Page 111 - Engineering

‫ﺍﳍﻨﺪﺳﺔ )ﺍﳉﺰء ﺍﻷﻭﻝ(‬ ‫‪٩٨‬‬

‫)‪ (٢٨‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ CD = 6 ،AC = 9 ،ABD = DAC ،‬و‬
‫‪ .[BCA] = 18‬ﻣﺎ ﻗﻴﻤﺔ ]‪ [ABD‬؟‬

‫‪B‬‬

‫‪D‬‬
‫‪6‬‬

‫)د( ‪10‬‬ ‫‪A‬‬ ‫‪) C‬ب( ‪9 8‬‬ ‫)أ( ‪6‬‬

‫)ج( ‪9‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )د(‪ :‬اﳌﺜﻠﺜﺎن ‪ △ACD‬و ‪ △BCA‬ﻣﺘﺸﺎ‪‬ﺎن ﻷن‬

‫‪2‬‬ ‫‪َ DBA = DAC‬و ‪.C = C‬‬

‫أن‬ ‫أي‬ ‫]‪. [ACD‬‬ ‫=‬ ‫‪‬‬ ‫‪2‬‬ ‫‪=4‬‬ ‫ﻓﺈن‬ ‫و‪‬ﺬا‬ ‫‪.CD‬‬ ‫=‬ ‫‪6‬‬ ‫=‬ ‫‪2‬‬ ‫أن‬ ‫ﻧﺮى‬ ‫ذﻟﻚ‬ ‫ﻣﻦ‬
‫‪3‬‬ ‫‪9‬‬ ‫‪9‬‬ ‫‪3‬‬
‫]‪[BCA‬‬ ‫‪CA‬‬

‫إذن‪،‬‬ ‫]‪.[ACD‬‬ ‫=‬ ‫]‪4 [BCA‬‬ ‫=‬ ‫‪4‬‬ ‫‪× 18‬‬ ‫=‬ ‫‪8‬‬
‫‪9‬‬ ‫‪9‬‬

‫‪.[ABD] = [BCA] − [ACD] = 18 − 8 = 10‬‬

‫)‪ (٢٩‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪ AD ،‬و ‪ BD‬ﻣﻨﺼﻔﺎن ﻟﻠﺰاوﻳﺘﲔ ‪ BAC‬و ‪ ABC‬ﻋﻠﻰ‬
‫اﻟﺘﻮاﱄ‪ .‬ﻣﺎ ﻗﻴﻤﺔ ‪ y‬ﺑﺪﻻﻟﺔ ‪ x‬؟‬

‫‪A‬‬

‫‪x‬‬
‫‪D‬‬

‫‪B‬‬ ‫‪y‬‬
‫‪C‬‬
‫)ب( ‪y = 2x − 180°‬‬
‫)د( ‪y = 2x − 90°‬‬ ‫)أ( ‪y = x − 180°‬‬
‫)ج( ‪y = 2x‬‬