《2025年5年高考3年模拟高中数学全一册人教B版》

7.(2023全国甲文,16,5分,中)在正方体$ABCD - A_1B_1C_1D_1$中,$AB = 4$,$O$为$AC_1$的中点,若该正方体的棱与球$O$的球面有公共点,则球$O$的半径的取值范围是________.

答案： 答案$[2\sqrt{2},2\sqrt{3}]$
解析 由该正方体的棱与球$O$的球面有公共点，可知最小球为与棱相切的球，最大球为正方体的外接球.当球与棱相切时，设球的半径为$R_{1}$，有$2R_{1}=\sqrt{2}\times4$，$\therefore R_{1}=2\sqrt{2}$，当球为正方体的外接球时，设球的半径为$R_{2}$，有$(2R_{2})^{2}=4^{2}+4^{2}+4^{2}$，$\therefore R_{2}=2\sqrt{3}$，所以球$O$的半径的取值范围是$[2\sqrt{2},2\sqrt{3}]$.

8.(2020新高考Ⅰ,16,5分,难)已知直四棱柱$ABCD - A_1B_1C_1D_1$的棱长均为2,$\angle BAD = 60^{\circ}$.以$D_1$为球心,$\sqrt{5}$为半径的球面与侧面$BCC_1B_1$的交线长为________.

答案：
答案$\frac{\sqrt{2}\pi}{2}$
解析 易知四边形$A_{1}B_{1}C_{1}D_{1}$为菱形，$\angle B_{1}A_{1}D_{1}=60^{\circ}$，连接$B_{1}D_{1}$，则$\triangle B_{1}C_{1}D_{1}$为正三角形，取$B_{1}C_{1}$的中点$O$，连接$D_{1}O$，易得$D_{1}O\perp B_{1}C_{1}$，又$D_{1}O\perp BB_{1}$，$BB_{1}\cap B_{1}C_{1}=B_{1}$，$\therefore D_{1}O\perp$平面$BCC_{1}B_{1}$，取$BB_{1}$的中点$E$，$CC_{1}$的中点$F$，连接$D_{1}E,D_{1}F,OE,OF,EF$，易知$D_{1}E = D_{1}F=\sqrt{5}$，易知以$D_{1}$为球心，$\sqrt{5}$为半径的球面与侧面$BCC_{1}B_{1}$的交线是以$O$为圆心，$OE$为半径的$\overset{\frown}{EF}$(难点：确定交线)，$\because B_{1}E = B_{1}O = 1$，$\therefore OE=\sqrt{2}$，同理$OF=\sqrt{2}$，易知$EF = 2$，$\therefore\angle EOF = 90^{\circ}$，$\therefore\overset{\frown}{EF}$的长$=\frac{1}{4}\times(2\pi\times\sqrt{2})=\frac{\sqrt{2}\pi}{2}$.

1.(2021新高考Ⅱ,5,5分,易)正四棱台的上、下底面的边长为2,4,侧棱长为2,则四棱台的体积为（ ）
A.56
B.$28\sqrt{2}$
C.$\frac{56}{3}$
D.$\frac{28\sqrt{2}}{3}$

答案：
D 如图，正四棱台$ABCD - A_{1}B_{1}C_{1}D_{1}$，$O$与$O_{1}$分别为上、下底面的中心，则四边形$OO_{1}B_{1}B$为直角梯形，且$OB=\sqrt{2}$，$O_{1}B_{1}=2\sqrt{2}$，$BB_{1}=2$，过点$B$作$BM\perp O_{1}B_{1}$，垂足为$M$，可得$BM=\sqrt{2}$，即正四棱台的高为$\sqrt{2}$.故正四棱台的体积$V=\frac{1}{3}\times\sqrt{2}\times(4 + 8+16)=\frac{28\sqrt{2}}{3}$.选D.

