Page 39 - Engineering

‫ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ(‬ ‫‪٢٦‬‬

‫‪5x + 7 = 3x + 23‬‬

‫‪2x = 16‬‬

‫‪x = 8°‬‬

‫إذن‪ .EAD + BAC = (5 × 8 + 7) + (3 × 8 + 23) = 94° ،‬و‪‬ﺬا ﻳﻜﻮن‬
‫‪.DAC = 180° − 94° = 86°‬‬

‫)‪ (٧‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ EOF ،‬ﻣﻨﺼﻒ ﻋﻤﻮدي ﻟﻠﻘﻄﻌﺔ ‪،OC = 18 .BC‬‬
‫‪ .BO = 4x − 6 ،CD = 3x − 7 ،AB = 2x + 1‬ﻣﺎ ﻃﻮل اﻟﻘﻄﻌﺔ‬

‫‪ AD‬؟‬

‫‪F‬‬

‫‪AB‬‬ ‫‪O CD‬‬

‫)د( ‪50‬‬ ‫)ج( ‪54‬‬ ‫‪E‬‬ ‫)أ( ‪60‬‬

‫)ب( ‪58‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )أ(‪ :‬ﲟﺎ أن ‪ EOF‬اﳌﻨﺼﻒ اﻟﻌﻤﻮدي ﻟﻠﻘﻄﻌﺔ ‪ BC‬ﻓﺈن‬

‫‪BO = OC‬‬

‫‪4x − 6 = 18‬‬

‫‪x =6‬‬

‫إذن‪ .CD = 3x − 7 = 11 ،AB = 2x + 1 = 13 ،‬وﻟﺬا ﻓﺈن‬

‫‪AD = AB + BO + OC + CD = 13 + 18 + 18 + 11 = 60 .‬‬

‫)‪ (٨‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ ‪ DBG ،CBF ،ABE‬ﺛﻼﺛﺔ ﻣﺴﺘﻘﻴﻤﺎت ﺗﺘﻘﺎﻃﻊ ﰲ‬
‫اﻟﻨﻘﻄﺔ ‪ .B‬ﻣﺎ ﻗﻴﺎس اﻟﺰاوﻳﺔ ‪ ABC‬؟‬