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W1ac+cours
1.
ﺑﺴﻢ ﺍﷲ ﺍﻟﺮﺣﻤﻦ ﺍﻟﺮﺣﻴﻢ ﻭﺍﻟﺼﻼﺓ ﻭﺍﻟﺴﻼﻡ ﻋﻠﻰ ﺃﺷﺮﻑ ﺍﻟﻤﺨﻠﻮﻗﻴﻦ ﻣﺤﻤﺪ ﺳﻴﺪ ﺍﻟﻤﺮﺳﻠﻴﻦ ﻭﻋﻠﻰ ﺁﻟﻪ ﻭﺻﺤﺒﻪ ﺃﺟﻤﻌﻴﻦ
ﺃﻣﺎ ﺑﻌﺪ ٬ ﻳﺴﺮﻧﻲ ﺃﻥ ﺃﻗﺪﻡ ﻟﻜﻢ ﻫﺬﺍ ﺍﻟﻌﻤﻞ ﺍﻟﻤﺘﻮﺍﺿﻊ ﻭﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻠﻰ ﺟﻤﻴﻊ ﺩﺭﻭﺱ ﻭﻟﻰ ﺛﺎﻧﻮﻱ ﺇﻋﺪﺍﺩﻱ ﻣﺠﻤﻌﺔ ﻓﻲ ﻛﺘﺎﺏ ﻭﺍﺣﺪ ﻣﻔﻬﺮﺱ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﻟﻤﺴﺘﻮﻯ ﺍﻷ ﺟﻤﻌﺖ ﻣﻦ ﻣﻮﻗﻊ www.anissmaths.ift.cx ﻟﻸﺳﺘﺎﺫ ﺍﻟ ﻬﺪﻱ ﻋﻨﻴﺲ ﻤ ﻟﺘﺼﻔﺢ ﺃﻱ ﺩﺭﺱ ﺃﺿﻐﻂ ﻋﻠﻰ ﻋﻨﻮﺍﻧﻪ ﻓﻲ ﺍﻟﻔﻬﺮﺱ ﻭﻛﺬﻟﻚ ﺍﻟﺘﻤﺎﺭﻳﻦ ﻭﻟﻠﺮﺟﻮﻉ ﺇﻟﻰ ﺍﻟﻔﻬﺮﺱ ﺇﺿﻐﻂ ﻋﻠﻰR ﺗﺠﻤﻴﻊ ﻭﺗﺮﺗﻴﺐ ﻭﻓﻬﺮﺳﺖ ALMOHANNAD
2.
ﺍﻟﻌﻤﻠﻴﺎﺕ ﻋﻠﻰ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ ﻭ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ
ﺍﻟﻤﺴﺘﻘﻴﻢ ﻭ ﺃﺟــﺰﺍﺅﻩ ﺴﺮﻳﺔ + ﺍﻟﻜﺘﺎﺑﺎﺕ ﺍﻟﻜﺴﺮﻳﺔ ﻣﻘﺎﺭﻧﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻜ ﺍﻟﻌﻤﻠﻴﺎﺕ ﻋﻠﻰ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻜﺴﺮﻳﺔ ﻭﺍﺳــﻂ ﻗﻄﻌﺔ + ﺍﻟﻤﺘﻔﺎﻭﺗﺔ ﺍﻟﻤﺜﻠﺜﻴﺔ ﻣﻘﺎﺭﻧﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻭ ﺗﻘﺪﻳﻢ ﻣﺜﻠﺜﺎﺕ ﺧــﺎﺻﺔ + ﻣﺠﻤﻮﻉ ﻗﻴﺎﺳﺎﺕ ﺯﻭﺍﻳﺎ ﻣﺜﻠﺚ ﻃﺮﺡ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻭ ﺟﻤﻊ ﻗﺴﻤﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻭ ﺿﺮﺏ ﺍﻻﺭﺗﻔﺎﻋﺎﺕ ﻓﻲ ﻣﺜﻠﺚ ﻭ ﺍﻟﻤﻨﺼﻔﺎﺕ ﺍﻟـﻘــــﻮﻯ ﺍﻟﺘﻤﺎﺛﻞ ﺍﻟﻤﺮﻛــﺰﻱ ﺍﻟﺘﻌﻤﻴــﻞ ﻭ ﺍﻟﻨﺸــﺮ ﻣﺘــﻮﺍﺯﻱ ﺍﻷﺿــﻼﻉ ﺍﻟﻤـﻌــﺎﺩﻻﺕ ﻭ ﺍﻟﻤﺴــﺎﺋﻞ ﺍﻟﺮﺑﺎﻋﻴــﺎﺕ ﺍﻟﺨــﺎﺻﺔ ﺍﻟﺰﻭﺍﻳﺎ ﺍﻟﻤﻜﻮﻧﺔ ﻣﻦ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻭ ﻗــﺎﻃﻊ ﺍﻟـﺘـﻨــــﺎﺳﺒﻴــﺔ ﺍﻟﻤﻌﻠــﻢ ﻓﻲ ﺍﻟﻤﺴﺘــﻮﻯ + ﺍﻟﻤﺴﺘﻘﻴــﻢ ﺍﻟﻤــﺪﺭﺝ ﺍﻟـــــﺪﺍﺋــﺮﺓ ﺍﻷﺳﻄــﻮﺍﻧﺔ ﺍﻟﻘــﺎﺋﻤﺔ ﻭ ﺍﻟﻤﻮﺷــﻮﺭ ﺍﻟﻘــﺎﺋﻢ ﺍﻟﺤﺠــﻮﻡ ﻭ ﺍﻟﻤﺴــﺎﺣﺎﺕ ﻭ ﺍﻟﻤﺤﻴـﻄــﺎﺕ ﺍﻹﺣﺼــــﺎء
3.
ﻷﻋﺪﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ ﺍﻟﻄﺒﻴﻌﻴﺔ ﻭ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ
ﺍﻟﻌﻤﻠﻴﺎﺕ ﻋﻠﻰ ﺍ 1( – ﺣﺴﺎﺏ ﺳﻠﺴﻠﺔ ﻣﻦ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺑﺪﻭﻥ ﺃﻗﻮﺍﺱ : ( ﻗﺎﻋﺪﺓ 1 : ﺃ ﻟﺤﺴﺎﺏ ﺗﻌﺒﻴﺮ ﺟﺒﺮﻱ ﻣﻜﻮﻥ ﻣﻦ ﺳﻠﺴﻠﺔ ﻣﻦ ﻋﻤﻠﻴﺘﻲ ﺍﻟﺠﻤﻊ ﻭ ﺍﻟﻄﺮﺡ ﻓﻘﻂ ﺃﻭ ﺍﻟﻀﺮﺏ ﻭ ﺍﻟﻘﺴﻤﺔ ﻓﻘﻂ ﻭ ﺑﺪﻭﻥ ﺃﻗﻮﺍﺱ , ﻧﻨﺠﺰ ﺍﻟﻌﻤﻠﻴﺎﺕ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﺇﻟﻰ ﺍﻟﻴﻤﻴﻦ ﺣﺴﺐ ﺍﻟﺘﺮﺗﻴﺐ . * ﻣﺜﺎﻝ : 5,1 – 9 – 7,3 + 5,0 + 5,3 – 11 + 5,2 = A 5,1 – 9 – 7,3 + 5,0 + 5,3 – 5,31 = 5,1 – 9 – 7,3 + 5,0 + 5,3 – 01 = 5,1 – 9 – 7,3 + 5,0 + 5,7 = 5,1 – 9 – 7,3 + 8 = 5,1 – 9 – 7,11 = 5,1 – 7,2 = 2,1 = ( ﻗﺎﻋﺪﺓ 2 : ﺏ ﻟﺤﺴﺎﺏ ﺗﻌﺒﻴﺮ ﺟﺒﺮﻱ ﻳﺘﻜﻮﻥ ﻣﻦ ﺳﻠﺴﻠﺔ ﻣﻦ ﺍﻟﻌﻤﻠﻴﺎﺕ ﻭﺑﺪﻭﻥ ﺃﻗﻮﺍﺱ ‘ ﻧﻨﺠﺰ ﻋﻤﻠﻴﺘﻲ ﺍﻟﻀﺮﺏ ﻭ ﺍﻟﻘﺴﻤﺔ ﻗﺒﻞ ﻋﻤﻠﻴﺘﻲ ﺍﻟﺠﻤﻊ ﻭ ﺍﻟﻄﺮﺡ ﺛﻢ ﻧﻄﺒﻖ ﺍﻟﻘﺎﻋﺪﺓ 1 . * ﻣﺜﺎﻝ : 5,1 – 4 : 6,8 + 11 – 2 B = 22 – 2,5 + 7 x 5,1 – 51,2 + 11 – 41 + 5,2 – 22 = 5,1 – 51,2 + 11 – 41 + 5,02 = 5,1 – 51,2 + 11 – 5,43 = 5,1 – 51,2 + 11 – 5,32 = 5,1 – 51,2 + 5,21 = 5,1 – 56,41 = 51,31 =
4.
2( – ﺣﺴﺎﺏ ﺳﻠﺴﻠﺔ ﻣﻦ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺑﺄﻗــﻮﺍﺱ :
( ﻗﺎﻋﺪﺓ 3 : ﺝ ﻌﺒﻴﺮ ﺟﺒﺮﻱ ﻣﻜﻮﻥ ﻣﻦ ﺳﻠﺴﻠﺔ ﻣﻦ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺑﺄﻗﻮﺍﺱ ﻟﺤﺴﺎﺏ ﺗ ﻧﺤﺴﺐ ﺃﻭﻻ ﻣﺎ ﺑﻴﻦ ﻗﻮﺳﻴﻦ ﺛﻢ ﻧﻨﺠﺰ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻷﺧﺮﻯ . * ﻣﺜﺎﻝ : 2,3 – ) 4 – 8,5 ( C = 3,5 + [ 14 – ( 1,5 + 3 ) ] x 2 – 0,5 x 2,3 – 8,1 = 3,5 + [ 14 – 4,5 ] x 2 – 0,5 x 2,3 – 8,1 = 3,5 + 9,5 x 2 – 0,5 x 2,3 – 9,0 – 91 + 5,3 = 2,3 – 9,0 – 5,22 = 2,3 – 4,12 = 2,81 = 3( – ﺗﻮﺯﻳﻌﻴﺔ ﺍﻟﻀﺮﺏ ﻋﻠﻰ ﺍﻟﺠﻤﻊ ﻭ ﺍﻟﻄﺮﺡ : ( ﻗﺎﻋﺪﺓ 4 : ﺩ aﻭ bﻭ kﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ . k x ( a + b ) = a x k + b x k ; k x ( a – b ) = a x k – b x k ( a + b ) x k = a x k + b x k ; ( a – b ) x k = a x k – b x k * ﻣﺜﺎﻝ : ) 5,5 – 11 ( D = 2,5 x ( 4 + 7,2 ) E = 3 x 5,5 = 2,5 x 4 + 2,5 x 7,2 = 3 x 11 – 3 x ,61 – 33 = 81 + 01 = 71 = 82 = 5,1 F = ( 6,5 + 1 ) x 5 G = ( 13 – 9,2 ) x 2,9 = 5 x 6,5 + 5 x 1 = 1,5 x 13 – 1,5 x 8,31 – 5,91 = 5 + 5,23 = 5,73 = 7,5 =
5.
ﺍﻟﻤﺴﺘﻘﻴــﻢ ﻭ ﺃﺟــﺰﺍﺅﻩ
– Iﺍﻟﻤﺴﺘﻘﻴﻢ ﺍﻟﻨﻘــﻂ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﺔ . 1( – ﺗﻌﺮﻳﻒ : ﻭﺩ , ﻭ ﻫﻮ ﻏﻴﺮ ﻣﺤﺪ ﺍﻟﻤﺴﺘﻘﻴﻢ ﻫﻮ ﻣﺠﻤﻮﻋﺔ ﻣﻦ ﻧﻘﻂ ﺍﻟﻤﺴﺘﻮﻯ * ﻣﺜﺎﻝ : ﺍﻟﺸﻜﻞ ﺍﻟﺘﺎﻟﻲ ﻳﻤﺜﻞ ﻣﺴﺘﻘﻴﻤﺎ ﻭ ﻗﺪ ﺭﻣﺰﻧﺎ ﻟﻪ ﺑﺎﻟﺮﻣﺰ : ). (D )(D 2( – ﺍﻟﻤﺴﺘﻘﻴﻢ ﺍﻟﻤﺎﺭ ﻣﻦ ﻧﻘﻄﺘﻴﻦ : * ﺧﺎﺻﻴﺔ : ﻣﻦ ﻧﻘﻄﺘﻴﻦ ﻣﺨﺘﻠﻔﺘﻴﻦ ﻳﻤﺮ ﻣﺴﺘﻘﻴﻢ ﻭﺣﻴـــﺪ * ﻣﺜﺎﻝ : ﻧﺮﻣﺰﻟﻬﺬﺍ ﺍﻟﻤﺴﺘﻘﻴﻢ ﺑﺎﻟﺮﻣﺰ : ). (AB * ﻣﻼﺣــﻈـﺔ ﻫﺎﻣــﺔ : ﻣﻦ ﻧﻘﻄﺔ ﻭﺍﺣﺪﺓ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ ﺗﻤﺮ ﻋــﺪﺓ ﻣﺴﺘﻘﻴﻤﺎﺕ
6.
3( – ﺍﻟﻨﻘﻂ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﺔ :
ﺗﻜﻮﻥ ﻧﻘﻂ ﻣﺴﺘﻘﻴﻤﻴﺔ ﺇﺫﺍ ﻛﺎﻧﺖ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﻧﻔﺲ ﺍﻟﻤﺴﺘﻘﻴﻢ * ﺗﻌﺮﻳﻒ : * ﻣﺜﺎﻝ : ﻧﻘﻮﻝ ﺃﻥ ﺍﻟﻨﻘﻂ Aﻭ Bﻭ Cﻭ Dﻣﺴﺘﻘﻴﻤﻴﺔ . ﻧﻘﻮﻝ ﺃﻥ ﺍﻟﻨﻘﻂ Eﻭ Fﻭ Gﻏﻴﺮ ﻣﺴﺘﻘﻴﻤﻴﺔ . _ IIﺍﻷﻭﺿﺎﻉ ﺍﻟﻨﺴﺒﻴﺔ ﻟﻤﺴﺘﻘﻴﻤﻴﻦ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ : 1( – ﺍﻟﻤﺴﺘﻘﻴﻤﺎﻥ ﺍﻟﻤﺘﻘﺎﻃﻌﺎﻥ : * ﺗﻌﺮﻳﻒ : ﻳﻜﻮﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻘﺎﻃﻌﻴﻦ ﺇﺫﺍ ﻛﺎﻧﺎ ﻳﺸﺘﺮﻛﺎﻥ ﻓﻲ ﻧﻘﻄﺔ ﻭﺍﺣﺪﺓ * ﻣﺜﺎﻝ : ﻧﻘﻮﻝ ﺃﻥ ) (Dﻭ ) (Lﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻘﺎﻃﻌﺎﻥ . 2( ﺴﺘﻘﻴﻤﺎﻥ ﺍﻟﻤﻨﻄﺒﻘﺎﻥ : ﺍﻟﻤ * ﺗﻌﺮﻳﻒ : ﻳﻜﻮﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﻨﻄﺒﻘﻴﻦ ﺇﺫﺍ ﻛﺎﻧﺎ ﻳﺸﺘﺮﻛﺎﻥ ﻓﻲ ﺃﻛﺜﺮ ﻣﻦ ﻧﻘﻄﺔ ﻭﺍﺣﺪﺓ . * ﻣﺜﺎﻝ : ﻧﻘﻮﻝ ﺃﻥ ) (Lﻭ ) (Kﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﻨﻄﺒﻘﺎﻥ .
7.
3( – ﺍﻟﻤﺴﺘﻘﻴﻤﺎﻥ ﺍﻟﻤﺘﻮﺍﺯﻳﺎﻥ ﻗﻄﻌﺎ :
ﻳﻜﻮﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻗﻄﻌﺎ ﺇﺫﺍ ﻛﺎﻧﺎ ﻻ ﻳﺸﺘﺮﻛﺎﻥ ﻓﻲ ﺃﻳﺔ ﻧﻘﻄﺔ * ﺗﻌﺮﻳﻒ : * ﻣﺜﺎﻝ : ﻧﻘﻮﻝ ﺃﻥ ) (Dﻭ ) (Lﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﺎﻥ ﻗﻄﻌﺎ ﻭ ﻧﻜـــﺘﺐ : )(D) // (L ﻭ ﻧﻘﺮﺃ : ) (Dﻳﻮﺍﺯﻱ ) (Lﻭ ) (Lﻳﻮﺍﺯﻱ . ﺃ _ IIIﺍﻟﻤﺴﺘﻘﻴﻤﺎﻥ ﺍﻟﻤﺘﻌﺎﻣﺪﺍﻥ : 1( – ﺗﻌﺮﻳﻒ : ﻳﻜﻮﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻌﺎﻣﺪﻳﻦ ﺇﺫﺍ ﻛﺎﻧﺎ ﻳﺤﺪﺩﺍﻥ ﺯﺍﻭﻳﺔ ﻗﺎﺋﻤﺔ * ﻣﺜﺎﻝ : ﻧﻘﻮﻝ ﺃﻥ ﺍﻟﻤﺴﺘﻘﻴﻢ ) (Dﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺍﻟﻤﺴﺘﻘﻴﻢ ) ( Rﻭ ﻧﻜــﺘﺐ : )( R ) ^ (D ﻭ ﻧﻘﺮﺃ : ) (Dﻋﻤﻮﺩﻱ ﻋﻠﻰ ) ( Rﺃﻭ ) ( Rﻋﻤﻮﺩﻱ ﻋﻠﻰ )(D 2( – ﺧـﺎﺻﻴﺔ : ﻣﻦ ﻧﻘﻄﺔ ﻣﻌﻠﻮﻣﺔ ﻳﻤﺮ ﻣﺴﺘﻘﻴﻢ ﻭﺣﻴــﺪ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﻣﺴﺘﻨﻘﻴﻢ ﻣﻌﻠﻮﻡ _ IVﻧﺼﻒ ﻣﺴﺘﻘﻴﻢ : 1( – ﻣﺜﺎﻝ : ﺟﺰء ﺍﻟﻤﺴﺘﻘﻴﻢ ) (Dﺍﻟﻤﻠﻮﻥ ﺑﺎﻷﺣﻤﺮ ﻳﺴﻤﻰ : ﻧﺼﻒ ﻣﺴﺘﻘﻴﻢ ﺃﺻﻠﻪ Aﻭ ﻳﻤﺮ ﻣﻦ . B ﻭ ﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺮﻣﺰ : ). [AB ﻧﺴﻤﻲ ﺍﻟﻤﺴﺘﻘﻴﻢ ) : (Dﺣــﺎﻣﻞ ﻧﺼﻒ ﺍﻟﻤﺴﺘﻘﻴﻢ ). [AB
8.
2( – ﻧﺼﻔﺎ ﺍﻟﻤﺴﺘﻘﻴﻢ ﺍﻟﻤﺘﻘﺎﺑﻼﻥ :
* ﺗﻌﺮﻳﻒ : ﻳﻜﻮﻥ ﻧﺼﻔﺎ ﻣﺴﺘﻘﻴﻢ ﻣﺘﻘﺎﺑﻠﻴﻦ ﺇﺫﺍ ﻛﺎﻧﺎ ﻣﺨﺘﻠﻔﻴﻦ ﻭ ﻛﺎﻥ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻷﺻﻞ ﻭ ﻧﻔﺲ ﺍﻟﺤــﺎﻣﻞ . * ﻣﺜﺎﻝ : ﻧﻼﺣﻆ ﺃﻥ ﻧﺼﻔﻲ ﺍﻟﻤﺘﻘﻴﻢ ) [ABﻭ ) [ACﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﺮﺃﺱ Aﻭ ﻧﻔﺲ ﺍﻟﺤﺎﻣﻞ ). (D ﻧﻘﻮﻝ ﺃﻥ ) [ABﻭ ) [ACﻧﺼﻔﺎ ﻣﺴﺘﻘﻴﻢ ﻣﺘﻘﺎﺑﻠﻴﻦ . 3( – ﺍﻟﻤﺴﻘﻂ ﺍﻟﻌﻤﻮﺩﻱ ﻟﻨﻘﻄﺔ ﻋﻠﻰ ﻣﺴﺘﻘﻴﻢ : * ﺗﻌﺮﻳﻒ : ﺍﻟﻤﺴﻘﻂ ﺍﻟﻌﻤﻮﺩﻱ ﻟﻨﻘﻄﺔ Eﻋﻠﻰ ﻣﺴﺘﻘﻴﻢ ) (Dﻫﻲ Hﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ )(D ﻭ ﺍﻟﻤﺴﺘﻘﻴﻢ ﺍﻟﻌﻤﻮﺩﻱ ﻋﻠﻴﻪ ﻓﻲ . H * ﻣﺜﺎﻝ : ﺍﻟﻤﺴﺎﻓﺔ EHﺗﺴﻤﻰ : ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﺍﻟﻨﻘﻄﺔ Eﻭ ﺍﻟﻤﺴﺘﻘﻴﻢ )(D _ Vﺍﻟﻘﻄﻌﺔ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﺔ : 1( – ﻣﺜﺎﻝ : ﻧﺴﻤﻲ ﻫﺬﺍ ﺍﻟﺸﻜﻞ : ﻗـﻄــﻌـﺔ ﻣﺴﺘﻘﻴـﻤﻴــﺔ . ﻭ ﻧﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ : ]. [AB Aﻭ Bﻳﺴﻤﻴﺎﻥ : ﻃﺮﻓﻲ ﻄﻌﺔ ]. [AB ﺍﻟﻘ ﺍﻟﻤﺴﺘﻘﻴﻢ ) (ABﻳﺴﻤﻰ ﺣﺎﻣﻞ ﺍﻟﻘﻄﻌﺔ ][AB 2( – ﻣﻨﺘﺼﻒ ﻗﻄﻌﺔ : * ﺗﻌﺮﻳﻒ : ﻣﻨﺘﺼﻒ ﻗﻄﻌﺔ ﻫﻮ ﻧﻘﻄﺔ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻘﻄﻌﺔ ﻭ ﻣﺘﺴﺎﻭﻳﺔ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻦ ﻃﺮﻓﻲ ﻫﺬﻩ ﺍﻟﻘﻄﻌــﺔ .
9.
* ﻣﺜﺎﻝ :
ﻧﺴﻤﻲ ﺍﻟﻨﻘﻄﺔ Mﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌــﺔ ]. [AB ﻭ MA = MB * ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : Mﻣﻨﺼﻒ ﺍﻟﻘﻄﻌﺔ ] [ABﻳﻌﻨﻲ ﺃﻥ : ]M Î [AB 3( – ﺍﻟﻘﻄﻌﺘﺎﻥ ﺍﻟﻤﺘﻘﺎﻳﺴﺘﺎﻥ : * ﺗﻌﺮﻳﻒ : ﺗﻜﻮﻥ ﻗﻄﻌﺘﺎﻥ ﻣﺘﻘﺎﻳﺴﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻄـــﻮﻝ * ﻣﺜﺎﻝ : ﻧﻘﻮﻝ ﺃﻥ ] [ABﻭ ] [CDﻗﻄﻌﺘﺎﻥ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ , ﻭ ﻧﻜــﺘﺐ : AB = CD
10.
ﺍﻟﻜﺘﺎﺑﺎﺕ ﺍﻟﻜﺴﺮﻳﺔ ﻭ ﻣﻘﺎﺭﻧﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻜﺴﺮﻳﺔ
1( – ﺍﻟﻜﺘﺎﺑﺎﺕ ﺍﻟﻜﺴﺮﻳﺔ ﻟﻌﺪﺩ ﻛﺴﺮﻱ : * ﻗﺎﻋﺪﺓ 1 : aﻭ bﻋﺪﺩﺍﻥ ﻋﺸﺮﻳﺎﻥ ﻏﻴﺮ ﻣﻨﻌﺪﻣﻴﻦ . ﻳﻤﻜﻦ ﺇﻳﺠﺎﺩ ﻛﺘﺎﺑﺎﺕ ﻛﺴﺮﻳﺔ ﺍﻋﺪﺩ ﻛﺴﺮﻱ ﻭ ﺫﻟﻚ ﺑﻀﺮﺏ ﺃﻭ ﻌﺪﺩ ﺍﻟﻜﺴﺮﻱ ﻋﻠﻰ ﻧﻔﺲ ﺍﻟﻌﺪﺩ ﺍﻟﻐﻴﺮ ﺍﻟﻤﻨﻌﺪﻡ . ﻗﺴﻤﺔ ﺣﺪﻱ ﻫﺬﺍ ﺍﻟ aﻭ bﻭ mﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﺑﺤﻴﺚ : m aﻏﻴﺮ ﻣﻨﻌﺪﻣﻴﻦ . ﻭ ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : a ´ m a a : m a ;; = = b ´ m b b : m b 6 2 : 21 21 51 3 ´ 5 5 = ;; = = = * ﺃﻣﺜﻠﺔ : 7 2 : 41 41 72 3 ´ 9 9 2( – ﺟﻌﻞ ﻣﻘﺎﻡ ﻋﺸﺮﻱ ﻟﻜﺘﺎﺑﺔ ﻛﺴﺮﻳﺔ ﻋﺪﺩﺍ ﺻﺤﻴﺤﺎ : * ﻗﺎﻋﺪﺓ 2 : ﻟﺠﻌﻞ ﻣﻘﺎﻡ ﻋﺪﺩ ﻛﺴﺮﻱ ﻋﺪﺩﺍ ﺻﺤﻴﺤﺎ , ﻧﻀﺮﺏ ﺣﺪﻱ ﻫﺬﺍ ﺍﻟﻌﺪﺩ ﺍﻟﻜﺴﺮﻱ ﻓﻲ : ﺃﻭ 001 0001 ﺃﻭ ....... ﺃﻭ 01 31 00031 0001 ´ 31 7 007 001´ 7 011 01 11 11 ´ = = ;; = = ;; = = * ﺃﻣﺜﻠﺔ : 2101 0001 ´ 210 1 210 1 , , 2 001´ 20 0 20 0 , , 53 01´ 5 3 5 3 , , 3( – ﻣﻘﺎﺭﻧﺔ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻤﻘﺎﻡ : * ﻗﺎﻋﺪﺓ 3 : , ﻓﺈﻥ ﺃﻛﺒﺮﻫﻤﺎ ﻫﻮ ﺍﻟﺬﻱ ﻟﻪ ﺃﻛﺒﺮ ﺑﺴﻂ ﺇﺫﺍ ﻛﺎﻥ ﻟﻌﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻧﻔﺲ ﺍﻟﻤﻘﺎﻡ
11.
