题目内容

已知△ABC内接于以O为圆心,1为半径的圆,且3
OA
+4
OB
+5
OC
=
0

(1)求数量积,
OA
OB
OB
OC
OC
OA

(2)求△ABC的面积.
(1)∵3
OA
+4
OB
+5
OC
=
0
,且外接圆的半径r=1,
(3
OA
+4
OB
)
2
=(-5
OC
)
2
=25.
9+16+24
OA
OB
=25

OA
OB
=0

同理可得,
OB
OC
=-
4
5
OA
OC
=-
3
5

(2)设C(m,n)则3(1,0)+4(0,1)+5(m,n)=0.
m=-
3
5
,n=-
4
5

S=S△AOB+S△AOC+S△BOC=
6
5
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