So there's an altitude going from the vertex adjacent to the side 8cm long to a side measuring 90 degrees.

The altitude creates 2 right triangles within and so they share one side.
Here's a image I drew up in paint to help show the problem.

http://img521.imageshack.us/img521/2083/isos.png

I'm trying to figure out what "y" is but I figure I'd need "x" to do it. I set up 2 equations using Pythagoras theorem.

10^2=(10-x)^2+Y^2 and 8^2=Y^2+X^2.

Solving it left me with -64=20x. It being a negative probably means I'm wrong. I've checked it over and didn't find mistakes so maybe I'm just approaching it wrong.

4^2 +x^2 = 10^2

16 + x^2 = 100

x^2 = 100 - 16

x^2 = 84

x = sq root of 84

x = 9.165 approx.

http://img521.imageshack.us/img521/2083/isos.png

Let's do it again.

10^2=(10-x)^2+y^2... a

8^2=y^2+x^2 ... b

Expand the first equation.

10^2=(10-x)^2+y^2

10^2=100 - 20x + x^2 +y^2... c

Subtract b from c.

\begin{array}{rcr}10^2 &=& 100 - 20x + x^2 +y^2 \\

8^2 &=&x^2+y^2 \\

36 &=& 100 - 20x\end{array}

This gives x = \frac{36-100}{-20} = 3.2

Can you find the rest now? :)