71 15
17 31 7 3 > 15 ﻷﻥ > ;; ﻷﻥ 17 < 31 < ;; ﻷﻥ 3 > 7 > * ﺃﻣﺜﻠﺔ : 2 2 9 9 11 11 71 4( – ﺭﻧﺔ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﺒﺴﻂ : ﻣﻘﺎ * ﻗﺎﻋﺪﺓ 4 : , ﻓﺈﻥ ﺃﻛﺒﺮﻫﻤﺎ ﻫﻮ ﺍﻟﺬﻱ ﻟﻪ ﺃﺻﻐﺮ ﻣﻘﺎﻡ ﺇﺫﺍ ﻛﺎﻥ ﻟﻌﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻧﻔﺲ ﺍﻟﺒﺴﻂ 71 71 7 7 3 3 ﻷﻥ 22 < 9 > ﻷﻥ 31 > 14 ;; > ﻷﻥ 13 < 11 ;; > * ﺃﻣﺜﻠﺔ : 22 9 31 14 13 11 5( – ﺭﻧﺔ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻣﻘﺎﻡ ﺃﺣﺪﻫﻤﺎ ﻣﻀﺎﻋﻒ ﻟﻤﻘﺎﻡ ﺍﻵﺧﺮ : ﻣﻘﺎ * ﻗﺎﻋﺪﺓ 4 : ﻟﻤﻘﺎﺭﻧﺔ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻣﻘﺎﻡ ﺃﺣﺪﻫﻤﺎ ﻣﻀﺎﻋﻒ ﻟﻤﻘﺎﻡ ﺍﻵﺧﺮ , ﻧﻮﺣﺪ ﻣﻘﺎﻣﻴﻬﻤﺎ ﺛﻢ ﻧﻄﺒﻖ ﺍﻟﻘﺎﻋﺪﺓ 3 * ﻣﺜﺎﻝ : 7 5 ﻭ ﻟﻨﻘﺎﺭﻥ ﺍﻟﻌﺪﺩﻳﻦ : 4 61 82 4 ´ 7 7 5 5 = = ﻭ = ﻟﺪﻳﻨﺎ : 61 4 ´ 4 4 61 61 82 5 ﻷﻥ 82 < 5 < ﻭﺑﻤﺎ ﺃﻥ 61 61 7 5 < ﻓﺈﻥ 4 61 6( – ﺭﻧﺔ ﻋﺪﺩ ﻛﺴﺮﻱ ﻭ 1 : ﻣﻘﺎ 5 : * ﻗﺎﻋﺪﺓ ﻳﻜﻮﻥ ﻋﺪﺩ ﻛﺴﺮﻱ ﺃﻛﺒﺮ ﻣﻦ 1 ﺇﺫﺍ ﻛﺎﻥ ﺑﺴﻄﻪ ﺃﻛﺒﺮ ﻣﻦ ﻣﻘﺎﻣﻪ , ﻭ ﻳﻜﻮﻥ ﺃﺻﻐﺮ ﻣﻦ 1 ﺇﺫﺍ ﻛﺎﻥ ﺑﺴﻄﻪ ﺃﺻﻐﺮ ﻣﻦ ﻣﻘﺎﻣﻪ . 5 17 ﻷﻥ 3 < 5 1 < ;; 25 > 17 ﻷﻥ 1 > * ﻣﺜﺎﻝ : 73 25
12.
ﺍﻟﻌﻤﻠﻴﺎﺕ ﻋﻠﻰ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻜﺴــﺮﻳﺔ
1( – ﺟﺪﺍء ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ : * ﻗﺎﻋﺪﺓ 1 : ﻘﺎﻡ ﻓﻲ ﺍﻟﻤﻘﺎﻡ . ﻟﺤﺴﺎﺏ ﺟﺪﺍء ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻧﻀﺮﺏ ﺍﻟﺒﺴﻂ ﻓﻲ ﺍﻟﺒﺴﻂ ﻭ ﺍﻟﻤ a c a ´ c c a = ´ ﻋﺪﺩﺍﻥ ﻛﺴﺮﻳﺎﻥ : ﻭ b d b ´ d d b 54 3 ´ 51 3 51 3 31 711 9 ´ 31 77 7 ´ 11 7 11 = ´ = ´ 5,1 = ;; = 9 ´ = ;; = ´ = * ﺃﻣﺜﻠﺔ : 07 7 ´ 01 7 01 7 22 22 1 ´ 22 01 2 ´ 5 2 5 ﺎﻡ : 2( – ﻣﺠﻤﻮﻉ ﻭ ﻓﺮﻕ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻤﻘ * ﻗﺎﻋﺪﺓ 2 : ﻭ ﻓﺮﻕ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻤﻘﺎﻡ : ﻧﺤﺘﻔﻆ ﺑﻨﻔﺲ ﻟﺤﺴﺎﺏ ﻣﺠﻤﻮﻉ ﺃ . ﺍﻟﻤﻘﺎﻡ ﺛﻢ ﻧﺤﺴﺐ ﻣﺠﻤﻮﻉ ﺃﻭ ﻓﺮﻕ ﺍﻟﺒﺴﻄﻴﻦ a c a - c a c a + c c a = - = + ﻭ ( > c ) a ﻭ ﻋﺪﺩﺍﻥ ﻛﺴﺮﻳﺎﻥ : b b b b b b b b 8 91 - 72 91 72 81 7 + 11 7 11 = - = ;; = + = * ﺃﻣﺜﻠﺔ : 9 9 9 9 5 5 5 5 3( – ﻣﺠﻤﻮﻉ ﻭ ﻓﺮﻕ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻣﻘﺎﻡ ﺃﺣﺪﻫﻤﺎ ﻣﻀﺎﻋﻒ ﻣﻘﺎﻡ ﺍﻵﺧﺮ : * ﻗﺎﻋﺪﺓ 3 : ﻟﺤﺴﺎﺏ ﻣﺠﻤﻮﻉ ﺃﻭ ﻓﺮﻕ ﻋﺪﺩﻳﻦ ﻛﺴﺮﻳﻴﻦ ﻣﻘﺎﻡ ﺃﺣﺪﻫﻤﺎ ﻣﻀﺎﻋﻒ ﻟﻤﻘﺎﻡ ﺍﻵﺧﺮ , ﻧﻮﺣﺪ ﻣﻘﺎﻣﻴﻬﻤﺎ ﺛﻢ ﻧﻄﺒﻖ ﺍﻟﻘﺎﻋﺪﺓ 2 . 23 7 - 93 7 93 7 31 62 11 + 51 11 51 11 5 - = - = = ;; = + + = = * ﺃﻣﺜﻠﺔ : 3 9 9 9 9 9 12 12 12 7 12 12
13.
ﺣﺎﻻﺕ ﺧـــﺎﺻﺔ :
8 83 2 - 04 2 04 1 93 12 + 81 12 81 7 9 = - - = = ;; = + + = = 02 02 01 4 02 02 42 42 8 21 42 42 93 33 27 3 21 59 53 + 06 53 06 5 51 = - - = ;; = + + = = 66 66 66 6 11 82 82 4 7 82 82 ﺗﻘﻨﻴﺎﺕ ﻭ ﻣﻬﺎﺭﺍﺕ 7 1 1 31 7 5 + + ,1 = B 5 ;; + 11 = A + + + ﻟﻨﺤﺴﺐ ﺍﻟﻤﺠﻤﻮﻋﻴﻦ Aﻭ Bﺑﺄﺑﺴﻂ ﻃﺮﻳﻘﺔ : 02 5 9 6 3 6 7 1 1 7 31 5 + + ,1 = B 5 ) + ( + ) + ( + 11 = A 02 5 6 6 9 3 7 1 51 1 12 31 + 5 = B + + + 11 = A ) + ( + 6 9 9 02 5 02 22 81 1 7 51 + + 11 = A + ) + ( = B 9 6 5 02 02 22 1 22 + 3 + 11 = A = B + 9 5 02 22 + 41 = A 1 11 9 + = B 5 01 22 621 = A + 2 11 9 9 + = B 841 01 01 = A 31 9 = B 01
14.
ﺍﻟﻤﺘﻔﺎﻭﺗﺔ ﺍﻟﻤﺜﻠﺜﻴﺔ ﻭ ﻭﺍﺳــﻂ ﻗﻄﻌﺔ
1( – ﺍﻟﻤﺘﻔﺎﻭﺗﺔ ﺍﻟﻤﺜﻠﺜﻴﺔ : * ﺧﺎﺻﻴﺔ 1 : Aﻭ Bﻭ Cﺛﻼﺙ ﻧﻘﻂ ﻣﺨﺘﻠﻔــﺔ ﺇﺫﺍ ﻛﺎﻧﺖ Cﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻘﻄﻌﺔ ] [ABﻓﺈﻥ : AB = AC + BC ﺇﺫﺍ ﻛﺎﻧﺖ Cﻻ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻘﻄﻌﺔ ] [ABﻓﺈﻥ : AB < AC + BC * ﻣﺜﺎﻝ : AB = AC + BC AB < AC + BCﻭ ﻛﺬﻟﻚ : AC < AB + BCﻭ BC < AB + AC ﻭ ﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ ﻣﺎ ﻳﻠﻲ : ﻲ ﻣﺜﻠﺚ ﻃﻮﻝ ﺃﻱ ﺿﻠﻊ ﻣﻦ ﺃﺿﻼﻋﻪ ﺃﺻﻐﺮ ﻣﻦ ﻣﺠﻤﻮﻉ ﻃﻮﻟﻲ ﻓ ﺍﻟﻀﻠﻌﻴﻦ ﺍﻵﺧــــﺮﻳﻦ . ﺗﻄﺒﻴﻖ : ﻫﻞ ﻳﻤﻜﻦ ﺭﺳﻢ ﺍﻟﻤﺜﻠﺚ ABCﺑﺤﻴﺚ : AB = 7cmﻭ AC = 17cmﻭ BC = 5 cm؟ ﻧﻼﺣﻆ ﺃﻥ : 21 = 7 + 5 ﻭ ﺃﻥ 21 > 71 ﺃﻱ ﺃﻥ AC > AB + BC ﺇﺫﻥ : ﻻ ﻳﻤﻜﻦ ﺭﺳﻢ ﺍﻟﻤﺜﻠﺚ . ABC 2( – ﻭﺍﺳـــﻂ ﻗـﻄــﻌــﺔ : * ﺗﻌــﺮﻳﻒ : ﻭﺍﺳﻂ ﻗﻄﻌﺔ ﻫﻮ ﻣﺴﺘﻘﻴﻢ ﻳﻤﺮ ﻣﻦ ﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌــﺔ ﻭ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺣﺎﻣﻠﻬﺎ
15.
* ﻣﺜﺎﻝ :
ﻟﻨﺮﺳﻢ ﻗﻄﻌﺔ ] [ABﻗﻄﻌﺔ ﻭ ) (Dﻭﺍﺳﻄﻬﺎ * ﺧﺎﺻﻴﺔ 2 : ﻛﻞ ﻧﻘﻄﺔ ﺗﻨﺘﻤﻲ ﺇﺍﻟﻰ ﻭﺍﺳﻂ ﻗﻄﻌﺔ ﺗﻜﻮﻥ ﻣﺘﺴﺎﻭﻳﺔ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻦ ﻃﺮﻓﻴﻬﺎ * ﺑﺘﻌﺒﻴﺮ ﺁﺧــﺮ : ] [ABﻗﻄﻌﺔ ﻭ )∆( ﻭﺍﺳﻄﻬﺎ ﻭ Mﻧﻘﻄﺔ ﻣﻦ ﺍﻣﺴﺘﻮﻯ . ) M Î (Dﻳﻌﻨﻲ ﺃﻥ MA = MB * ﺧﺎﺻﻴﺔ 3 : ﻛﻞ ﻧﻘﻄﺔ ﻣﺘﺴﺎﻭﻳﺔ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻦ ﻃﺮﻓﻲ ﻗﻄﻌﺔ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﻭﺍﺳﻂ ﻫﺬﻩ ﺍﻟﻘﻄﻌﺔ * ﺑﺘﻌﺒﻴﺮ ﺁﺧــﺮ : ] [ABﻗﻄﻌﺔ ﻭ )∆( ﻭﺍﺳﻄﻬﺎ ﻭ Mﻧﻘﻄﺔ ﻣﻦ ﺍﻣﺴﺘﻮﻯ . MA = MBﻳﻌﻨﻲ ﺃﻥ )M Î (D
16.
3( – ﻭﺍﺳﻄﺎﺕ ﻣﺜﻠﺚ :
* ﺗﻌﺮﻳﻒ 2 : ﻭﺍﺳﻂ ﻣﺜﻠﺚ ﻫﻮ ﻭﺍﺳﻂ ﻛﻞ ﺿـــﻠﻊ ﻣﻦ ﺃﺿــــﻼﻋــﻪ ﻣﺜﺎﻝ : ABCﻣﺜﻠﺚ ﻭ (Dﻭﺍﺳﻂ ﺍﻟﻀﻠﻊ ]. [BC ) ﻧﺴﻤﻲ ﺍﻟﻤﺴﺘﻘﻴﻢ (Dﻭﺍﺳﻂ ﺍﻟﻤﺜﻠﺚ ABC ) ﺧﺎﺻﻴﺔ 4 : * ﻭﺍﺳﻄﺎﺕ ﻣﺜﻠﺚ ﺗﺘﻼﻗﻰ ﻓﻲ ﻧﻘﻄﺔ ﻭﺍﺣﺪﺓ ﺗﺴﻤﻰ ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﻤﺤﻴﻄﺔ ﺑﻬﺬﺍ ﺍﻟﻤﺜﻠﺚ ﻣﺜﺎﻝ :
17.
ﺗﻘﺪﻳﻢ ﻭ ﻣﻘﺎﺭﻧﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ
_Iﺗ ﻘ ﺪﻳﻢ . ـ ـــ 1( – ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻤﻮﺟﺒﺔ ﻭ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﺴﺎﻟﺒﺔ : * ﺗﻌﺮﻳﻒ 1 : ﺍﻷﻋﺪﺍﺩ ﻣﺜﻞ : 0 ; 1 ; 2 , 41 ; 41,3 ; 11 ; 5,2 ﺗﺴﻤﻰ ﺃﻋﺪﺍﺩﺍ ﻋﺸﺮﺑﺔ ﻣﻮﺟﺒﺔ . ﺍﻷﻋﺪﺍﺩ ﻣﺜﻞ : 0 ; 2 ; 1 ; 44,0 ; 21 ; 5,2 ﺗﺴﻤﻰ ﺃﻋﺪﺍﺩﺍ ﻋﺸﺮﻳﺔ ﺳﺎﻟﺒﺔ . ﺍﻟﻌﺪﺩ 0 ﻫﻮ ﻋﺪﺩ ﻋﺸﺮﻱ ﻣﻮﺟﺐ ﻭ ﺳﺎﻟﺐ ﻓﻲ ﺁﻥ ﻭﺍﺣﺪ . * ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﺒﻴﺔ : 2( – ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴ * ﺗﻌﺮﻳﻒ 2 : ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻤﻮﺟﺒﺔ ﻭ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﺴﺎﻟﺒﺔ ﺗﻜﻮﻥ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ * ﻣﻼﺣﻈﺔ ﺭﻫﺎﻣﺔ : ﺍﻷﻋﺪﺍﺩ ﻣﺜﻞ : 0 ; 1 ; 8 , 2 ; 41 ; 1 ; 5 ; 15 ; 11 .... ﺗﺴﻤﻰ ﺃﻋﺪﺍﺩﺍ ﺻﺤﻴﺤﺔ ﻧﺴﺒﻴﺔ . ﻛﻞ ﻋﺪﺩ ﺻﺤﻴﺢ ﻧﺴﺒﻲ ﻫﻮ ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ . ﺍﻟﻌﺪﺩ ﻣﺜﻞ : 21,41 ﺃﻭ 5,2 ﻫﻮ ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﻭ ﻟﻴﺲ ﺑﻌﺪﺩ ﺻﺤﻴﺢ ﻧﺴﺒﻲ . 3( – ﺍﻟﻤﺴﺘﻘﻴﻢ ﺍﻟﻤﺪﺭﺝ : ﻧﻌﺘﺒﺮ ) (Dﻣﺴﺘﻘﻴﻤﺎ ﻭ Oﻭ Iﻧﻘﻄﺘﻴﻦ ﻣﺨﺘﻠﻔﺘﻴﻦ ﻣﻦ ) . (Dﻟﻨﺪﺭﺝ ﺍﻟﻤﺴﺘﻘﻴﻢ ) (Dﺑﻮﺍﺳﻄﺔ ﺍﻟﻘﻄﻌ ﺔ ][OI ) ﺃﻇﺮ ﺍﻟﺸﻜــﻞ ﺃﺳﻔﻠﻪ ( . )(D E F O I A B , , , , , , , , , , , , , , , 0 1 ﺍﻷﻋﺪﺍﺩ ﺍﻟﺴﺎﻟﺒﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻤﻮﺟﺒﺔ ﻛﻞ ﻧﻘﻄﺔ ﻣﻦ ﺍﻟﻤﺴﺘﻘﻴﻢ ) (Dﻣﺮﺗﺒﻄﺔ ﺑﻌﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﻳﺴﻤﻰ ﺃﻓﺼﻮﻝ ﻫﺬﻩ ﺍﻟﻨﻘﻄﺔ . ﺍﻟﻨﻘﻄﺔ Oﺗﺴﻤﻰ ﺃﺻﻞ ﺍﻟﻤﺴﺘﻘﻴﻢ ﺍﻟﻤﺪﺭﺝ ). (D ﻃﻮﻝ ﺍﻟﻘﻄﻌﺔ ] [OIﻳﺴﻤﻰ ﻭﺣــﺪﺓ ﺍﻟﺘﺪﺭﻳﺞ . ﺍﻟﻨﻘﻄﺔ Eﺃﻓﺼﻮﻟﻬﺎ 4 ﺍﻟﻨﻘﻄﺔ Aﺃﻓﺼﻮﻟﻬﺎ 3 ﺍﻟﻨﻘﻄﺔ Oﺃﻓﺼﻮﻟﻬﺎ 0 ﺍﻟﻨﻘﻄﺔ Fﺃﻓﺼﻮﻟﻬﺎ 5,3 ﺍﻟﻨﻘﻄﺔ Bﺃﻓﺼﻮﻟﻬﺎ 5,3 Iﺃﻓﺼﻮﻟﻬﺎ 1 ﺍﻟﻨﻘﻄﺔ 4( – ﻣﺴﺎﻓﺔ ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﻋﻦ ﺍﻟﺼﻔﺮ : * ﺗﻌﺮﻳﻒ 3 : ﺪﺭﺟﺎ ﺃﺻﻠﻪ Oﻭ Mﻧﻘﻄﺔ ﻣﻦ ) (Dﺃﻓﺼﻮﻟﻬﺎ ﺍﻟﻌﺪﺩ . a ﻧﻌﺘﺒﺮ ) (Dﻣﺴﺘﻘﻴﻤﺎ ﻣ ﻣﺴﺎﻓﺔ ﺍﻟﻌﺪﺩ aﻋﻦ ﺍﻟﺼﻔﺮ ﻫﻮ ﻃﻮﻝ ﺍﻟﻘﻄﻌﺔ ]. [OM
18.
5( – ﻣﻘﺎﺑﻞ ﻋﺪﺩ ﻋﺸﺮﻱ :
* ﺗﻌﺮﻳﻒ 4 : ﻳﻜﻮﻥ ﻋﺪﺩﺍﻥ ﻣﺘﻘﺎﺑﻠﻴﻦ ﺇﺫﺍ ﻛﺎﻧﺖ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻦ ﺍﻟﺼﻔﺮ ﻭ ﺇﺷﺎﺭﺗﺎﻫﻤﺎ ﻣﺨﺘﻠﻔﺘﻴﻦ . 11 ﻭ 11 ﻋﺪﺩﺍﻥ ﻣﺘﻘﺎﺑﻼﻥ ;; 2,1 ﻭ 2,1 ﻋﺪﺩﺍﻥ ﻣﺘﻘﺎﺑﻼﻥ * ﺃﻣﺜﻠﺔ : 3 ﻭ 3 ﻋﺪﺩﺍﻥ ﻣﺘﻘﺎﺑﻼﻥ 23,0 ﻭ 23,0 ﻋﺪﺩﺍﻥ ﻣﺘﻘﺎﺑﻼﻥ ;; ﻣﻘﺎﺑﻞ ﺍﻟﻌﺪﺩ 0 ﻫﻮ ﺍﻟﻌﺪﺩ 0 _ IIﻟﻤﻘ ﺎﺭﻧــﺔ : ﺍ ـــــ 1( – ﻣﻘﺎﺭﻧﺔ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻓﻲ ﺍﻹﺷﺎﺭﺓ : * ﻗﺎﻋﺪﺓ 1 : ﻛﻞ ﻋﺪﺩ ﻋﺸﺮﻱ ﻣﻮﺟﺐ ﺃﻛﺒﺮ ﻣﻦ ﻛﻞ ﻋﺪﺩ ﻋﺸﺮﻱ ﺳﺎﻟﺐ ﻏﻴﺮ ﻣﻨﻌﺪﻡ ;; 7,41 > 22 0 < 21,33 ;; 0 > 44,52 ;; 5,1 < 54,0 * ﺃﻣﺜﻠﺔ : 2( – ﻣﻘﺎﺭﻧﺔ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﺳﺎﻟﺒﻴﻦ : * ﻗﺎﻋﺪﺓ 2 : ﺇﺫﺍ ﻛﺎﻥ ﻋﺪﺩﺍﻥ ﻋﺸﺮﻳﺎﻥ ﺳﺎﻟﺒﺎﻳﻦ ﻓﺈﻥ ﺃﻛﺒﺮﻫﻤﺎ ﻫﻮ ﺍﻷﻗﺮﺏ ﻣﻦ ﺍﻟﺼﻔﺮ ;; 1 < 5,2 3522 > 0 ;; 63 > 1,0 * ﺃﻣﺜﻠﺔ : ﺍﻟﻌﺪﺩ 0 ﻫﻮ ﺃﻛﺒﺮ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺴﺎﻟﺒﺔ ﻭ ﺃﺻﻐﺮ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻤﻮﺟﺒﺔ * ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﻭ £ . 3( – ﺍﻟﺮﻣﺰﺍﻥ : ³ ﻭ 33 ³ 33 ﺍﻟﺮﻣﺰ ³ ﻳﻘﺮﺃ : ﺃﻛﺒﺮ ﻣﻦ ﺃﻭ ﻳﺴﺎﻭﻱ ﻭ ﻳﺴﺘﻌﻤﻞ ﻓﻲ ﺣﺎﻟﺘﻴﻦ ﻣﺜﻞ : 32 ³ 3,11 ﺴﺎﻭﻱ ﻭ ﻳﺴﺘﻌﻤﻞ ﻓﻲ ﺣﺎﻟﺘﻴﻦ ﻣﺜﻞ : 5,1 5,73 £ ﻭ 6,7 – 6,7 £ ﺍﻟﺮﻣﺰ £ ﻳﻘﺮﺃ : ﺃﺻﻐﺮ ﻣﻦ ﺃﻭ ﻳ ﺗﻘﻨﻴﺎﺕ : ﻟﺘﺮﺗﻴﺐ ﻋﺪﺓ ﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ ﻧﺮﺗﺐ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺴﺎﻟﺒﺔ ﻓﻴﻤﺎ ﺑﻴﻨﻬﺎ ﺛﻢ ﻧﺮﺗﺐ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻤﻮﺟﺒﺔ ﻓﻴﻤﺎ ﺑﻴﻨﻬﺎ ﺛﻢ ﻧﺮﺗﺐ ﺍﻟﻜﻞ ﻣﺜﺎﻝ : ﻟﻨﺮﺗﺐ ﺍﻷﻋﺪﺍﺩ : 6,41 ;; 11 ;; 55,8 ;; 9,5 ;; 6 ;; 5,1 ;; 52 ;; 0 ﻟﺪﻳﻨﺎ : 0 < 5,1 < 6 < 55,8 < 6,41 ﻭ 52 < 11 < 9,5 < 0 52 < 11 < 9,5 < 0 < 5,1 < 6 < 55,8 < 6,41 ﺇﺫﻥ
19.
