Page 108 - Engineering

‫ﺍﳌﺜﻠﺜﺎﺕ ‪٩٥‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )ب(‪ :‬ﰲ اﳌﺜﻠﺚ ‪ △PQR‬ﻟﺪﻳﻨﺎ‬

‫)‪QPR = UPV = 180 − (3x + 4x‬‬
‫)‪QRP = XRY = 180 − (6x + 7x‬‬

‫إذن‪،‬‬

‫‪180 − 7x + 180 − 13x + 5x = 180‬‬

‫أي أن ‪ .15x = 180‬و‪‬ﺬا ﻓﺈن ‪.x = 12‬‬

‫)‪ [Aust.MC 1991] (٢٤‬ﰲ اﻟﺸﻜﻞ اﳌﺮﻓﻖ‪،PR = QR = 12 ،‬‬

‫‪ .RS = RT = 8‬ﻣﺴﺎﺣﺔ اﻟﺸﻜﻞ ‪ RSXT‬ﺗﺴﺎوي ‪ 8‬وﺣﺪات ﻣﺮﺑﻌﺔ‪.‬‬

‫ﻣﺴﺎﺣﺔ ‪ △PRQ‬ﺑﺎﻟﻮﺣﺪات اﳌﺮﺑﻌﺔ ﺗﺴﺎوي‬

‫)د( ‪18‬‬ ‫)ج( ‪17‬‬ ‫)ب( ‪16‬‬ ‫)أ( ‪15‬‬

‫‪Q‬‬
‫‪T‬‬

‫‪X‬‬

‫‪R SP‬‬

‫اﻟﺤﻞ‪ :‬اﻹﺟﺎﺑﺔ ﻫﻲ )أ(‪ :‬ﻻﺣﻆ أن ‪ .△PXS ≡ △QXT‬وﻟﺬا ﻓﺈن‬

‫] ‪ .[PXS ] = [QXT‬ﻧﻔﺮض أن ﻫﺬﻩ اﳌﺴﺎﺣﺔ ﻫﻲ ‪ x‬وﻟﻨﻔﺮض أن ]‪.y = [PXQ‬‬

‫ﻋﻨﺪﺋﺬ ‪ .[RXS ] = 4‬ﲟﺎ أن ‪ RS = 8‬و ‪ PS = 4‬ﻓﺈن ‪ .[RXS ] = 8‬إذن‪،‬‬
‫‪[PXS ] 4‬‬ ‫‪[PXS ] x‬‬

‫‪ . 4 = 8‬وﻣﻦ ﰒ ﻓﺈن ‪ .x = 2‬وﺑﺎﳌﺜﻞ‪،‬‬

‫‪x4‬‬
‫‪8 = [PTR] = 8 + x = 10‬‬
‫‪4 [PTQ] y + x y + 2‬‬

‫و‪‬ﺬا ﻓﺈن ‪ .y = 3‬إذن‬
‫‪[PRQ] = 3 + 2 + 2 + 8 = 15 .‬‬