过P点作PE⊥AC，PF⊥BD，∵四边形ABCD是矩形，
∴AD⊥CD，
∴△PEA∽△CDA，
∴
| PE |
| CD |
| PA |
| CA |
∵AC=BD=
| 32+42 |
∴
| PE |
| 3 |
| PA |
| 5 |
同理：△PFD∽△BAD，
∴
| PF |
| AB |
| PD |
| BD |
∴
| PF |
| 3 |
| PD |
| 5 |
∴①+②得：
| PE+PF |
| 3 |
| PA+PD |
| 5 |
| AD |
| 5 |
| 4 |
| 5 |
∴PE+PF=
| 12 |
| 5 |
即点P到矩形的两条对角线AC和BD的距离之和是：
| 12 |
| 5 |
故答案为：
| 12 |
| 5 |
过P点作PE⊥AC，PF⊥BD，| PE |
| CD |
| PA |
| CA |
| 32+42 |
| PE |
| 3 |
| PA |
| 5 |
| PF |
| AB |
| PD |
| BD |
| PF |
| 3 |
| PD |
| 5 |
| PE+PF |
| 3 |
| PA+PD |
| 5 |
| AD |
| 5 |
| 4 |
| 5 |
| 12 |
| 5 |
| 12 |
| 5 |
| 12 |
| 5 |