Question

AB为半圆O的直径,AB=2,P为AB延长线上一点,且OP=2

Answer 1

设∠QOP=a

则：QO=1,

PQ^2=OQ^2+OP^2-2OP*OQcosa=1+4-2*2cosa=5-4cosa

SOPRQ=S△OPQ+S△PQR

=1/2*OP*OQsina+1/2*PQ*√3PQ/2

=sina+√3PQ^2/4

=sina+√3(5-4cosa)/4

=sina-√3cosa+5√3/4

=2sin(a-π/3)+5√3/4

所以,a-π/3=π/2,即：a=5π/6时

四边形OPRQ的面积有最大值=2+5√3/4

这时,∠QOP=a=5π/6=150°