高雄市明誠中學高二數學平時測驗日期

(1)

高雄市明誠中學高二數學平時測驗日期：99.10.21 範

圍

2-1

空間基本概念班級二年____班姓座號

名

一、填充題

(每題10

分 )

1.ABCD為四面體﹐已知AB垂直平面BCD﹐又BD⊥CD﹐CD=3﹒BC=5﹐AB=12﹐ 則AD長為____________﹒

解答 4 10 解析

由圖﹕BD=4﹐∴AD= 12²+4² = 160=4 10﹒

2.金字塔每一個稜長均為a﹐底部正方形﹐側邊正三角形如圖所示﹐θ 為ABP平面與

CBP平面所夾之二面角﹐則cosθ=____________﹒

解答 1

−3 解析

取BP之中點M ﹐則 3 AM =CM = 2 a﹐

又AC= 2a﹐∴cosθ =cos∠AMC

2 2 2 2

2

3 3 1

2 1

4 4 2

3 3

2 2 2 2

a a a a

a a a

+ − −

= = = −

⋅ ⋅

﹒

3.正四面體ABCD一邊AB長為4公分﹐M 為AD的中點﹐N為BC的中點﹐則

(1)MN=____________﹔(2)相鄰二側面形成之兩面角角度為θ ﹐cosθ =_______﹒ 解答 (1) 2 2 ;(2)1

3 解析

(1)△AND中﹐AN =DN﹐ 中線NM ⊥AD﹐ △ANM 中﹐AM =2﹐AN=2 3﹐

∴NM² =AN²−AM²=12− =4 8﹐ ∴NM = 8 =2 2﹒ (2)△BCM中﹐

2 2 2

cos 2

BM CM BC BM CM

θ = ⁺ ⁻

( ) ( )

^{2 3} ² ^{2 3} ² ⁴² 1 2 2 3 2 3 3

+ −

= =

⋅ ⋅ ﹒

4. 如圖﹐ABCD為四面體﹐已知AD垂直於平面BCD﹐BC⊥BD﹐BC=7﹐ 24

AB= ﹐AD=15﹐則(1)AC之長為____________﹔

(2)若平面ABD與平面ACD之夾角為θ ﹐則sinθ之值為____________﹒

(2)

解答 (1)25;(2) 7 20

解析 (1)AD⊥平面BCD﹐BC⊥BD﹐由三垂線定理知AB⊥BC﹐ 故AC= AB²+BC² = 24²+7² =25﹒

(2)∵AD⊥平面BCD﹐∴BD⊥AD且CD⊥AD﹐

故∠BDC為平面ABD與平面ACD之夾角﹐ △BCD中﹐

2 2

7 7

sin 25 15 20

BC

θ =CD= =

− ﹒

5.如圖﹐設A﹐B﹐C在平面E上﹐且AB⊥BC﹐另設PA ⊥平面E於A﹐已知

8

PA= ﹐AB=6﹐BC=24﹐求PC=____________﹒

解答 26

解析 △PAB中﹐PB= 8²+6² =10﹐

∵PA⊥E﹐AB⊥BC﹐由三垂線定理知PB⊥BC﹐

△PBC中﹐PC= PB²+BC² = 10²+24² =26﹒

6.△ABC是邊長為4的正三角形﹐D﹐E﹐F為三邊中點﹐沿DE﹐EF﹐FD上摺﹐使A﹐B﹐C三點重疊在P點成為一個正四面體PDEF﹐則此四面體頂點P 與底面DEF的高度為__﹒

解答 2 6 3

解析 2 2 3

3 3

DH = DM= ﹐∴ 4

4 3

PH= − 8 2 6

3 3

= = ﹒

7.設正四面體ABCD﹐其每一稜長均為a﹐已知AB之中點為M ﹐CD之中點為N﹐求

(1)AB AN

 

⋅ =____________﹔(2)MN 之長為____________﹒

解答 (1)

2

a ;(2) 2 2 a 解析

(1) 1 1

2 2

AB AN⋅ =AB⋅ AC+ AD

    

¹ ¹

2AB AC 2AB AD

=

   

⋅ + ⋅ ¹

(

^{cos 60}

)

¹

(

^{cos 60}

)

