Page 130 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪١١٧‬‬

‫‪[ABC ] = 4[APN ] = 4 × 2 = 8‬‬

‫اﳊﻞ اﻟﺜﺎﱐ‬

‫‪) △APN ∼ △ABC‬ﻛﻤﺎ ﰲ اﳊﻞ اﻷول(‪ .‬إذن‪ .N = C = 90° ،‬ﺑﺎﳌﺜﻞ‬

‫‪ .△PMB ∼ △ACB‬إذن‪ .M = C = 90° ،‬إذن‪،‬‬

‫‪ AN = NC = PM = 1 AC‬و ‪.NP = CM = MB = 1CB‬‬

‫‪22‬‬

‫ﻣﻦ ذﻟﻚ ﻳﻜﻮن ‪ .△PMB ≡ △ANP‬إذن‪ .[PMB] = 2 ،‬وﻣﻦ اﻟﻮاﺿﺢ أن‬

‫‪ .[NPMC ] = 2[ANP ] = 4‬إذن‪،‬‬

‫‪[ABC ] = 2 + 2 + 4 = 8 .‬‬

‫اﳊﻞ اﻟﺜﺎﻟﺚ‬

‫ﺻﻞ ﺑﲔ اﻟﻨﻘﻄﺘﲔ ‪ C‬و ‪ .P‬ﻋﻨﺪﺋﺬ‪ .△CPN ≡ △PCM ،‬ﲟﺎ أن ‪AN = NC‬‬

‫وأن ‪ PN‬ارﺗﻔﺎع ﻟﻜﻞ ﻣﻦ اﳌﺜﻠﺜﲔ ‪ △PNA‬و ‪ △PNC‬ﻓﺈن ] ‪.[PNA] = [PNC‬‬

‫وﺑﺎﳌﺜﻞ‪ .[PMC ] = [PMB] ،‬وﺑﺎﻟﺘﺎﱄ ﻓﻤﺴﺎﺣﺔ اﳌﺜﻠﺜﺎت اﻷرﺑﻌﺔ اﻟﺼﻐﲑة ﻣﺘﺴﺎوﻳﺔ‬

‫وﺗﺴﺎوي ‪ 2‬إذن‪،‬‬

‫‪[ABC ] = 2 + 2 + 2 + 2 = 8‬‬

‫)‪ [MAΘ 2011] (٥٣‬ﰲ اﳌﺜﻠﺚ ‪ .AC = 5 ،BC = 3 ،△ABC‬ﻣﺎ ﻋﺪد‬
‫اﻟﻘﻴﻢ اﳌﻤﻜﻨﺔ ﻟﻠﻄﻮل ‪ AB‬ﻟﻜﻲ ﻳﻜﻮن ‪ △ABC‬ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ؟‬

‫)أ( ‪) 0‬ب( ‪) 1‬ج( ‪) 2‬د( ‪3‬‬
‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ج(‪ :‬ﻫﻨﺎك ﺧﻴﺎران ﻟﻄﻮل ‪ AB‬اﻷول ﻣﻨﻬﻤﺎ ﻫﻮ أن ﻳﻜﻮن‬
‫‪ AC‬ﻫﻮ اﻟﻮﺗﺮ )اﻷﻃﻮل(‪ .‬ﰲ ﻫﺬﻩ اﳊﺎﻟﺔ ‪ .AB = 52 − 32 = 4‬وأﻣﺎ اﳋﻴﺎر‬

‫اﻟﺜﺎﱐ ﻓﻬﻮ أن ﻳﻜﻮن ‪ AB‬ﻫﻮ اﻟﻮﺗﺮ‪ .‬ﰲ ﻫﺬﻩ اﳊﺎﻟﺔ ‪.AB = 52 + 32 = 34‬‬