题目内容
如图,已知边长为a的正方形ABCD,点E在AB上,点F在BC的延长线上,EF与AC交于点O,且AE=CF.
(1)若a=4,则四边形EBFD的面积为______;
(2)若AE=
AB,求四边形ACFD与四边形EBFD面积的比;
(3)设BE=m,用含m的式子表示△AOE与△COF面积的差.
(1)若a=4,则四边形EBFD的面积为______;
(2)若AE=
1 |
3 |
(3)设BE=m,用含m的式子表示△AOE与△COF面积的差.
(1)∵AE=CF,∠EAD=∠FCD,AD=CD,
∴△DAE≌△DCF,
∴四边形EBFD的面积=正方形ABC的面积=42=16;
(2)CF=AE=
AB=
,
∵四边形ABCD为正方形,
∴BC=CD=AD=AB=a,∠ABC=∠BCD=∠CDA=∠DAB=90°,AD∥BC,
∴S四边形ACFD=
=
=
,
S四边形EBFD=S四边形EBCD+S△CFD=S四边形EBCD+S△AED=S正方形ABCD=a2,
∴S四边形ACFD:S四边形EBFD=
:a2=2:3;
(3)CF=AE=a-m,FB=a+a-m=2a-m,
由(2)知∠ABC=90°,AB=BC,可得,
S△AOE+S四边形EOCB=S△ABC=
=
,
S△COF+S四边形EOCB=S△EBF=
=
=
,
∴S△AOE+S四边形EOCB-(S△COF+S四边形EOCB)=
-
=
,
即S△AOE-S△COF=
.
∴△DAE≌△DCF,
∴四边形EBFD的面积=正方形ABC的面积=42=16;
(2)CF=AE=
1 |
3 |
a |
3 |
∵四边形ABCD为正方形,
∴BC=CD=AD=AB=a,∠ABC=∠BCD=∠CDA=∠DAB=90°,AD∥BC,
∴S四边形ACFD=
(CF+AD)CD |
2 |
(
| ||
2 |
2a2 |
3 |
S四边形EBFD=S四边形EBCD+S△CFD=S四边形EBCD+S△AED=S正方形ABCD=a2,
∴S四边形ACFD:S四边形EBFD=
2a2 |
3 |
(3)CF=AE=a-m,FB=a+a-m=2a-m,
由(2)知∠ABC=90°,AB=BC,可得,
S△AOE+S四边形EOCB=S△ABC=
AB2 |
2 |
a2 |
2 |
S△COF+S四边形EOCB=S△EBF=
EB•FB |
2 |
m(2a-m) |
2 |
2am-m2 |
2 |
∴S△AOE+S四边形EOCB-(S△COF+S四边形EOCB)=
a2 |
2 |
2am-m2 |
2 |
a2-2am+m2 |
2 |
即S△AOE-S△COF=
a2-2am+m2 |
2 |
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