ﻤﻮﻉ ﻗﻴﺎﺳﺎﺕ ﺯﻭﺍﻳﺎ ﻣﺜﻠﺚ / ﻣﺜﻠﺜﺎﺕ ﺧﺎﺻﺔ
ﻣﺠ _Iﻣﺠﻤﻮﻉ ﻗﻴﺎﺳﺎﺕ ﺯﻭﺍﻳﺎ ﻣﺜﻠﺚ . 1( – ﺍﻟﺰﻭﺍﻳﺎ : ﺗﻌﺎﺭﻳﻒ ﻭ ﻣﻔﺮﺩﺍﺕ : Tﺍﻟﺸﻜﻞ ﺟﺎﻧﺒﻪ ﻳﺴﻤﻰ : ﺯﺍﻭﻳﺔ . ﻳﺮﻣﺰ ﻟﻬﺬﻩ ﺍﻟﺰﺍﻭﻳﺔ ﺑﺎﻟﺮﻣﺰ : A ˆ B O ﺍﻟﻨﻘﻄﺔ Oﺗﺴﻤﻰ ﺭﺃﺱ ﻫﺬﻩ ﺍﻟﺰﺍﻭﻳﺔ . ﻧﺼﻔﺎ ﺍﻟﻤﺴﺘﻘﻴﻢ ) [OAﻭ ) [OBﻳﺴﻤﻴﺎﻥ : ﺿﻠﻌﻲ ﻫﺬﻩ ﺍﻟﺰﺍﻭﻳﺔ . Tﺯﻭﺍﻳﺎ ﺧﺎﺻﺔ : ± ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﻨﻌﺪﻣﺔ : ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﻨﻌﺪﻣﺔ ﻫﻲ ﺯﺍﻭﻳﺔ ﻗﻴﺎﺳﻬﺎ °0 . ± ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺤﺎﺩﺓ : ﺎ ﻣﺤﺼﻮﺭ ﺑﻴﻦ °0 ﻭ °09 . ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﺤﺎﺩﺓ ﻫﻲ ﺯﺍﻭﻳﺔ ﻗﻴﺎﺳﻬ ± ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻘﺎﺋﻤﺔ : ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻘﺎﺋﻤﺔ ﻫﻲ ﺯﺍﻭﻳﺔ ﻗﻴﺎﺳﻬﺎ °09 .
20.
± ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﻨﻔﺮﺟﺔ : ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﻨﻔﺮﺟﺔ ﻫﻲ ﺯﺍﻭﻳﺔ ﻗﻴﺎﺳﻬﺎ ﻣﺤﺼﻮﺭ ﺑﻴﻦ °09 ﻭ
°081 . ± ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﺔ : ﺍﻭﻳﺔ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﺔ ﻫﻲ ﺯﺍﻭﻳﺔ ﻗﻴﺎﺳﻬﺎ °081 ﺍﻟﺰ ± ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﻠﻴــﺌﺔ : ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﻠﻴﺌﺔ ﻫﻲ ﺯﺍﻭﻳﺔ ﻗﻴﺎﺳــﻬﺎ °063 . Tﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﻘﺎﻳﺴﺘﺎﻥ : ﺗﻜﻮﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻘﺎﻳﺴﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻘﻴﺎﺱ . Tﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﺤﺎﺫﻳﺘﺎﻥ : ﺗﻜﻮﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺤﺎﺫﻳﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ : ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﺮﺃﺱ . ﻟﻬﻤﺎ ﺿﻠﻊ ﻣﺸﺘﺮﻙ . ﻭ ﻳﺘﻘﺎﻃﻌﺎﻥ ﻓﻲ ﺍﻟﻀﻠﻊ ﺍﻟﻤﺸﺘﺮﻙ . Tﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﺘﺎﻣﺘﺎﻥ : ﺗﻜﻮﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺘﺎﻣﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ ﻣﺠﻤﻮﻉ ﻗﻴﺎﺳﻬﻤﺎ ﻳﺴﺎﻭﻱ °09 Tﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﻜﺎﻣﻠﺘﺎﻥ : ﺗﻜﻮﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻜﺎﻣﻠﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ ﻣﺠﻤﻮﻉ ﻗﻴﺎﺳﻬﻤﺎ ﻳﺴﺎﻭﻱ °081 2( – ﻣﺠﻤﻮﻉ ﻗﻴﺎﺳﺎﺕ ﺯﻭﺍﻳﺎ ﻣﺜﻠﺚ : * ﺧﺎﺻﻴﺔ 1 : ﻣﺠﻤﻮﻉ ﻗﻴﺎﺳﺎﺕ ﺯﻭﺍﻳﺎ ﻣﺜﻠﺚ ﻳﺴﺎﻭﻱ °081 ABCﻣﺜﻠﺚ
21.
3( – ﻣﺜﻠﺜﺎﺕ ﺧـــﺎﺻﺔ :
± ﺍﻟﻤﺜﻠﺚ ﺍﻟﻘﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ : * ﺗﻌﺮﻳﻒ 1 : ﻛﻞ ﻣﺜﻠﺚ ﻟﻪ ﺯﺍﻭﻳﺔ ﻗﺎﺋﻤﺔ ﻳﺴﻤﻰ ﻣﺜﻠﺚ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﺜﻠﺚ ﺍﻟﻘﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﻫﻮ ﻣﺜﻠﺚ ﻟﻪ ﺯﺍﻭﻳﺔ ﻗﺎﺋﻤﺔ * ﻣﺜﺎﻝ : ABCﻣﺜﺎﺙ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﻓﻲ . A * ﺧﺎﺻﻴﺔ 2 : ﺇﺫﺍ ﻛﺎﻥ ﻣﺜﻠﺚ ﻗﺎﺋﻢ ﺍﺯﺍﻭﻳﺔ ﻓﺈﻥ ﺯﺍﻭﻳﺘﺎﻩ ﺍﻟﺤﺎﺩﺗﻴﻦ ﻣﺘﺘﺎﻣﺘﻴﻦ * ﺧﺎﺻﻴﺔ 3 : ﺇﺫﺍ ﻛﺎﻥ ﻟﻤﺜﻠﺚ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺘﺎﻣﺘﺎﻥ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ± ﺍﻟﻤﺜﻠﺚ ﺍﻟﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ : * ﺗﻌﺮﻳﻒ 2 : ﻳﻜﻮﻥ ﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﺇﺫﺍ ﻛﺎﻥ ﻟﻪ ﺿﻠﻌﺎﻥ ﻣﺘﻘﺎﻳﺴﺎﻥ * ﻣﺜﺎﻝ : ABCﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﺭﺃﺳﻪ A
22.
* ﺧﺎﺻﻴﺔ :
4 ﺇﺫﺍ ﻛﺎﻥ ﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴ ﻴﻦ ﻓﺈﻥ ﺯﺍﻭﺗﻲ ﺍﻟﻘﺎﻋﺪﺓ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ ﺎﻗ ˆ ˆ ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : ABCﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﺭﺃﺳﻪ Aﻳﻌﻨﻲ ﺃﻥ : B = C * ﺧﺎﺻﻴﺔ : 5 ﺇﺫﺍ ﻛﺎﻥ ﻟﻤﺜﻠﺚ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻘﻠﻴﺴﺘﺎﻥ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ˆ ˆ ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : ABCﻣﺜﻠﺚ ﺑﺤﻴﺚ B = Cﻳﻌﻨﻲ ﺃﻥ : ABCﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﺭﺃﺳﻪ . A ± ﺍﻟﻤﺜﻠﺚ ﺍﻟﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﻭ ﺍﻟﻘﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ : * ﺗﻌﺮﻳﻒ 3 : ﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﻭ ﺍﻟﻘﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﻫﻮ ﻣﺜﻠﺚ ﻟﻪ ﺿﻠﻌﺎﻥ ﻣﺘﻘﺎﻳﺴﺎﻥ ﻭ ﺯﺍﻭﻳﺔ ﻗﺎﺋﻤﺔ ﺍﻟﻤﺜﻠﺚ ﺍﻟﻤﺘﺴ C ﻦ ﻭ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﻓﻲ . A * ﻣﺜﺎﻝ : ABCﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴ A B * ﺧﺎﺻﻴﺔ : 6 ﺇﺫﺍ ﻛﺎﻥ ﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﻭ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﻓﺈﻥ ﺯﺍﻭﻳﺘﻲ ﺍﻟﻘﺎﻋﺪﺓ ﻣﺘﻘﺎ ﻳﺴﺘﺎﻥ ﻭ ﻗﻴﺎﺳﻬﻤﺎ °54 ˆ ˆ ° 54 = ABC = ACB * ﻣﺜﺎﻝ : ABCﻣﺜﻠﺚ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﻭ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﻓﻲ Aﺇﺫﻥ :
23.
± ﺍﻟﻤﺜﻠﺚ ﺍﻟﻤﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ :
* ﺗﻌﺮﻳﻒ 4 : ﺍﻟﻤﺜﻠﺚ ﺍﻟﻤﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ ﻫﻮ ﻣﺜﻠﺚ ﺟﻤﻴﻊ ﺃﺿﻼﻋﻪ ﻣﺘﻘﺎﻳﺴﺔ * ﻣﺜﺎﻝ : ABCﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ . * ﺧﺎﺻﻴﺔ : 7 ﺇﺫﺍ ﻛﺎﻥ ﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ ﻓﺈﻥ ﺟﻤﻴﻊ ﺯﻭﺍﻳﺎﻩ ﻣﺘﻘﺎﻳﺴﺔ ﻭ ﻗﻴﺎﺱ ﻛﻞ ﻣﻨﻬﺎ °06 * ﺧﺎﺻﻴﺔ : 8 ﺇﺫﺍ ﻛﺎﻧﺖ ﺯﻭﺍﻳﺎ ﻣﺜﻠﺚ ﻣﺘﻘﺎﻳﺴﺔ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ
24.
ﺟﻤﻊ ﻭ ﻃﺮﺡ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ
1( – ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ : ( ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻹﺷﺎﺭﺓ : ﺃ * ﻗﺎﻋﺪﺓ 1 : ﻟﺤﺴﺎﺏ ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻹﺷﺎﺭﺓ ﻧﺤﺘﻔﻆ ﺑﺎﻹﺷﺎﺭﺓ ﺛﻢ ﻧﺠﻤﻊ ﻣﺴﺎﻓﺘﻴﻬﻤﺎ ﻋﻦ ﺍﻟﺼﻔﺮ . ;; 9,32 = 5,1 + 4,22 * ﺃﻣﺜﻠﺔ : 5,21 = ) 7 + 5,5( – = ) 7 –( + 5,5 – ;; 51,071 = 51,85 + 211 522,175 – = ) 75 + 522,,415 ( – = ) 75 –( + 522,415 – ( ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻓﻲ ﺍﻹﺷﺎﺭﺓ : ﺏ * ﻗﺎﻋﺪﺓ 2 : ﻟﺤﺴﺎﺏ ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻓﻲ ﺍﻹﺷﺎﺭﺓ ﻧﺄﺧﺬ ﺇﺷﺎﺭﺓ ﺍﻟﻌﺪﺩ ﺍﻷﺑﻌﺪ ﻋﻦ ﺍﻟﺼﻔﺮ ﺛﻢ ﻧﺤﺴﺐ ﻓﺮﻕ ﻣﺴﺎﻓﺘﻴﻬﻤﺎ ﻋﻦ ﺍﻟﺼﻔﺮ . 62,31 – = ) 41,21 – 4,52( – = ) 4,52 –( + 41,21 * ﺃﻣﺜﻠﺔ : 98,12 = ) 11,41 – 63 ( + = 63 + 11,41 – 5,97 = ) 5,54 – 521 ( + = ) 5,54 –( + 521 51,02 – = ) 5,11 – 56,13 ( – = 5,11 + 56,13 – ﺎﺑﻠﻴﻦ : ( ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻣﺘﻘ ﺝ * ﻗﺎﻋﺪﺓ 3 : ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻳﻜﻮ ﺩﺍﺋﻤﺎ ﻣﻨﻌﺪﻣﺎ ) ﺃﻱ ﻳﺴﺎﻭﻱ ﺻﻔﺮ ( . 0 = ) a + ( aﻭ 0 = a + a aﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ . ;; 0 = ) 88,521 –( + 88,521 * ﺃﻣﺜﻠﺔ : 0 = 7633 + 7633 – 0 = ) 85211 –( + 85211 ;; 0 = 7,953 + 7,953 – 2( – ﻓﺮﻕ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ : * ﻗﺎﻋﺪﺓ 4 : ﻟﺤﺴﺎﺏ ﻓﺮﻕ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻧﻀﻴﻒ ﺇﻟﻰ ﺍﻟﺤﺪ ﺍﻷﻭﻝ ﻣﻘﺎﺑﻞ ﺍﻟﺤﺪ ﺍﻟﺜﺎﻧﻲ . aﻭ bﻋﺪﺩﺍﻥ ﻋﺸﺮﻳﺎﻥ ﻧﺴﺒﻴﺎﻥ : ) a – b = a + ( b * ﺃﻣﺜﻠﺔ : 57,9 = ) 5,11 – 52,12 ( + = ) 5,11 –( + 52,12 = 5,11 – 52,12 55,52 = 21 + 55,31 = ) 21 ( – 55,31 05 = ) 61 + 43( – = ) 61 –( + 43 – = 61 – 43 – 41,54 = ) 02 – 41,56 ( – = 02 + 41,56 – = ) 02 –( – 41,56 –
25.
ﺗﻘﻨﻴﺎﺕ 1( ﻹﺯﺍﻟﺔ ﺍﻷﻗﻮﺍﺱ ﺍﻟﻤﺴﺒﻮﻗﺔ ﺑﻌﻼﻣﺔ
+ : ﻧﺰﻳﻞ ﻋﻼﻣﺔ + ﻭ ﻧﺤﺪﻑ ﺍﻷﻗﻮﺍﻕ ﺑﺪﻭﻥ ﺗﻐﻴﻴﺮ ﺇﺷﺎﺭﺓ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺘﻲ ﺑﺪﺍﺧﻠﻬﺎ . ﻹﺯﺍﻟﺔ ﺍﻷﻗﻮﺍﺱ ﺍﻟﻤﺴﺒﻮﻗﺔ ﺑﻌﻼﻣﺔ – : ﻧﺰﻳﻞ ﻋﻼﻣﺔ – ﻭ ﻧﺤﺪﻑ ﺍﻷﻗﻤﺎﺱ ﻣﻊ ﺗﻐﻴﻴﺮ ﺇﺷﺎﺭﺓ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺘﻲ ﺑﺪﺍﺧﻠﻬﺎ . )2 + 11 – 45 ( + )5,1 – 33 + 5,2 –( + 11 = A * ﺃﻣﺜﻠﺔ : 2 + 11 – 45 + 5,1 – 33 + 5,2 – 11 = ) 66,42 + 5,1 – 25 ( – ) 1+ 85 – 44,21 + 55 –( – 6,2 = B 66,42 – 5,1 + 25 – 1 – 85 + 44,21 – 55 + 6,2 = 2( ﺣﺴﺎﺏ ﺗﻌﺒﻴﺮ ﺟﺒﺮﻱ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺃﻗﻮﺍﺱ ﻭ ﻣﻌﻘﻮﻓﺎﺕ ﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻘﺎﻋﺪﺓ ﺃﻋﻼﻩ . 1( – ﻧﺰﻳﻞ ﺍﻷﻗﻮﺍﺱ ﻭ ﺍﻟﻤﻌﻘﻮﻓﺎﺕ ﺑﺪﺃ ﺑﺎﻷﻗﻮﺍﺱ ﺍﻟﺪﺍﺧﻠﻴﺔ ﻣﻊ ﺗﻄﺒﻴﻖ ﺍﻟﻘﺎﻋﺪﺓ ﺃﻋﻼﻩ . 2( – ﻧﺠﻤﻊ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻤﺘﻘﺎﺑﻠﺔ ﻓﻴﻤﺎ ﺑﻴﻨﻬﺎ ﺛﻢ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻤﻮﺟﺒﺔ ﻭ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺴﺎﻟﺒﺔ 7 – ) 5,2 + 41 –( – ) 1+ 5,11 –( + 5,2 = A * ﻠﺔ : ﺃﻣﺜ 7 – 5,2 – 41 + 1 + 5,11 – 5,2 = 7 – 5,11 – 41 + 1 + 5,2 – 5,2 = 5,71 – 51 + 0 = ) 51 – 5,71 ( – = 5,2 – = ) 3 + 5,5–( – 22 + ] 1 – ) 7 – 5,3 ( + 5,11 –[ – ) 1 – 5,3 ( = B 3 – 5,5 + 22 + ] 1 – 7 – 5,3 + 5,11–[ – 1 – 5,3 = 3 – 5,5 + 22 + 1 – 7 – 5,3 – 5,11 + 1 – 5,3 = 3 – 7 – 5,5 + 22 + 5,11 + 1 – 1 + 5,3 – 5,3 = 01 – 93 + 0 + 0 = 01 – 93 = 92 =
26.
ﺿﺮﺏ ﻭ ﻗﺴﻤﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ
1( – ﺿﺮﺏ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ : ( ﺟﺪﺍء ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻹﺷﺎﺭﺓ : ﺃ * ﻗﺎﻋﺪﺓ 1 : ﺟﺪﺍء ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻹﺷﺎﺭﺓ ﻫﻮ ﻋﺪﺩ ﻋﺸﺮﻱ ﻣﻮﺟﺐ 5,0 = ) 01–( – 21 x (–5 ) = 105 ;; 0,05 x * ﺃﻣﺜﻠﺔ : 0 = ) 621–( –125,89 x 0 = 0 ;; 0 x ( ﺟﺪﺍء ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻓﻲ ﺍﻹﺷﺎﺭﺓ : ﺏ * ﻗﺎﻋﺪﺓ 2 : ﺟﺪﺍء ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻓﻲ ﺍﻹﺷﺎﺭﺓ ﻫﻮ ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﺳﺎﻟﺐ ;; 15– = ) 2–( 25,5 x * ﺃﻣﺜﻠﺔ : 575– = 05 –11,5 x ;; 011– = ) 5–( 22 x 057 = 01 –75 x ( ﺟﺪﺍء ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﻓﻲ : 1 ﻭ 1 : ﺝ * ﻗﺎﻋﺪﺓ 3 : ﻣﺠﻤﻮﻉ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻳﻜﻮ ﺩﺍﺋﻤﺎ ﻣﻨﻌﺪﻣﺎ ) ﺃﻱ ﻳﺴﺎﻭﻱ ﺻﻔﺮ ( . aﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ . a + ( 1 ) = aﻭ 1 + a = a 0 = ) a + ( aﻭ ﻭ 1 x a = a a + a = 0 a x 1 = a aﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ . * ﺃﻣﺜﻠﺔ : 7633 = 1 1 x (– 125,88 ) = –125,88 ;; 3367 x 35211– = 85211 – 359,7 x (–1 ) = 359,7 ;; – 1 x ( ﺟﺪﺍء ﻋﺪﺓ ﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ : ﺩ * ﻗﺎﻋﺪﺓ 4 : ﺮﻳﺔ ﻧﺴﺒﻴﺔ ﻳﻜﻮﻥ : ﺟﺪﺍء ﻋﺪﺓ ﺃﻋﺪﺍﺩ ﻋﺸ . ﻣﻮﺟﺒﺎ : ﺇﺫﺍ ﻋﺪﺩ ﻋﻮﺍﻣﻠﻪ ﺍﻟﺴﺎﻟﺒﺔ ﺯﻭﺟﻴﺎ ﺳﺎﻟﺒﺎ : ﺇﺫﺍ ﻛﺎﻥ ﻋﺪﺩ ﻋﻮﺍﻣﻠﻪ ﺍﻟﺴﺎﻟﺒﺔ ﻓﺮﺩﻳﺎ . ) 5–( A = –5 x 1,3 x (–7 ) x (–25 ) x 1 x * ﺃﻣﺜﻠﺔ : 7,1 B = 11 x (–25,4 ) x 14 x (–1 ) x (–0,5 ) x * ﻟﺪﻳﻨﺎ ﺍﻟﺠﺪﺍء Aﻋﺪﺩ ﻋﻮﺍﻣﻠﻪ ﺍﻟﺴﺎﻟﺒﺔ ﻫﻮ 4 ﻭ ﻫﻮ ﻋﺪﺩ ﺯﻭﺟﻲ , ﺇﺫﻥ Aﻋﺪﺩ ﻣﻮﺟﺐ . * ﻟﺪﻳﻨﺎ ﺍﻟﺠﺪﺍء Bﻋﺪﺩ ﻋﻮﺍﻣﻠﻪ ﺍﻟﺴﺎﻟﺒﺔ ﻫﻮ 3 ﻭ ﻫﻮ ﻋﺪﺩ ﻓﺮﺩﻱ , ﺇﺫﻥ Bﻋﺪﺩ ﺳﺎﻟﺐ .
27.
* ﻗﺎﻋﺪﺓ 5 :
ﻻ ﻳﺘﻐﻴﺮ ﺟﺪﺍء ﻋﺪﺓ ﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ ﺇﺫﺍ ﻏﻴﺮﻧﺎ ﺗﺮﺗﻴﺐ ﻋﻮﺍﻣﻠﻪ ﺃﻭ ﻋﻮﺿﻨﺎ ﺑﻌﻀﺎ ﻣﻨﻬﺎ ﺑﺠﺪﺍﺋﻬﺎ . ) 5,1–( A = (–2 ) x 5,5 x 50 x * ﻣﺜﺎﻝ : ) ) 5,1–( = ( –2 x 50 ) x ( 5,5 x ) 52,8–( = –100 x 528 = ﺗﻘﻨﻴﺎﺕ ﻟﺤﺴﺎﺏ ﺟﺪﺍء ﻋﺪﺓ ﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ ﻧﺤﺪﺩ ﺃﻭﻻ ﺇﺷﺎﺭﺓ ﻫﺬﺍ ﺍﻟﺠﺪﺍء ﺛﻢ ﻧﻄﺒﻖ ﺍﻟﻘﺎﻋﺪﺓ 4 . ﺃﻣﺜﻠﺔ : 5,6 A = (–7,5 ) x 25 x –4 ) x ) 5,6 = + ( 7,5 x 25 x 4 x ) 5,6 = + ( ( 25 x 5 ) x ( 7,5 x 57,84 = 100 x 5784 = 5,7 B = –6 x 5 x (–1,5 ) x (–1 ) x ) 5,7 = – ( 6 x 5 x 1 x ) 5,7 = – ( (6 x 5 x 1 ) x ( 1,5 x ) 52,11 = – ( 30 x 5,733– = 2( – ﻗﺴﻤﺔ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ : ( ﺧﺎﺭﺝ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻹﺷﺎﺭﺓ : ﺃ * ﻗﺎﻋﺪﺓ 6 : ﺧﺎﺭﺝ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻹﺷﺎﺭﺓ ﻫﻮ ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﻣﻮﺟﺐ ;; 51,26 = ) 31 –( : 59,708 – 011 = 1,7 : 187 * ﺃﻣﺜﻠﺔ : ( ﺧﺎﺭﺝ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻓﻲ ﺍﻹﺷﺎﺭﺓ : ﺏ * ﻗﺎﻋﺪﺓ 7 : ﺧﺎﺭﺝ ﻋﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ ﻣﺨﺘﻠﻔﻴﻦ ﻓﻲ ﺍﻹﺷﺎﺭﺓ ﻫﻮ ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﺳﺎﻟﺐ ;; 51,26 – = ) 31–( : 59,708 011 – = 1,7 : 187 – * ﺃﻣﺜﻠﺔ :
28.
- a a
- a a a = ﻭ = -= * ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : - b b b - b b ( ﺍﻟﺨﺎﺭﺝ ﺍﻟﻤﻘﺮﺏ ﻭ ﺍﻟﺘﺄﻃﻴﺮ : ﺝ 1( – ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺨﺎﺭﺝ ﻣﻮﺟﺒﺎ : 22 22 7 ﻧﻌﺘﺒﺮ ﺍﻟﺨﺎﺭﺝ * ﻣﺜﺎﻝ : 7 01 41,3 03 02 22 ﺇﻟﻰ 1 ﻧﺘﻔﺮﻳﻂ ﻫﻲ : 3 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 ﺇﻟﻰ 1 ﺑﺈﻓﺮﺍﻁ ﻫﻲ : 4 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 22 < 3 ﺇﻟﻰ 1 ﻫﻮ : 4 < ﺇﺫﻥ ﺗﺄﻃﻴﺮ ﺍﻟﻌﺪﺩ 7 7 22 ﺇﻟﻰ 1,0 ﻧﺘﻔﺮﻳﻂ ﻫﻲ : 1,3 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 ﺇﻟﻰ 1,0 ﺑﺈﻓﺮﺍﻁ ﻫﻲ : 2,3 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 22 < 1,3 2,3 < ﺇﻟﻰ 1,0 ﻫﻮ : ﺇﺫﻥ ﺗﺄﻃﻴﺮ ﺍﻟﻌﺪﺩ 7 7 2( – ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺨﺎﺭﺝ ﺳﺎﻟﺒﺎ : 22 - * ﻣﺜﺎﻝ : ﻧﻌﺘﺒﺮ ﺍﻟﺨﺎﺭﺝ 7 22 - ﺇﻟﻰ 1 ﻧﺘﻔﺮﻳﻂ ﻫﻲ : 4 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 - ﺇﻟﻰ 1 ﺑﺈﻓﺮﺍﻁ ﻫﻲ : 3 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 22 - ﺇﻟﻰ 1 ﻫﻮ : 3 < - < 4 ﺇﺫﻥ ﺗﺄﻃﻴﺮ ﺍﻟﻌﺪﺩ 7 7 22 - ﺇﻟﻰ 1,0 ﻧﺘﻔﺮﻳﻂ ﻫﻲ : 2,3 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 - ﺇﻟﻰ 1,0 ﺑﺈﻓﺮﺍﻁ ﻫﻲ : 1,3 . * ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻘﺮﺑﺔ ﻟﻠﻌﺪﺩ 7 22 22 - ﺇﻟﻰ 1,0 ﻫﻮ : 1,3 < - < 2,3 ﺇﺫﻥ ﺗﺄﻃﻴﺮ ﺍﻟﻌﺪﺩ 7 7
29.
