The first volume of the eighth grade is the fourth chapter of mathematics. Quadrilateral natural exploration questions? Have the answer! High score!

Exploration of quadrilateral properties

First, multiple choice questions

1, as shown in the figure, the quadrangle represented by the shaded part is ().

A. rectangle B. parallelogram

C. diamond-shaped square

2. (Haidian, Beijing, 2002) A school plans to build a flower bed with both central symmetry and axial symmetry. The design schemes collected from students include isosceles triangle, regular triangle, isosceles trapezoid, diamond and so on. What do you think meets the requirements is ().

A. isosceles triangle B. regular triangle

C. isosceles trapezoid D. rhombus

3. (Shanxi, 2002) A, B, C and D are coplanar, from ① AB ‖ CD; ②AB = CD； ③ BC ‖ AD; (4) Choosing two of the four conditions BC = AD can make the quadrilateral ABCD a parallelogram.

A.3 kinds B. 4 kinds

C.5 species D. 6 species

4. (Nanjing, 2002) Among the following figures, the one with the largest symmetry axis is ().

A. Circle B. Square

C. isosceles triangle D. line segment

5. The students played with a kaleidoscope, which is surrounded by three pieces of glass with the same length and width, as shown in the figure. The small triangles in the figure are all congruent equilateral triangles, and the diamond AEFG can be regarded as taking the diamond ABCD as the center ().

A. rotate 60 degrees clockwise to obtain B. rotate clockwise 120 degrees to obtain.

C. rotate 60 degrees counterclockwise to get d. rotate counterclockwise 120 degrees to get d.

6, the following modes are both central symmetry and axial symmetry is ()

A B C D

7. As shown in Figure E, F, G and H are the midpoint of BD, AC, AD and BC respectively, and AB=DC. The following conclusions are drawn: ①GH divides EF equally in L; ②ef⊥gh； ③GFHE is rectangular; ④GH bisects ∠ EGF; The correct one is ()

A.① and ② B.② and ③

C.①②④ D.②③④

8. As shown in the figure, in the square ABCD, AO⊥BD, OE, FG and HI are all perpendicular to AD, and EF, GH and IJ are all perpendicular to AO. If S△AIJ= 1, then s squared ABCD= ().

248 B.252

C.254 D.256

9. If the diamond rotates around its center on the plane so that it coincides with the original diamond, the rotation angle should be at least ().

A. 180

From 270 to 360 A.D.

10. If each interior angle of a polygon is equal to 144, then the sum of its interior angles is ().

1260

c 1620d 1800

Volume b

Second, comprehensive questions

1 1. Cut and weld the existing isosceles right-angled triangular iron plate into a parallelogram with an angle of 45. Please design a simplest scheme and prove that your scheme does get a parallelogram that meets the requirements.

12, (1), there are two squares ABCD and OPQS with side length a, and the vertex o of the square OPQS is the center of the square ABCD. Try to judge the size relationship between AP and BS and prove it.

(2) If the side lengths of the square ABCD and OPQS are A and B (A

13. It is known that in quadrilateral ABCD, AB=DC, AC=BD, AD≠BC. It is proved that the quadrilateral ABCD is an isosceles trapezoid.

14, in the isosceles trapezoid ABCD, AD‖BC, AB=DC, the point P is the point on the side of BC, PE⊥AB, PF⊥CD, BG⊥CD, and the steps are E, F and G respectively. Verification: PE+PF = BG.

The correct answer is D.

The correct answer is B.

The correct answer is a.

The correct answer is D.

The correct answer is C.

The correct answer is D.

The correct answer is a.

The correct answer is B.

Analysis of 1 1;

Take points E and D of AC and BC, cut along ED with DE, make point E unchanged, and point C coincide with point A, then welding is simple.

Proof tip:

According to the conditions, △CED is an isosceles right triangle, so ∠ CED = ∠ CDE = 45 = ∠ AEF.