2 a a 2 a a

= ⋅ ⋅ ° + ⋅ ⋅ ° ²

2

=a ﹒

(2) 3

AN= 2 a﹐∴ 3 ² 1 ² 2

4 4 2

MN= a − a = a﹒

8. 如右圖﹐其中底面為邊長是2的正方形﹐四個側面是腰為3的等腰三角形﹐若底面與相鄰兩側面所形成的二面角角度為θ﹐則cosθ =____________﹒

解答 2 4

(3)

解析

取M 為BC中點﹐A在BCDE平面之投影H ﹐則∠AMH=θ﹐ 又AM = 3²−1² =2 2﹐ MH =1﹐∴ 1 2

cosθ=2 2 = 4 ﹒

9. 如右圖﹐正立方體ABCD−EFGH﹐則

(1)將正立方體的十二個稜延長為直線﹐則與AC



互為歪斜線的直線有______條﹔

(2)若平面ACH和平面ACD的夾角為θ﹐求cosθ=____________﹒

解答 (1)6;(2) 3 3

解析 (1)BF



_﹐_DH



_﹐_EF



_﹐_HG



_﹐_EH



_﹐_FG



_共₆_條﹒

(2)

2

1 3

cos 2

6 3 3

2 a a

θ = = = ﹒

10.有一側稜長均為8的金字塔形﹐其側面為四個等腰三角形﹐底面是邊長為6的

正方形﹐若底面與側面之夾角為α﹐則cosα =____________﹒

解答 3 55 解析

64 9 55

AM = − = ﹐

( ) ( )

² ² ^{55 9} ⁴⁶

AH = AM − HM = − = ﹐∴ 3

cos 55

HM AM

α = = ﹒

11.設點O在平面E上之投影點為A﹐L為平面E上不過A的直線﹐A在L上之投影

點為B﹐點C在L上﹐BC=12﹐OC=13﹐AB=4﹐則O至平面E之距離為______﹒

解答 3

解析 ^{d O E}

(

^,

)

⁼^OA⁼³^﹒

12.已知四面體ABCD中﹐AD垂直平面BCD於D﹐BD⊥BC於B﹐AD=4﹐BC=3﹐

5

BD= ﹐則AC=____________﹒

解答 5 2 解析

(4)

由三垂線定理﹐∵AD⊥平面BCD﹐ DB⊥BC﹐∴AB⊥BC﹐

∴AB= 4²+5² = 41﹐ ^AC⁼

( )

⁴¹ ²⁺³² ⁼ ⁵⁰⁼^{5 2}^﹒

13.正四面體OABC的各稜長為6﹐點O在底面ABC上的正射影為H ﹐則

(1)OH 長為____________﹔(2)四面體體積為____________﹒

解答 (1) 2 6 ;(2)18 2 解析

(1) 2 2

3 3 2 3

3 3

AH= AM = ⋅ = ﹐ 故^OH ⁼ ⁶²⁻

( )

^{2 3} ² ⁼^{2 6}^﹒

(2)體積 1 3 ²

6 2 6

3 4

 

=  ⋅ ⋅

  18 2= ﹒

14.A﹐B﹐C﹐D為一個正四面體之四頂點﹐θ 為AB



_與_CD



之夾角﹐則cosθ=____﹒

解答 0

解析 AB CD

    

⋅ =AB⋅AD AC−  =AB AD AB AC

   

⋅ − ⋅ =0﹐∴cos AB CD 0 AB CD θ=

 

^⋅ =

 

^﹒

15.設OA⊥平面E於A﹐直線L在E上﹐AB⊥L於B﹐C為L上一點﹐若OA=4﹐

3

AB= ﹐∠BOC= °30 ﹐則BC=____________﹒

解答 5 3 3 解析

OA⊥E﹐AB⊥ ⇒L OB⊥L﹐∴OB= 4²+3² =5 1 5 3

5 3 3

⇒BC= ⋅ = ﹒

16.三角錐（四面體）ABCD﹐頂點A﹐底面為△BCD﹐已知AB=AC=AD= 21﹐

底邊BC=CD=DB=6﹐求

(1)平面ABC與底面BCD的銳夾角為___________﹔

(2)若AH垂直於底面BCD於H ﹐則高AH的長為__________﹒

解答 (1) 60°;(2)3 解析

(1)AM = 21 9− =2 3﹐ DM =3 3﹐

∴ 12 27 21 18 1

cosθ =2 2 3 3 3⁺ ⁻ =36=2

⋅ ⋅ ﹐ ∴θ =60°﹒

(2) 2

3 2 3

DH = DM = ﹐∴AH = 21 12− =3﹒

17.有一正四角錐﹐底面是邊長10公分的正方形﹐側面是腰長13公分的等腰三角

θ θ

(5)

