帮忙解下两道数学题!要求步骤!!
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1、证明:过点A作AE⊥BD于E,过点C作CF⊥BD于F
∵S△AOB
(1/2)OB●AE
OB
-----------
=
---------------
=
-----
S△AOD
(1/2)OD●AE
OD
S△COB
(1/2)OB●CF
OB
-----------
=
---------------
=
-----
S△COD
(1/2)OD●CF
OD
∴S△AOB
S△COB
-----------
=
-----------
S△AOD
S△COD
2、解:(1)过点C作CE⊥AB于E,过点D作DF⊥AB于F
∵AB//CD,∴CE=DF
∴△ABD和△ABC同底等高
∴S△ABD=S△ABC
又∵S△BOC=S△ABC-S△AOB
S△AOD=S△ABD-S△AOB
∴S△BOC=S△AOD=6平方厘米
(2)过点A作AH⊥BD于H
∵S△AOD=(1/2)OD●AH
S△AOB=(1/2)OB●AH
∴OD
S△AOD
6
2
----
=
----------
=
--
=
--
OB
S△AOB
9
3
同理,OC
S△BOC
6
2
----
=
----------
=
--
=
--
OA
S△BOA
9
3
∵S△AOB
(1/2)OB●AE
OB
-----------
=
---------------
=
-----
S△AOD
(1/2)OD●AE
OD
S△COB
(1/2)OB●CF
OB
-----------
=
---------------
=
-----
S△COD
(1/2)OD●CF
OD
∴S△AOB
S△COB
-----------
=
-----------
S△AOD
S△COD
2、解:(1)过点C作CE⊥AB于E,过点D作DF⊥AB于F
∵AB//CD,∴CE=DF
∴△ABD和△ABC同底等高
∴S△ABD=S△ABC
又∵S△BOC=S△ABC-S△AOB
S△AOD=S△ABD-S△AOB
∴S△BOC=S△AOD=6平方厘米
(2)过点A作AH⊥BD于H
∵S△AOD=(1/2)OD●AH
S△AOB=(1/2)OB●AH
∴OD
S△AOD
6
2
----
=
----------
=
--
=
--
OB
S△AOB
9
3
同理,OC
S△BOC
6
2
----
=
----------
=
--
=
--
OA
S△BOA
9
3
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