I need help with this Geometry question asap. It's on special triangles.


Answer:
4√5
Step-by-step explanation:
the rule is: AB/n=n/BC.
from the rule above n²=AB*BC and AB=sqrt(144-n²).
After substitution n²=10*√(144-n²).
In order to solve this equation:
n⁴=100(144-n²); ⇔ n⁴+100n²-14400=0;
[tex]\left[\begin{array}{ccc}n^2=-180\\\ n^2=80\end{array}\right] \ => \ \left[\begin{array}{ccc}n=\sqrt{-80} \\\ n=\sqrt{80} \end{array}\right] \ => \ n=\sqrt{80}.[/tex]
P.S. sqrt{80}=4√5.