空间几何体的外接球与内切球练习题(9)

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所属分类:空间几何体

1.若三棱锥$S - ABC$的三条侧棱两两垂直,且$SA = 2$,$SB = SC = 4$,则该三棱锥的外接球半径为( )

A. $3$         B. $6$         C. $36$         D. $9$

解:${(2R)^2} = \sqrt {4 + 16 + 16} = 6$,$R = 3$

【三棱锥有一侧棱垂直于底面,且底面是直角三角形】【共两种】

2.三棱锥$S - ABC$中,侧棱$SA \perp $平面$ABC$,底面$ABC$是边长为$\sqrt 3 $的正三角形,$SA = 2\sqrt 3 $,则该三棱锥的外接球体积等于________.

解析:$2r = \dfrac{{\sqrt 3 }}{{\sin {{60}^ \circ }}} = 2$,${(2R)^2} = 4 + 12 = 16$,${R^2} = 4$,$R = 2$,

外接球体积$\dfrac{4}{3}\pi \cdot 8 = \dfrac{{32\pi }}{3}$

【外心法(加中垂线)找球心;正弦定理求球小圆半径】

3.正三棱锥$S - ABC$中,底面$ABC$是边长为$\sqrt 3 $的正三角形,侧棱长为$2$,则该三棱锥的外接球体积等于________.

解析:$\triangle ABC$外接圆的半径为$1$,三棱锥$S - ABC$的直径为$2R = \dfrac{2}{{\sin {{60}^ \circ }}} = \dfrac{4}{{\sqrt 3 }}$,外接球半径$R = \dfrac{2}{{\sqrt 3 }}$,

或${R^2} = {(R - \sqrt 3 )^2} + 1$,$R = \dfrac{2}{{\sqrt 3 }}$,外接球体积$V = \dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi \cdot \dfrac{8}{{3\sqrt 3 }} = \dfrac{{32\sqrt 3 \pi }}{{27}}$,

4.三棱锥$P - ABC$中,平面$PAC \perp $平面$ABC$,$\triangle PAC$边长为$2$的正三角形,$AB \perp BC$,则三棱锥$P - ABC$外接球的半径为________.

解析:$\triangle {PAC}$的外接圆是大圆,$2R = \dfrac{2}{{\sin {{60}^ \circ }}} = \dfrac{4}{{\sqrt 3 }}$,$R = \dfrac{2}{{\sqrt 3 }}$,

5.三棱锥$P - ABC$中,平面$PAC \perp $平面$ABC$,$AC = 2$,$PA = PC = 3$,$AB \perp BC$,则三棱锥$P - ABC$外接球的半径为________.

解析:$\cos \angle P = \dfrac{{P{A^2} + P{C^2} - A{C^2}}}{{2PA \cdot PC}} = \dfrac{{9 + 9 - 4}}{{2 \cdot 3 \cdot 3}} = \dfrac{7}{9}$,

${\sin ^2}\angle P = 1 - {(\dfrac{7}{9})^2} = \dfrac{{16 \cdot 2}}{{81}}$,$\sin \angle P = \dfrac{{4\sqrt 2 }}{9}$,

$2R = \dfrac{2}{{\dfrac{{4\sqrt 2 }}{9}}} = \dfrac{9}{{2\sqrt 2 }} = \dfrac{{9\sqrt 2 }}{4}$,$R = \dfrac{{9\sqrt 2 }}{8}$

6.三棱锥$P - ABC$中,平面$PAC \perp $平面$ABC$,$AC = 2$,$PA \perp PC$,$AB \perp BC$,则三棱锥$P - ABC$外接球的半径为________.

解:$AC$是公共的斜边,$AC$的中点是球心$O$,球半径为$R = 1.$

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