2.(2022新高考Ⅰ,4,5分,易)南水北调工程缓解了北方一些地区水资源短缺问题,其中一部分水蓄入某水库.已知该水库水位为海拔148.5 m时,相应水面的面积为140.0 km²；水位为海拔157.5 m时,相应水面的面积为180.0 km².将该水库在这两个水位间的形状看作一个棱台,则该水库水位从海拔148.5 m上升到157.5 m时,增加的水量约为（$\sqrt{7}\approx2.65$）（ ）
A.$1.0\times10^9 m^3$
B.$1.2\times10^9 m^3$
C.$1.4\times10^9 m^3$
D.$1.6\times10^9 m^3$

答案： C 由该水库在这两个水位间的形状看作一个棱台，增加的水量即为棱台的体积.$140km^{2}=140\times10^{6}m^{2}$，$180km^{2}=180\times10^{6}m^{2}$，由棱台的体积公式$V=\frac{1}{3}(S+\sqrt{SS'}+S')h$，可得$V_{增加水量}=\frac{140\times10^{6}+180\times10^{6}+\sqrt{140\times10^{6}\times180\times10^{6}}}{3}\times(157.5 - 148.5)=\frac{(140 + 180+60\sqrt{7})\times10^{6}\times9}{3}\approx(320 + 60\times2.65)\times10^{6}\times3=1437\times10^{6}\approx1.4\times10^{9}m^{3}$.故选C.
名师点睛 水库水位变化即为棱台的高.

3.(2023全国甲文,10,5分,中)在三棱锥$P - ABC$中,$\triangle ABC$是边长为2的等边三角形,$PA = PB = 2$,$PC = \sqrt{6}$,则该棱锥的体积为（ ）
A.1
B.$\sqrt{3}$
C.2
D.3

答案：
A 取$AB$的中点$D$，连接$CD,PD$，$\because\triangle ABC,\triangle PAB$为等边三角形，$\therefore CD\perp AB,PD\perp AB$，$\because\triangle ABC$的边长为$2$，$PA = PB = 2$，$\therefore PD = CD=\sqrt{3}$，又$\because PC=\sqrt{6}$，$\therefore PD^{2}+CD^{2}=PC^{2}$，$\therefore PD\perp CD$，又$\because CD\cap AB = D$，$CD,AB\subset$平面$ABC$，$\therefore PD\perp$平面$ABC$，$\therefore PD$为三棱锥$P - ABC$的高，$\therefore V_{P - ABC}=\frac{1}{3}S_{\triangle ABC}\cdot PD=\frac{1}{3}\times\frac{1}{2}\times2\times\sqrt{3}\times\sqrt{3}=1$，故选A.
一题多解
(分割求和法)取$AB$的中点$D$，连接$CD,PD$，$\because\triangle ABC,\triangle PAB$为等边三角形，$\therefore CD\perp AB,PD\perp AB$.又$\because CD\cap PD = D$，$\therefore AB\perp$平面$PCD$，$\because\triangle ABC$的边长为$2$，$PA = PB = 2$，$\therefore PD = CD=\sqrt{3}$，又$\because PC=\sqrt{6}$，$\therefore PD^{2}+CD^{2}=PC^{2}$，$\therefore PD\perp CD$，$\therefore S_{\triangle PCD}=\frac{1}{2}PD\cdot CD=\frac{1}{2}\times(\sqrt{3})^{2}=\frac{3}{2}$，$\therefore V_{P - ABC}=V_{A - PCD}+V_{B - PCD}=\frac{1}{3}S_{\triangle PCD}\cdot(AD + BD)=\frac{1}{3}\times\frac{3}{2}\times2 = 1$，故选A.

4.(2022新高考Ⅱ,7,5分,难)已知正三棱台的高为1,上、下底面边长分别为$3\sqrt{3}$和$4\sqrt{3}$,其顶点都在同一球面上,则该球的表面积为（ ）
A.100π
B.128π
C.144π
D.192π

答案：
A 设正三棱台为$A'B'C'-ABC$，$\triangle A'B'C'$，$\triangle ABC$的外心分别为$D',D$，则$A'D' = 3$，$AD = 4$，又知$D'D = 1$，所以正三棱台的外接球球心在线段$D'D$的延长线上，设球心为$O$，半径为$R$，如图所示，在$Rt\triangle A'D'O$中，$R^{2}=3^{2}+(DO + 1)^{2}$①，在$Rt\triangle ADO$中，$R^{2}=4^{2}+DO^{2}$②，由①②得$R = 5$，所以该球的表面积为$4\pi\times5^{2}=100\pi$，故选A.

