1. ## Geometry(?) Problem

I'm pretty sure I've done this in grade 8, but sadly I threw out last year's Math notes... Reminder anybody? Please and thank you's.

In the diagram, triangle XYZ is right-angled at X, with YX = 60 and XZ = 80. W is the point on YZ so that WX is perpendicular to YZ. Determine the length of WZ.

2. Hello, anh1tran!

In the diagram, triangle $XYZ$ is right-angled at $X$,
with $YX = 60$ and $XZ = 80$.
$XW$ is perpendicular to $YZ$.
Determine the length of $WZ$.
Code:
            X
*
*|   *
60 * |       *   80
*  |           *
*   |               *
*    |                   *
*  *  *  *  *  *  *  *  *  *  *
Y     W                       Z
: - - - - - 100 - - - - - - - :

Using Pythagorus: . $YZ^2 \:=\:60^2+80^2 \:=\:100,000\quad\Rightarrow\quad YZ = 100$

Since $\Delta WXZ \sim \Delta XYZ$, we have: . $\frac{WZ}{XZ} \:=\:\frac{XZ}{YZ}$

Hence: . $\frac{WZ}{80}\:=\:\frac{80}{100}\quad\Rightarrow\q uad WZ \:=\:\frac{6400}{100} \:=\:\boxed{64}$

3. Oh my.. Thank you so much!