题目内容
已知在等腰梯形
D中,
∥
.
(1)若
,
,梯形的高是4,求梯形的周长;
(2)若
,
,梯形的高是h,梯形的周长为c,请用
表示c;
(3)若
,
,
.求证:
⊥
.
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(1)若
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(2)若
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(3)若
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(1)26 (2)c=
(3)见解析
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(1) 解:如图,作DE∥AB,DF⊥BC.
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因为AD∥BC ,所以四边形ABED是平行四边形,
所以AB=DE,AD=BE.
因为AB=CD,所以DE=DC.
又DF⊥BC,所以EF=FC.
因为AD=5,BC=11, 梯形的高是4,
所以EC=BC-AD=6,EF=FC=3,DF=4,
从而
,
梯形的周长为AB+BC+CD+AD=5+11+5+5=26.
(2) 解:若AD=a,BC=b,梯形的高是h,则DF=h,EF=FC=
(b-a),
.
所以梯形的周长c=AB+BC+CD+AD=
.
(3)证明:如图,过点D作AC的平行线,交BC的延长线于点E.

由等腰梯形的性质得AC=BD.因为AD∥BC, ED∥AC,
所以四边形ACED是平行四边形,
所以AD=CE,AC=DE,从而BD=DE=
.
又BE=BC+CE=BC+AD=10,
所以
,
所以DE⊥BD,即AC⊥BD.

因为AD∥BC ,所以四边形ABED是平行四边形,
所以AB=DE,AD=BE.
因为AB=CD,所以DE=DC.
又DF⊥BC,所以EF=FC.
因为AD=5,BC=11, 梯形的高是4,
所以EC=BC-AD=6,EF=FC=3,DF=4,
从而

梯形的周长为AB+BC+CD+AD=5+11+5+5=26.
(2) 解:若AD=a,BC=b,梯形的高是h,则DF=h,EF=FC=

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所以梯形的周长c=AB+BC+CD+AD=

(3)证明:如图,过点D作AC的平行线,交BC的延长线于点E.

由等腰梯形的性质得AC=BD.因为AD∥BC, ED∥AC,
所以四边形ACED是平行四边形,
所以AD=CE,AC=DE,从而BD=DE=

又BE=BC+CE=BC+AD=10,
所以

所以DE⊥BD,即AC⊥BD.
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