A multiple-choice question (this big question *** 24 points)

1. Take the following groups as three sides of a triangle, in which () can form a right triangle.

(A) 17， 15，8 (B) 1/3， 1/4， 1/5

2. If the degree of one angle of a triangle is equal to the sum of the degrees of the other two angles, then the triangle must be ().

(a) acute triangle (b) right triangle (c) obtuse triangle (d) isosceles triangle

3. The following groups of line segments, can form a triangle is ().

(A)5 12 13(B)5 12 7(C)8 18 7(D)3 4 8

4. As shown in the figure: in Rt△ABC, ∠ C = 90, AD bisects ∠BAC, AE=AC and connects DE, then the following conclusion is incorrect ().

(ADC = Germany (B)≈ADC =≈ Ade (C)≈ Debu = 90(D)≈BDE =≈ Dee.

5. The lengths of the three sides of a triangle are 15, 20 and 25 respectively, so the height of its largest side is ().

12 (B) 10 (C) 8 (D) 5

6. The following statement is incorrect ()

(a) The corresponding angles of congruent triangles are equal.

(b) The bisectors of the corresponding angles of congruent triangles are equal.

(c) Triangles with equal bisectors must be congruent.

(d) The bisector of an angle is the set of all points with equal distance to both sides of the angle.

7. A triangle with 2 and 8 sides and an integer third side has ().

(a) Three (b) Four (c) Five (d) Countless

8. In the following figure, the figure that is not axisymmetric is ().

(a) line segment MN (B) equilateral triangle (c) right triangle (d) obtuse angle ∠AOB

9. As shown in the figure, in △ABC, AB=AC, BE=CF, and AD⊥BC is in D, and congruent triangles in this figure * * * has ().

(A)2 pairs (B)3 pairs (C)4 pairs (D)5 pairs.

10. The obtuse angle between bisectors of two acute angles of a right triangle is ().

(A) 125(B) 135(C) 145(D) 150

1 1. The obtuse angle between bisectors of two acute angles of a right triangle is ().

(A) 125(B) 135(C) 145(D) 150

12. as shown in the figure: ∠A=∠D, ∠C=∠F, if △ ABC △ def, the condition to be given is ().

(A)AC = DE(B)AB = DF(C)BF = CE(D)∠ABC =∠DEF

2. Fill in the blanks (40 points for this big question * * *)

1. In Rt△ABC, ∠ c = 90, if AB= 13, BC= 12, then AC =；; If AB= 10 and AC: BC= 3: 4, then BC=

2. If two sides of a triangle are 5 and 9 respectively, the range of the third side X is.

There is a triangle whose two sides are 3 and 5 respectively. To make this triangle a right triangle, its third side is equal to.

4. As shown in the figure, in isosceles △ABC, AB=AC, ∠ A = 50, BO and CO are bisectors of ∠ABC and ∠ACB respectively, and BO and CO intersect at O, then: ∠BOC=

5. Let α be the base angle of isosceles triangle, and the range of α is ().

(A)0 & lt； α& lt； 90(B)α& lt； 90(C)0 & lt； α≤90(D)0≤α& lt； 90

6. As shown in the figure, △ ABC △ DBE, △ A = 50 and △ E = 30.

Then ∠ADB= degrees, ∠DBC= degrees.

7. In △ABC, the following reasoning process is correct ()

(a) If ∠A=∠B, then AB=AC.

(b) If ∠A=∠B, then AB=BC.

(c) If CA=CB, then ∠ A = ∠ B.

(d) If AB=BC, then ∠ B = ∠ A.

8. If the outer angle of a triangle is smaller than its adjacent inner angle, then the triangle must be a triangle.

9. In isosceles △ABC, AB=2BC, and its circumference is 45, then the length of AB is

10. The inverse proposition of the proposition "A triangle with equal angles is congruent triangles" is:

Among them, the original proposition is a proposition and the inverse proposition is a proposition.

1 1. As shown in the figure, AB‖DC, AD‖BC, AC, BD and EF intersect at O, AE=CF, and in the figure △ AOE△, △ ABC△, congruent triangles * * pair.

12. As shown in the figure, in Rt△ABC and Rt△DEF,

AB = DE (known)

= (known)

∴Rt△ABC≌Rt△DEF (________)

13. If the outer angle of a triangle is smaller than its adjacent inner angle, then the triangle must be a triangle.

14. As shown in the figure, BO and CO are bisectors of ∠ABC and ∠ACB respectively, and ∠ BOC = 136, then = degrees.

15. If the outer angle of an isosceles triangle is 80 degrees, then its base angle is 10 degrees.

16. In isosceles Rt△ABC, CD is the center line of the bottom edge, and AD= 1, then AC=. If the length of an equilateral triangle is 2, then its height is 0.