5.(2023全国乙理,8,5分,中)已知圆锥$PO$的底面半径为$\sqrt{3}$,$O$为底面圆心,$PA$,$PB$为圆锥的母线,$\angle AOB = 120^{\circ}$,若$\triangle PAB$的面积等于$\frac{9\sqrt{3}}{4}$,则该圆锥的体积为（ ）
A.π
B.$\sqrt{6}\pi$
C.3π
D.$3\sqrt{6}\pi$

答案： B 设圆锥的母线长为$l$，底面半径为$r$，则$r=\sqrt{3}$.在$\triangle AOB$中，由$\angle AOB = 120^{\circ}$，$OA = OB=\sqrt{3}$，得$AB = 3$.$S_{\triangle PAB}=\frac{1}{2}AB\cdot\sqrt{l^{2}-(\frac{AB}{2})^{2}}=\frac{1}{2}\times3\times\sqrt{l^{2}-\frac{9}{4}}=\frac{9\sqrt{3}}{4}$，所以$l = 3$，则圆锥的高$h=\sqrt{l^{2}-r^{2}}=\sqrt{9 - 3}=\sqrt{6}$，故$V_{圆锥}=\frac{1}{3}S_{底}\cdot h=\frac{1}{3}\pi(\sqrt{3})^{2}\cdot\sqrt{6}=\sqrt{6}\pi$，故选B.

6.(2022新高考Ⅰ,8,5分,难)已知正四棱锥的侧棱长为$l$,其各顶点都在同一球面上.若该球的体积为36π,且$3\leq l\leq3\sqrt{3}$,则该正四棱锥体积的取值范围是（ ）
A.$\left[18,\frac{81}{4}\right]$
B.$\left[\frac{27}{4},\frac{81}{4}\right]$
C.$\left[\frac{27}{4},\frac{64}{3}\right]$
D.$[18,27]$

答案： C 因为球的体积为$36\pi$，所以球的半径$R = 3$，设正四棱锥的底面边长为$2a$，高为$h$，则$l^{2}=2a^{2}+h^{2}$，$3^{2}=2a^{2}+(h - 3)^{2}$，所以正四棱锥的体积$V=\frac{1}{3}Sh=\frac{1}{3}\times4a^{2}\times h=\frac{2}{3}\times(l^{2}-\frac{l^{4}}{36})\times\frac{l^{2}}{6}=\frac{1}{9}(l^{4}-\frac{l^{6}}{36})$，所以$V'=\frac{1}{9}(4l^{3}-\frac{l^{5}}{6})=\frac{1}{9}l^{3}(\frac{24 - l^{2}}{6})$，当$3\leq l＜2\sqrt{6}$时，$V'＞0$，当$2\sqrt{6}＜l\leq3\sqrt{3}$时，$V'＜0$，所以当$l = 2\sqrt{6}$时，正四棱锥的体积$V$取最大值，最大值为$\frac{64}{3}$，当$l = 3$时，$V=\frac{27}{4}$，当$l = 3\sqrt{3}$时，$V=\frac{81}{4}$，所以正四棱锥的体积$V$的最小值为$\frac{27}{4}$，所以该正四棱锥体积的取值范围是$[\frac{27}{4},\frac{64}{3}]$.故选C.

7.(多选)(2023新课标Ⅱ,9,5分,中)已知圆锥的顶点为$P$,底面圆心为$O$,$AB$为底面直径,$\angle APB = 120^{\circ}$,$PA = 2$,点$C$在底面圆周上,且二面角$P - AC - O$为$45^{\circ}$,则（ ）
A.该圆锥的体积为π
B.该圆锥的侧面积为$4\sqrt{3}\pi$
C.$AC = 2\sqrt{2}$
D.$\triangle PAC$的面积为$\sqrt{3}$

