(2014?拱墅区二模)已知,如图,双曲线y=4x(x>0)与直线EF交于点A,点B,且AE=AB=BF,连结AO,BO,它
(2014?拱墅区二模)已知,如图,双曲线y=4x(x>0)与直线EF交于点A,点B,且AE=AB=BF,连结AO,BO,它们分别与双曲线y=2x(x>0)交于点C,点D...
(2014?拱墅区二模)已知,如图,双曲线y=4x(x>0)与直线EF交于点A,点B,且AE=AB=BF,连结AO,BO,它们分别与双曲线y=2x(x>0)交于点C,点D,则:(1)AB与CD的位置关系是______;(2)四边形ABDC的面积为______.
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解:(1)如图,过点A作AM⊥x轴于点M,过点D作DH⊥x轴于点H,过点B作BN⊥x轴于点N,
∴AM∥DH∥BN∥y轴,
设点A的坐标为:(m,
),
∵AE=AB=BF,
∴OM=MN=NF,
∴点B的坐标为:(2m,
),
∴S△OAB=S△OAM+S梯形AMNB-S△OBN=2+
×(
+
)×(2m-m)-2=3,
∵DH∥BN,
∴△ODH∽△OBN,
∴
=
=
,
∵DH?OH=2,BN?ON=4,
∴(
)2=
=
,
同理:(
)2=
,
∴
=
,
∴AB∥CD
故答案为:AB∥CD
(2)∵
=
,∠COD=∠AOB,
∴△COD∽△AOB,
∴
=(
)2=
,
∴S△COD=
,
∴S四边形ABDC=
.
故答案为:
.
∴AM∥DH∥BN∥y轴,
设点A的坐标为:(m,
4 |
m |
∵AE=AB=BF,
∴OM=MN=NF,
∴点B的坐标为:(2m,
2 |
m |
∴S△OAB=S△OAM+S梯形AMNB-S△OBN=2+
1 |
2 |
2 |
m |
4 |
m |
∵DH∥BN,
∴△ODH∽△OBN,
∴
OD |
OB |
DH |
BN |
OH |
ON |
∵DH?OH=2,BN?ON=4,
∴(
OD |
OB |
2 |
4 |
1 |
2 |
同理:(
OC |
OA |
1 |
2 |
∴
OC |
OA |
OD |
OB |
∴AB∥CD
故答案为:AB∥CD
(2)∵
OC |
OA |
OD |
OB |
∴△COD∽△AOB,
∴
S△COD |
S△AOB |
OD |
OB |
1 |
2 |
∴S△COD=
3 |
2 |
∴S四边形ABDC=
3 |
2 |
故答案为:
3 |
2 |
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