Find the altitude (length of a segment perpendicular to both bases) of the isosceles trapezoid shown.

SOLUTION
The trapezoid can be re-drawn for a better understanding as follows, since it is an isosceles trapezoid
So, we can use either side to find the altitude, which I have labelled as x.
Now, we can see two right-angle triangles which are the same, so we can use anyone. From Pythagoras, we have that
[tex]\begin{gathered} \text{hyp}^2=opp^2+adj^2 \\ 9^2=3^2+x^2 \\ 81=9+x^2 \\ x^2=81-9 \\ x^2=72 \\ x=\sqrt[]{72} \\ x=8.485 \end{gathered}[/tex]hence the answer is 8.5 to the nearest tenth