Question

求解答，数学题，初三要过程

Answer 1

作BE⊥OC,DF⊥OC
∴RT△BOE≌RT△DOF(ASA)
∴BE=DF,OE=OD
∴B和D关于原点O对称
∴lxbl=lxdl,lyal=lydl,
OA=OC=OB=OD=2 ,y=kx

∵x^2+y^2=4,y=kx
x=2/√(1+k^2),,y=2k/√(1+k^2)
Sabcd=S△AOC+S△BOC+S△AOD

=1/2*OC*OA+1/2*OC*lxbl+1/2*OA*lydl
=1/2*2*2+1/2*2*lxbl+1/2*2*lydl

=2+lxbl+lydl
=2+lxbl+lybl
=2+lxbl+lkxbl
=2+2/√(1+k^2)+2k/√(1+k^2)
=2+2[(1+k)/√(1+k^2)]
令[(1+k)/√(1+k^2)]=a
(1+k)^2=a^2(1+k^2)
(a^2-1)*k^2-2k+a^2-1=0
判别式=4-4(a^2-1)^2>=0
(a^2-1)^2<=1
-1<=a^2-1<=1
0<=a^2<=2
0<=a<=√2
所以,(1+k)/√(1+k^2)式子最大值为√2
即原式最大值为=2+2√2
所以,Sabcd最大值=2+2√2

追问

答案是四倍根号二

一次函数在二四象限