Junior high school geometry comprehensive examination questions and answers (time 120, full mark 100) 1. Fill in the blanks (22 points for this question * * *, 2 points for each blank) 1. The two sides of a triangle are 9 and 2 respectively, and the third side is odd, so the third side is .2. △ ABC。 If the maximum side length of similar △ a ′ b ′ c ′ is 10, the area of △ a ′ b ′ c ′ is .4. Chords AC and BD intersect at e in a circle, and ∠ BEC = 130, then ∠ ABCD =. The area of delta △AOB is 0.6. The lengths of two right angles of a right triangle are 5cm and 12cm respectively, and the median length on the hypotenuse is 0.7. The upper base length of the trapezoid is 2, the middle length is 5, and the lower base length of the trapezoid is 0.9. As shown in the figure, the two groups of sides of quadrilateral ABCD extend to satisfy e and f respectively, and if DF=2DA, 65438+. If BC=a and ∠ B = 30, then AD is equal to. 2. Multiple choice questions (44 points for this question, 4 points for each small question) 1. The complementary angle of an angle and its complementary angle are complementary. Then this angle is [] a.30 b.45 c.60 d.752. The quadrilateral obtained by connecting the midpoints of the sides of the isosceles trapezoid in turn is [] A. Rectangular B. Square C. Diamond D. Trapezoidal 3. As shown in the figure, DF∨EG∨BC, AD=DE=EB. △ABC is divided into three parts, and the area ratio is [] A.1:2: 3b.1:1:kloc-0/:4: 9d.1:3: 54. If the radius of both circles is 4. Then the positional relationship between these two circles is [] A. Intersection B. Cut C. External D. External 5. Given that the central angle of the sector is 120 and the radius is 3cm, the area of the sector is [] 6. Given that the hypotenuse of Rt△ABC is 10 and the radius of inscribed circle is 2, the length of two right-angled sides is [B a straight line between two parallel lines. A parallel line whose distance from two parallel lines is equal to 2 cm. A parallel line whose distance from these two parallel lines is equal to 1 cm. 8. Secant PBC makes a circle at a point outside the circle, intersecting with point B and point C. Tangents PM and M are tangent points. If PB=2 and BC=3, then the length of PM is [] 9. Known: ABC, EF∨CD, and ∠ ABC = 20, ∨. Then the degree of ∠BCF is [] A.160 b.150 c.70d.5010. As shown in figure OA=OB, point C is on OA, point D is on OB, OC=OD, and AD and BC intersect at E. In the figure, congruent triangles * * has [] A.2 to B.3 to C.4 to D.5 to 1 1. The figure with both axial symmetry and central symmetry is [] A. isosceles triangle B. isosceles trapezoid C. parallelogram D. Line segment 3. Calculation problem (7 points for each small question in this question * * * 65438) I first saw the ship at 30 southwest of B, and after half an hour, I saw the ship at 60 southwest of B, and asked for the speed of the ship. 2. It is known that the radius of ⊙O is 2cm, the secant of PAB is ⊙O, Pb = 4 cm, PA = 3 cm, and PC is ⊙. 4 points for each small question) 1. As shown in the figure, in △ABC, BF⊥AC,CG⊥AD,F and G are vertical rulers, and D and E are the midpoint of BC and FG respectively. Evidence: DE⊥FG 2. As shown in the figure, AE∨BC, D, AF=CE, FG in the parallelogram. Ed intersects AC at Q, and the extension line of ED intersects AB at P: PDQE=PEQD 4. As shown in the figure, in the trapezoidal abcd, ABCD, AD=BC, the circle O with the diameter of AD intersects AB at point E, and the tangent EF of the circle O intersects BC at point F. It is proved that: (1) ∠ def = (2)EF⊥BC 5. As shown in the figure, the chord AC of ⊙O, BD and F intersect .. Fill in the blanks (4 points for each small question *)1.b2.c3.c4.b5.a6.c7.d8.c9.d10.c1.d3. (This question * */ AB= ∴MN=20 (km), that is, the ship sails for 20 km in half an hour, the ship speed is 40 km/h, PC is the tangent of O, CD and OP, RT △ OCD, RT △ OPC IV. D is the midpoint of ∴GD=FD BC, △GDF is an isosceles triangle, ∵E is the midpoint of GF ∴DE⊥GF 2. It is proved that ∵ quadrilateral ABCD is a parallelogram ∴ad∨BC∠ 1 =∞. Ad∨BC ∴fg∥eh∴ quadrilateral FHEG is a parallelogram, and GH and EF are the diagonal lines of the parallelogram ∴GH and EF are bisected. 3. Proof: ∫AE∨BC∴∠ 1 =∞. ∠ 2 =∠ 3 ∴△ AQE ∽△ CQD: AE ∨ BC: BD = CD ∴ that is, PDQE=PEQD 4. Proof: (1) In trapezoidal ABCD, DC \. ∠ Deb = 90, that is ∠ DEF+∠ BEF = 90 and ∠ DEF = ∠ B ∠ B+∠ BEF = 90 ∴∠ EFB.