分析:由矩形ABCD可得:S△AOD= S矩形ABCD,又由AB=8,BC=15,可求得AC的长,则可求得OA与OD的长,又由S△AOD=S△APO+S△DPO= OA?PE+ OD?PF,代入数值即可求得结果.
解:过点P作PE⊥AC于E,PF⊥BD与F,连接OP, ∵四边形ABCD是矩形, ∴AC=BD,OA=OC=AC,OB=OD=BD,∠ABC=90°, S△AOD=S矩形ABCD, ∴OA=OD=AC, ∵AB=8,BC=15, ∴AC===17,S△AOD=S矩形ABCD=30, ∴OA=OD=, ∴S△AOD=S△APO+S△DPO=OA?PE+OD?PF=OA?(PE+PF)=×(PE+PF)=30, ∴PE+PF=. ∴点P到矩形的两条对角线AC和BD的距离之和是. 故答案为:. |