17. If the waist length of an isosceles triangle is 4 and the waist height is 2, then the vertex angle of the isosceles triangle is ().

(a) 30 (b) 120 (c) 40 (d) 30 or 150.

18. As shown in the figure, AD is the symmetry axis of △ABC. If ∠DAC=30? , DC=4cm, then the circumference of △ABC is cm.

19. As shown in the figure, in △ABC, AB=AC, the middle vertical line DE of AB passes through AC to E, and the vertical foot is D. If ∠A=40? Then ∠ bec =; If the circumference of △BEC is 20cm, then the bottom BC=.

20. As shown in the figure, in Rt△ABC, ∠ ACB = 90, DE is the middle vertical line of BC, AB intersects with E, and the vertical foot is D. If AC = ∠ 3 and BC = 3, ∠A= degrees. △ The circumference of △CDE is.

Three. True or false (5 points for this big question)

1. Two equilateral triangles have one side corresponding to the congruence of the same triangle. ( )

2. Regarding the symmetry of two triangles, the one with the same area is ().

3. Two triangles with one angle and two equal sides are congruent. ( )

4. The condition of a triangle with lines A, B and C as sides is A+B >; c()

5. The median line of two sides and one of them corresponds to the congruence of two triangles. ( )

Four. Calculation problem (this big problem *** 5 points)

1. As shown in the figure, in △ABC, ∠ B = 40, ∠ C = 62, AD is the height on the side of BC, and AE is the bisector of ∠BAC.

Find: ∠ the degree of ∠DAE.

Verb (abbreviation of verb) drawing problem (6 points for this big problem)

1. As shown in figure △ABC, draw the bisector of ∠A with a scale and a protractor; The center line of AC side; Height on the side of AB.

2. As shown in the figure: ∠ α and line segment α. Solution: isosceles △ABC, so that ∠ AD=α ∠ α, AB = AC, and the height on the side of BC = α.

There are two factories on the same side of the railway, A and B. It is necessary to build a warehouse next to the railway, make the distance between the two factories equal, and draw the location of the warehouse.

Six. Answer the question (this big question *** 5 points)

1. As shown in the figure, in rtδABC, c = 90, DE⊥AB in D, BC= 1, AC=AD= 1. Find: the length of gain and quilt.

Seven. Proof problem (this big problem * *15 points)

1. If the three sides of ABC are m2-n2, m2+n2 and 2mn respectively. (m & gtn & gt0)

Prove that ABC is a right triangle.

2. As shown in the figure, in △ABC, BC=2AB, and D and E are the midpoint of BC and BD respectively.

Verification: AC=2AE

3. As shown in the figure, in △ABC, the bisector of ∠ABC and the bisector of ∠ACB intersect at D, DE‖BC at E and AC at F. ..

Verification: BE=EF+CF

Second grade geometry-triangle-answer

A multiple-choice question (this big question *** 24 points)

1.:A

2.:B

3.:A

4.:D

5.:A

6.:C

7.:A

8.:C

9.:C

10.:B

1 1.:B

12.:C

2. Fill in the blanks (40 points for this big question * * *)

1.:5,8

2.:4 & ltx & lt 14

3.:4 or √34

4.: 1 15

5.:A

6.:50,20

7.:C

8.: obtuse angle

9.: 18

10.: The angles corresponding to congruent triangles are equal. Fake, really.

1 1.:COF，CDA，6

12.:AC=DF，SAS

13.: obtuse angle

14.:92

15.:40

16.:√2,√3

17.:D

18.:24

19.:30? , 8 cm

20.:60? , 1/2(3√3+3)

Three. True or false (5 points for this big question)

1.:√

2.:√

3.:×

4.:×

5.:√

Four. Calculation problem (this big problem *** 5 points)

1.: solution: ∵AD⊥BC (known)

∴∠ CAD+∠ C = 90 (the two acute angles of a right triangle are complementary).

∠CAD=90 -62 =28

And ∵∠ BAC+∠ B+∠ C = 180 (triangle interior angle theorem)

∴∠bac= 180-∠b-∠c = 180-40-62 = 78

And AE equally divided ∠BAC, ∴∠ CAE = ∠ BAC = 39.

∠DAE =∠CAE-∠CAD = 39-28 = 1 1

Verb (abbreviation of verb) drawing problem (6 points for this big problem)

1.: Sketch.

2. Practice: (1) as ∠A=∠α,

(2) Let ∠A bisector AD, on which the intercept AD = α.

(3) The vertical line that intersects D as AD intersects ∠A on both sides of B and C..

△ABC is the isosceles triangle to be made.

