1. The space geometry formed by the right triangle rotating around the hypotenuse is ().
A. a cone B. a cone and a cylinder C. two cones D. a cone and a frustum.
2. If,, equals ............... ()
A.B. C. D。
3. In the following propositions: ① If A α, B α, then AB α; (2) If α and β are α, α and β must intersect in a straight line, and let M and A M3 pass through three points and have only one plane; 4 if a? b，c？ B, then a//c/C. The number of correct propositions ()
A. 1 B. 2 C. 3 D. 4
4. As shown in the figure, the area of the plane figure is ()
a4 b . 4 c . 2d . 8
5. If yes, then = () College Entrance Examination Resource Network
A.0 B. 1 C.2 D.3
6. The vertices of the cube are all on the sphere, and its side length is, so the radius of the sphere is () cm.
BC 1 year
7. Let the domain of even function f(x) be r, and when x is increasing function, the relationship between the sizes of f (-2), f () and f (-3) is ().
a . f()& gt； f(-3)>f(-2) B.f()>f(-2)>f(-3)
c . f()& lt； f(-3)& lt； f(-2)d . f()& lt； f(-2)& lt； f(-3)
8. The following proposition is wrong ()
A. If, then there must be a straight line parallel to the plane.
B. If, then all straight lines are perpendicular to the plane.
C If the plane is not perpendicular to the plane, then there must be no straight line perpendicular to the plane.
D. If, then
9. If the sides of the three cones P-ABC are equal, then the projection o of point P on the bottom surface is △ABC's ().
A. inside B. outside C. hanging down D. center of gravity
10. Let the function pair be satisfied arbitrarily, and then = ()
A.- 2 BC
Fill in the blanks (4 points for each small question, *** 16 points)
1 1. If the side of a cylinder is made of rectangular hard paper with the length and width of 3 and 3 respectively, then the radius of the bottom of the cylinder is _ _ _ _ _ _.
12. In a cube whose midpoints are respectively, the angles formed by straight lines on different planes are _ _ _ _ _ _ _.
13. If the function decreases in the interval, the real number range is.
14. It is known that m and n are different straight lines and non-coincident planes, and the following proposition is given:
(1) If it is parallel to any straight line in the plane.
(2) if, then
(3) if, then
(4) If, then
Among the above propositions, the serial number of true propositions is _ _ _ _ _ _ (write the serial numbers of all true propositions).
Third, answer questions:
15. (Full score for this small question 10)
Calculation: log2.56.25+LG+ln ()+log2 (log216)
16. (The full score of this small question is 12)
The picture on the right is a three-view view of space geometry, according to
Size in the figure (unit:), find the geometric surface area.
And volume.
17. (Full score for this small question 10)
As shown in the figure, in the cube ABCD-a1b1c1d1,e and f are edges AD and AB.
midpoint
(1) verification: EF‖ plane CB1d1; +0;
(2) verification: plane CAA 1C 1⊥ plane CB 1D 1. +0.
18. (Full score for this small question 10)
As shown in the figure, in the cone,, are the two diameters of the bottom circle,
And yes.
(1) Find the surface area of the cone;
(2) Find the tangent of the angle formed by straight lines on different planes.
19. (The full score of this small question is 12)
As shown in the figure, ABCD is a square, O is the center of the square,
The bottom of PO ABCD, e is the midpoint of PC.
Verification: (1)PA‖ plane BDE
(2) Plane packaging plane BDE
(3) Find the size of dihedral angle e-bd-a. ..
20. (Full score for this small question 10)
As shown in the figure, plane ABCD⊥ plane ABEF, ABCD is a square and ABEF is a rectangle.
G is the midpoint of EF,
(1) Verify the plane AGC⊥ plane bgc；;
(2) Find the sine of the included angle between GB and plane AGC.
Senior one mathematics final examination paper reference answer.
1. Multiple choice questions: (4 points for each small question, * * * 40 points)
The title is 1 23455 6789 10.
Answer C, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B. B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A B A b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a
Fill in the blanks: (4 points for each small question, *** 16 points)
1 1. or12.13.14.3④
Third, answer questions:
15, (10) Original formula =2-2+ =
16.( 12 minutes) Solution: According to the three views, the space geometry is a regular triangular prism with a base length of 2 and a side length of 3.
Its bottom area is:, and lateral area is:
Its total area is:
Its volume is: (m3)
17.( 10 score)
The solution (1) connects BD, so BDD 1B 1 is a parallelogram, ∴BD //B 1D 1
* ef//BD ∴ef//b 1d 1
EF surface CB 1D 1
B 1D 1 surface CB 1D 1
EF// plane CB 1D 1
(2)∵b 1d 1⊥a 1c 1⊥aa 1b 1d 1 ⊥.
b 1d 1 face c 1b 1d 1
∴ aircraft CAA 1C 1⊥ aircraft C 1B 1D 1.
18.( 10)
Solution: (1),
, ,
.
(2) This is the angle formed by straight lines in different planes.
, ,
In,,,
,
The tangent of the nonplanar line to the angle is.
19 and (12) prove that (1)∵O is the midpoint of AC, e is the midpoint of PC, ∴OE‖AP,
There are also: OE plane BDE, PA plane BDE, ∴ PA plane BDE.
(2) abcd at the bottom of Po, ∴PO BD,
Ac bd and AC PO=O∴BD plane PAC,
And BD plane BDE, ∴ plane packaging plane BDE.
(3) From (2), we can know that the BD plane includes PAC, ∴BD OE, BD OC,
∠EOC is the plane angle of dihedral angle e-BD-C.
(∠EOA is the plane angle of dihedral angle E-BD-A)
In RT△POC, OC= and PC=2 can be obtained.
In △EOC, OC=, CE= 1, OE= PA= 1.
∴∠ EOC = 45 ∴∠ EOA = 135, that is, dihedral angle E-BD-A is 135.
20.( 10) (1) It is proved that the square abcd surface ABCD⊥ surface ABEF satisfies AB,
∴ CB ⊥ face abef: Ag, UK Faceabef, ∴CB⊥AG, CB⊥BG
And AD=2a, AF= a, ABEF is a rectangle, g is the midpoint of EF,
∴AG=BG=, AB=2a, AB2=AG2+BG2, ∴ ag: CG ∩ BG = b ∴ ag ∴ plane CBG and ag plane AGC, so plane AGC⊥ plane BGC.
(2) Solution: As shown in the figure, surface AGC⊥ surface BGC can be seen from (i) that it intersects GC. If BH⊥GC is made on the plane BGC and the vertical foot is H, then BH⊥ plane AGC,
∴∠BGH is the included angle formed by GB and plane AGC.
∴, BG= in Rt△CBG,
∴
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