解答 5 12

解析由圖﹕AM =12﹐ MH=5﹐∴ 5

cos 12

MH

θ= AM = ﹒

18. 有一個底為正方形的直角錐﹐每一稜長都是10﹐設□ABCD為其底﹐O為其錐頂﹐(1)求此直角錐之高=____________﹔

(2)若兩側面之夾角為θ﹐則cosθ=____________﹒

解答 (1) 5 2 ;(2) 1

−3 解析

(1) 1

2 5 2

CH = AC= ﹐OC=10

2 2

OH OC CH

⇒ = − = 100 50− =5 2﹒

(2)取OB中點M﹐連接AM﹐CM﹐

△AMC中﹐AM =CM =5 3﹐AC=10 2﹐

( ) ( ) (

^{5 3} ² ^{5 3} ² ^{10 2}

)

²

cosθ= 2 5 3 5 3⁺ ⁻

⋅ ⋅

1

= −3﹒

19.圖為一單位正立方體ABCD-EFGH（即稜長1）﹐則四面體A-CFH的表面

積為____________﹒

解答 12

解析如圖﹐四面體ACHF之四面均為邊長是 2的正三角形﹐

所以表面積為⁴^⋅ ₄³^⋅

( )

² ²⁼^{2 3}⁼ ¹²^﹒

20.正八面體各稜長均為a﹐相鄰二側面之夾角為θ﹐求cosθ=____________﹒

解答 1

−3 解析

取AC中點M ﹐∵各面均為正△﹐∴BM ⊥AC﹐DM ⊥AC﹐

∴∠BMD為二側面之夾角﹒

△BDM 中﹐ 3

BM = 2 a﹐ 3

DM = 2 a﹐BD= 2a﹐

(6)

∴

2 2 2

cos 2

BM DM BD BM DM

θ= ⁺ ⁻

⋅ ⋅

2 2

( )

3 3 2

2 2 2 1

3 3 3

2 2 2

a a a

a a

   

+ −

   

   

= = −

⋅ ⋅

﹒

21.若2θ為正八面體相鄰二面所成之二面角﹐則cosθ=____________﹒

解答 3 3 解析

設正八面體之中心為O﹐CD中點為M ﹐稜長為a﹐

△PCD為正△﹐∴ 3 PM = 2 a﹐

2 OM =a﹐

∴cos 2 1

3 3

2 a OM

PM a

θ = = = 3

= 3 ﹒

22.不共面三射線OX



_﹐OY



_﹐OZ



_互成_30°_角﹐_P_∈_OX



_﹐_OP₌₂_﹐_P_至平面_YOZ_之投影為_Q_﹐Q_至

OY



_之垂足為_R_﹐又_QR



_交_OZ



_於_S_﹐求_PS²₊_OR²₌____________﹒

解答 11 4 3− 解析

∵PQ⊥平面OYZ ﹐QR⊥OY﹐∴PR⊥OY﹐

△OPR中﹐OR=OP⋅cos 30° 3

2 3

= ⋅ 2 = ﹐

△OSR中﹐ 2

sec30 3 2

OS=OR⋅ ° = ⋅ 3 = ﹐

△OPS中﹐由餘弦定理PS² =OP²+OS²−2OP OS⋅ ⋅cos 30°

2 2 3

2 2 2 2 2 8 4 3

= + − ⋅ ⋅ ⋅ 2 = − ﹐ ∴^PS²⁺^OR²^{= −}⁸ ^{4 3}⁺

( )

