题目如图OP=1OP=1OP=1,过点PPP作PP1⊥OPPP_{1}\bot OPPP1⊥OP且PP1=1PP_{1}=1PP1=1,得OP1=2OP_{1}=\sqrt{2}OP1=2;再过点P1P_{1}P1作P1P2⊥OP1P_{1}P_{2}\bot OP_{1}P1P2⊥OP1且P1P2=1P_{1}P_{2}=1P1P2=1,得OP2=3OP_{2}=\sqrt{3}OP2=3;又过点P2P_{2}P2作P2P3⊥OP2P_{2}P_{3}\bot OP_{2}P2P3⊥OP2且P2P3=1P_{2}P_{3}=1P2P3=1,得OP3=2…OP_{3}=2\ldotsOP3=2…依此法继续作下去,得OP4019OP_{4019}OP4019等于( )A.4017\sqrt{4017}4017B.4018\sqrt{4018}4018C.4019\sqrt{4019}4019D.4020\sqrt{4020}4020知识点:勾股定理章节:图形的性质 / 三角形