Answer:
by using pythagoras theorem
[tex] {h}^{2} = {p}^{2} + {b }^{2} \\ {28}^{2} = {17}^{2} + {es}^{2} \\ {28}^{2} - {17}^{2} = {es}^{2} \\ {es}^{2} = 495 \\ es = \sqrt{495} \\ es = 3 \sqrt{55} [/tex]
