Page 120 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪١٠٧‬‬

‫إذن‪،‬‬

‫‪AF = FG = AG = 3‬‬
‫‪AE EC AC 4‬‬

‫ﻣﻦ ذﻟﻚ ﻧﺮى أن ‪ .h = FG = 3 EC‬اﻵن‪،‬‬

‫‪4‬‬

‫] ‪[ABF‬‬ ‫=‬ ‫‪1 × AB‬‬ ‫‪×h‬‬ ‫=‬ ‫‪1‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪AC‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪EC‬‬ ‫‪‬‬
‫‪2‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪4‬‬

‫=‬ ‫‪3‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪× AC‬‬ ‫‪× EC‬‬ ‫‪‬‬ ‫=‬ ‫] ‪3 [ACE‬‬
‫‪16‬‬ ‫‪2‬‬ ‫‪16‬‬

‫وﺑﺎﳌﺜﻞ‪ ،‬ﻧﺮى أن ] ‪ .[BCD] = [DEF ] = 3 [ACE‬إذن‪،‬‬

‫‪16‬‬

‫] ‪[DBF ] = [ACE ] − 3 × 3 [ACE ] = 7 [ACE‬‬
‫‪16 16‬‬

‫‪.‬‬ ‫‪[DBF‬‬ ‫]‬ ‫=‬ ‫‪7‬‬ ‫و‪‬ﺬا ﻳﻜﻮن‬
‫‪[ACE‬‬ ‫]‬ ‫‪16‬‬

‫)‪ [AMC10A 2008] (٤١‬ﻣﺜﻠﺚ ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ﳏﻴﻄﻪ ‪ 32‬وﻣﺴﺎﺣﺘﻪ ‪ .20‬ﻣﺎ ﻃﻮل‬

‫وﺗﺮﻩ ؟‬

‫‪63‬‬ ‫)د(‬ ‫‪61‬‬ ‫)ج(‬ ‫‪59‬‬ ‫)ب(‬ ‫‪57‬‬ ‫)أ(‬
‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ب(‪:‬‬

‫‪z‬‬
‫‪x‬‬

‫‪y‬‬

‫ﻟﺪﻳﻨﺎ ‪ x + y + z = 32‬و ‪ . 1 xy = 20‬ﻣﻦ ﻣﱪﻫﻨﺔ ﻓﻴﺜﺎﻏﻮرس ﻟﺪﻳﻨﺎ‬

‫‪2‬‬
‫‪z = x2 + y2‬‬