ﺍﻟﻤﻨﺼﻔــﺎﺕ ﻭ ﺍﻻﺭﺗﻔــﺎﻋــﺎﺕ ﻓﻲ ﻣﺜﻠﺚ
1( – ﺍﻟﻤﻨﺼﻔﺎﺕ ﻓﻲ ﻣﺜﻠﺚ : ( ﻣﻨﺼﻒ ﺯﺍﻭﻳﺔ : ﺃ * ﺗﻌﺮﻳﻒ 1 : ﻣﻨﺼﻒ ﺯﺍﻭﻳﺔ ﻫﻮ ﻧﺼﻒ ﻣﺴﺘﻘﻴﻢ ﺃﺻﻠﻪ ﺭﺃﺱ ﺍﻟﺰﺍﻭﻳﺔ , ﻳﻮﺟﺪ ﺑﺪﺍﺧﻠﻬﺎ ﻭ ﻳﻘﺴﻤﻬﺎ ﺇﻟﻰ ﻴﻦ ﺯﺍﻭﻳﺘﻴﻦ ﻣﺘﻘﺎﻳﺴﺘ * ﻣﺜﺎﻝ : ﻧﻌﺘﺒﺮ A ˆ Bﺯﺍﻭﻳﺔ ﻭ ) [OMﻣﻨﺼﻔﻬﺎ . O ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﻤﻴﺰﺓ ﻟﻤﻨﺼﻒ ﺯﺍﻭﻳﺔ : ﺏ * ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷــﺮﺓ : ﻛﻞ ﻧﻘﻄﺔ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﻣﻨﺼﻒ ﺯﺍﻭﻳﺔ ﺗﺒﻌﺪ ﺑﻨﻔﺲ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻦ ﺿﻠﻌﻲ ﻫﺬﻩ ﺍﻟﺰﺍﻭﻳﺔ ﺎ : EK = EL ﺳﻴﻜﻮﻥ ﻟﺪﻳﻨ * ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ : ﺔ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﻣﻨﺼﻒ ﻫﺬﻩ ﺍﻟﺰﺍﻭﻳﺔ ﻛﻞ ﻧﻘﻄﺔ ﺗﺒﻌﺪ ﺑﻨﻔﺲ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻦ ﺿﻠﻌﻲ ﺯﺍﻭﻳ * ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﻤﻴﺰﺓ : ﻣﻨﺼﻒ ﺯﺍﻭﻳﺔ ﻫﻮ ﻣﺠﻤﻮﻋﺔ ﻣﻦ ﻧﻘﻂ ﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﺘﺴﺎﻭﻳﺔ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻦ ﺿﻠﻌﻴﻬﺎ ( ﻣﻨﺼﻔﺎﺕ ﻣﺜﻠﺚ : ﺝ * ﺗﻌﺮﻳﻒ 2 : ﻣﻨﺼﻒ ﻣﺜﻠﺚ ﻫﻮ ﻣﻨﺼﻒ ﺇﺣﺪﻯ ﺯﻭﺍﻳﺎﻩ
30.
* ﻣﺜﺎﻝ :
ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﻟﻠﻤﺜﻠﺚ ﺛﻼﺙ ﻣﻨﺼﻔﺎﺕ . * ﺧﺎﺻﻴـــﺔ : ﻣﻨﺼﻔﺎﺕ ﻣﺜﻠﺚ ﺗﺘﻼﻗﻰ ﻓﻲ ﻧﻘﻄﺔ ﻭﺍﺣﺪﺓ ﺗﺴﻤﻰ ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﻤﺤﺎﻃﺔ ﺑﻬﺬﺍ ﺍﻟﻤﺜﻠﺚ * ﻣﺜﺎﻝ : ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﻹﻳﺠﺎﺩ ﻣﺮﻛﺰ ﺩﺍﺋﺮﺓ ﻣﺤﺎﻃﺔ ﺑﻤﺜﻠﺚ ﻳﻜﻔﻲ ﺭﺳﻢ ﻣﻨﺼﻔﻴﻦ ﻓﻘﻂ ﻣﻦ ﻣﻨﺼﻔﺎﺕ ﻫﺬﺍ ﺍﻟﻤﺜﻠﺚ . 2( – ﺍﻻﺭﺗﻔﺎﻋﺎﺕ ﻓﻲ ﻣﺜﻠﺚ : ( ﺟﺪﺍء ﻋﺪﺓ ﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ : ﺃ * ﺗﻌﺮﻳﻒ 3 : ﺍﺭﺗﻔﺎﻉ ﻣﺜﻠﺚ ﻫﻮ ﻣﺴﺘﻘﻴﻢ ﻳﻤﺮ ﻣﻦ ﺃﺣﺪ ﺭﺅﻭﺱ ﺍﻟﻤﺜﻠﺚ ﻭ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺣﺎﻣﻞ ﺍﻟﻀﻠﻊ ﺍﻟﻤﻘﺎﺑﻞ ﻟﻬﺬﺍ ﺍﻟﺮﺃﺱ . * ﻣﺜﺎﻝ : ABCﻣﺜﻠﺚ ﻭ ) (AHﺍﻻﺭﺗﻔﺎﻉ ﺍﻟﻤﻮﺍﻓﻖ ﻟﻠﻀﻠﻊ ]. [BC
31.
· ﺣﺎﻟﺔ ﺧﺎﺻ
ﺔ : ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﻟﻠﻤﺜﻠﺚ ﺛﻼﺙ ﺍﺭﺗﻔﺎﻋﺎﺕ . * ﺧﺎﺻﻴﺔ : ﺍﺭﺗﻔﺎﻋﺎﺕ ﻣﺜﻠﺚ ﺗﺘﻼﻗﻰ ﻓﻲ ﻧﻘﻄﺔ ﻭﺍﺣﺪﺓ ﺗﺴﻤﻰ ﻣﺮﻛﺰ ﺗﻌﺎﻣــﺪ ﻫﺬﺍ ﺍﻟﻤﺜﻠﺚ . * ﻣﺜﺎﻝ : 02 ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﻟﺮﺳﻢ ﻣﺮﻛﺰ ﺗﻌﺎﻣـــﺪ ﻣﺜﻠﺚ ﻳﻜﻔﻲ ﺭﺳﻢ ﺍﺭﺗﻔﺎﻋﻴﻦ ﻓﻘﻂ ﻣﻦ ﺍﺭﺗﻔﺎﻋﺎﺕ ﻫﺬﺍ ﺍﻟﻤﺜﻠﺚ .
32.
ﺍﻟـﻘــــــــــــــــﻮﻯ
1( – ﻗﻮﺓ ﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ : ( ﻣﺜﺎﻝ : ﺃ ﻧﻌﺘﺒﺮ ﺍﻟﺠﺪﺍء ﺍﻵﺗﻲ : 5,2 A = 2,5 x 2,5 x 2,5 x 2,5 x ﻳﺘﻜﻮﻥ ﻫﺬﺍﺍﻟﺠﺪﺍء ﻣﻦ ﺧﻤﺴﺔ ﻋﻮﺍﻣﻞ ﻣﺴﺎﻭﻳﺔ ﻟﻠﻌﺪﺩ 5,2 . ﻧﺴﻤﻲ ﺇﺫﻥ ﻫﺬﺍ ﺍﻟﺠﺪﺍء : ﺍﻟﻘﻮﺓ ﺍﻟﺨﺎﻣﺴﺔ ﻟﻠﻌﺪﺩ 5,2 . 5 .. ﺘﺐ : )5,2( ﻭ ﻧﻘــﺮﺃ : ﺇﺛﻨﺎﻥ ﺃﺱ ﺧﻤﺴــﺔ ﻭ ﻧﻜ 5 5 ﺍﻟﻌﺪﺩ 5,2 ﻳﺴﻤﻰ : ﺃﺳﺎﺱ ﺍﻟﻘﻮﺓ )5,2( ﻭ ﺍﻟﻌﺪﺩ 5 ﻳﺴﻤﻰ : ﺃﺱ ﺍﻟﻘﻮﺓ )5,2( . ( ﺗﻌﺮﻳﻒ : ﺏ . aﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ ﺃ ﺒﺮ ﻣﻦ 1 ﻭ nﻋﺪﺩ ﺻﺤﻴﺢ ﻃﺒﻴﻌﻲ ﻏﻴﺮ ﻣﻨﻌﺪﻡ ﻛ a n = a ´ a ´ a ´ a ´ a ´ ...... ´ a 44442 44414 3 ) nﻣﻦ ﺍﻟﻌﻮﺍﻣﻞ ( ﻣﻼﺣﻈﺎﺕ ﻫﺎﻣﺔ : 0 0 1 ﺓ 0 ﻻ ﻣﻌﻨﻰ ﻟﻬﺎ . ﺍﻟﻘﻮ , , 1 = ( 0 ¹ a ) a a = a ﻣﻔــــﺮﺩﺍﺕ : n · ﻧﺴﻤﻲ aﺃﺳــﺎﺱ ﺍﻟﻘﻮﺓ . a n · ﻧﺴﻤﻲ nﺃﺱ ﺍﻟﻘﻮﺓ . a ( ﺇﺷﺎﺭﺓ ﻗـــﻮﺓ ﺃﺳﺎﺳﻬﺎ ﺳـــﺎﻟﺐ : ﺝ * ﺧــﺎﺻﻴﺔ 1 : ﺗﻜﻮﻥ ﻗــﻮﺓ ﺃﺳﺎﺳﻬﺎ ﺳﺎﻟﺐ : · ﻣﻮﺟﺒﺔ : ﺇﺫﺍ ﻛﺎﻥ ﺃﺳﻬﺎ ﻋﺪﺩﺍ ﺯﻭﺟﻴﺎ . · ﺳﺎﻟﺒﺔ : ﺇﺫﺍ ﻛﺎﻥ ﺃﺳﻬﺎ ﻋﺪﺩﺍ ﻓﺮﺩﻳﺎ . 61 )11 ( ﻋﺪﺩ ﻣﻮﺟﺐ , ﻷﻥ ﺃﺳﻬﺎ ﻫﻮ 61 ﻭﻫﻮ ﻋﺪﺩ ﺯﻭﺟﻲ . ﺍﻟﻘﻮﺓ * ﻣﺜﺎﻝ : 12 ﺍﻟﻘﻮﺓ )9,5 ( ﻋﺪﺩ ﺳﺎﻟﺐ , ﻷﻥ ﺃﺳﻬﺎ ﻫﻮ 12 ﻭ ﻫﻮ ﻋﺪﺩ ﻓﺮﺩﻱ . 8 8 * ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﺍﻟﻘﻮﺓ )5 ( ﺗﺨﺘﻠﻒ ﻋﻦ ﺍﻟﻘﻮﺓ 5 ﻷﻥ : 8 ) 5 ( ﺃﺳﺎﺳﻬﺎ ﻫﻮ ) 5 ( ﻭﺣﺴﺐ ﺍﻟﺨﺎﺻﻴﺔ 1 ﻓﻬﻲ ﻣﻮﺟﺒﺔ . 8 5 ﺃﺳﺎﺳﻬﺎ ﻫﻮ 5 ﻭ ﻫﻲ ﺳﺎﻟﺒﺔ ﻷﻧﻬﺎ ﻻﺗﺨﻀﻊ ﻟﻠﺨﺎﺻﻴﺔ 1 .
33.
2( – ﺧـﺼـــﺎﺋـــﺺ ﺍﻟﻘــﻮﻯ :
aﻭ bﻋﺪﺩﺍﻥ ﻋﺸﺮﻳﺎﻥ ﻧﺴﺒﻴﺎﻥ ﻏﻴﺮ ﻣﻨﻌﺪﻣﻴــﻦ . mﻭ nﻋﺪﺩﺍﻥ ﺻﺤﻴﺤﺎﻥ ﻃﺒﻴﻴﻌﻴﺎﻥ . a m ´ a n = a m + n m - n a m = æ a ö ) (m > n ÷ ç a n è a ø n ) ( a m = a m ´n m ) ´a m ´ b m = ( a b m a m = æ a ö ÷ ç b m è b ø * ﺃﻣﺜﻠــﺔ : 62 a ´ a = a + 14 = a 21 41 21 42 a 5 ´ a ´ a 7 ´ a = a 5 + 11 + 7 + 1 = a 11 32 a 23 ´ b 23 = (a ´ b ) 72 = 51 - 24 = 24 a 51 a a a 54 (a 9 ) 5 = a 9 ´ 5 = a 11a = æ a ö 11 ÷ ç ÷ 11 ç a è b ø 3( – ﻗــــﻮﻯ ﺍﻟـﻌـــﺪﺩ 01 : * ﺧــﺎﺻﻴﺔ 2 : nﻋﺪﺩ ﺻﺤﻴﺢ ﻃﺒﻴﻌﻲ ﻏﻴﺮ ﻣﻨﻌﺪﻡ : 0.............0000001 = 10 n 3442441 ) nﻣﻦ ﺍﻷﺻﻔﺎﺭ ( * ﺃﻣﺜﻠــﺔ : 5 000001 = 01 11 000000000001 = 01 22 00000000000000000000001 = 01
34.
ﺍﻟﺘﻤــﺎﺛﻞ ﺍﻟﻤــﺮﻛــﺰﻱ
1( – ﻣﻤﺎﺛﻠﺔ ﻧﻘﻄﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻘﻄﺔ : ( ﻣﺜﺎﻝ : ﺃ Aﻭ Oﻧﻘﻄﺘﺎﻥ ﻣﺨﺘﻠﻔﺘﺎﻥ ﻣﻦ ﺍﻟﻤﺴﺘﻮﻯ . ﻟﻨﻨﺸﺊ ' Aﺑﺤﻴﺚ ﺗﻜﻮﻥ Oﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ]'. [AA ﻧﺴﻤﻲ ' Aﻣﻤﺎﺛﻠﺔ Aﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . Oﻭ ﻧﻘﻮﻝ ﻛﺬﻟﻚ : ' Aﻲ ﻣﻤﺎﺛﻠﺔ Aﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺘﻤﺎﺛﻞ ﺍﻟﻤﺮﻛﺰﻱ ﺍﻟﺬﻱ ﻣﺮﻛﺰﻩ . o ﻫ ﻧﻼﺣﻆ ﺃﻥ Aﻫﻲ ﻛﺬﻟﻚ ﻣﻤﺎﺛﻠﺔ ' Aﺑﺎﻟﻨﺒﺔ ﻟﻠﻨﻘﻄﺔ . Oﻧﻘﻮﻝ ﺇﺫﻥ : Aﻭ ' Aﻣﺘﻤﺎﺛﻠﺘﺎﻥ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . O ( ﺗﻌﺮﻳﻒ : ﺏ ﺗﻜﻮﻥ Aﻭ ' Aﻄﺘﻴﻦ ﻣﺘﻤﺎﺛﻠﺘﻴﻦ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻘﻄﺔ Oﺇﺫﺍ ﻛﺎﻧﺖ Oﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ]'[AA ﻧﻘ * ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﻣﻤﺎﺛﻠﺔ ﺍﻟﻨﻘﻄﺔ Oﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Oﻫﻲ Oﻧﻔﺴﻬﺎ . 2( – ﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ ﺍﻟﻤﺴﺎﻓﺔ : ( ﻣﺜﺎﻝ : ﺃ Aﻭ Bﻧﻘﻄﺘﺎﻥ ﻣﺨﺘﻠﻔﺘﺎﻥ ﺑﺤﻴﺚ AB = 4 cmﻭ Oﻧﻘﻄﺔ ﺧﺎﺭﺝ ﺍﻟﻤﺴﺘﻘﻴﻢ ). (AB ﻟﻨﻨﺸﺊ ' Aﻭ ' Bﻣﻤﺎﺛﻠﺘﻲ Aﻭ Bﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . O ﻟﻨﺤﺴﺐ ' A'Bﺑﺎﺳﺘﻌﻤﺎﻝ ﺍﻟﻤﺴﻄﺮﺓ . ﻧﻼﺣﻆ ﺃﻥ . A'B' = 4 cmﺇﺫﻥ : '. AB = A'B ( ﺧﺎﺻﻴــﺔ : ﺏ ﺍﻟﺘﻤﺎﺛﻞ ﺍﻟﻤﺮﻛﺰﻱ ﻳﺤﺎﻓﻆ ﻋﻠﻰ ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﻧﻘﻄﺘﻴﻦ
35.
3( – ﻣﻤﺎﺛﻼﺕ ﺑﻌﺾ ﺍﻷﺷﻜﺎﻝ :
( ﻣﻤﺎﺛﻼﺕ ﻧﻘﻂ ﻣﺴﺘﻘﻴﻤﻴﺔ : ﺃ · ﻣﺜﺎﻝ : Aﻭ Bﻭ Cﻧﻘﻂ ﻣﺴﺘﻘﻴﻤﻴﺔ ﻭ Oﻧﻘﻄﺔ ﺧﺎﺭﺝ ﺍﻟﻤﺴﺘﻘﻴﻢ ). (AC ﻟﻨﻨﺸﺊ ﺍﻟﻨﻘﻂ ' Aﻭ ' Bﻭ ' Cﻣﻤﺎﺛﻼﺕ ﺍﻟﻨﻘﻂ Aﻭ Bﻭ Cﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ O ﺣﻆ ﺃﻥ ' Aﻭ ' Bﻭ ' Cﻫﻲ ﻛﺬﻟﻚ ﻧﻘﻂ ﻣﺴﺘﻘﻴﻤﻴﺔ . ﻧﻼ · ﺧﺎﺻﻴﺔ : ﺍﻟﺘﻤﺎﺛﻞ ﺍﻟﻤﺮﻛﺰﻱ ﻳﺤﺎﻓﻆ ﻋﻠﻰ ﺍﺳﺘﻘﺎﻣﻴﺔ ﺍﻟﻨﻘﻂ ( ﻣﻤﺎﺛﻞ ﻣﺴﺘﻘﻴﻢ : ﺏ · ﻣﺜﺎﻝ : ) (Dﻣﺴﺘﻘﻴﻢ ﻭ Eﻧﻘﻄﺔ ﻻ ﺗﻨﺘﻤﻲ ﺇﻟﻴﻪ . ﻟﻨﻨﺸﺊ )' (Dﻣﻤﺎﺛﻞ ﺍﻟﻤﺴﺘﻘﻴﻢ ) (Dﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻄﺔ . E ﻣﻦ ﺃﺟﻞ ﻫﺬﺍ ﺳﻨﺄﺧﺬ ﻧﻘﻄﺘﻴﻦ ﻣﺨﺘﻠﻔﺘﻴﻦ ﺗﻨﺘﻤﻴﺎﻥ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻢ )(D ﺛﻢ ﻧﻨﺸﺊ ﻣﻤﺎﺛﻠﺘﻴﻬﻤﺎ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . E ﻧﻼﺣﻆ ﺃﻥ ﺍﻟﻤﺴﺘﻘﻴﻢ )' (Dﻳﻮﺍﺯﻱ ﺍﻟﻤﺴﺘﻘﻴﻢ ). (D · ﺧﺎﺻﻴﺔ : ﻣﻤﺎﺛﻞ ﻣﺴﺘﻘﻴﻢ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻘﻄﺔ ﻫﻮ ﻣﺴﺘﻘﻴﻢ ﻳﻮﺍﺯﻳﻪ ( ﻣﻤﺎﺛﻞ ﻧﺼﻒ ﻣﺴﺘﻘﻴﻢ : ﺝ · ﻣﺜﺎﻝ : ) [ABﻧﺼﻒ ﻣﺴﺘﻘﻴﻢ ﻭ Iﻧﻘﻄﺔ ﻻ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻢ )(AB . ﻟﻨﻨﺸﺊ ﻧﺼﻒ ﺍﻟﻤﺴﺘﻘﻴﻢ )' [A'Bﻣﻤﺎﺛﻞ ) [ABﺑﺎﻟﻨﺒﺔ ﻟﻠﻨﻘﻄﺔ I . ﻣﻦ ﺃﺟﻞ ﻫﺬﺍ ﺳﻨﻨﺸﺊ ' Aﻭ ' Bﻣﻤﺎﺛﻠﺘﻲ Aﻭ Bﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻋ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . I · ﺧﺎﺻﻴﺔ : ﻣﻤﺎﺛﻞ ﻧﺼﻒ ﻣﺴﺘﻘﻴﻢ ) [ABﺑﺎﻟﻨﺒﺔ ﻟﻨﻘﻄﺔ Oﻫﻮ ﻧﺼﻒ ﺍﻟﻤﺴﺘﻘﻴﻢ )' [A'Bﺑﺤﻴﺚ ' B' Aﻣﻤﺎﺛﻠﺘﻲ B Aﻋﻠﻰ ﻭ ﻭ ﺍﻟﺘﻮﺍﻟﻲ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . O ( ﻣﻤﺎﺛﻠﺔ ﻗﻄﻌﺔ : ﺩ · ﻣﺜﺎﻝ : ] [ABﻗﻄﻌﺔ ﻭ Mﻧﻘﻄﺔ ﺧﺎﺭﺝ ﺍﻟﻤﺴﺘﻘﻴﻢ ). (AB ﻟﻨﻨﺸﺊ ﺍﻟﻘﻄﻌﺔ ]' [A'Bﻣﻤﺎﺛﻠﺔ ﺍﻟﻘﻄﻌﺔ ] [ABﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . M
36.
ﻣﻦ ﺃﺟﻞ ﻫﺬﺍ ﺳﻨﻨﺸﺊ ' B' Aﻣﻤﺎﺛﻠﺘﻲ B Aﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﺑﺎﻟﻨﺴﺒﺔ
ﻭ ﻭ ﻟﻠﻨﻘﻄﺔ . M ( ﻭ ﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ ﺳﻴﻜﻮﻥ ﻟﺪﻳﻨﺎ ' ) AB = A'Bﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ ﺍﻟﻤﺴﺎﻓﺔ ﺃﻥ ﺍﻟﻘﻄﻌﺘﻴﻦ ] [A'"B'] [ABﻣﺘﻘﺎﻳﺴﺘﺎﻥ . ﻭ · ﺧﺎﺻﻴﺔ : ﻣﻤﺎﺛﻠﺔ ﻗﻄﻌﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻘﻄﺔ ﻫﻲ ﻗﻄﻌﺔ ﺗﻘﺎﻳﺴﻬﺎ ( ﻣﻤﺎﺛﻠﺔ ﺯﺍﻭﻳﺔ : ﻩ · ﻣﺜﺎﻝ : A ˆ Bﺯﺍﻭﻳﺔ ﻭ Eﻧﻘﻄﺔ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ . O ﻟﻨﻨﺸﺊ ﺍﻟﺰﺍﻭﻳﺔ ' A O Bﻣﻤﺎﺛﻠﺔ ﺍﻟﺰﺍﻭﻳﺔ A ˆ Bﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . E O ' ˆ ' ﻣﻦ ﺃﺟﻞ ﻫﺬﺍ ﺳﻨﻨﺸﺊ ' B' O' Aﻣﻤﺎﺛﻼﺕ B O Aﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻭ ﻭ ﻭ ﻭ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . E ' A ˆ B = A O B O ' ˆ ' ﻧﻼﺣﻆ ﺃﻥ : · ﺧﺎﺻﻴﺔ : ﻣﻤﺎﺛﻠﺔ ﺯﺍﻭﻳﺔ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻘﻄﺔ ﻫﻲ ﺯﺍﻭﻳﺔ ﺗﻘﺎﻳﺴﻬﺎ ( ﻣﻤﺎﺛﻠﺔ ﺩﺍﺋﺮﺓ : ﻭ · ﻣﺜﺎﻝ : ) (Cﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ Oﻭ ﺷﻌﺎﻋﻬﺎ rﻭ Eﻧﻘﻄﺔ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ . ﻟﻨﻨﺸﺊ ﺍﻟﺪﺍﺋﺮﺓ )' (Cﻣﻤﺎﺛﻠﺔ ) (Cﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻄﺔ . E ﻣﻦ ﺃﺟﻞ ﻫﺬﺍ ﺳﻨﺄﺧﺬ ﻧﻘﻄﺔ Aﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﺪﺍﺋﺮﺓ )(C ﺛﻢ ﻧﻨﺸﺊ ' A' Oﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . Eﻭ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﺘﻲ ﻣﺮﻛﺰﻫﺎ ﻭ ' Oﻭ ﺗﻤﺮ ﻣﻦ ' Aﻫﻲ ﻣﻤﺎﺛﻠﺔ ) (Cﺑﺎﻟﻨﺒﺔ ﻟﻠﻨﻘﻄﺔ . E ﻟﻨﺒﻴﻦ ﺃﻥ ﺍﻟﺪﺍﺋﺮﺗﻴﻦ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﺸﻌﺎﻉ . r ﻟﺪﻳﻨﺎ : ' Oﻣﻤﺎﺛﻠﺔ Oﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . E ' Aﻣﻤﺎﺛﻠﺔ Aﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . E ﺇﺫﻥ : ' ) OA = O'Aﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ ﺍﻟﻤﺴﺎﻓﺔ ( . ﻭ ﺑﻤﺎ ﺃﻥ : OA = rﻓﺈﻥ O'A' = r ﻭ ﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ ﺃﻥ ﻟﻠﺪﺍﺋﺮﺗﻴﻦ ) (C') (Cﻧﻔﺲ ﺍﻟﺸﻌﺎﻉ . r ﻭ · ﺧﺎﺻﻴﺔ : ﻬﺎ rﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻘﻄﺔ Eﻫﻲ ﻣﻤﺎﺛﻠﺔ ﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ Oﻭ ﺷﻌﺎﻋ ﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ ' Oﻣﻤﺎﺛﻞ Oﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Eﻭ ﺷﻌﺎﻋﻬﺎ r ﻣﻮﻗﻊ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺑﺎﻟﺜﺎﻧﻮﻱ ﺍﻹﻋﺪﺍﺩﻱ ﻟﻸﺳﺘﺎﺫ ﺍﻟﻤﻬﺪﻱ ﻋﻨﻴﺲ / ﺃﺳﺘﺎﺫ ﺑﺎﻟ _ww
37.
