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(1)见解析 (2)
(1)证明:在△ABC中,EF是等腰直角△ABC的中位线,
∴EF⊥AC,
在四棱锥A′
![](https://img.yulucn.com/upload/6/54/654524cf18da1b427704f2259543f148_thumb.jpg)
BCEF中,EF⊥A′E,EF⊥EC,
又EC∩A′E=E,∴EF⊥平面A′EC,
又A′C⊂平面A′EC,
∴EF⊥A′C.
(2)解:在直角梯形BCEF中,EC=2,BC=4,
∴S
△FBC =
![](https://img.yulucn.com/upload/4/b0/4b0acbbdd2325abb49e6dab7a57fb095_thumb.jpg)
BC·EC=4,
∵A′O⊥平面BCEF,
∴A′O⊥EC,
又∵O为EC的中点,
∴△A′EC为正三角形,边长为2,
∴A′O=
![](https://img.yulucn.com/upload/b/38/b387cceb7248525b58b6161b132b90a3_thumb.jpg)
,
∴
![](https://img.yulucn.com/upload/7/ba/7bab4f2696e3833d56b39f41af67e1e5_thumb.jpg)
=
![](https://img.yulucn.com/upload/9/8b/98b0e1836d27937b0042f19f05be7337_thumb.jpg)
=
![](https://img.yulucn.com/upload/7/c3/7c348d577cacb4da59d8eaa0681af591_thumb.jpg)
S
△FBC ·A′O=
![](https://img.yulucn.com/upload/7/c3/7c348d577cacb4da59d8eaa0681af591_thumb.jpg)
×4×
![](https://img.yulucn.com/upload/b/38/b387cceb7248525b58b6161b132b90a3_thumb.jpg)
=
![](https://img.yulucn.com/upload/2/1b/21b5d21a3d4f73da872d8779a410f452_thumb.jpg)
.
1年前
5