Question

如图,已知四边形ABCD内接于直径为3的圆O

Answer 1

AC=3,PC=0.6,
∴AP=2.4,
设BP=x,PD=y,则AB=BP=x+y,
由相交弦定理,xy=1.44,y=1.44/x,①
由△PAB∽△PDC得AB/DC=PA/PD,
∴DC=AB*PD/PA=(x+y)y/2.4,
AC是直径,
∴BC=√[9-(x+y)^2],AD=√{9-[(x+y)y/2.4]^2},
仿上,BC/AD=BP/AP,
∴2.4√[9-(x+y)^2]=x√{9-[(x+y)y/2.4]^2},
平方得5.76[9-(x+y)^2]=x^2*{9-y^2*(x+y)^2/5.76},②
把①代入②,得5.76[9-(x+1.44/x)^2]=9x^2-0.36(x+1.44/x)^2,
9(5.76-x^2)=5.4(x+1.44/x)^2,
5(5.76-x^2)=3(x^2+2.88+1.44^2/x^2),
8x^2-20.16+6.2208/x^2=0,
x^4-2.52x^2+0.7776=0,
x^2=2.16或0.36,
∴x=0.6√6或0.6（舍）,
代入①,y=0.4√6.
AB=x+y=√6,BC=√[9-(x+y)^2]=√3,DC=(x+y)y/2.4=1,AD=√{9-[(x+y)y/2.4]^2}=2√2,
∴四边形ABCD的周长=√6+√3+1+2√2.