· ﺗﻘﻨﻴﺎﺕ : ﻟﺮﺳﻢ ﻣﻤﺎﺛﻠﺔ ﺩﺍﺋﺮﺓ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻨﻘﻄﺔ ﻧﺮﺳﻢ ﻣﻤﺎﺛﻞ ﺍﻟﻤﺮﻛﺰ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻬﺬﻩ ﺍﻟﻨﻘﻄﺔ ﺛﻢ ﻧﺤﺘﻔﻆ ﺑﻨﻔﺲ ﺍﻟﺸﻌﺎﻉ .
( ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﺷﻜﻞ : ﺯ · ﺧﺎﺻﻴﺔ : ﻧﺴﻤﻲ ﻧﻘﻄﺔ Oﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﺷﻜﻞ Fﺇﺫﺍ ﻛﺎﻥ ﻣﻤﺎﺛﻞ ﻫﺬﺍ ﺍﻟﺸﻜﻞ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Oﻫﻮ ﺍﻟﺸﻜﻞ Fﻧﻔﺴﻪ . · ﺜﺎﻝ : ﻣ 2( – ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﻗﻄﻌﺔ : 1( – ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﺩﺍﺋﺮﺓ : ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﺩﺍﺋﺮﺓ ﻫﻮ ﻣﺮﻛﺰﻫﺎ ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﻗﻄﻌﺔ ﻫﻮ ﻣﻨﺘﺼﻔﻬﺎ
38.
ﺍﻟـﻨـﺸــــــــــــﺮ ﻭ ﺍﻟـﺘﻌـﻤﻴـــــــــــــــﻞ
_ Iﺍﻟﻨﺸﺮ : 1( – ﺟﺪﺍء ﻋﺪﺩ ﻓﻲ ﻣﺠﻤﻮﻉ ﻭﺟﺪﺍء ﻋﺪﺩ ﻓﻲ ﻓﺮﻕ : * ﻣﺜﺎﻝ 1 : ﻟﻨﺤﺴﺐ ﺛﻢ ﻧﻘﺎﺭﻥ ﺍﻟﻌﺪﺩﻳﻦ : ) 3,7 + 5,5( A = 2 xﻭ 3,7 B = 2 x 5,5 + 2 x 3,7 B = 2 x 5,5 + 2 x ) 3,7 + 5,5 ( A = 2 x ﻟﺪﻳﻨﺎ : 8,21 = 11 + 14,6 = 2 x 6,52 = 6,52 = ﻧﻼﺣﻆ ﺃﻥ : A = Bﺃﻱ ﺃﻥ : 3,7 2 x ( 5,5 + 7,3 ) = 2 x 5,5 + 2x ﻧﻘﻮﻝ ﺃﻧﻨﺎ ﻗﺪ ﻧﺸﺮﻧﺎ ﺍﻟﺠﺪﺍء ) 3,7 + 5,5 ( 2 x * ﻣﺜﺎﻝ 2 : ﻟﻨﺤﺴﺐ ﺛﻢ ﻧﻘﺎﺭﻥ ﺍﻟﻌﺪﺩﻳﻦ : ) 5,7 3 ( C = 6,5 xﻭ 5,7 D = 6,5 x 3 6,5 x 5,7 D = 6,5 x 3 6,5 x ) 5,7 – 3 ( C = 6,5 x ﻟﺪﻳﻨﺎ : 57,84 5,91 = ) 5,4 ( = 6,5 x 52,92 = 52,92 = 5,7 6,5 x ( 3 7,5 ) = 6,5 x 3 6,5 x ﻧﻼﺣﻆ ﺃﻥ : C = Dﺃﻱ ﺃﻥ : * ﻗﺎﻋﺪﺓ 1 : aﻭ xﻭ yﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ . a ( x - y ) = a.x ﻭ a ( x + y ) = a.x + a.y - a.y ﺍﻟﻨﺸﺮ ﻫﻮ ﻛﺘﺎﺑﺔ ﺟﺪﺍء ﻋﻠﻰ ﺷﻜﻞ ﻣﺠﻤﻮﻉ ﺃﻭ ﻓﺮﻕ . ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : * ﺑﺼﻔﺔ ﻋﺎﻣﺔ : n c b aﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ . ﻭ ﻭ ﻭ n ( a + b – c ) = na + nb nc
39.
* ﺍﻟﻤﺠﻤﻮﻉ : a + a + a + a + a +........+ a
Aﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ . a + a + a + a +........+a = n.a nﻣﻦ ﺍﻟﺤﺪﻭﺩ * ﻣﺜﺎﻝ : 51 = 3 3 + 3 + 3 + 3 + 3 + 3 = 6 x ) 5 ( ( 5 ) + ( 5 ) + ( 5 ) + ( 5 ) = 4 x */ ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ 1 : aﻋﺪﺩ ﻋﺸﺮﻱ ﻧﺴﺒﻲ . ﺑﺴﻂ ﻣﺎﻳﻠﻲ : A = 3a + 25a ;; B = 3,5a + 2a + a C = 7a 3,5a ;; D = 11,5a 2,5a a ;; E = 21a – 3,5a + 5,5a ﺍﻟﺤﻞ : A = 3a + 25a = 28a ;; B = 3,5a + 2a + a = 6,5a C = 7a 3,5a = 3,5a ;; D = 11,5a 2,5a a = 15a E = 21a 3,5a + 5,5a = 21a + 5,5a 3,5a = 26,5a 3,5a = 23a */ ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ 2 : bﻭ bﻋﺪﺩﺍﻥ ﻋﺸﺮﻳﺎﻥ ﻧﺴﺒﻴﺎﻥ . ﺑﺴﻂ ﻣﺎﻳﻠﻲ : 11 A = 2a + 3b 5,5a + 7b ;; B = a + b 11b + 0,5a ﺍﻟﺤﻞ : A = 2a + 3b 5,5a + 7b 11 B = a + b 11b + 0,5a = 2a 5,5a + 3b + 7b 11 = a + 0,5a + b 11b = 3,5a + 10b 11 = 1,5a 10b 2( – ﺍء ﻣﺠﻤــﻮﻋﻴــﻦ : ﺟﺪ * ﻗﺎﻋﺪﺓ 2 : aﻭ y x bﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ : ﻭ ﻭ ( a + b ) ( x + y ) = ax + ay + bx + by ) ( a + b ) ( x + y ) = a ( x + y ) + b ( x + y ﺗﻘﻨﻴﺎﺕ : ) = ( ax + ay ) + ( bx + by = ax + ay + bx + by
40.
* ﻣﺜﺎﻝ : ) 6 + A = ( 2 + x ) ( x
) B = ( 3 + x ) ( 1 x ) 6 + = 2 ( x + 6 ) + x ( x ) = 3 ( 1 x ) + x ( 1 x 2 2 ) = ( 2x + 12 ) + ( x + 6x ) = ( 3 3x ) + ( x x 2 2 = 2x + 12 + x + 6x = 3 + 3x + x x 2 2 21 + = 2x + 6x + x 3 = 3x + x x 2 2 21 + = 8x + x 3 = 4x x ) 5 C = ( 2x 3 ) ( x ) 5 = 2x ( x 5 ) 3 ( x 2 ) 51 = ( 2x 10x ) ( 3x 2 51 + = 2x + 10x + 3x 2 51 + = 2x + 13x _ IIﺍﻟﺘﻌﻤﻴــﻞ : 1( – ﺗﻌﺮﻳــﻒ : ﺍﻟﺘﻌﻤﻴﻞ ﻫﻮ ﻛﺘﺎﺑﺔ ﻣﺠﻤﻮﻉ ﺃﻭ ﻓﺮﻕ ﻋﻠﻰ ﺷﻜﻞ ﺟــﺪﺍء · ﻣﻼﺣﻈﺔ ﻫﺎﻣــﺔ : ﻗﺒﻞ ﺃﻥ ﻧﻌﻤﻞ ﻧﺒﺤﺚ ﻋﻦ ﺍﻟﻌﺎﻣﻞ ﺍﻟﻤﺸﺘﺮﻙ 2( – ﻣﺜﺎﻝ : 2 A = 6a + ax ;; B = a + 3a ;; C = 2ax 4ay ) = a ( 6 + x ) = a ( a + 3 ) = 2a ( x 2y 3 2 ) D = 5x + 10x 20x ;; E = 2abx + 6aby 2ab ;; F = 2x ( 1 + x ) + 7 ( 1 + x 2 ) 7 + = 5x ( x + 2x 4 ) = 2ab ( x + 3y 1 ) = ( 1 + x ) ( 2x _ IIIﺍﻟﻨﺸﺮ ﻭ ﺍﻟﺘﻌﻤﻴﻞ ﻭ ﺍﻟﻤﺘﻄﺎﺑﻘﺎﺕ ﺍﻟﻬﺎﻣﺔ : 1( – ﺧﺎﺻﻴﺔ : ) ﺍﻟﻤﺘﻄﺎﺑﻘﺎﺕ ﺍﻟﻬﺎﻣﺔ ( aﻭ bﻋﺪﺩﺍﻥ ﻋﺸﺮﻳﺎﻥ ﻧﺴﺒﻴﺎﻥ : 2 2 2 ( a + b ) = a + 2ab + b 2 2 2 ( a b ) = a 2ab + b 2 2 ( a b ) ( a + b ) = a b
41.
: 2( – ﺗﻄﺒﻴـﻘـــــﺎﺕ
: ( ﻟﻨﻨﺸﺮ ﺛﻢ ﻧﺒﺴﻂ ﻣﺎ ﻳﻠﻲ ﺃ 2 2 2 ( a + 3 ) ;; ( x + y ) ;; ( 2x + 1 ) 2 2 2 ( x 5 ) ;; ( 2 y ) ;; ( 3x 7 ) ( x 2 ) ( x + 2 ) ;; ( 6x + 1 ) ( 6x 1 ) 2 2 2 2 2 2 ( a + 3 ) = a + 2 . a . 3 + 3 ;; ( x + y ) = x + 2 . x . y + y 2 2 2 = a + 6a + 9 = x + 2xy + y 2 2 2 2 2 2 ( 2x + 1 ) = ( 2x ) + 2 . 2x . 1 + 1 ;; ( x 5 ) = x 2 . x . 5 + 5 2 2 = 4x + 4x + 1 = x 10x + 25 2 2 2 2 2 2 ( 2 y ) = 2 2 . 2 . y + y ;; ( 3x 7 ) = ( 3x ) 2 . 3x . 7 + 7 2 2 = 4 4y + y = 9x 42x + 49 2 2 2 2 ( x 2 ) ( x + 2 ) = x 2 ;; ( 6x + 1 ) ( 6x 1 ) = ( 6x ) 1 2 2 = x 4 = 36x 1 : ( ﻟﻨﻌﻤﻞ ﻣﺎ ﻳﻠﻲ ﺏ 2 2 2 A = x + 4x + 1 ;; B = 25x + 30x + 9 ;; C = 49 28x + 4x 2 2 2 D = 9x 30x + 25 ;; E = 4 x ;; F = 81x 121 2 2 2 A = x + 4x + 1 ;; B = 25x + 30x + 9 ;; C = 49 28x + 4x 2 2 2 2 2 2 = x + 2 . x . 2 + 2 = ( 5x ) + 2 . 5x . 3 + 3 = 7 2 . 7 . 2x + ( 2x ) 2 2 2 = ( x + 2 ) = ( 5x + 3 ) = ( 7 2x ) 2 2 2 D = 9x 30x + 25 ;; E = 4 x ;; F = 81x 121 2 2 2 2 2 2 = ( 3x ) 2 . 3x . 5 + 5 = 2 x = ( 9x ) 11 2 = ( 3x 5 ) = ( 2 x ) ( 2 + x ) = ( 9x 11 ) ( 9x + 11 )
42.
ﻣﺘـــﻮﺍﺯﻱ ﺍﻷﺿــــــﻼﻉ
_ Iﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ : 1( – ﻣﺜﺎﻝ : )1 (Dﻭ )2 (Dﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﺎﻥ . )1 (Lﻭ )2 (Lﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﺎﻥ ﻳﻘﻄﻌﺎﻥ )1 (Dﻭ )2 (Dﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻓﻲ : Aﻭ Bﻭ Cﻭ . D ﻧﺴﻤﻲ ﺍﻟﺮﺑﺎﻋﻲ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ 2( – ﺗﻌﺮﻳﻒ : ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻫﻮ ﺭﺑﺎﻋﻲ ﺣﺎﻣﻼ ﻛﻞ ﺿﻠﻌﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻓﻴﻪ ﻣﺘﻮﺍﺯﻳﻴﻦ _ IIﺧﺼﺎﺋــﺺ : 1( – ﺧﺎﺻﻴﺔ ﺍﻟﻘﻄﺮﻳﻴﻦ : ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ : ﺃ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻗﻄﺮﺍﻩ ﻳﺘﻘﺎﻃﻌﺎﻥ ﻓﻲ . O ﻧﻼﺣﻆ ﺃﻥ Oﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﺮﻳﻴﻦ ] [ACﻭ ]. [BD ﻧﻘــﻮﻝ ﺇﺫﻥ : ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻓﺈﻥ ﻟﻘﻄﺮﻳﻪ ﻧﻔﺲ ﺍﻟﻤﻨﺘﺼﻒ * ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﻧﺴﻤﻲ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﻗﻄﺮﻱ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻣﺮﻛﺰﻩ .
43.
( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ :
ﺏ ﻨﺘﺼﻒ Oﻭ ﺣﺎﻣﻼﻫﻤﺎ ﻏﻴﺮ ﻣﺘﻌﺎﻣﺪﻳﻦ : Aﻭ Bﻭ Cﻭ Dﻧﻘﻂ ﺑﺤﻴﺚ ] [ACﻭ ] [BDﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻤ ﻟﻨﺒﺮﻫﻦ ﺃﻥ ﺍﻟﺮﺑﺎﻋﻲ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ . ﻣﻦ ﺃﺟﻞ ﻫﺬﺍ ﺳﻨﺒﺮﻫﻦ ﺃﻥ ) (ABﻳﻮﺍﺯﻱ ) (CDﻭ ﺃﻥ ) (ADﻳﻮﺍﺯﻱ ): (BC ﻧﻌﻠﻢ ﺃﻥ Oﻣﻨﺘﺼﻒ ] [ACﻭ ] [BDﺇﺫﻥ : Aﻭ Cﻣﺘﻤﺎﺛﻠﺘﻴﻦ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . O ﻴﻦ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . O Bﻭ Dﻣﺘﻤﺎﺛﻠﺘ ﺇﺫﻥ : ﺍﻟﻤﺴﺘﻘﻴﻤﻴﻦ ) (ABﻭ ) (CDﻣﺘﻤﺎﺛﻠﻴﻦ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Oﻭ ﻛﺬﻟﻚ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﻦ ) (ADﻭ ). (BC ﻭ ﻣﻨﻪ ﻓﺈﻥ ) (CD) // (ABﻭ )(BC) // (AD ﻭ ﺑﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ) ﺣﺴﺐ ﺍﻟﺘﻌﺮﻳﻒ ( ﻣﺮﻛﺰﻩ ﺍﻟﻨﻘﻄﺔ . O ﻧﻘــﻮﻝ ﺇﺫﻥ : ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻗﻄﺮﺍﻩ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻤﻨﺘﺼﻒ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ * ﺗﻤﺮﻳﻦ ﺗﻄﺒﻴﻘﻲ : ABCﻣﺜﻠﺚ ﻭ Iﻣﻨﺘﺼﻒ ]. [AC 1( – ﺃﻧﺸﺊ Dﻣﻤﺎﺛﻠﺔ Bﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . I 2( – ﺃﺛﺒﺖ ﺃﻥ ﺍﻟﺮﺑﺎﻋﻲ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿــﻼﻉ . ﺍﻟﺤــــﻞ : 1( – ﺍﻟﺸﻜـــــﻞ : 2( – ﻟﻨﺜﺒﺖ ﺃﻥ ﺍﻟﺮﺑﺎﻋﻲ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿـــﻼﻉ : ﻧﻌﻠﻢ ﺃﻥ : Iﻣﻨﺘﺼﻒ ](1) . [AC ﻭ ﻟﺪﻳﻨﺎ Dﻣﻤﺎﺛﻠﺔ Bﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ . I ﺇﺫﻥ : Iﻣﻨﺘﺼﻒ ](2) . [BD ﻉ ) ﺣﺴﺐ ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﻘﻄﺮﻳﻦ ( . ﻣﻦ )1( ﻭ )2( ﻧﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﺮﺑﺎﻋﻲ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿـــﻼ .
44.
2( – ﺧﺎﺻﻴﺔ ﺍﻷﺿﻼﻉ ﺍﻟﻤﺘﻘﺎﺑﻠﺔ :
( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ : ﺃ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻣﺮﻛﺰﻩ . O ﻟﻨﺒﻴﻦ : AB = CDﻭ AD = BC ﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ . ABCD ﻧﻌﻠﻢ ﺃﻥ Oﻣﺮﻛﺰ ﻣ ﺇﺫﻥ Oﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﺮﻳﻦ ]. [BD] [AC ﻭ ﻭ ﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ ﺃﻥ : Aﻭ Cﻣﺘﻤﺎﺛﻠﺘﻴﻦ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Oﻭ ﻛﺬﻟﻚ Bﻭ . D ( . ﻭ ﺑﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ : AB = CDﻭ ) AD = BCﺣﺴﺐ ﺧﺎﺻﻴﺔ ﺍﻟﺤﻔﺎﻅ ﻋﻠﻰ ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﻧﻘﻄﺘﻴﻦ ﻧﻘــﻮﻝ ﺇﺫﻥ : ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻓﺈﻥ ﻛﻞ ﺿﻠﻌﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻓﻴﻪ ﻣﺘﻘﺎﻳﺴﺎﻥ ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ : ﺏ ﺇﺫﺍ ﻛﺎﻥ ﻟﺮﺑﺎﻋﻲ ﻛﻞ ﺿﻠﻌﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻓﻴﻪ ﻣﺘﻘﺎﻳﺴﺎﻥ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ 3( – ﺧﺎﺻﻴﺔ ﺍﻟﺰﻭﺍﻳﺎ ﺍﻟﻤﺘﻘﺎﺑﻠﺔ : ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ : ﺃ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻣﺮﻛﺰﻩ . O ABCD ﻟﻨﺒﻴﻦ ﺃﻥ A ˆ C = A ˆ Cﻭ ﺃﻥ . B ˆ C = B ˆ D A C B D
45.
ﻧﻌﻠﻢ ﺃﻥ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻣﺮﻛﺰﻩ . O
ﺇﺫﻥ : Oﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﺮﻳﻦ ] [ACﻭ ]. [BD ﻭ ﻣﻨﻪ ﻓﺈﻥ : Aﻭ Cﻣﺘﻤﺎﺛﻠﺘﻴﻦ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Oﻭ ﻛﺬﻟﻚ Bﻭ . D ﺇﺫﻥ ﺍﻟﺰﺍﻭﻳﺘﺎﻥ A ˆ Cﻭ A ˆ Cﻣﺘﻤﺎﺛﻠﺘﺎﻥ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻨﻘﻄﺔ Oﻭ ﻛﺬﻟﻚ ﺍﻟﺰﺍﻭﻳﺘﻴﻦ B ˆ Dﻭ B ˆ D C A D B ˆ D ˆ ˆ ˆ C ﻭ ﺑﺎﻟﺘﺎﻟﻲ ﻓﺈﻥ : A B C = A Dﻭ B C = B A D ﻧﻘــﻮﻝ ﺇﺫﻥ : ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻓﺈﻥ ﻛﻞ ﺯﺍﻭﻳﺘﻴﻦ ﻣﺘﻘﺎﺑﻠﺘﻴﻦ ﻓﻴﻪ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ ﻴﺔ : ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴ ﺏ ﺇﺫﺍ ﻛﺎﻥ ﻟﺮﺑﺎﻋﻲ ﻛﻞ ﺯﺍﻭﻳﺘﻴﻦ ﻣﺘﻘﺎﺑﺎﺗﻴﻦ ﻓﻴﻪ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ 4( – ﺍﺭﺗﻔﺎﻉ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿـــﻼﻉ : ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻭ Hﺍﻟﻤﺴﻘﻂ ﺍﻟﻌﻤﻮﺩﻱ ﻟﻠﻨﻘﻄﺔ Aﻋﻠﻰ ﺍﻟﻤﺴﺘﻘﻴﻢ ). (CD ﻧﺴﻤﻲ AHﺍﺭﺗﺘﻔﺎﻉ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ . ABCD 5( – ﺧﺎﺻﻴﺔ ﺇﺿــﺎﻓﻴﺔ : ﺇﺫﺍ ﻛﺎﻥ ﻟﺮﺑﺎﻋﻲ ﺿﻠﻌﺎﻥ ﻣﺘﻘﺎﺑﻼﻥ ﻭ ﺣﺎﻣﻼﻫﻤﺎ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿـــﻼﻉ
46.
ﺍﻟـﻤــﻌــــﺎﺩﻻﺕ
_ Iﺍﻟﻤﻌﺎﺩﻻﺕ : 1( – ﺗﻌﺮﻳﻒ : ﺎﻥ : aﻭ bﻋﺪﺩﺍﻥ ﻋﺸﺮﻳﺎﻥ ﻧﺴﺒﻴ ﻧﺴﻤﻲ ﻣﻌﺎﺩﻟﺔ ﻣﻦ ﺍﻟﺪﺭﺟﺔ ﺍﻷﻭﻟﻰ ﺑﻤﺠﻬﻮﻝ ﻭﺍﺣﺪ ﻛﻞ ﻣﻦ ﺍﻟﻜﺘﺎﺑﺘﻴﻦ : a + x = bﻭ (a ¹ 0) ax = b 2( – ﺃﻣﺜﻠﺔ : 5 = 11 + x = 22 ;; 5 + x = 10 ;; x 2 = 8 ;; 6 x = 2,6 ;; 1 x 41 = 3x = 12 ;; 7x = 21 ;; 4x = 16 ;; 5x = 0 ;; 2,5x = 1 ;; 7x _ IIﺣﻞ ﻣﻌﺎﺩﻟﺔ : 1( – ﺗﻌﺮﻳﻒ : ﺣﻞ ﻣﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﺒﺤﺚ ﻋﻦ ﺍﻟﻤﺠﻬﻮﻝ . x 2( – ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : a + x = b ( ﻗﺎﻋــﺪﺓ : ﺍ ﺣﻞ ﻣﻌﺎﺩﻟﺔ a + x = bﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ x = b a ( ﺃﻣﺜﻠﺔ : ﺏ ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : 11 = 6 + xﻫﻮ ﺍﻟﻌﺪ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ : 5 = 6 11 = . x ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = 2 + xﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ : 2 = 2 + 0 = ) 2 ( 0 = . x ﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ : 4 = 5,2 5,1 = . x ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : 5,1 = 2,5 xﻫﻮ ﺍﻟﻌﺪ ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : 1 = 5 xﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ : 4 = 5 + 1 = . x 2( – ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : (a ¹ 0 ) ax = b ( ﻗﺎﻋــﺪﺓ : ﺍ x = b/a ﺣﻞ ﻣﻌﺎﺩﻟﺔ (a ¹ 0) ax = bﻫﻮ ﺍﻟ ﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ
47.
