Question

what is the value of y?
3√3 units
6√3 units
9√3 units
12√3 units

Explanation:

Step1: Identify similar triangles

Triangles $\triangle NTU$ and $\triangle NMT$ are similar (right triangles sharing $\angle N$).

Step2: Set up proportion

Corresponding sides: $\frac{y}{NM} = \frac{NM}{y + TM}$
$NM = 9 + 3 = 12$, $TM = 6$, so:
$\frac{y}{12} = \frac{12}{y + 6}$

Step3: Cross-multiply to solve for $y$

$y(y + 6) = 12^2$
$y^2 + 6y - 144 = 0$
Use quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=6$, $c=-144$:

$$\begin{align*} y &= \frac{-6 \pm \sqrt{6^2 - 4(1)(-144)}}{2(1)}\\ &= \frac{-6 \pm \sqrt{36 + 576}}{2}\\ &= \frac{-6 \pm \sqrt{612}}{2}\\ &= \frac{-6 \pm 6\sqrt{17}}{2}\\ &= -3 \pm 3\sqrt{17} \end{align*}$$

Since length is positive, $y = -3 + 3\sqrt{17}$ is not an option, so use geometric mean altitude theorem for right triangle:
First, find altitude $TU$: $TU = \sqrt{6 \times 3} = 3\sqrt{2}$
Then use Pythagoras in $\triangle NTU$: $y^2 + (3\sqrt{2})^2 = 9^2$
$y^2 + 18 = 81$
$y^2 = 63$
$y = 3\sqrt{7}$ (incorrect, re-identify similar triangles correctly: $\triangle NTM \sim \triangle TUM$)
Correct proportion: $\frac{y}{9} = \frac{12}{y}$
$y^2 = 108$
$y = 6\sqrt{3}$

Answer:

$6\sqrt{3}$ units