3. Practice: The midline railway with line segment AB is located at point C, and point C is the warehouse location.

Six. Answer the question (this big question *** 5 points)

1.: solution: ∫BC = AC = 1

∠ c = 90, then: ∠ b = 45.

AB2=BC2+AC2=2，AB=√2

And ∵DE⊥AB, ∠ b = 45.

∴DE=DB=AB-AD=√2- 1

∴BE=√2DE=√2(√2- 1)=2-√2

Seven. Proof problem (this big problem * *15 points)

1.: Proof: ∫ (m2-N2)+(2mn) 2 = M4-2m2n2+N4+4m2n2.

=m4+2m2n2+n4

=(m2+n2)

∴δabc is a right triangle.

2. Prove: extend AE to F, make AE=EF, connect DF, in △ABE and △FDE.

BE=DE，

∠AEB =∠ Federal Reserve

AE=EF

∴△abe?△FDE(SAS)

∴∠B=∠FDE，

DF=AB

∴D is the midpoint of BC, and BC =2AB.

∴DF=AB= BC=DC

And: BD= BC=AB, ∴∠BAD=∠BDA.

∠ADC=∠BAC+∠B，∠ADF=∠BDA+∠FDE

∴∠ADC=∠ADF

DF=DC (certification) ∴△ ADF △ ACD (SAS)

∠ADF=∠ADC (authentication)

AD=AD (male side)

∴AF=AC ∴AC=2AE

3. Proof: BC

DB share ∠ABC, CD share ∠ACM

∴∠EBD=∠DBC=∠BDE，

∠ACD=∠DCM=∠FDC

∴BE=DE,CF=DF

And: BE=EF+DF

∴BE=EF+CF

I. Fill in the blanks:

1. As shown in the figure, make a circle with the origin of rectangular coordinate system as the center and 4 as the diameter. The straight line L passes through the origin, and the sector areas sandwiched in the positive direction of X axis are P and Q respectively. Try to write the resolution function of p about Q _ _ _ _ _ _ _ _ _

2. In the plane rectangular coordinate system, the origin of the center of the square ABCD with a side length of 2, four sides are perpendicular to the coordinate axis, and point P is a point on the X axis, so the coordinate of the vertex P of a regular triangle with P as the vertex and the side of the square as the edge is _ _ _ _ _ _ _ _ _ _ _.

3 As shown in the figure, take a rectangular piece of paper, length AB= 10cm, width BC = 5 * root (3) cm, fold it in half with dotted line CE (point E on AD) as the crease, so that point D falls on the AB side, AE =-cm, ∠ DCE =-degree.

4. In the rectangular coordinate system, the circle O and the straight line y= -4x/3 +4 are tangent to the point C, so the coordinate of the point C is _ _ _ _ _

2. Calculation problems:

1. Known in ⊿ABC, AB=8, AC=6, D is a point above BC, BD: DC = 2: 3.

Find out the scope of the advertisement.

2. As shown in the figure, square ABCD, points M and N are on BC and CD respectively, so that MN=BM+DN can find the size of ∠MAN.

Three. Comprehensive question:

1, it is known that the quadratic function y=-x2+8x- 12 image intersects the X axis at points A and B, and the linear function image passes through points A and C (3,3).

(1) solution: analytical formula of linear function.

(2) When x is what value, the value of the linear function is less than that of the quadratic function.

(3) Can we find a point P on the symmetry axis of quadratic function image to minimize the value of PA+PC? Please explain the reason.

2. As shown in the figure, in the plane rectangular coordinate system, the straight line AB intersects the X axis and the Y axis at points A and B respectively, OA = OB = B. Draw a circle with point O as the center and A (A < B) as the radius, which intersects the straight line AB at points C and D respectively, also called CF⊥OA,CE⊥OB, and the vertical feet are points F and E respectively.

(1) Find the resolution function of straight line AB;

(2) Find the perimeter of rectangular OFCE (expressed by an algebraic expression containing a and b);

(3) Let point P be an arbitrary moving point on straight line AB, and then let point P be PF⊥x axis, PE⊥y axis, and the vertical foot be F point and E point, and try to explore whether the rectangular perimeter of OFPE is constant. And explain why.

3. As shown in Figure 3, in the plane rectangular coordinate system, the straight line AB intersects with the X axis and the Y axis at points A and B respectively, OA = 6 and OB = 8. Draw a circle with O point as the center and 5cm length as the radius, intersect with straight line AB at C point and D point, and intersect with the negative semi-axis of X axis at M point.

(1) Find the analytical formula of straight line AB; (2) Find the coordinates of point C;

(3) Find the analytical formula of parabola passing through points A, C and M;

(4) Is there a point P on the parabola of (3) that makes the area of △PAM 1 1? If it exists, request the coordinates of point P, if it does not exist, please explain the reason.

I hope it helps you.