（本小题满分 13 分） 图5所示的几何体是将高为 2，底…——2011 高考数学第 17 题答案解析

$\because O_{1}^{\prime} H^{\prime}=A^{\prime} O_{1}^{\prime}$
$\therefore O_{1}^{\prime} H^{\prime} / / O_{2}^{\prime} B^{\prime}$ ，四边形 $O_{1}^{\prime} O_{2}^{\prime} B^{\prime} H^{\prime}$ 是平行四边形

$\therefore O_{1}^{\prime} O_{2}^{\prime} / / H^{\prime} B^{\prime}$
$\because O_{1}^{\prime} O_{2}^{\prime} \perp O_{2} O_{2}^{\prime}, \quad O_{1}^{\prime} O_{2}^{\prime} \perp B^{\prime} O_{2}^{\prime}, \quad O_{2} O_{2}^{\prime} \cap B^{\prime} O_{2}^{\prime}=O_{2}^{\prime}$
$\therefore O_{1}^{\prime} O_{2}^{\prime} \perp$ 面 $O_{2} O_{2}^{\prime} B^{\prime} B$
$\therefore H^{\prime} B^{\prime} \perp$ 面 $O_{2} O_{2}{ }^{\prime} B^{\prime} B, \quad B O_{2}{ }^{\prime} \subset$ 面 $O_{2} O_{2}{ }^{\prime} B^{\prime} B$
$\therefore B O_{2}{ }^{\prime} \perp H^{\prime} B^{\prime}$
易知四边形 $A A^{\prime} H^{\prime} H$ 是正方形，且边长 $A A^{\prime}=2$
$\because \tan \angle H O_{1}^{\prime} H^{\prime}=\frac{H H^{\prime}}{O_{1}^{\prime} H^{\prime}}=2, \tan \angle A^{\prime} H^{\prime} G=\frac{A^{\prime} G}{A^{\prime} H^{\prime}}=\frac{1}{2}$
$\therefore \tan \angle H O_{1}^{\prime} H^{\prime} \cdot \tan \angle A^{\prime} H^{\prime} G=1$
$\therefore \angle H O_{1}^{\prime} H^{\prime}+\angle A^{\prime} H^{\prime} G=90^{\circ}$
$\therefore H O_{1}^{\prime} \perp H^{\prime} G$
易知 $O_{1}^{\prime} O_{2}^{\prime} / / H B$ ，四边形 $O_{1}^{\prime} O_{2}^{\prime} B H$ 是平行四边形
$\therefore B O_{2}{ }^{\prime} / / H O_{1}{ }^{\prime}$
$\therefore B O_{2}{ }^{\prime} \perp H^{\prime} G, \quad H^{\prime} G \bigcap H^{\prime} B^{\prime}=H^{\prime}$
$\therefore B O_{2}{ }^{\prime} \perp$ 平面 $H^{\prime} B^{\prime} G$.