答案：
AC 对于A，连接$PO$，$\because PA = PB = 2$，$\angle APB = 120^{\circ}$，$\therefore AB = 2\sqrt{3}$，$PO = 1$，$\therefore$圆锥的体积$V=\frac{1}{3}\pi\times(\sqrt{3})^{2}\times1=\pi$，故A正确.对于B，$S_{侧}=\frac{1}{2}\times2\pi\times\sqrt{3}\times2=2\sqrt{3}\pi$，故B错误.对于C，取$AC$的中点$D$，连接$PD,OD$.$\because OA = OC$，$PA = PC$，$D$为$AC$的中点，$\therefore OD\perp AC$，$PD\perp AC$.$\therefore\angle PDO$即为二面角$P - AC - O$的平面角，$\therefore\angle PDO = 45^{\circ}$，又$PO\perp$底面圆，$OD\subset$底面圆，$\therefore PO\perp OD$，$\therefore PO = DO = 1$，$\therefore PD=\sqrt{2}$，$AC = 2AD = 2\sqrt{OA^{2}-OD^{2}}=2\sqrt{(\sqrt{3})^{2}-1^{2}}=2\sqrt{2}$，故C正确.对于D，$S_{\triangle PAC}=\frac{1}{2}AC\cdot PD=\frac{1}{2}\times2\sqrt{2}\times\sqrt{2}=2$，故D错误.故选AC.

8.(多选)(2022新高考Ⅱ,11,5分,中)如图,四边形$ABCD$为正方形,$ED\perp$平面$ABCD$,$FB// ED$,$AB = ED = 2FB$.记三棱锥$E - ACD$,$F - ABC$,$F - ACE$的体积分别为$V_1$,$V_2$,$V_3$,则（ ）

A.$V_3 = 2V_2$
B.$V_3 = V_1$
C.$V_3 = V_1 + V_2$
D.$2V_3 = 3V_1$

答案：
CD 解法一：因为$ED\perp$平面$ABCD$，且$FB// ED$，所以$FB\perp$平面$ABCD$.设$AB = ED = 2FB = 2a$，则$FB = a$，则$V_{1}=\frac{1}{3}S_{\triangle ACD}\cdot ED=\frac{1}{3}\times\frac{1}{2}\cdot2a\cdot2a\cdot2a=\frac{4}{3}a^{3}$，所以$V_{2}=\frac{1}{2}V_{1}=\frac{2}{3}a^{3}$.如图，连接$BD$，交$AC$于$O$，连接$OE,OF$.易证$AC\perp$平面$BDEF$.$S_{\triangle EOF}=S_{梯形BDEF}-S_{\triangle ODE}-S_{\triangle OBF}=\frac{1}{2}(a + 2a)\times2\sqrt{2}a-\frac{1}{2}\times2a\times\sqrt{2}a-\frac{1}{2}\times a\times\sqrt{2}a=\frac{3\sqrt{2}}{2}a^{2}$，故$V_{3}=\frac{1}{3}S_{\triangle EOF}\cdot AO\times2=\frac{1}{3}\times\frac{3\sqrt{2}}{2}a^{2}\times\sqrt{2}a\times2=2a^{3}$，故$V_{1}+V_{2}=V_{3}$，$2V_{3}=3V_{1}$成立，故选CD.

解法二：所设及$V_{1},V_{2}$的计算同解法一.将原几何体“补成”如图所示的正方体.

则$V_{3}=8a^{3}-V_{1}-V_{2}-V_{E - AA_{1}B_{1}F}-V_{E - CC_{1}B_{1}F}=2a^{3}$，所以$V_{3}=3V_{2}$，$2V_{3}=3V_{1}$，$V_{3}=V_{1}+V_{2}$.所以选CD.

9.(2020新高考Ⅱ,13,5分,易)棱长为2的正方体$ABCD - A_1B_1C_1D_1$中,$M$,$N$分别为棱$BB_1$,$AB$的中点,则三棱锥$A_1 - D_1MN$的体积为________.

