giai bai tap hinh hoc lop 10 sgk bai 3

... www.Vuihoc24h.vn – Kênh học tập Online Page 1 ÔN TẬP HÌNH HỌC LỚP 9 100 BÀI TẬP HÌNH HỌC CÓ LỜI GIẢI PHẦN 1 Bài 1: Cho ABC có các ... tiếp tam giác tại hai điểm M và N. 1. Chứng minh:BEDC nội tiếp. 2. Chứng minh: góc DEA=ACB. 3. Chứng minh: DE // với tiếp tuyến tai A của đường tròn ngoại tiếp tam giác. 4. Gọi O là tâm đường...

... tam giác ABC vôùi A( -5 ; 6) ; B (-4 ; -1) và C(4 ; 3) . Tìm D để ABCD là hình bình haønh a) D (3 ; 10) b) D (3 ; -10) c) D( -3 ; 10) d) D( -3 ; -10) Caâu 25 : Gọi I là trung điểm của đoạn AB. Câu ... 5 ; 2) vaø C(1 ; -3) có tâm đường tròn ngoại tiếp I là a) I = (3 ; 1 2 − ) b)I = (3 ; -1) c) I = ( -3 ; 1 2 − ) d) I = (3 ; 1 2 ) Câu 23 : Cho a r =( 1 ; 2) và b r = (3 ; 4) ; cho c r ... C(m -3; 4). Tìm giá trị của m để A ; B ; C thẳng hàng a) m = 2 b) m = 3 c) m = -2 d) m = 1 Câu 18 : Cho tam giác ABC với A ( 3; -1) ; B(-4;2) ; C(4; 3) . Tìm D để ABDC laø hbh a) D( 3; 6) b) D( -3; 6)...

... 10 Trang 17 Câu 1: Giá trị của 0 0 sin60 cos30+ bằng bao nhiêu? I. 3 2 II. 3 III. 3 3 IV. 1 Câu 2: Giá trị của 0 0 tan30 cot30+ bằng bao nhiêu? I. 4 3 II. 1 3 3 + III. 2 3 ... Câu 10: Cho ba ñiểm (1;2), ( 2; 1), (3; 2) A B C − − − a) T ọa ñộ véc tơ AB  là I. ( 3; 3) − − II. (3; 3) III. (3; 3) − IV. ( 3; 3)− b) T ọa ñộ trung ñiểm BC là I. 5 3 ( ; ... Tọa ñộ của véc tơ 2 3a i j= +    là I. (2; -3) II. (3; -2) III. (2 ;3) IV. (3; 2) Câu 2: T ọa ñộ của véc tơ 3a j= −   là I. ( -3; 0) II. (0; -3) III. (0 ;3) IV. (3; 0) Câu 3: T ọa ñộ của véc...

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... kính R = 3 37 . Tính độ dài các cạnh BC, AC và diện tích ΔABC. <1=18> Cho ΔABC với A = 60 o (B > C), bán kính các đường tròn ngoại tiếp, nội tiếp: R = 3 3 13 cm , r = 2 33 cm. Tính ... 2 37 , nội tiếp trong đường tròn bán kính R = 3 3 13 . Tính độ dài các cạnh và diện tích ΔABC. * - 4 - Vectơ <27> Trên đường thẳng  cho 3 điểm P, Q, R và trên đường thẳng m cho 3 ... những điểm M sao cho MA.MB = 2MI 2 trong đó A(0;5), B(– 4 ;3) và I là trung điểm đoạn AB. <58> Cho 3 im A (3; 5), B (3; 3), C(1;2). ơ. Chứng minh rằng A, B, C là các đỉnh của 1 tam...

... GV: TRƯƠNG VĂN THANH. ĐT: 0974 8109 57.http:violet.vn/truongthanh85 Bài toán: CÂN BẰNG CỦA VẬT RẮN KHÔNNG CÓ TRỤC QUAY CỐ ĐỊNH Xác ... quy cần tịnh tiến để các lực đồng quy để giải. Bước 2 :Phân tích các lực tác dụng lên vật. Bước 3 Áp dụng định luật II Newton cho vật cân bằng : ∑ = 0F (1) Bước 4: Giải phương trình vecto (1):có ... dây BC và lực nén lên thanh AB. Đáp số :T=N=5N Bài 2: THE METHODS FOR SOLVING PROLEM OF PHYSICS 10 1 1 l 2 l F 2 F 1 F 2 F 1 l 2 l 1 F F ...

