1. the value of BD is the same to AD so is 16
3. for the third we can calculate DN by pythagoras
NA=8, AD=16 and DN=?
but by pythagoras
[tex]\begin{gathered} NA^2+DN^{2^{^{}}}=AD^2 \\ 8^2+DN^2=16^2 \\ DN^2=16^2-8^2 \\ DN=\sqrt[]{16^2-8^2} \\ DN=8\sqrt[]{3}\approx13.86 \end{gathered}[/tex]2.