答案：
答案$1$
解析 解法一：如图，易知$MN=\sqrt{2}$，连接$A_{1}B$交$MN$于点$O$，则$A_{1}O=\frac{3\sqrt{2}}{2}$，$\therefore V_{A_{1}-DMN}=V_{D_{1}-A_{1}MN}=\frac{1}{3}\times\frac{1}{2}\times\sqrt{2}\times\frac{3\sqrt{2}}{2}\times2 = 1$.
解法二：易知$A_{1}D_{1}\perp$平面$A_{1}MN$，则$A_{1}D_{1}$是三棱锥$D_{1}-A_{1}MN$的高.$\because S_{\triangle A_{1}MN}=S_{正方形AA_{1}B_{1}B}-S_{\triangle AA_{1}N}-S_{\triangle A_{1}B_{1}M}-S_{\triangle BMN}=2\times2-\frac{1}{2}\times2\times1-\frac{1}{2}\times2\times1-\frac{1}{2}\times1\times1=\frac{3}{2}$，$\therefore V_{A_{1}-DMN}=V_{D_{1}-A_{1}MN}=\frac{1}{3}\times2\times\frac{3}{2}=1$.

10.(2023新课标Ⅱ,14,5分,易)底面边长为4的正四棱锥被平行于其底面的平面所截,截去一个底面边长为2,高为3的正四棱锥,所得棱台的体积为________.

答案： 答案$28$
解析 棱台的两底面边长分别为$2$与$4$，高为$3$(由上、下底面边长可知棱台的高与截去的棱锥的高相等)，所以棱台的体积$V=\frac{1}{3}\times(2^{2}+4^{2}+\sqrt{2^{2}\times4^{2}})\times3=28$.

11.(2023新课标Ⅰ,14,5分,易)在正四棱台$ABCD - A_1B_1C_1D_1$中,$AB = 2$,$A_1B_1 = 1$,$AA_1 = \sqrt{2}$,则该棱台的体积为________.

答案：
答案$\frac{7\sqrt{6}}{6}$
解析 解法一：正四棱台$ABCD - A_{1}B_{1}C_{1}D_{1}$如图所示.设其上、下底面中心分别为$O_{1},O$，连接$O_{1}O,O_{1}A_{1},OA$，由正四棱台的定义可知$O_{1}O\perp$平面$ABCD$，$O_{1}O\perp$平面$A_{1}B_{1}C_{1}D_{1}$，四边形$ABCD$，$A_{1}B_{1}C_{1}D_{1}$均为正方形.$\because AB = 2$，$A_{1}B_{1}=1$，$\therefore AO=\sqrt{2}$，$A_{1}O_{1}=\frac{\sqrt{2}}{2}$，易知$AO// A_{1}O_{1}$，又$\because AA_{1}=\sqrt{2}$，$\therefore$在直角梯形$A_{1}O_{1}OA$中，$O_{1}O=\sqrt{(\sqrt{2})^{2}-(\sqrt{2}-\frac{\sqrt{2}}{2})^{2}}=\frac{\sqrt{6}}{2}$，$\therefore V_{正四棱台ABCD - A_{1}B_{1}C_{1}D_{1}}=\frac{1}{3}\times(1 + 4+\sqrt{1\times4})\times\frac{\sqrt{6}}{2}=\frac{7\sqrt{6}}{6}$.

解法二：将正四棱台补形为正四棱锥$P - ABCD$，如图，因为$AB = 2$，$A_{1}B_{1}=1$，所以$A_{1},B_{1},C_{1},D_{1}$分别为$PA,PB,PC,PD$的中点，又$AA_{1}=\sqrt{2}$，所以$PA = PB = PC = PD = 2\sqrt{2}$，过$P$作$PO\perp$平面$ABCD$，连接$OA$，易知$O$为正方形$ABCD$的中心，又因为$AB = 2$，所以$OA=\sqrt{2}$，所以$PO=\sqrt{PA^{2}-OA^{2}}=\sqrt{6}$，所以四棱锥$P - ABCD$的体积为$\frac{1}{3}PO\cdot S_{四边形ABCD}=\frac{1}{3}\times\sqrt{6}\times4=\frac{4\sqrt{6}}{3}$，同理，四棱锥$P - A_{1}B_{1}C_{1}D_{1}$的体积为$\frac{\sqrt{6}}{6}$，所以正四棱台$ABCD - A_{1}B_{1}C_{1}D_{1}$的体积为$\frac{4\sqrt{6}}{3}-\frac{\sqrt{6}}{6}=\frac{7\sqrt{6}}{6}$.