( ﺃﻣﺜﻠﺔ :
ﺏ 5 = . x ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : 5 = 2xﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ : 5 2 = , 2 3 = . x ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : 3 = 5xﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ : 6 0- = , 5 - 0 = . = 0 x ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : 0 = 7xﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ : 7 - _ IIﺧﺼﺎﺋﺺ : 1( – ﺍﻟﻘﺎﻋﺪﺓ 1 : . ﺇﺫﺍ ﺃﺿﻔﻨﺎ ﺃﻭ ﻃﺮﺣﻨﺎ ﻧﻔﺲ ﺍﻟﻌﺪﺩ ﺍﻟﻨﺴﺒﻲ ﺇﻟﻰ ﻃﺮﻓﻲ ﻣﺘﺴﺎﻭﻳﺔ ﻓﺈﻥ ﺍﻟﻤﺘﺴﺎﻭﻳﺔ ﻻ ﺗﺘﻐﻴﺮ ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : aﻭ bﻭ kﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ . a = bﻳﻌﻨﻲ : a + k = b + kﻭ a – k = b – k ﺪﺓ 2 : 2( – ﺍﻟﻘﺎﻋ ﺇﺫﺍ ﺿﺮﺑﻨﺎ ﻓﻲ ﻧﻔﺲ ﺍﻟﻌﺪﺩ ﺃﻭ ﻗﺴﻤﻨﺎ ﻋﻠﻰ ﻧﻔﺲ ﺍﻟﻌﺪﺩ ﺍﻟﻐﻴﺮ ﺍﻟﻤﻨﻌﺪﻡ ﻃﺮﻓﻲ ﻣﺘﺴﺎﻭﻳﺔ ﻓﺈﻥ ﺍﻟﻤﺘﺴﺎﻭﻳﺔ ﻻ ﺗﺘﻐﻴﺮ ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : aﻭ bﻭ kﻭ ' (k '¹ 0 kﺃﻋﺪﺍﺩ ﻋﺸﺮﻳﺔ ﻧﺴﺒﻴﺔ . ) a = bﻳﻌﻨﻲ : a x k = b x kﻭ 'a : k' = b : k ﺗﻘﻨﻴﺎﺕ : ﻷﻋﺪﺍﺩ ﺍﻟﺘﻲ ﻻﺗﺤﺘﻮﻱ ﻋﻠﻰ ﺍﻟﻌﺪﺩ ﺍﻟﻤﺠﻬﻮﻝ xﻣﻦ ﺍﻟﻄﺮﻑ ﺍﻷﻳﺴﺮ ﻟﻠﻤﻌﺎﺩﻟﺔ ﻭ ﺍﻷﻋﺪﺍﺩ ﺍﻟﺘﻲ ﺗﺤﺘﻮﻱ ﻋﻠﻰ 1 ﻧﺰﻳﻞ ﺍ ﺍﻟﻌﺪﺩ ﺍﻟﻤﺠﻬﻮﻝ xﻣﻦ ﺍﻟﻄﺮﻑ ﺍﻟﻸﻳﻤﻦ ﻟﻠﻤﻌﺎﺩﻟﺔ . 2 ﻋﻨﺪ ﺇﺯﺍﻟﺔ ﻋﺪﺩ ﻣﻦ ﻃﺮﻑ ﻣﻌﺎﺩﻟﺔ ﻧﻀﻴﻒ ﻣﻘﺎﺑﻠﻪ ﺇﻟﻰ ﺍﻟﻄﺮﻑ ﺍﻵﺧﺮ . ﺗﻄﺒﻴﻘﺎﺕ : ¤ ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ 7 = . 5 + x ﻟﺪﻳﻨﺎ : 5 – 7 = x 21 = ﺇﺫﻥ ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ 21 – . ¤ ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ 2 = . 5x 2 = = 0 4 x , ﻟﺪﻳﻨﺎ : 5 ﺇﺫﻥ ﺣﻞ ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ 4,0 .
48.
¤ ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ 1
. 3x + 5 = x ﻟﺪﻳﻨﺎ 5 1 = 3x x 6 = 2x 6 - = = -3 x 2 ﺇﺫﻥ ﺣﻞ ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ 3 - . ¤ ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ 5 . 2( x + 1) = x ﻟﺪﻳﻨﺎ 1 2x + 2 = x 2 1 = 2x x 3 = x ﺇﺫﻥ ﺣﻞ ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ 3 - . 2 + 2 x 1 - x . = + x ¤ ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ 4 2 + = ) ( 1 - 2 x + 2 4 x 2 x ﻟﺪﻳﻨﺎ 4 4 4 ) ( 1 - 2 x + 2 + 4 x = 2 x 2 - 2 x + 2 + 4 x = 2 x 2 - 2- = 2 x + 4 x - 2 x 4- = 4 x 4- = x 4 1- = x ﺇﺫﻥ ﺣﻞ ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ 1 - . ﺣﺎﻻﺕ ﺧﺎﺻﺔ : )0 ¹ (b ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ 0x = b ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻟﻴﺲ ﻟﻬﺎ ﺣﻼ )0 ¹ (a ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ 0 = ax ﺣﻞ ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﻌﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ 0 ﺣﻞ ﺍﻟﻌﺎﺩﻟﺔ 0 = 0x ﺟﻤﻴﻊ ﺍﻷﻋﺪﺍﺩ ﺍﻟﻌﺸﺸﺮﻳﺔ ﺍﻟﻨﺴﺒﻴﺔ ﺣﻞ ﻟﻬﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ
49.
_ IIﺣﻞ ﻣﺴــﺎﺋﻞ :
1( – ﻗﺎﻋﺪﺓ : ﻟﺤﻞ ﻣﺴﺄﻟﺔ ﻧﺘﺒﻊ ﺍﻟﻤﺮﺍﺣﻞ ﺍﻵﺗﻴ ﺔ : ﺭ ﺍﻟﻤﺠﻬﻮﻝ . 1 – ﺍﺧﺘﻴﺎ . 2 – ﺻﻴﺎﻏﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ . 3 – ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ – ﺍﻟﺨﻼﺻﺔ. 2( – ﻣﺜﺎﻝ : . ﺇﺫﺍ ﻋﻠﻤﺖ ﺃﻥ ﺍﻹﺭﺙ ﻳﺮﺟﻊ ﺇﻟﻰ ﺃﻭﻻﺩﻩ ﺍﻷﺭﺑﻌﺔ : ﺗﻮﻓﻲ ﺭﺟﻞ ﻭ ﺗﺮﻙ ﻣﺎﻻ ﻗﺪﺭﻩ 00006 ﺩﺭﻫﻤﺎ ﺇﺑﻦ ﻭ ﺛﻼﺙ ﺑﻨﺎﺕ ﻭ ﺃﻥ ﻟﻠﺬﻛﺮ ﺣﻆ ﺍﻷﻧﺜﻴﻴﻦ , ﻓﻜﻴﻒ ﺳﻴﻘﺴﻢ ﻫﺬﺍ ﺍﻹﺭﺙ ﻋﻠﻰ ﺍﻷﻭﻻﺩ ﺍﻷﺭﺑﻌﺔ ؟ ﺤــﻞ : ﺍﻟ 1( – ﺍﺧﺘﻴﺎﺭ ﺍﻟﻤﺠﻬﻮﻝ : ﻧﻌﺘﺒﺮ xﺣﻆ ﺑﻨﺖ . 2( – ﺻﻴﺎﻏﺔ ﺍﻟﻤﻌﺎﺩﻟﺔ : ﻮﻟﺪ ﻫﻮ . 2 x ﺇﺫﺍ ﻛﺎﻥ ﺣﻆ ﺑﻨﺖ ﻫﻮ xﻓﺈﻧﻪ ﺣﻆ ﺍﻟﺒﻨﺎﺕ ﺍﻟﺜﻼﺛﺔ ﻫﻮ 3 ﻭ ﺣﻆ ﺍﻟ x ﺇﺫﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ : 00006 = . 3 x + 2 x 3( – ﺣﻞ ﺍﻟﻤﻌﺎﺩﻟﺔ : ﻟﺪﻳﻨﺎ 00006 = 5 x 00006 = x 5 00021 = x ﺪﺩ ﺍﻟﻌﺸﺮﻱ ﺍﻟﻨﺴﺒﻲ 00021 . ﺇﺫﻥ ﺣﻞ ﻫﺬﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻟﻌ 4( – ﺣﻞ ﺍﻟﺨﻼﺻﺔ : . ﺣﻆ ﺍﻟﺒﻨﺎﺕ ﻫﻮ 00021 ﺩﺭﻫﻤﺎ ﻟﻜﻞ ﻭﺍﺣﺪﺓ ﺣﻆ ﺍﻟﻮﻟﺪ ﻫﻮ 00042 ﺩﺭﻫﻤﺎ .
50.
ﺍﻟــﺮﺑﺎﻋﻴــــــــــــﺎﺕ ﺍﻟـﺨـﺎﺻــــــــــــــﺔ
_ Iﺍﻟﻤﺴﺘﻄﻴﻞ : 1( – ﺗﻌﺮﻳﻒ : ﺍﻟﻤﺴﺘﻄﻴﻞ ﻫﻮ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻟﻪ ﺯﺍﻭﻳﺔ ﻗﺎﺋﻤﺔ 2( – ﻣﺜﺎﻝ : ABCDﻣﺴﺘﻄﻴﻞ . * ﻣﻼﺣﻈﺎﺕ ﻫﺎﻣﺔ : 1( – ﺟﻤﻴﻊ ﺯﻭﺍﻳﺎ ﺍﻟﻤﺴﺘﻄﻴﻞ ﻗﺎﺋﻤﺔ . 2( – ﻟﻠﻤﺴﺘﻄﻴﻞ ﺑﻌﺪﻳﻦ ﻫﻤﺎ : ﺍﻟﻄﻮﻝ ﻭ ﺍﻟﻌﺮﺽ . 3( – ﺍﻟﻤﺴﺘﻄﻴﻞ ﻟﻪ ﺟﻤﻴﻊ ﺧﺎﺻﻴﺎﺕ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ . 3( – ﺧﺎﺻﻴﺔ ﺍﻟﻘﻄﺮﻳﻦ : ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ : ﺃ ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﺴﺘﻄﻴﻼ ﻓﺈﻥ ﻟﻘﻄﺮﻳﻪ ﻧﻔﺲ ﺍﻟﻄﻮﻝ ABCDﻣﺴﺘﻄﻴﻞ ﻳﻌﻨﻲ ﺃﻥ : AC = BD ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ : ﺏ ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻗﻄﺮﺍﻩ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﻄﻮﻝ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺴﺘﻄﻴﻼ 4( – ﻣﺤﺎﻭﺭ ﻭﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﺍﻟﻤﺴﺘﻄﻴﻞ : ﻟﻠﻤﺴﺘﻄﻴﻞ ﻣﺤﻮﺭﺍ ﺗﻤﺎﺛﻞ ﻫﻤﺎ ﻭﺍﺳﻄﺎ ﻛﻞ ﺿﻠﻌﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻓﻴﻪ ﻭ ﻟﻪ ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﻭﺍﺣﺪ ﻫﻮ ﺗﻘﺎﻃﻊ ﻗﻄﺮﻳﻪ
51.
_ IIﺍﻟﻤﻌﻴـﻦ :
1( – ﺗﻌﺮﻳﻒ : ﺍﻟﻤﻌﻴﻦ ﻫﻮ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻟﻪ ﺿﻠﻌﺎﻥ ﻣﺘﺘﺎﺑﻌﺎﻥ ﻣﺘﻘﺎﻳﺴﺎﻥ 2( – ﻣﺜﺎﻝ : ABCDﻣﻌﻴﻦ . ﻣﻼﺣﻈﺎﺕ ﻫﺎﻣﺔ : * 1( – ﺟﻤﻴﻊ ﺃﺿﻼﻉ ﺍﻟﻤﻌﻴﻦ ﻣﺘﻘﺎﻳﺴﺔ . 2( – ﺍﻟﻤﻌﻴﻦ ﻟﻪ ﺟﻤﻴﻊ ﺧﺎﺻﻴﺎﺕ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ . 3( – ﺧﺎﺻﻴﺔ ﺍﻟﻘﻄﺮﻳﻦ : ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ : ﺃ ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﻌﻴﻨﺎ ﻓﺈﻥ ﺣﺎﻣﻼ ﻗﻄﺮﻳﻪ ﻣﺘﻌﺎﻣﺪﺍﻥ ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ : ﺏ ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻗﻄﺮﺍﻩ ﻣﺘﻌﺎﻣﺪﺍﻥ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﻌﻴﻨﺎ 4( – ﻣﺤﺎﻭﺭ ﻭﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﺍﻟﻤﻌﻴﻦ : ﻟﻠﻤﻌﻴﻦ ﻣﺤﻮﺭﺍ ﺗﻤﺎﺛﻞ ﻫﻤﺎ ﻭﺍﺳﻄﺎ ﻛﻞ ﺿﻠﻌﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻓﻴﻪ ﻭ ﻟﻪ ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﻭﺍﺣﺪ ﻫﻮ ﺗﻘﺎﻃﻊ ﻗﻄﺮﻳﻪ
52.
_ IIﺍﻟﻤﺮﺑــﻊ :
1( – ﺗﻌﺮﻳﻒ : ﺍﻟﻤﺮﺑﻊ ﻫﻮ ﻣﻌﻴﻦ ﻟﻪ ﺯﺍﻭﻳﺔ ﻗﺎﺋﻤﺔ 2( – ﻣﺜﺎﻝ : ABCDﻣﺮﺑﻊ . ﻣﻼﺣﻈﺎﺕ ﻫﺎﻣﺔ : * ﺑﻊ ﻗﺎﺋﻤﺔ . 1( – ﺟﻤﻴﻊ ﺯﻭﺍﻳﺎ ﺍﻟﻤﺮ 2( – ﺟﻤﻴﻊ ﺃﺿﻼﻉ ﺍﻟﻤﺮﺑﻊ . 3( – ﺍﻟﻤﺮﺑﻊ ﻟﻪ ﺟﻤﻴﻊ ﺧﺎﺻﻴﺎﺕ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ . 4( – ﺍﻟﻤﺮﺑﻊ ﻫﻮ ﻣﺴﺘﻄﻴﻞ ﻃﻮﻟﻪ ﻳﺴﺎﻭﻱ ﻋﺮﺿﻪ . 3( – ﺧﺎﺻﻴﺔ ﺍﻟﻘﻄﺮﻳﻦ : ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ : ﺃ ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﺮﺑﻌﺎ ﻓﺈﻥ ﻟﻘﻄﺮﻳﻪ ﻧﻔﺲ ﺍﻟﻄﻮﻝ ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ : ﺏ ﺇﺫﺍ ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﻣﻌﻴﻨﺎ ﻗﻄﺮﺍﻩ ﻣﻘﺎﻳﺴﺎﻥ ﻓﺈﻧﻪ ﻳﻜﻮﻥ ﻣﺮﺑﻌﺎ 4( – ﻣﺤﺎﻭﺭ ﻭﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﺍﻟﻤﺮﺑﻊ : ﻟﻠﻤﺮﺑﻊ ﺃﺭﺑﻌﺔ ﻣﺤﺎﻭﺭ ﺗﻤﺎﺛﻞ ﻫﻲ ﻭﺍﺳﻄﺎ ﻛﻞ ﺿﻠﻌﻴﻦ ﻣﺘﻘﺎﺑﻠﻴﻦ ﻓﻴﻪ ﻭ ﺣﺎﻣﻼ ﻗﻄﺮﻳﻪ ﻭ ﻟﻪ ﻣﺮﻛﺰ ﺗﻤﺎﺛﻞ ﻭﺍﺣﺪ ﻫﻮ ﺗﻘﺎﻃﻊ ﻗﻄﺮﻳﻪ
53.
ـﻮﻧﺔ ﻣﻦ ﻣـﺘـــــﻮﺍﺯﻳﻴﻦ ﻭ ﻗــــــــــﺎﻃﻊ
ﺍﻟﺰﻭﺍﻳﺎ ﺍﻟﻤـﻜـ _ Iﺗﺬﻛﻴــﺮ : 1( – ﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﺘﺎﻣﺘﺎﻥ ﻭﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﻜﺎﻣﻠﺘﺎﻥ : ¤ ﺗﻜﻮﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺘﺎﻣﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ ﻣﺠﻤﻮﻉ ﻗﻴﺎﺳﻬﻤﺎ °09 . ﻮﻉ ﻗﻴﺎﺳﻬﻤﺎ °081 . ¤ ﺗﻜﻮﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻜﺎﻣﻠﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ ﻣﺠﻤ 2( – ﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﺤﺎﺫﻳﺘﺎﻥ : ﺗﻜﻮﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺤﺎﺫﻳﺘﻴﻦ ﺇﺫﺍ ﻛﺎﻥ : ¤ ﻟﻬﻤﺎ ﻧﻔﺲ ﺍﻟﺮﺃﺱ . . ¤ ﻟﻬﻤﺎ ﺿﻠﻊ ﻣﺸﺘﺮﻙ ¤ ﺗﻘﺎﻃﻌﻬﻤﺎ ﻫﻮ ﺍﻟﻀﻠﻊ ﺍﻟﻤﺸﺘﺮﻙ . * ﻣﺜﺎﻝ : A B A ˆ Bﻭ B ˆ Cﺯﺍﻭﻳﺘﻨﺎ ﻣﺘﺤﺎﺫﻳﺘﺎﻥ O O O C _ IIﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﻘﺎﺑﻠﺘﺎﻥ ﺑﺎﻟﺮﺃﺱ : )2(D 1( – ﻣﺜﺎﻝ : D )1(D A ﻧﺴﻤﻲ ﺍﻟﺰﺍﻭﻳﺘﻴﻦ A ˆ Cﻭ : B ˆ D O O O O ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻘﺎﺑﻠﺘﺎﻥ ﺑﺎﻟﺮﺃﺱ B ﻭ ﻛﺬﻟﻚ ﺍﻟﺰﺍﻭﻳﺘﻴﻦ : A ˆ Dﻭ ˆ C O O B C
54.
2( – ﺧﺎﺻﻴﺔ :
ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻘﺎﺑﻠﺘﺎﻥ ﺑﺎﻟﺮﺃﺱ ﺗﻜﻮﻧﺎﻥ ﻣﺘﻘﺎﻳﺴﺘﻴﻦ _ IIIﺍﻟﺰﻭﺍﻳﺎ ﺍﻟﻤﻜﻮﻧﺔ ﻣﻦ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻭﻗﺎﻃﻊ : 1( – ﺗﻌﺎﺭﻳﻒ : ( ﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﺒﺎﺩﻟﺘﺎﻥ ﺩﺍﺧﻠﻴﺎ : ﺃ )1 (D2) (Dﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻘﺎﻃﻌﺎﻥ ﻭ ) (Lﻗﺎﻃﻊ ﻟﻬﻤﺎ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻓﻲ . B A ﻭ ﻭ )(L )1(D A E ﻧﺴﻤﻲ ﺍﻟﺰﺍﻭﻳﺘﻴﻦ E ˆ Bﻭ : A ˆ F B A ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺒﺎﺩﻟﺘﺎﻥ ﺩﺍﺧﻠﻴﺎ )2(D B F ( ﺍﻟﺰﺍﻭﻳﺘﺎﻥ ﺍﻟﻤﺘﻨﺎﻇﺮﺗﺎﻥ : ﺏ )(L )1 (D2) (Dﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻘﺎﻃﻌﺎﻥ ﻭ ) (Lﻗﺎﻃﻊ ﻟﻬﻤﺎ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻓﻲ . B A ﻭ ﻭ )1(D A E ﻧﺴﻤﻲ ﺍﻟﺰﺍﻭﻳﺘﻴﻦ E ˆ Bﻭ : F ˆ C B A ﻥ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻨﺎﻇﺮﺗﺎ )2(D F B 2( – ﺧﺼــﺎﺋــﺺ : C ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ ﻟﻠﺰﺍﻭﻳﺘﻴﻦ ﺍﻟﻤﺘﺒﺎﺩﻟﺘﻴﻦ ﺩﺍﺧﻠﻴﺎ : ﺃ )1 (D2) (Dﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﺎﻥ ﻭ ) (Lﻗﺎﻃﻊ ﻟﻬﻤﺎ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻓﻲ . B A ﻭ ﻭ )(L E A )1(D )2(D F B
55.
E ˆ B = F ˆ A
A B ﻧﻼﺣــﻆ ﺃﻥ : ﻧﻘﻮﻝ ﺇﺫﻥ : ﺇﺫﺍ ﻛﺎﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻓﺈﻧﻬﻤﺎ ﻳﺤﺪﺩﺍﻥ ﻣﻊ ﻛﻞ ﻗﺎﻃﻊ ﻟﻬﻤﺎ ﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺒﺎﺩﻟﺘﺎﻥ ﺩﺍﺧﻠﻴﺎ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ﻭ Mﻧﻘﻄﺔ ﻣﻦ ﻧﺼﻒ ﺍﻟﻤﺴﺘﻘﻴﻢ ) [CDﺧﺎﺭﺝ ﺍﻟﻘﻄﻌﺔ ]. [CD * ﻣﺜﺎﻝ : ﻟﻨﺒﻴﻦ ﺃﻥ : . B ˆ D = A ˆ M A D A B ﻧﻌﺘﺒﺮ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﻦ ) (CD) (ABﻭ ﺍﻟﻘﺎﻃﻊ ﻟﻬﻤﺎ ). (AD ﻭ ﻟﺪﻳﻨﺎ : A ˆ Mﻭ B ˆ Dﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺒﺎﺩﻟﺘﺎﻥ ﺩﺍﺧﻠﻴﺎ . A D ﻭ ﻧﻌﻠﻢ ﺃﻥ ﺍﻟﺮﺑﺎﻋﻲ ABCDﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ , ﺇﺫﻥ : M ) ) (CD) // (ABﺣﺴﺐ ﺍﻟﺘﻌﺮﻳﻒ ( . D C ﻭ ﻣﻨﻪ ﻓﺈﻥ : B ˆ D = A ˆ M A D ( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻤﺒﺎﺷﺮﺓ ﻟﻠﺰﺍﻭﻳﺘﻴﻦ ﺍﻟﻤﺘﻨﺎﻇﺮﺗﻴﻦ : ﺏ )1 (D2) (Dﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﺎﻥ ﻭ ) (Lﻗﺎﻃﻊ ﻟﻬﻤﺎ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﻓﻲ . B A ﻭ ﻭ )(L )1(D E A ﻧﻼﺣﻆ ﺃﻥ : E ˆ B = F ˆ G A B )2(D F B G ﻧﻘﻮﻝ ﺇﺫﻥ : ﻭﻳﺘﺎﻥ ﻣﺘﻨﺎﻇﺮﺗﺎﻥ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ ﺇﺫﺍ ﻛﺎﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻓﺈﻧﻬﻤﺎ ﻳﺤﺪﺩﺍﻥ ﻣﻊ ﻛﻞ ﻗﺎﻃﻊ ﻟﻬﻤﺎ ﺯﺍ ﻤﺮ ﻣﻦ Aﻭ ﻳﻮﺍﺯﻱ ﺍﻟﻤﺴﺘﻘﻴﻢ ). (BC ABCﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ ﻭ ) (AFﻣﺴﺘﻘﻴﻢ ﻳ * ﻣﺜﺎﻝ : ﻭ Eﻧﻘﻄﺔ ) [BAﺧﺎﺭﺝ ]. [AB E ﻟﻨﺤﺴﺐ . E ˆ F A F ﻧﻌﺘﺒﺮ ﺍﻟﻤﺘﻘﻴﻤﻴﻦ ) (BCﻭ ) (AFﻭ ﺍﻟﻘﺎﻃﻊ ﻟﻬﻤﺎ ). (EB A ﻟﺪﻳﻨﺎ : E ˆ Fﻭ A ˆ Cﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﻨﺎﻇﺮﺗﺎﻥ . B A ﻭ ﺑﻤﺎ ﺃﻥ ) (BC) // (AFﻓﺈﻥ : = . A ˆ C E ˆ F B A ﻥ ﺍﻟﻤﺜﻠﺚ ABCﻣﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ , ﺇﺫﻥ : °06 = . A ˆ C B ﻭﻧﻌﻠﻢ ﺃ ﻭ ﻣﻨﻪ ﻓﺈﻥ : °06 = . E ˆ F A C B
56.
( ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﻌﻜﺴﻴﺔ ﻟﻠﺰﺍﻭﻳﺘﻴﻦ ﺍﻟﻤﺘﺒﺎﺩﻟﺘﻴﻦ ﺩﺍﺧﻠﻴﺎ ﻭ ﺍﻟﺰﺍﻭﻳﺘﻴﻦ ﺍﻟﻤﺘﻨﺎﻇﺮﺗﻴﻦ :
ﺝ ﺇﺫﺍ ﺣﺪﺩ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﻊ ﻗﺎﻃﻊ ﻟﻬﻤﺎ ﺯﺍﻭﻳﺘﻴﻦ ﻣﺘﺒﺎﺩﻟﺘﻴﻦ ﺩﺍﺧﻠﻴﺎ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ ﺃﻭ ﺯﺍﻭﻳﺘﻴﻦ ﻣﺘﻨﺎﻇﺮﺗﻴﻦ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ ﻓﺈﻧﻬﻤﺎ ﻳﻜﻮﻧﺎﻥ ﻣﺘﻮﺍﺯﻳﻴﻦ ﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﺭﺃﺳﻪ Aﺑﺤﻴﺚ 08 = . B ˆ C A ° ABCﻣﺜﻠﺚ ﻣ * ﻣﺜﺎﻝ : ) [AEﻧﺼﻒ ﻣﺴﺘﻘﻴﻢ ﺑﺤﻴﺚ C ˆ Bﻭ B ˆ Eﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺤﺎﺫﻳﺘﺎﻥ ﻭ 05 = . BAE ° A A ﻟﻨﺒﻴﻦ ﺃﻥ ). (AE) // (BC ﻟﺪﻳﻨﺎ ABCﻣﺜﻠﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﺭﺃﺳﻪ . A 08 - 081 ° ° = A ˆ C = A ˆ B B C ﺇﺫﻥ : 05 = ° 2 ﻧﻌﺘﺒﺮ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﻦ ) (BC) (EAﻭ ﺍﻟﻘﺎﻃﻊ ﻟﻬﻤﺎ ). (AB ﻭ ﻟﺪﻳﻨﺎ : B ˆ Eﻭ A ˆ Cﺯﺍﻭﻳﺘﺎﻥ ﻣﺘﺒﺎﺩﻟﺘﺎﻥ ﺩﺍﺧﻠﻴﺎ . B A ﻧﻌﻠﻢ ﺃﻥ 05 = . B ˆ Eﻭ ﺑﻤﺎ ﺃﻥ 05 = A ˆ Cﻓﺈﻥ : B ° A ° . B ˆ E = A ˆ C A B ﻭﻣﻨﻪ ﻓﺈﻥ : )(BC) // (AE _ IVﺧﺎﺻﻴﺎﺕ ﺍﻟﺘﻮﺍﺯﻱ ﻭ ﺍﻟﺘﻌﺎﻣﺪ : 1( – ﺍﻟﺨﺎﺻﻴﺔ ﺍﻷﻭﻟﻰ : ﺇﺫﺍ ﻛﺎﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻓﺈﻥ ﻛﻞ ﻣﺴﺘﻘﻴﻢ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺃﺣﺪﻫﻤﺎ ﻳﻜﻮﻥ ﻋﻤﻮﺩﻳﺎ ﻋﻠﻰ ﺍﻵﺧﺮ )1(D2) // (D : )(D2) ┴ (L ﻓﺈﻥ ﺇﺫﺍ ﻛﺎﻥ ﻭ * ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : )(D1) ┴ (L
57.
2( – ﺍﻟﺨﺎﺻﻴﺔ ﺍﻟﺜﺎﻧﻴﺔ : ﺇﺫﺍ ﻛﺎﻥ ﻣﺴﺘﻘﻴﻤﺎﻥ ﻣﺘﻌﺎﻣﺪﻳﻦ ﻓﺈﻥ ﻛﻞ ﻣﺴﺘﻘﻴﻢ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺃﺣﺪﻫﻤﺎ
ﻳﻜﻮﻥ ﻣﻮﺍﺯﻳﺎ ﻟﻶﺧﺮ . )1(D2) ┴ (D : )(D2) // (L ﺇﺫﺍ ﻛﺎﻥ ﻭ ﻓﺈﻥ * ﺑﺘﻌﺒﻴﺮ ﺁﺧﺮ : )(D1) ┴ (L
58.
ﺍﻟﺘـﻨــﺎﺳﺒﻴـــــــــــــــــﺔ
_ Iﺍﻟﺘﻨﺎﺳﺒﻴﺔ : 1( – ﺟﺪﻭﻝ ﺍﻟﺘﻨﺎﺳﺒﻴﺔ : ( ﻣﺜﺎﻝ 1 : ﺃ 7 4 4,5 11 5,2 ﻻﺣﻆ ﺍﻟﺠﺪﻭﻝ ﺍﻵﺗﻲ : 3 X 12 21 2,61 33 5,7 . ﻧﻼﺣﻆ ﺃﻧﻨﺎ ﺿﺮﺑﻨﺎ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻷﻭﻝ ﻓﻲ ﻧﻔﺲ ﺍﻟﻌﺪﺩ 3 ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻟﺜﺎﻧﻲ ﻧﺴﻤﻲ ﺍﻟﻌﺪﺩ 3 : ﻌﺎﻣﻞ ﺍﻟﺘﻨﺎﺳﺐ . ﻣ ﻧﻘﻮﻝ ﺇﺫﻥ : nﻫﺬﺍ ﺍﻟﺠﺪﻭﻝ ﻳﺤﻘﻖ ﻭﺿﻌﻴﺔ ﺍﻟﺘﻨﺎﺳﺒﻴﺔ . 12 21 2 61 33 ,7 5 , = = = = nﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻟﺜﺎﻧﻲ ﻣﺘﻨﺎﺳﺒﺔ ﻣﻊ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻷﻭﻝ . ﻭ ﻧﻜﺘﺐ : 3 = 7 4 4 5 11 5 2 , , ( ﻣﺜﺎﻝ 2 : ﺏ 4 5,5 5,7 6 5 ﻻﺣﻆ ﺍﻟﺠﺪﻭﻝ ﺍﻵﺗﻲ : ? 61 11 03 21 01 ﻧﻼﺣﻆ ﺃﻧﻨﺎ ﻟﻢ ﻧﻀﺮﺏ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻷﻭﻝ ﻓﻲ ﻧﻔﺲ ﺍﻟﻌﺪﺩ ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻟﺜﺎﻧﻲ . ﻧﻘﻮﻝ ﺇﺫﻥ : nﻫﺬﺍ ﺍﻟﺠﺪﻭﻝ ﻻﻳﺤﻘﻖ ﻭﺿﻌﻴﺔ ﺍﻟﺘﻨﺎﺳﺒﻴﺔ . 61 03 11 21 01 ﻭ 4 = = = = nﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻟﺜﺎﻧﻲ ﻏﻴﺮ ﻣﺘﻨﺎﺳﺒﺔ ﻣﻊ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻷﻭﻝ . ﻭ ﻧﻜﺘﺐ : 2 = 4 5 7 , 5 5 6 5 , 2( – ﻣﺒﻴﺎﻥ ﺍﻟﺘﻨﺎﺳﺒﻴﺔ : ( ﻣﺜﺎﻝ 1 : ﺃ ﻻﺣﻆ ﺍﻟﻤﺒﻴﺎﻥ ﺍﻵﺗﻲ : ﻧﻼﺣﻆ ﺃﻥ ﺟﻤﻴﻊ ﻧﻘﻄﻪ ﻣﺴﺘﻘﻴﻤﻴﺔ ﻣﻊ ﺃﺻﻞ ﺍﻟﻤﻌﻠﻢ . ﻧﻘﻮﻝ ﺇﺫﻥ : ¤ ﻫﺬﺍ ﺍﻟﻤﺒﻴﺎﻥ ﻳﺤﻘﻖ ﻭﺿﻌﻴﺔ ﺍﻟﺘﻨﺎﺳﺒﻴﺔ .
59.
( ﻣﺜﺎﻝ 2 :
ﺏ ﻻﺣﻆ ﺍﻟﻤﺒﻴﺎﻥ ﺍﻵﺗﻲ : ﻧﻼﺣﻆ ﺃﻥ ﺣﺠﻤﻴﻊ ﻧﻘﻄﻪ ﻏﻴﺮ ﻣﺴﺘﻘﻴﻤﻴﺔ ﻣﻊ ﺃﺻﻞ ﺍﻟﻤﻌﻠﻢ . ﻧﻘﻮﻝ ﺇﺫﻥ : ¤ ﻫﺬﺍ ﺍﻟﻤﺒﻴﺎﻥ ﻻ ﻳﺤﻘﻖ ﻭﺿﻌﻴﺔ ﺍﻟﺘﻨﺎﺳﺒﻴﺔ . 3( – ﺍﻟﺮﺍﺑﻊ ﺍﻟﻤﺘﻨﺎﺳﺐ : ( ﺗﻌﺮﻳﻒ : ﺃ c a ﻗﻴﻤﺔ ﺍﻟﻌﺪﺩ xﺑﺎﻟﺠﺪﻭﻝ ﺟﺎﻧﺒﻪ ﺗﺴﻤﻰ ﺍﻟﺮﺍﺑﻊ ﺍﻟﻤﺘﻨﺎﺳﺐ x b ﻊ ﺍﻟﻤﺘﻨﺎﺳﺐ ﺑﺎﺳﺘﻌﻤﺎﻝ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻨﺎﺳﺐ . ( ﻣﺜﺎﻝ 1 : ﺣﺴﺎﺏ ﺍﻟﺮﺍﺑ ﺏ 52 5,41 ﻧﻌﺘﺒﺮ ﺟﺪﻭﻝ ﺍﻟﺘﻨﺎﺳﺐ ﺍﻵﺗﻲ : x 5 5 . ﻟﺪﻳﻨﺎ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻨﺎﺳﺐ ﻫﻮ : 2 0 = , 52 9,2 = x ﺃﻱ ﺇﺫﻥ : 2,0 x = 14,5 x ﺐ . ( ﻣﺜﺎﻝ 2 : ﺇﺗﻤﺎﻡ ﺟــﺪﻭﻝ ﺍﻟﺘﻨﺎﺳ ﺝ ﻓﻲ ﺍﻟﺠﺪﻭﻝ ﺍﻵﺗﻲ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻟﺜﺎﻧﻲ ﻣﺘﻨﺎﺳﺒﺔ ﻣﻊ ﺃﻋﺪﺍﺩ ﺍﻟﺴﻄﺮ ﺍﻷﻭﻝ . 2 7 y 11 t x 19 742 z 889 ﻟﻨﺤﺴﺐ : xﻭ yﻭ zﻭ . t 19 ﻟﺪﻳﻨﺎ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻨﺎﺳﺐ ﻫﻮ : 31 = . 7 ;; 62 = 31 x = 2 x ﺇﺫﻥ : 91 = 31 : 742 = y 67 = 31 : 889 = z = 11 x 13 = 143 ;; t
60.
_ IIﺳﻠﻢ ﺍﻟﺘﺼﻤﻴﻤﺎﺕ ﻭ ﺍﻟﺨﺮﺍﺋﻂ :
1( – ﺗﻌﺮﻳــﻒ : ﺍﻟﺴﻠﻢ ﻫﻮ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻨﺎﺳﺐ ﺑﻴﻦ ﺍﻟﻘﻴﺎﺳﺎﺕ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻟﺸﻴﺊ ﻭ ﺍﻟﻘﻴﺎﺳﺎﺕ ـﻢ ﺑﺎﻟﺮﻣﺰ : e ﻋﻠﻰ ﺗﺼﻤﻴﻢ ﺃﻭ ﺧﺮﻳﻄﺔ ﻟﻬﺬﺍ ﺍﻟﺸﻴﺊ . ﻳﺮﻣﺰ ﻟﻠﺴﻠــ ﻠﻰ ﺍﻟﺘﺼﻤﻴﻢ ﺍﻟﻘﻴﺎﺱ ﻋ = e ﻣﻼﺣﻈﺔ ﻫﺎﻣــﺔ : ﺍﻟﻘﻴﺎﺱ ﺍﻟﺤﻘﻴﻘﻲ ﺍﻟﻤﺴﺎﻓﺔ ﻋﻠﻰ ﺍﻟﺨﺮﻳﻄﺔ )y 125 (cm 2( – ﻣﺜــﺎﻝ : ﺮ ﺍﻟﺠﺪﻭﻝ ﺍﻵﺗﻲ : ﻧﻌﺘﺒ 75 x ﺍﻟﻤﺴﺎﻓﺔ ﺍﻟﺤﻘﻴﻘﻴﺔ )(km 1 . ﻟﻨﺤﺴﺐ xﻭ yﻋﻠﻤﺎ ﺃﻥ ﺍﻟﺴﻠــﻢ ﻫﻮ : 000052 1 . ﻭ ﻫﺬﺍ ﻳﻌﻨﻲ ﺃﻥ 1 cmﻋﻠﻰ ﺍﻟﺨﺮﻳﻄﺔ ﻳﻤﺜﻞ 250000 cmﻓﻲ ﺍﻟﺤﻘﻴﻘﺔ . ﻟﺪﻳﻨﺎ ﻣﻌﺎﻣﻞ ﺍﻟﺘﻨﺎﺳﺐ ﻫﻮ : 000052 ﻭ ﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻘﻴﺎﺳﺎﺕ ﻋﻠﻰ ﺍﻟﺨﺮﻳﻄﺔ ﻣﺘﻨﺎﺳﺒﺔ ﻣﻊ ﺍﻟﻘﻴﺎﺳﺎﺕ ﺍﻟﺤﻘﻴﻘﻴﺔ . 1 : 521 = x ﺇﺫﻥ : 000052 000052 x = 125 xﺃﻱ x = 31250000 cm = 312,5 km 1 y = 30 cm ﺃﻱ y = 7500000 x 000052 _ IIIﺍﻟﺤﺮﻛﺔ ﺍﻟﻤﻨﺘﻈﻤﺔ : 1( – ﺗﻌﺮﻳــﻒ : ﻳﻜﻮﻥ ﺟﺴﻢ ﻓﻲ ﺣﺮﻛﺔ ﻣﻨﺘﻈﻤﺔ ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﻤﺴﺎﻓﺎﺕ ﺍﻟﺘﻲ ﻳﻘﻄﻌﻬﺎ ﻣﺘﻨﺎﺳﺒﺔ ﻣﻊ ﺍﻟﻤﺪﺩ ﺍﻟﺰﻣﻨﻴﺔ ﺍﻟﻤﻮﺍﻓﻘﺔ ﻟﻬﺎ . 2( – ﻣﺜــﺎﻝ 1 : ﻲ ﺗﺴﺘﻐﺮﻗﻬﺎ ﺳﻴﺎﺭﺓ ﻟﻘﻄﻊ ﻣﺴﺎﻓﺎﺕ . ﺍﻟﺠﺪﻭﻝ ﺍﻵﺗﻲ ﻳﺒﻴﻦ ﺍﻟﻤﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ ﺍﻟﺘ 6 5 ﺍﻟﻤﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ )(h ﺍﻟﻤﺴﺎﻓﺔ ﺍﻟﻤﻘﻄﻮﻋﺔ )480 400 (km ﻟﺪﻳﻨﺎ : 084 004 08 = ﻭ 08 = 6 5 084 004 = 08 = ﻧﻼﺣﻆ ﺃﻥ : 5 6 ﻧﻘﻮﻝ ﺇﺫﻥ : ﻫﺬﻩ ﺍﻟﺴﻴﺎﺭﺓ ﻓﻲ ﺣﺮﻛﺔ ﻣﻨﺘﻈﻤﺔ .
61.
3( – ﻣﺜــﺎﻝ 2 : ﺍﻟﺠﺪﻭﻝ ﺍﻵﺗﻲ ﻳﺒﻴﻦ ﺍﻟﻤﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ ﺍﻟﺘﻲ ﻳﺴﺘﻐﺮﻗﻬﺎ ﻗﻄﺎﺭ ﻟﻘﻄﻊ ﻣﺴﺎﻓﺎﺕ .
7 6 5 ﺍﻟﻤﺪﺓ ﺍﻟﺰﻣﻨﻴﺔ )(h ﺍﻟﻤﺴﺎﻓﺔ ﺍﻟﻤﻘﻄﻮﻋﺔ 077 055 63 )(km ﻟﺪﻳﻨﺎ : 077 063 055 011 = ﻭ 06 = ﻭ 011 = 7 6 5 063 077 055 06 = ﻭ = 011 = ﻧﻼﺣﻆ ﺃﻥ : 6 5 7 063 077 055 = ¹ ﺃﻱ ﺃﻥ : 5 7 6 ﻫﺬﺍ ﺍﻟﻘﻄﺎﺭ ﻟﻴﺲ ﻓﻲ ﺣﺮﻛﺔ ﻣﻨﺘﻈﻤﺔ . ﻧﻘﻮﻝ ﺇﺫﻥ :
62.
ﺍﻟﻤﺴﺘﻘﻴــﻢ ﺍﻟﻤــﺪﺭﺝ ﻭ ﺍﻟﻤـﻌــﻠﻢ ﻓﻲ ﺍﻟﻤـﺴـﺘـــﻮﻯ
_ Iﺍﻟﻤﺴﺘﻘـﻴﻢ ﺍﻟـﻤــﺪﺭﺝ : : 1( – ﺗﺬﻛﻴــﺮ . ﻧﻌﺘﺒﺮ ﻣﺴﺘﻘﻴﻤﺎ ) (Dﻣﺪﺭﺝ , ﺑﺤﻴﺚ ] [OIﻫﻲ ﻭﺣﺪﺓ ﺍﻟﺘﺪﺭﻳﺞ ﻧﺴﻤﻲ ﺍﻟﻌﺪﺩ 0 ﺃﻓﺼــﻮﻝ ﺍﻟﻨﻘﻄﺔ Oﻭ ﺍﻟﻌﺪﺩ 1 ﺃﻓﺼــﻮﻝ ﺍﻟﻨﻘﻄﺔ . I ﺃﻓﺼــﻮﻝ ﺍﻟﻨﻘﻄﺔ Aﻫﻮ ﺍﻟﻌﺪﺩ 4 . ﻭ ﻧﻜﺘﺐ : )4( Aﺃﻭ 4 = . xA 3 = . xB ﻓﺼــﻮﻝ ﺍﻟﻨﻘﻄﺔ Bﻫﻮ ﺍﻟﻌﺪﺩ 3 – . ﻭ ﻧﻜﺘﺐ : )3 ( Bﺃﻭ ﺃ : 2( – ﺍﻷﻓﺼﻮﻝ ﻭ ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﻧﻘﻄﺘﻴﻦ * ﺗﻌﺮﻳﻒ : ﻟﺤﺴﺎﺏ ﺍﻟﻤﺴﺎﻓﺔ ﺑﻴﻦ ﻧﻘﻄﺘﻴﻦ ﻧﻄﺮﺡ ﻣﻦ ﺍﻷﻓﺼﻮﻝ ﺍﻟﻜﺒﻴﺮ ﺍﻷﻓﺼﻮﻝ ﺍﻟﺼﻐﻴﺮ * : ﻣﺜﺎﻝ )2( C( 1,5) B( 5) Aﻧﻘﻂ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﻣﺴﺘﻘﻴﻢ ﻣﺪﺭﺝ . ﻭ ﻭ ﻟﻨﺤﺴﺐ ﺍﻟﻤﺴﺎﻓﺎﺕ ABﻭ BCﻭ . AC ﻟﺪﻳﻨﺎ : AC = xA xC BC = xC xB AB = xA xB )5,1 ( 2 = )5 ( 5,1 = = 5,1 + 2 = 5 + 5,1 = 5 + 2 = 5,3 = 5,3 = 7 = : 3( – ﺃﻓﺼﻮﻝ ﻣﻨﺘﺼﻒ ﻭﻗﻄﻌﺔ * ﺗﻌﺮﻳﻒ : ﺃﻓﺼﻮﻝ ﻣﻨﺘﺼﻒ ﻗﻄﻌﺔ ﻫﻮ ﻧﺼﻒ ﻣﺠﻤﻮﻉ ﺃﻓﺼﻮﻟﻲ ﻃﺮﻓﻴﻬﺎ * ﻣﺜﺎﻝ : . ) 5 ( Aﻭ ) 4 ( Bﻧﻘﻄﺘﺎﻥ ﻣﻦ ﻣﺴﺘﻘﻴﻢ ﻣﺪﺭﺝ ﻟﻨﺤﺴﺐ ﺃﻓﺼﻮﻝ Eﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ]. [AB
63.
= x A + x B
x ﻟﺪﻳﻨﺎ : E 2 1 - 4 + 5 - = = 5 0- = , 2 2 ﺇﺫﻥ : )5,0 (E _ IIﺍﻟﻤﻌﻠﻢ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ : : 1( – ﺇﻧﺸﺎء ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ ﻧﻌﺘﺒﺮ ) (Dﻭ ) (Δﻣﺴﺘﻘﻴﻤﻴﻦ ﻣﺪﺭﺟﻴﻦ ﻋﻠﻰ ﺍﻟﺘﻮﺍﻟﻲ ﺑﻮﺍﺳﻄﺔ ] [OJ] [OIﻭ ﻣﺘﻌﺎﻣﺪﻳﻦ ﻓﻲ ﺍﻟﻨﻘﻄﺔ . O ﻭ * ﻣﻼﺣﻈﺔ ﻫﺎﻣﺔ : ﺇﺫﺍ ﻛﺎﻥ OI = OJﻧﻘﻮﻝ ﺃﻥ ﺍﻟﻤﺴﺘﻮﻯ ﻣﻨﺴﻮﺏ ﺇﻟﻰ ﻣﻌــﻠﻢ ﻣﻤﻨﻈﻢ ﻭ ﻣﺘﻌﺎﻣﺪ . * ﻣﻔــﺮﺩﺍﺕ : ﻮﺭ ﺍﻷﻓﺎﺻﻴﻞ . ﻧﺴﻤﻲ ﺍﻟﻤﺴﺘﻘﻴﻢ ) : (OIﻣﺤــ ﻧﺴﻤﻲ ﺍﻟﻤﺴﺘﻘﻴﻢ ) : (OJﻣﺤــﻮﺭ ﺍﻷﺭﺍﺗﻴﺐ . ﻧﺮﻣﺰ ﻟﻤﻌﻠﻢ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ ﺑﺎﻟﺮﻣــﺰ : ) . ( O ; I ; J : 2( – ﺇﺣﺪﺍﺛﻴﺘﺎ ﻧﻘﻄﺔ * ﺗﻌﺮﻳﻒ : ﻛﻞ ﻧﻘﻄﺔ Mﻣﻦ ﺍﻟﻤﺴﺘﻮﻯ ﻣﺮﺗﺒﻄﺔ ﺑﻌﺪﺩﻳﻦ ﻋﺸﺮﻳﻴﻦ ﻧﺴﺒﻴﻴﻦ xMﻭ yM ﻳﺴﻤﻴﺎﻥ ﺇﺣﺪﺍﺛﻴﺘﻲ ﺍﻟﻨﻘﻄﺔ . Mﻭ ﻧﻜﺘﺐ : ) . M( xM ; yM
64.
* ﻣﺜﺎﻝ :
ﻧﻌﺘﺒﺮ ﺍﻟﻤﺴﺘﻮﻯ ﻣﻨﺴﻮﺑﺎ ﺇﻟﻰ ﻣﻌﻠﻢ ﻣﻤﻨﻈﻢ ﻭ ﻣﺘﻌﺎﻣﺪ ) . (O ; I ; J ﻟﻨﻨﺸﺊ ﺍﻟﻨﻘﻂ : ) 2 ; 1 ( Aﻭ ) 2 ; 3 ( Bﻭ ) 4 ; 5,0 ( Cﻭ ) 3 ; 0 ( Dﻭ ) 0 ; 2 ( Eﻭ ) 5 ; 5 (F : 3( – ﺇﺣﺪﺍﺛﻴﺘﺎ ﻣﻨﺘﺼﻒ ﻧﻘﻄﺔ * ﺗﻌﺮﻳﻒ : ] [ABﻗﻄﻌﺔ ﻭ Eﻣﻨﺘﺼﻔﻬﺎ . y A + y B + = y E x E = x A x Bﻭ 2 2 * ﻣﺜﺎﻝ : ) 5 ; 2 ( Aﻭ ) 6 ; 4 ( Bﻧﻘﻄﺘﺎﻥ ﻣﻦ ﺍﻟﻤﺴﺘﻮﻯ ﻣﻨﺴﻮﺏ ﺇﻟﻰ ﻣﻌﻠﻢ ﻣﻤﻨﻈﻢ ﻣﺘﻌﺎﻣﺪ . ﻟﻨﺤﺴﺐ ﺇﺣﺪﺍﺛﻴﺜﻲ Eﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ]. [AB 11 - 6 ( + 5 - y A + y B ) - 1- = 2 - = 4 ( + 2 = x A + x B ) - = y E = = 5 5- = , ﻭ = x E ﻟﺪﻳﻨﺎ : 2 2 2 2 2 2 ﺇﺫﻥ : ) 5,5 ; 1 (E
65.
ﺍﻟــﺪﺍﺋـــــــــــــــــــــــﺮﺓ
_ Iﺍﻟﺪﺍﺋﺮﺓ : 1( – ﻣﺜــﺎﻝ : ﻧﻌﺘﺒﺮ ) ( Cﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ Oﻭﺷﻌﺎﻋﻬﺎ . 2 cm ﻟﺘﻜﻦ Aﻭ D C Bﻧﻘﻂ ﻣﺨﺘﻠﻔﺔ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﺪﺍﺋﺮﺓ ) . ( C ﻭ ﻭ OA = 2 cmﻭ OB = 2 cmﻭ OC = 2 cmﻭ . OD = 2 cm ﻟﺪﻳﻨﺎ : ﻧﻞ ﺍﺣﻆ ﺃﻥ ﺍﻟﻨﻘﻂ Aﻭ Bﻭ Cﻭ Dﺗﺒﻌﺪ ﺑﻨﻔﺲ ﺍﻟﻤﺴﺎﻓﺔ 2 cmﻋﻦ ﺍﻟﻤﺮﻛﺰ . O 2( – ﺗﻌﺮﻳﻒ : ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﺘﻲ ﻣﺮﻛﺰﻫﺎ Oﻭﺷﻌﺎﻋﻬﺎ rﻫﻲ ﻣﺠﻤﻮﻋﺔ ﺍﻟﻨﻘﻂ ﺍﻟﺘﻲ ﻣﺴﺎﻓﺘﻬﺎ ﻋﻦ ﺍﻟﻤﺮﻛﺰ Oﻫﻲ . r 3( – ﻣﻔﺮﺩﺍﺕ : ﻭﺗﺮ ﺩﺍﺋﺮﺓ ﻫﻮ ﻗﻄﻌﺔ ﻃﺮﻓﻴﻬﺎ ﻳﻨﺘﻤﻴﺎﻥ ﺇﻟﻰ ﺍﻟﺪﺍﺋﺮﺓ * ﺍﻟﻮﺗﺮ : ) ( Cﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ Oﻭ ﺷﻌﺎﻋﻬﺎ [AB] . rﻭ ] [EFﻗﻄﻌﺘﺎﻥ ﻃﺮﻓﻴﻬﻤﺎ ﻳﻨﺘﻤﻴﺎﻥ ﺇﻟﻰ ﺍﻟﺪﺍﺋﺮﺓ . ﻧﺴﻤﻲ ﻛﻼ ﻣﻦ ] [ABﻭ ] [EFﻭﺗﺮ ﻟﻠﺪﺍﺋﺮﺓ ) . ( C ﻗﻄﺮ ﺩﺍﺋﺮﺓ ﻫﻮ ﻭﺗﺮ ﻳﻤﺮ ﻣﻦ ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ : * ﺍﻟﻘﻄﺮ ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺃﻋﻼﻩ ﻧﺴﻤﻲ ] [EFﻗﻄﺮ ﻟﻠﺪﺍﺋﺮﺓ ) . ( C
66.
