It is known that in the right triangle ABC, C = 90 °, AC = 3cm, BC = 4cm, the midpoint of BC is O, Po ⊥ plane ABC, and Po = 5cm. Please tell me the detailed steps. I just learned solid geometry

It is known that in the right triangle ABC, C = 90 °, AC = 3cm, BC = 4cm, the midpoint of BC is O, Po ⊥ plane ABC, and Po = 5cm. Please tell me the detailed steps. I just learned solid geometry

In the triangle, make od perpendicular to point D, and the line PD is the distance from point P to ab
From the plane geometry knowledge, it is easy to know: ab = 5, OB = 2, OD = 6 / 5
PD^2=PO^2+OD^2=25+36/25=661/25
PD = radical (661) / 5