Question

（本小題滿分12分）

         已知四棱錐的三視圖如圖所示，為正三角形．

         （Ⅰ）在平面中作一條與底面平行的直線,并說明理由；

         （Ⅱ）求證：平面；

         （Ⅲ）求三棱錐的高．

 

Answer 1

【答案】

（Ⅰ）分別取中點(diǎn)，連結(jié)，則即為所求，下證之：·········· 1分

∵　分別為中點(diǎn)，

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

∵　平面，平面，··· 3分

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

（作法不唯一）

（Ⅱ）見解析；（Ⅲ）　．

【解析】（I）在平面PCD內(nèi)作一條與CD平行的直線即可.可以考慮作三角形PCD的中位線.

（II）由于PA垂直AC，所以只須證AC垂直AB即可.可以利用勾股定理進(jìn)行證明.

（III）求三棱錐的高可以考慮其特殊性，采用換底的方法利用體積法求解是一條比較好的求解方法.本小題可以考慮利用進(jìn)行求解.

（Ⅰ）分別取中點(diǎn)，連結(jié)，則即為所求，下證之：·········· 1分

∵　分別為中點(diǎn)，

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

∵　平面，平面，··· 3分

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

（作法不唯一）

（Ⅱ）由三視圖可知，平面，，四邊形為直角梯形．

過點(diǎn)作于，則,．

∴　,，

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

∵　平面,平面,

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

∵　，

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

（Ⅲ）∵　為正三角形，

∴　．

在中，．

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

（其中為三棱錐的高）．

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

∵　，

∴　．        12分