2. Analysis because BC is tangent to ⊙O, BC⊥AB. According to Pythagorean theorem, AC = AB2+BC2 = 22+22 = 22. C.
3.B
4. Analyze that the length of the bus is the length of the hypotenuse BC of Rt△ABC, the radius of the bottom circle is the length of AB, and the perimeter of the bottom circle is 2AB? π=2? 3 π = 6 π. According to the formula, sector area = 12× sector arc length× sector radius =12× 6 π× BC =12× 6 π× 5 =15 π, so D is selected.
5. Analyze the chord of ∵AB⊥CD in E, where AB is ⊙O in diameter and CD is ⊙ O, and CE = de can be obtained according to the vertical diameter theorem.
6. As shown in the figure, after point A, do AM⊥BC in M, followed by OB.
At Rt△ABC, ab = ac, AM⊥BC in m, BC = 6, ∴ BM = CM = 12bc = 3, ∠ ABM = 45. ∴ in Rt△ABM, BM = AM =
∴ OM = 2。 In Rt△BOM, OM = 2 and BM = 3. According to Pythagorean Theorem, Bo = 13.
7. Analysis in RT △ABC, ∠ A+∠ C = 90, S shadow = SRT △ ABC-14s ⊙ A =12× 6× 8-14×π× (60.
8. By analyzing the link OC, oc⊥cd∫OA = oc, ∴∠ A = ∠ OCA = 30, ∴∠ Doc = 60, ∴∠ D = 30.
9. Analysis: PA and PB are tangents of ⊙O, and tangents are A, B, ∴OA⊥AP, ob:BP. ∵∠ P = 60，∴∠ AOB = 65436。
10. Analysis BC = 23cm. Shaded area in the figure = sector AOB area+triangle BOC area = (16π 3+23) cm2.
1 1. The area of the whole graph can be regarded as a sector with a radius of 6 and a central angle of 60 and a semicircle with a diameter of 6. The area of the shaded part can be regarded as the area of the whole graph minus the area of the semicircle with a diameter of AB, that is, S shadow = S sector Bab' = 60 π× 62360 = 6 π, so a is selected.
12. Analysis △ABC is inscribed in ⊙O, D is the midpoint of line segment AB, and OD⊥AB can be known from the vertical diameter theorem, from which the midpoint of point E can be deduced, thus proving that ①AB⊥DE, ② AE = BE, and ⑤ = 12 hold.
13. As shown in the analysis, O passing through the center of the circle is OC⊥AB, and the vertical foot is E. From the vertical diameter theorem, OC bisects AB and AE = be. ∫ab = 1.6，∴ AE = 0.8。
14. analysis because Tan α = AOOB = 43, AO = 8, so OB = 6, then the bottom area S = 36 π (square meter).
15. Analysis: AB‖OC, ∴∠a =∠o .∠o = 2∠b = 2×22 = 44, ∴∠.
16. If the link is analyzed, then ∠ FOD = ∠ EOD = 40. And OC = of, ∴∠ FCD = 12 ∠ FOD = 20.
17.( 1) Proof: link BD, ∫ =, ∴∠ A = ∠ Abd, ∴ AD = BD.
∠∠A+∠C = 90，∠DBA+∠DBC=90，
∴∠ C =∠ DBC，∴ BD = DC。 Therefore, AD = DC.
(2) solution: connect OD and cross AB to F.
∵DE is the tangent of ⊙O, ∴OD⊥DE.
∫ =, OD passes through the center of the circle, ∴OD⊥AB.
And ∵AB⊥BC, ∴ quadrilateral FBED is a rectangle, ∴DE⊥BC.
That is, ∠ dec = 90, de = ec, ∴ c = 45.
∴sinC=sin45 =22。
18. Solution: (1) ∠ BOE = 60, ∴∠ A = 12 ∠ BOE = 30.
(2) In △ABC, ∫cosc = 12, ∴∠ C = 60.
And ∵∠a = 30°, ∴∠ABC = 90°, ∴AB⊥BC, ∴BC are tangents of ⊙ O. 。
(3) point m is the midpoint of ∴om⊥ae. AE.
In Rt△ABC, BC = 23,
∴AB=BC？ tan60 =23×3=6。
∴oa=ab2=3,∴od= 12oa=32,∴md=om-od=32.
19.∴DE=BE.
(2) It is proved that the connection OD is known from (1) ∠ DOE = ∠ BOE, and in △COD and △COB, CO = CO and OD = OB. ∴.∴△cod?△cob(SAS)
(3) Solution: In Rt△ADG, ∵ Sina = DGAD = 45, let DG = 4x, then AD =