...  ⇔   ∠ − ∠ = ∠ =     b) S vp = S qBOC - S V BOC = 2 2 0 . .120 1 . 3. 36 0 2 2 R R R π − = 2 2 2 . . 3 .(4 3 3) 3 4 12 R R R π = Bài 54 Cho tam giác ABC có ba góc nhän néi tiÕp (O; R), ... (2). Từ (1) và (2) => AMN là tam giác đều => S AMN = 2 3 3 4 R . => S = S (O) - S ∆ AMN = 2 R π - 2 3 3 4 R = 2 (4 3 3 4 R Bài 56 Cho tam giác ABC nội tiếp (O; R), tia phân giác ... tròn đờng kính OA cố định. 5. Ta cã AM. AN = 3R 2 , AN = R 3 . => AM =AN = R 3 => AMN cân tại A. (1) Xét ABN vuông tại N ta có AB = 2R; AN = R 3 => BN = R => ∠ABN = 60 0 . ABN =...

... ∆ABC ⇒OAN=OBM =30 o ; OA=OB=R và BM=AN(gt)⇒∆OMB=∆ONA ⇒OM=ON ⇒OMN cân ở O. 2/C/m OMAN nội tiếp: do ∆OBM=∆ONA(cmt)⇒BMO=ANO mà BMO+AMO=2v⇒ANO+AMO=2v. ⇒AMON nội tiếp. 3/ C/m BC 2 +DC 2 =3R 2 . Do BO ... điểm của AG. 4. Chứng tỏ EOFG nội tiếp. 37 Hình 38 1/C/m AHPK nội tiếp(sử dụng tổng hai góc đối) 2/C/m: HB.KP=HP.KC C/m hai ∆ vuông HPB và KPC đồng dạng. 3/ C/m HD=FE: Do FE//DO và DF//EP (FE ... DE.DB=DF.DA Xét hai tam giác vuông BDA và FDE có góc D chung. ⇒∆BDA∽∆FDE⇒đpcm. 3/ C/m IE là tâm đường tròn ngoại tiếp ∆FBC: Xem câu 3 bài 35 . 4/ C/m: BI 2 =BF.BC-IF.IC MỘT TRĂM BÀI TẬP HÌNH HỌC LỚP...

... DB= == oo AB n Sin AB 60sin2180 2 3 3a Do góc EDB=EFB=1v⇒EDFB nội tiếp trong đường tròn tâm O đường kính EB.Theo Pi Ta Go trong tam giác vuông EDB có:EB 2 =2ED 2 =2.( 3 3a ) 2 . ⇒EB= 3 6a ⇒OE= 6 6a 2/C/m ... ABD=ACD (Cuứng chaộn cung AD) ãDo MECD noọi tieỏp neõn MCD=MED (Cuứng chaộn cung MD) 2 Hinh3 Hinh4 3. c/m EC là phân giác của góc DAC. 4. C/m FD là đường trung trực của MB. 5. Chứng tỏ A;D;N ... góc đối. 2/C/m: NQ.NA=NH.NM. Xét hai ∆vuông NQM và ∆NAH đồng dạng. 3/ C/m MN là phân giác của góc BMQ. Có hai caùch: 5 Hinh8 Hinh9 4/Do OO’//BC và cung MC=MD ⇒O’M⊥BC⇒O’M⊥OO’⇒∆NO’M vuông ở O’...

... xyz 2 230 ++-= sao cho MA = MB = MC. ĐS: 1) xyz 2460 +-+= 2) M(2; 3; –7). Baøi 54. (ĐH 2008D) Trong không gian với hệ tọa độ Oxyz, cho bốn điểm A (3; 3; 0), B (3; 0; 3) , C(0; 3; 3) , D (3; 3; 3) . ... ĐS: 1) H (3; 0; 2), r = 4 2) M(0; 1; 3) , M 18 533 ;; 35 3 535 ổử ỗữ ốứ . Baứi 63. (H 2009B) 1. Trong không gian với hệ tọa độ Oxyz, cho tứ diện ABCD có các đỉnh A(1; 2; 1), B(–2; 1; 3) , C(2; ... trình: d 1 : xyz 31 1 23 -+ == - , d 2 : xyz 43 112 == . Đề thi Tốt nghiệp – Đại học Trần Sĩ Tùng Trang 96 ĐS: 1) (P): xyz 260 -+-= 2) M (1;1 ;3) - hoặc M 557 ;; 33 3 ổử ỗữ ốứ . Baứi...