Question

What is the value of y?

Answer 1

Apply Pythagorean theorem in ∆TMU

$\\ \rm\Rrightarrow UT^2=6^2-3^2=36-9=27$

$\\ \rm\Rrightarrow UT=3\sqrt{3}$

Now

• in ∆TNU

$\\ \rm\Rrightarrow y^2=(3\sqrt)^2+9^2$

$\\ \rm\Rrightarrow y^2=27+81$

$\\ \rm\Rrightarrow y^2=108$

$\\ \rm\Rrightarrow y=6\sqrt{3}$

Answer 2

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

$\implies \dfrac{y}{9}=\dfrac{6}{3\sqrt{3} }$

$\implies (3\sqrt{3})y=6 \times 9$

$\implies y=\dfrac{54}{3\sqrt{3}}$

$\implies y=6\sqrt{3}$