3( – ﺧـﺎﺻﻴﺔ :
) ( Cﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ Oﻭ ﺷﻌﺎﻋﻬﺎ rﻭ Mﻧﻘﻄﺔ ﻣﻦ ﺍﻟﻤﺴﺘﻮﻯ . ﻓﺈﻥ OM = r ¤ ﺇﺫﺍ ﻛﺎﻥ ) M Î (C ﻓﺈﻥ ) M Î (C ¤ ﺇﺫﺍ ﻛﺎﻥ OM = r _ Iﻣﻤﺎﺱ ﺍﻟﺪﺍﺋﺮﺓ : 1( – ﻣﺜــﺎﻝ : ) ( Cﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ Oﻭ ﺷﻌﺎﻋﻬﺎ . r Aﻧﻘﻄﺔ ﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﺪﺍﺋﺮﺓ ) ( Cﻭ ) (Lﻣﺴﺘﻘﻴﻢ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ) (OAﻓﻲ ﺍﻟﻨﻘﻄﺔ . A ﻧﺴﻤﻲ ﺍﻟﻤﺴﺘﻘﻴﻢ ) (Lﻣﻤﺎﺱ ﺍﻟﺪﺍﺋﺮﺓ ) ( Cﻓﻲ ﺍﻟﻨﻘﻄﺔ A 2( – ﺗﻌﺮﻳﻒ : ﻣﻤﺎﺱ ﺩﺍﺋﺮﺓ ﻓﻲ ﻧﻘﻄﺔ Mﺗﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﺪﺍﺋﺮﺓ ﻫﻮ ﻣﺴﺘﻘﻴﻢ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺣﺎﻣﻞ ﺍﻟﺸﻌﺎﻉ ﻓﻲ ﺍﻟﻨﻘﻄﺔ . M 3( – ﺧـﺎﺻﻴﺔ : ) ( Cﺩﺍﺋﺮﺓ ﻣﺮﻛﺰﻫﺎ Oﻭ ﺷﻌﺎﻋﻬﺎ M . rﻧﻘﻄﺔ ﻣﻦ ﺍﻟﻤﺴﺘﻮﻯ ﻭ ) (Lﺘﻘﻴﻢ . ﻣﺴ ) M Î (C ﻳﻌﻨﻲ ﺃﻥ ) ( Lﻣﻤﺎﺱ ﺍﻟﺪﺍﺋﺮﺓ ) ( Cﻓﻲ ﺍﻟﻨﻘﻄﺔ . M OM ( ) ^ ( L ) ) M Î (C ) ( Lﻣﻤﺎﺱ ﺍﻟﺪﺍﺋﺮﺓ ) ( Cﻓﻲ ﺍﻟﻨﻘﻄﺔ . Mﻳﻌﻨﻲ ﺃﻥ OM ( ) ^ ( L ) : ﻧﺮﻣﺰ ﻟﺪﺍﺋﺮﺓ ) ( Cﺮﻛﺰﻫﺎ Oﻭ ﺷﻌﺎﻋﻬﺎ rﺑﺎﻟﺮﻣﺰ : )C (O ; r ﻣ * ﺗﺮﻣﻴــﺰ
67.
ﺍﻟـﻤــﻮﺷــﻮﺭ ﺍﻟﻘــﺎﺋﻢ ﻭ ﺍﻷﺳـﻄـــﻮﺍﻧﺔ ﺍﻟـﻘــــــﺎﺋﻤﺔ
_ Iﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋﻢ : 1( – ﺗﻌﺮﻳﻒ : ﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋﻢ ﻫﻮ ﻣﺠﺴﻢ ﻳﺘﻜﻮﻥ ﻣﻦ : 1( – ﻭﺟﻬﻴﻦ ﻣﺘﻮﺍﺯﻳﻴﻦ ﻗﺎﺑﻠﻴﻦ ﻟﻠﺘﻄﺎﺑﻖ ﻫﻤﺎ : ﻗﺎﻋﺪﺗﺎﻥ ﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋﻢ . 2( – ﺃﺣﺮﻑ ﺟﺎﻧﺒﻴﺔ ﻣﺘﻘﺎﻳﺴﺔ ﻫﻲ : ﺍﺭﺗﻔﺎﻉ ﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋﻢ . 3( – ﺃﻭﺟﻪ ﺟﺎﻧﺒﻴﺔ ﻭ ﻫﻲ ﻋﻠﻰ ﺷﻜﻞ : ﻣﺴﺘﻄﻴﻼﺕ . * ﻣﻼﺣﻈﺎﺕ ﻫﺎﻣﺔ : 1( – ﻋﺪﺩ ﺍﻷﻭﺟﻪ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻟﻤﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻳﺴﺎﻭﻱ ﻋﺪﺩ ﺃﺿﻼﻉ ﻗﺎﻋﺪﺗﻪ . ﻞ ﻣﺜﻠﺚ ﺃﻭ ﻣﺮﺑﻊ ﺃﻭ ﻣﺴﺘﻄﻴﻞ ﺃﻭ ﻣﻀﻠﻊ ﺭﺑﺎﻋﻲ 2( – ﻗﺎﻋﺪﺗﺎ ﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋﻢ ﺇﻣﺎ ﺃﻥ ﺗﻜﻮﻧﺎ ﻋﻠﻰ ﺷﻜ ﺃﻭ ﻣﻀﻠﻊ ﺧﻤﺎﺳﻲ .......... 2( – ﺃﻣﺜﻠــﺔ : ( ﻣﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻗﺎﻋﺪﺗﺎﻩ ﻣﺜﻠﺚ : ﺃ ﺍﻟﻘﺎﻋﺪﺗﺎﻥ ﻫﻤﺎ : ﺍﻟﻤﺜﻠﺜﺎﻥ ABCﻭ . DEF ﺍﻷﺣﺮﻑ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ]. [CF] [BE] [AD ﻭ ﻭ ﺍﻷﻭﺟﻪ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ﺍﻟﻤﺴﺘﻄﻴﻼﺕ ADFC ﻭ ADEBﻭ . BEFC ( ﻣﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻗﺎﻋﺪﺗﺎﻩ ﻣﺴﺘﻄﻴﻞ : ﺏ ﺍﻟﻘﺎﻋﺪﺗﺎﻥ ﻫﻤﺎ : ﺍﻟﻤﺴﺘﻄﻴﻼﻥ ABCDﻭ . EFGH ﺍﻷﺣﺮﻑ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ][DH] [CG] [BF] [AE ﻭ ﻭ ﻭ . ﺍﻷﻭﺟﻪ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ﺍﻟﻤﺴﺘﻄﻴﻼﺕ DCGH AEHD ﻭ . ABFE BCGF ﻭ ﻭ ﻲ ﻫﺬﺍ ﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋﻢ : ﻣﺘﻮﺍﺯﻱ ﺍﻟﻤﺴﺘﻄﻴﻼﺕ ﺍﻟﻘﺎﺋﻢ . ﻧﺴﻤ
68.
( ﻣﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻗﺎﻋﺪﺗﺎﻩ ﻣﺮﺑﻊ :
ﺝ ﺍﻟﻘﺎﻋﺪﺗﺎﻥ ﻫﻤﺎ : ﺍﻟﻤﺮﺑﻌﺎﻥ . EFGH ABCD ﻭ ﺍﻷﺣﺮﻑ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ][DH] [CG] [BF] [AE ﻭ ﻭ ﻭ . ﻷﻭﺟﻪ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ﺍﻟﻤﺮﺑﻌﺎﺕ DCGH AEHD ﻭ ﺍ . ABFE BCGF ﻭ ﻭ ﻧﺴﻤﻲ ﻫﺬﺍ ﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋﻢ : ﻣـﻜـﻌـــﺐ . ( ﻣﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻗﺎﻋﺪﺗﺎﻩ ﻣﻀﻠﻊ ﺧﻤﺎﺳﻲ : ﺩ ﺍﻟﻘﺎﻋﺪﺗﺎﻥ ﻫﻤﺎ : ﺍﻟﻤﺮﺑﻌﺎﻥ . FGHMN ABCDE ﻭ ﺍﻷﺣﺮﻑ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ][CH] [BG] [AF ﻭ ﻭ ]. [DM] [EN ﻭ ﻭ ﺍﻷﻭﺟﻪ ﺍﻟﺠﺎﻧﺒﻴﺔ ﻫﻲ : ﺍﻟﻤﺴﺘﻄﻴﻼﺕ DMNE AENF ﻭ DCHM BCGFﻭ . BCHG ﻭ ﻭ _ IIﺍﻷﺳﻄﻮﺍﻧﺔ ﺍﻟﻘﺎﺋﻤﺔ : * ﻣﺜﺎﻝ : ﺎﻋﺪﺗﺎﻥ ﻫﻤﺎ : ﺍﻟﺪﺍﺋﺮﺗﺎﻥ ﺍﻟﻘ ﺍﻟﻠﺘﺎﻥ ﻣﺮﻛﺰﻫﻤﺎ Aﻭ . B ﺍﻻﺭﺗﻔﺎﻉ ﻫﻮ : . AB A D _ IIIﻧﺸﺮ ﺍﻟﻤﻮﺷﻮ ﺍﻟﻘﺎﺋﻢ ﻭ ﺍﻷﺳﻄﻮﺍﻧﺔ ﺍﻟﻘﺎﺋﻤﺔ : B E ﻢ : 1( – ﻧﺸﺮ ﺍﻟﻤﻮﺷﻮﺭ ﺍﻟﻘﺎﺋ A D ( ﻣﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻗﺎﻋﺪﺗﺎﻩ ﻣﺜﻠﺚ : ﺃ C F A D
69.
( ﻣﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻗﺎﻋﺪﺗﺎﻩ ﻣﺴﺘﻄﻴﻞ :
ﺏ A B A E F B A D H G C D D C ( ﻣﻮﺷﻮﺭ ﻗﺎﺋﻢ ﻗﺎﻋﺪﺗﺎﻩ ﻣﺮﺑﻊ : ﺝ A B B A E F B C D H G C D C 2( – ﻧﺸﺮ ﺍﻷﺳﻄﻮﺍﻧﺔ ﺍﻟﻘﺎﺋﻤﺔ : A B B
70.
ﺍﻟﻤﺤﻴـﻄـــــــــﺎﺕ ﻭ ﺍﻟﻤـﺴــــــــﺎﺣـﺎﺕ ﻭ ﺍﻟﺤﺠــــــــﻮﻡ
_ Iﺍﻟﻤﺤﻴﻄﺎﺕ ﻭ ﺍﻟﻤﺴﺎﺣﺎﺕ ﻓﻲ ﺍﻟﻤﺴﺘﻮﻯ : ﺍﻟﻤﺴﺎﺣﺔ ﺍﻟﻤﺤﻴﻂ ﺍﻟﺸﻜﻞ A ﺍﻟﻤﺜﻠﺚ : BC ´ AH = S P = AB + AC + BC 2 B H C A B ﺍﻟﻤﺮﺑﻊ : 2 p = 4 ´ AB S = AB D C A B ﺍﻟﻤﺴﺘﻄﻴﻞ : S = AB ´ BC ) P = 2 AB + BC ( D C A ﺍﻟﻤﻌﻴﻦ : B AC ´ BD =S P = 4 AB 2 D C
71.
ﺍﻟﻤﺴﺎﺣﺔ
ﺍﻟﻤﺤﻴﻂ ﺍﻟﺸﻜﻞ : ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ A S = AB ´ AH P = 2 AB + BC ) ( D B H C : ﺷﺒﻪ ﺍﻟﻤﻨﺤﺮﻑ ( AB = CD ´ AH ) A B S = p = AB = BC = CD = DA 2 D H C : ﺍﻟﺪﺍﺋﺮﺓ O r 2 S = π x r P = 2π x r
72.
_ IIﺍﻟﻤﺤﻴﻄﺎﺕ ﻭ ﺍﻟﻤﺴﺎﺣﺎﺕ ﻓﻲ ﺍﻟﻔﻀﺎء : ﺍﻟﺤﺠــﻢ
ﺍﻟﻤﺴﺎﺣﺔ ﺍﻟﻜﻠﻴﺔ ﺍﻟﻤﺴﺎﺣﺔ ﺍﻟﺠﺎﻧﺒﻴﺔ ﺍﻟﺸﻜﻞ ﻣﺘﻮﺍﺯﻱ ﺍﻟﻤﺴﺘﻄﻴﻼﺕ ﺍﻟﻘﺎﺋﻢ : A B V=AB.AD.AE ST=SL+2AB.AD )SL=2(AB.AE+AD.AE E F D C H G ﺍﻟﻤﻜﻌﺐ : A B D C 3 2 2 V= AB ST= 6AB SL= 4AB E F H G A ﺍﻟﻤﻮﺷﻮﺭ B ﺍﻟﻘﺎﺋﻢ : V = SB x AE ST = SL + 2SB E C SL= P x AE D = Pﻣﺤﻴﻂ ﺍﻟﻘﺎﻋﺪﺓ E F H = SBﻣﺴﺎﺣﺔ ﺍﻟﻘﺎﻋﺪﺓ G E ﺍﻷﺳﻄﻮﺍﻧﺔ A ﺍﻟﻘﺎﺋﻤﺔ : r 2 2 V = πr .h ST = SL + 2πr SL = 2π . r.h h B
73.
ﺍﻹﺣــﺼــــــــــــــــــﺎء
_ Iﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ : 1( – ﻗﺎﻋﺪﺓ 1 : x ´ n ﺗﻄﺒﻴﻖ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ % xﻋﻠﻰ ﺍﻟﻌﺪﺩ nﻫﻮ ﺣﺴﺎﺏ : 001 * ﻣﺜﺎﻝ 1 : ﺑﻘﺴﻢ ﻳﺤﺘﻮﻱ ﻋﻠﻰ 04 ﺗﻠﻤﻴﺬﺍ ﻳﻮﺟﺪ %06 ﻣﻦ ﺍﻹﻧﺎﺙ . . ﻟﻨﺤﺪﺩ ﻋﺪﺩ ﺍﻹﻧﺎﺙ ﻭ ﺍﻟﺬﻛﻮﺭ 0042 06 ´ 04 ﻭ ﻣﻨﻪ ﻓﺈﻥ : 61 = 42 – 04 = 42 = ﻟﺪﻳﻨﺎ : 001 001 ﺇﺫﻥ ﻋﺪﺩ ﺍﻹﻧﺎﺙ ﻫﻮ : 42 ﻭ ﻋﺪﺩ ﺍﻟﺬﻛﻮﺭ ﻫﻮ : 61 * ﻣﺜﺎﻝ 2 : ﺏ ﺛﻤﻨﻪ 24 DHﺇﺫﺍ ﻛﻨﺎ ﺳﻨﺴﺘﻔﻴﺪ ﻣﻦ ﺗﺨﻔﻴﺾ ﻗﺪﺭﻩ % 01 ؟ ﻛﻢ ﺳﻨﺪﻓﻊ ﻟﺸﺮﺍء ﻛﺘﺎ ﻟﻨﺤﺪﺩ ﻗﻴﻤﺔ ﺍﻟﺘﺨﻔﻴﺾ : 042 01 ´ 42 = 4 2 = , ﻟﺪﻳﻨﺎ : 001 001 ﺇﺫﻥ ﻗﻴﻤﺔ ﺍﻟﺘﺨﻔﻴﺾ ﻫﻲ : . 2,4 DH ﻣﺎ ﺳﻨﺪﻓﻌﻪ ﻟﺸﺮﺍء ﺍﻟﻜﺘﺎﺏ : ﻟﺪﻳﻨﺎ : 6,12 = 4,2 – 42 ﺇﺫﻥ ﺳﻨﺪﻓﻊ ﻟﺸﺮﺍء ﺍﻟﻜﺘﺎﺏ : . 21,6 DH 2( – ﻗﺎﻋﺪﺓ 2 : b ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﻌﺪﺩ bﻳﺸﻜﻞ % xﻣﻦ ﺍﻟﻌﺪﺩ aﻓﺈﻥ 001 ´ = x a * ﻣﺜﺎﻝ 1 : 2 2 ﻣﻨﺰﻝ ﻣﺴﺎﺣﺘﻪ . 90 mﺑﻪ ﺣﺠﺮﺓ ﻣﺴﺎﺣﺘﻬﺎ . 20 m ﻟﻨﺤﺪﺩ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺍﻟﺘﻲ ﺗﻤﺜﻠﻬﺨﺎ ﻣﺴﺎﺣﺔ ﺍﻟﺤﺠﺮﺓ ﻣﻦ ﻣﺴﺎﺣﺔ ﺍﻟﻤﻨﺰﻝ . 02 ﻟﺪﻳﻨﺎ : 22 22 = 001 ´ = x , 09 ﺇﺫﻥ : ﻣﺴﺎﺣﺔ ﺍﻟﺤﺠﺮﺓ ﺗﻤﺜﻞ % 22,22 ﻣﻦ ﻣﺴﺎﺣﺔ ﺍﻟﻤﻨﺰﻝ .
74.
* ﻣﺜﺎﻝ 2 :
125 kgﻣﻦ ﺍﻟﻘﻤﺢ ﺗﻌﻄﻲ 93,75 kgﻣﻦ ﺍﻟﺪﻗﻴﻖ . ﻟﻨﺤﺴﺐ ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺍﻟﺘﻲ ﻳﻌﻄﻴﻬﺎ ﺍﻟﻘﻤﺢ ﻣﻦ ﻛﺜﻠﺔ ﺍﻟﺪﻗﻴﻖ . 57 39 , = x ﻟﺪﻳﻨﺎ : 57 = 001 ´ 521 ﺇﺫﻥ : ﺍﻟﻨﺴﺒﺔ ﺍﻟﻤﺌﻮﻳﺔ ﺍﻟﺘﻲ ﻳﻌﻄﻴﻬﺎ ﺍﻟﻘﻤﺢ ﻣﻦ ﻛﺘﻠﺔ ﺍﻟﺪﻗﻴﻖ ﻫﻲ : % 57 . _ Iﺍﻹﺣﺼــﺎء : : 1( – ﺟﻤﻊ ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﻭ ﺗﻨﻈﻴﻤﻬﺎ ﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﺍﻵﺗﻴﺔ ﺗﺘﻌﻠﻖ ﺑﻨﻘﻂ ﻓﺮﺽ ﻣﺤﺮﻭﺱ ﻓﻲ ﻣﺎﺩﺓ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ . 6 31 4 11 6 01 9 7 01 9 7 41 11 5 31 5 41 8 21 01 21 01 8 6 01 11 8 01 9 21 7 41 11 4 8 21 6 : 2( – ﺍﻟﺠﺪﻭﻝ ﺍﻹﺣﺼﺎﺋﻲ 9 8 7 6 5 4 01 11 21 31 41 ( ﺍﻟﻨﻘﻂ ) ﺍﻟﻤﻴﺰﺓ 3 5 3 4 1 2 6 4 4 2 3 ( ﻋﺪﺩ ﺍﻟﺘﻼﻣﻴﺬ ) ﺍﻟﺤﺼﻴﺺ : * ﻣﻔــﺮﺩﺍﺕ nﺍﻟﺴﺎﻛﻨﺔ ﺍﻹﺣﺼﺎﺋﻴﺔ : ﻫﻲ ﺍﻷﺷﺨﺎﺹ ﺍﻟﺬﻳﻦ ﺃﺟﺮﻳﺖ ﻋﻠﻴﻬﻢ ﻋﻤﻠﻴﺔ ﺍﻹﺣﺼــﺎء . ) ﻓﻲ ﻫﺬﺍ ﺍﻟﻤ ﺜﺎﻝ ﺍﻟﺴﺎﻛﻨﺔ ﺍﻹﺣﺼﺎﺋﻴﺔ ﻫﻢ ﺍﻟﺘﻼﻣﻴﺬ ﺍﻟﺬﻳﻦ ﻗﺎﻣﻮﺍ ﺑﻺﻧﺠﺎﺯ ﺍﻟﻔﺮﺽ ﺍﻟﻤﺤﺮﻭﺱ ( . nﺍﻟـﻤﻴــﺰﺓ : ﻫﻲ ﻻﻟﺤﺎﻟﺔ ﺍﻟﻤﺪﺭﻭﺳﺔ . ) ﻓﻲ ﻫﺬﺍ ﺍﻟﻤﺜﺎﻝ ﺍﻟﻤﻴﺰﺓ ﻫﻲ ﻧﻘﻂ ﺍﻟﻔﺮﺽ ﺍﻟﻤﺤﺮﻭﺱ ( . nﺍﻟﺤﺼﻴﺺ : ﻋﺪﺩ ﺍﻟﺘﻼﻣﻴﺬ ﻟﻜﻞ ﻣﻴﺰﺓ . ) ﻓﻲ ﻫﺬﺍ ﺍﻟﻤﺜﺎﻝ : ﺣﺼﻴﺺ ﺍﻟﻤﻴﺰﺓ 21 ﻫﻮ 4 . ﺣﺼﻴﺺ ﺍﻟﻤﻴﺰﺓ 5 ﻫﻮ 1 ( . nﺍﻟﺤﺼﻴﺺ ﺍﻹﺟﻤﺎﻟﻲ : ﻋﺪﺩ ﺍﻷﺷﺨﺎﺹ ﺍﻟﺬﻳﻦ ﺃﺟﺮﻳﺖ ﻋﻠﻴﻬﻢ ﻋﻤﻠﻴﺔﺍﻹﺣﺼﺎء ) ﻓﻲ ﻫﺬﺍ ﺍﻟﻤﺜﺎﻝ ﺍﻟﺤﺼﺒﻴﺺ ﺍﻹﺟﻤﺎﻟﻲ ﻫﻮ . 73( 2( – ﺍﻟﺘﻤﺜﻴﻞ ﺍﻟﻤﺒﻴﺎﻧﻲ : ﺺ ) ﻋﺪﺩ ﺍﻟﺘﻼﻣﻴﺬ ( ﺍﻟﺤﺼﻴ ( ﺍﻟﻤﺨﻄﻂ ﺑﺎﻟﻘﻀﺒﺎﻥ : ﺃ ﺍﻟﻤﻴﺰﺓ ﻳﺔ ﺑﺔ ﻳﺔ ﺃﺳﺘﺎﺫ ﺜﺎﻧﻮﻳﺔ ﻸﺳ ﺎﺫ ﺜﺎﻧﻮ ﻣﻮﻗ
75.
) ﺍﻟﻨﻘﻂ (
( ﺍﻟﻤﺨﻄﻂ ﺍﻟﻘﻄﺎﻋﻲ : ﺏ * ﺟﺪﻭﻝ ﺍﻟﻔﺌﺎﺕ ﻭﺍﻟﻨﺴﺐ ﺍﻟﻤﺌﻮﻳﺔ ﻭﻗﻴﺎﺱ ﺍﻟﺰﻭﺍﻳﺎ : 7 < D :11 £ n £ 14 C : 9 £ n < 11 B : 7 £ n < 9 A : 4 £ n 4 ﺍﻟﻔﺌﺎﺕ 31 9 8 7 73 ﻋﺪﺩ ﺍﻟﺘﻼﻣﻴﺬ % 43 % 42 % 22 % 91 % 001 ﺍﻟﻨﺴﺐ ﺍﻟﻤﺌﻮﻳﺔ °621 °88 °87 °86 °063 ﻗﻴﺎﺱ ﺍﻟﺰﻭﺍﻳﺎ * ﺍﻟﻤﺨﻄﻂ ﺍﻟﻘﻄﺎﻋﻲ : 3( – ﺍﻟـﺘـــﺮﺩﺩ : ( ﺗﻌﺮﻳﻒ : ﺃ ﺗﺮﺩﺩ ﻣﻴﺰﺓ ﻫﻮ ﺧﺎﺭﺝ ﺣﺼﻴﺼﻬﺎ ﻋﻠﻰ ﺍﻟﺤﺼﻴﺺ ﺍﻹﺟﻤﺎﻟﻲ ( ﻣـﺜــﺎﻝ : ﺏ ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺃﻋﻼﻩ : 3 18 0 = , ﺗﺮﺩﺩ ﺍﻟﻤﻴﺰﺓ 41 ﻫﻮ : 73 6 861 0 = , : ﺗﺮﺩﺩ ﺍﻟﻤﻴﺰﺓ 01 ﻫﻮ 73
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