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在平行六面体ABCD-A1B1C1D1中,AB=1,AD=2,AA1=3,∠BAD=60°,∠BAA1=∠DAA1=90°,则AC1的长为()A.25B.4C.5D.26
题目详情
在平行六面体ABCD-A1B1C1D1中,AB=1,AD=2,AA1=3,∠BAD=60°,∠BAA1=∠DAA1=90°,则AC1的长为( )
A.2
B.4
C.5
D.2
11111111
2
B.4
C.5
D.2
5 5
2
6 6
A.2
5 |
B.4
C.5
D.2
6 |
2
5 |
B.4
C.5
D.2
6 |
5 |
2
6 |
6 |
▼优质解答
答案和解析
在平行六面体ABCD-A11B11C11D11中,
∵AB=1,AD=2,AA11=3∠BAD=60°,∠BAA11=∠DAA11=90°,
=
+
+
,
∴
2=(
+
+
)2
=
2+
2+
2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B.
AC1 AC1 AC11=
+
+
,
∴
2=(
+
+
)2
=
2+
2+
2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B.
AB AB AB+
BC BC BC+
CC1 CC1 CC11,
∴
2=(
+
+
)2
=
2+
2+
2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B.
AC1 AC1 AC1122=(
+
+
)2
=
2+
2+
2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B.
AB AB AB+
BC BC BC+
CC1 CC1 CC11)22
=
2+
2+
2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B.
AB AB AB2+
2+
2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B. 2+
BC BC BC2+
2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B. 2+
CC1 CC1 CC112+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B. 2+2
•
+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B. 2
AB AB AB•
BC BC BC+2
•
+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B.
AB AB AB•
CC1 CC1 CC11+2
•
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B.
BC BC BC•
CC1 CC1 CC11
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
|=4.
故选B. |
AC1 AC1 AC11|=4.
故选B.
![](http://hiphotos.baidu.com/zhidao/pic/item/bba1cd11728b471066632d7ac0cec3fdfc03233c.jpg)
∵AB=1,AD=2,AA11=3∠BAD=60°,∠BAA11=∠DAA11=90°,
AC1 |
AB |
BC |
CC1 |
∴
AC1 |
AB |
BC |
CC1 |
=
AB |
BC |
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B.
AC1 |
AB |
BC |
CC1 |
∴
AC1 |
AB |
BC |
CC1 |
=
AB |
BC |
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B.
AB |
BC |
CC1 |
∴
AC1 |
AB |
BC |
CC1 |
=
AB |
BC |
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B.
AC1 |
AB |
BC |
CC1 |
=
AB |
BC |
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B.
AB |
BC |
CC1 |
=
AB |
BC |
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B.
AB |
BC |
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B. 2+
BC |
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B. 2+
CC1 |
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B. 2+2
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B. 2
AB |
BC |
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B.
AB |
CC1 |
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B.
BC |
CC1 |
=1+4+9+2×1×2×cos60°+0+0
=16,
∴|
AC1 |
故选B. |
AC1 |
故选B.
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