³ ² ⁼^{11 4 3}⁻ ^﹒

23.一長方體ABCD−EFGH﹐如圖﹐EF =2﹐EH=3﹐EA=4﹐又AG與DF之

交角θ ﹐則sinθ=____________﹒

解答 12 5 29

解析 DF=AG= 4 9 16+ + = 29﹐AF= 4 16+ = 20﹐ 令DF交AG於O﹐△OAF中﹐

(7)

2 2 2 29 29 4 4 20

cos 2 2 1 29 1 29

2 2

OA OF AF OA OF

θ = ⁺ ⁻ = ⁺ ⁻

⋅ ⋅ ⋅

11

= −29 12 5 sinθ 29

⇒ = ﹒

24. 有公共底邊的兩個等腰三角形﹐它們所在的平面組成60°的二面角﹐公共邊長為16﹐一個三角形的腰長為17﹐另一個三角形的兩腰互相垂直﹐求這兩個等腰三角形頂點間的距離為____________﹒

解答 13

解析 AB=16﹐AC=BC=17﹐設AB中點為M﹐則∠CMD=60°﹐ 直角△ACM 中﹐ 1

2 8

AM = AB= ﹐∴CM = AC²−CM² = 17²−8² =15﹐

△ADM中﹐直角∆ADB的斜邊中線 1 2 8

DM = AB=AM= ﹐

△CMD中﹐餘弦定理CD²=DM²+CM²−2DM CM⋅ ⋅cos 60°

1 64 225 2 8 15 169

= + − ⋅ ⋅ ⋅ =2 ﹐CD=13﹒

25.正四面體ABCD﹐一稜長為10﹐有一隻螞蟻由點A﹐沿△ABC﹐△BCD﹐△ABD﹐△ACD之

順序﹐在側面上移動﹐終點為C﹐則移動之最短距離為____________﹒

解答 10 7 解析

將正四面體攤開如圖﹐其爬行之最短距離為

( )

²⁰ ² ¹⁰² 2 20 10 cos120 700

AC= + − ⋅ ⋅ ⋅ ° = =10 7﹒

26. 右圖是一個四角錐﹐它的底面ABCD是一個邊長為2的正方形﹐四個側面均是腰長為 3的等腰三角形﹐則

(1)此四角錐的體積為____________﹔

(2)相鄰兩側面的夾角為θ﹐sinθ=____________﹒

解答 (1)4

3;(2) 3 2 解析

(1)體積 1 ² 4

3 2 1 3

= ⋅ ⋅ = ﹒

(2)(i)△ 1

2 2 2

OAB= ⋅ ⋅2 = ﹐

又△OAB 1 1

2 3

2OB AH 2 AH

= ⋅ ⇒ = ⋅ ⋅ ﹐ ∴ 2 2 AH = 3 ﹒

(8)

(ii) 2 2

CH=AH = 3 ﹐AC=2 2﹐

∴

8 8 3 3 8

cos 2 2 2 2

2 3 3

θ= ^{+ −}

⋅ ⋅

8 3 1

16 2

3

−

= = − ﹐ 故 3 sinθ= 2 ﹒

27. 右圖是邊長為a的正立方體﹐求 (1)四面體ABCD的體積____________﹔

(2)AC



在平面ABD上的正射影長=____________﹔

(3)若平面ABD與平面BCD之夾角為θ ﹐求sinθ =____________﹒

解答 (1)1 ³ 6a ;(2) 6

3 a;(3) 6 3 解析

(1)所求 1 1 1 ³

3 2 a a a 6a

=  ⋅ ⋅  = ﹒

(2)設C到平面ABD的距離為h ABCD的體積 1

=3△ABD h⋅ 1 ³ 1 3 ² 6a 3  4 2a  h

⇒ = ⋅ ⋅ ⋅

 

3 h a

⇒ = ﹐

∴所求

2 2 2 2 2 6

3 3 3

AC h a a a a

= − = − = = ﹒

(3)設M 為BD中點 2

CM= 2 a﹐ 3 6

2 2 2

AM = ⋅ a= a﹐

∴

2 2

3 2

2 2 1 cos

2 6 3

2 2 2

a a a a a

θ = ⁺ ⁻ =

⋅ ⋅

2 6

sinθ 3 3

⇒ = = ﹒

28.四面體ABCD中﹐AB=AC=AD=9﹐BC=CD=BD=6﹐則直線AB與直線CD

的距離為____________﹒

解答 23 解析

(i)取CD之中點M ﹐作MN⊥AB﹒

(ii) 3

6 3 3 BM = 2 ⋅ = ﹐

2 2

3 3 2 3

BH = BM = ⋅ = ⇒ AH= 81 12− = 69﹒

(9)

