Answer :

Apply Pythagorean theorem in ∆TMU

[tex]\\ \rm\Rrightarrow UT^2=6^2-3^2=36-9=27[/tex]

[tex]\\ \rm\Rrightarrow UT=3\sqrt{3}[/tex]

Now

  • in ∆TNU

[tex]\\ \rm\Rrightarrow y^2=(3\sqrt)^2+9^2[/tex]

[tex]\\ \rm\Rrightarrow y^2=27+81[/tex]

[tex]\\ \rm\Rrightarrow y^2=108[/tex]

[tex]\\ \rm\Rrightarrow y=6\sqrt{3}[/tex]

semsee45

Answer:

6√3 units

Step-by-step explanation:

Similar right triangle theorem states that as ΔNUT ≅ ΔTUM then

NT : NU = TM : TU

First find the length of TU using Pythagoras' Theorem:  a² + b² = c²

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

⇒ UM² + TU² = TM²

⇒ 3² + TU² = 6²

⇒ TU² = 27

⇒ TU = √(27)

⇒ TU = 3√3

Given:

  • NT = y
  • NU = 9
  • TM = 6
  • TU = 3√3

NT : NU = TM : TU

⇒ y : 9 = 6 : 3√3

[tex]\implies \dfrac{y}{9}=\dfrac{6}{3\sqrt{3} }[/tex]

[tex]\implies (3\sqrt{3})y=6 \times 9[/tex]

[tex]\implies y=\dfrac{54}{3\sqrt{3}}[/tex]

[tex]\implies y=6\sqrt{3}[/tex]

Other Questions