
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
In right triangle MUT to find TU
TU² + 3² = 6², that is
TU² + 9 = 36 ( subtract 9 from both sides )
TU = 27 ( take the square root of both sides )
TU = [tex]\sqrt{27}[/tex] = [tex]\sqrt{9(3)}[/tex] = 3[tex]\sqrt{3}[/tex]
In right triangle NUT to find y
y² = 9² + (3[tex]\sqrt{3}[/tex] )² = 81 + 27 = 108 ( take the square root of both sides )
y = [tex]\sqrt{108}[/tex] = [tex]\sqrt{36(3)}[/tex] = 6[tex]\sqrt{3}[/tex]
