Question

（本小题满分12分）

         已知四棱锥的三视图如图所示，为正三角形．

         （Ⅰ）在平面中作一条与底面平行的直线,并说明理由；

         （Ⅱ）求证：平面；

         （Ⅲ）求三棱锥的高．

 

Answer 1

【答案】

（Ⅰ）分别取中点，连结，则即为所求，下证之：·········· 1分

∵　分别为中点，

∴　．················································ 2分

∵　平面，平面，··· 3分

∴　平面．··································· 4分

（作法不唯一）

（Ⅱ）见解析；（Ⅲ）　．

【解析】（I）在平面PCD内作一条与CD平行的直线即可.可以考虑作三角形PCD的中位线.

（II）由于PA垂直AC，所以只须证AC垂直AB即可.可以利用勾股定理进行证明.

（III）求三棱锥的高可以考虑其特殊性，采用换底的方法利用体积法求解是一条比较好的求解方法.本小题可以考虑利用进行求解.

（Ⅰ）分别取中点，连结，则即为所求，下证之：·········· 1分

∵　分别为中点，

∴　．················································ 2分

∵　平面，平面，··· 3分

∴　平面．··································· 4分

（作法不唯一）

（Ⅱ）由三视图可知，平面，，四边形为直角梯形．

过点作于，则,．

∴　,，

∴　，故．······························································· 6分

∵　平面,平面,

∴　.···································································································· 7分

∵　，

∴　平面．······················································································· 8分

（Ⅲ）∵　为正三角形，

∴　．

在中，．

∴　，··································· 10分

（其中为三棱锥的高）．

························································································································ 11分

∵　，

∴　．        12分