(iii)△ 1

ABM =2BM AH⋅ 1

2AB MN

= ⋅ ⇒BM AH⋅ =AB MN⋅

⇒3 3 ⋅ 69= ⋅9 MN⇒9 23=9MN﹐故MN= 23﹒

29. 四面體ABCD中﹐AB=AC=BC=BD=CD=4

(1) 當AD=2時﹐若相鄰兩面ABC與BCD的夾角為θ ﹐則cosθ =____________﹔

(2) 當四面體ABCD有最大體積時﹐AD的長為____________﹒

解答 (1)5

6;(2) 2 6 解析

(1)設BC的中點為M﹐ ∴ 3

4 2 3

DM = 2 ⋅ = =AM﹐

∴ 12 12 4 20 5

cosθ =2 2 3 2 3⁺ ⁻ =24=6

⋅ ⋅ ﹒

(2)∵△ 3 ²

4 4 3

BCD= 4 ⋅ = ﹐ ∴ABCD體積 1

= ⋅3 △BCD AH⋅ 4 3

3 AH

= ⋅ （∴AH越大越好）

（由三垂線定理中知H 落在DM



_上） ^{4 3}

3 AM

≤ ⋅

（即平面ABC⊥平面BCD

時，

AH =AM =2 3最大）﹐

∴AD= AM²+MD² = 12 12+ =2 6﹒

30.立方體ABCD−EFGH之稜長為a﹐GF﹐GH之中點為M ﹐N﹐則四面體AEMN

之體積為____________﹒

解答

3

8 a

解析 △EMN =□EFGH−△EFM −△EHN−△MGN

² 1 1 3 ²

2 2 2 2 2 2 8

a a a

a a a

= − ⋅ ⋅ ⋅ − ⋅ ⋅ = ﹐ 故四面體AEMN之體積為1

3⋅△

3

1 3 2

3 8 8

EMN AE⋅ = ⋅ a ⋅ =a a ﹒

31.一正立方體之稜長為a﹐共頂點的三稜為AB﹐AC﹐AD﹐則A到平面BCD的距離為__________﹒

解答 3 3 a 解析

設A到平面BCD之距離為h﹐則四面體ABCD體積=四面體CABD體積

(10)

( )

² ²

1 3 1

3 4 2 3 2

a h a a

⇒ ⋅ ⋅ = ⋅ ⋅ 3

h 3 a

⇒ = ﹐∴A到平面BCD之距離為 3 3 a﹒ 32. 如右圖﹐長方體ABCD−EFGH中﹐AE=1﹐AB=2﹐AD=3

(1)有一蜜蜂從A點飛到G點﹐其飛行的最短距離為____________﹔

(2)有一螞蟻從A點爬到G點﹐其爬行的最短距離為____________﹒

解答 (1) 14 ;(2) 18 解析

(1)長方體對角線長ÂG⁼ ÂE²⁺ÊG² ⁼ ÂE²⁺

(

^EF²⁺^FG²

)

⁼ ¹²⁺²²⁺³² ⁼ ¹⁴^﹒

(2)(i)將矩形DCGH 沿DH攤開﹐如左圖﹐ 此時^AG⁼

(

^{3 2}⁺

)

²⁺¹² ⁼ ²⁶^﹒

(ii)將矩形DCGH 沿DC攤開﹐如中圖﹐ 此時^AG⁼

(

^{3 1}⁺

)

²⁺²² ⁼ ²⁰^﹒

(iii)將矩形BCGF沿BC攤開﹐如右圖﹐此時^AG⁼ ³²⁺

(

^{2 1}⁺

)

² ⁼ ¹⁸^﹒

由(i)﹐(ii)﹐(iii)知爬行的最短距離為 18﹒

高雄市明誠中學高二數學平時測驗日期